Guy David

H ÖLDER REGULARITY OF TWO-DIMENSIONAL ALMOST-MINIMAL SETS IN R n

Keywords: Minimal sets, Almgren restricted sets, Almost-minimal or quasiminimal sets, Soap films, Hausdorff measure

On donne une démonstration différente et sans doute plus élémentaire d'une bonne partie du résultat de régularité de Jean Taylor sur les ensembles presque-minimaux d'Almgren. On en profite pour donner des précisions sur les ensembles presque minimaux, généraliser une partie du théorème de Taylor aux ensembles presque minimaux de dimension 2 dans R n , et donner la caractérisation attendue des ensembles fermés E de dimension 2 dans R 3 qui sont minimaux, au sens où H 2 (E \ F ) ≤ H 2 (F \ E) pour tout fermé F tel qu'il existe une partie bornée B telle que F = E hors de B et F sépare les points de R 3 \ B qui sont séparés par E.

Introduction

The core of this paper is a slightly different, probably more elementary, and very detailed proof of a good part of Jean Taylor's regularity theorem [Ta] for Almgren almostminimal sets of dimension 2 in R 3 .

Recall that this result says that these sets are locally C 1 -equivalent to minimal cones (i.e., planes or cones of type Y or T, as described near Figures 1.1 and 1.2). We shall only prove the biHölder equivalence in this text; the C 1 -equivalence needs a little bit of extra work, which is left for a next paper [D3].

The proof will extend to 2-dimensional sets in R n , and will also allow us to show that every M S-minimal set of dimension 2 in R 3 (where the definition of competitors uses the separation constraint in Definition 1.4 below) is a minimal cone. The issue shows up naturally in the study of the Mumford-Shah functional (what are the global Mumford-Shah minimizers (u, K) in R 3 for which u is locally constant), to the point that the author felt compelled to announce Theorem 1.9 in [D2].

The local biHölder equivalence to minimal cones can be useful as a first step to C 1 equivalence, as in [D3], but could also give enough geometric control on the almost-minimal 1 sets to allow one, perhaps, to prove existence results for functionals with a dominant area term under suitable topological constraints.

One of the main initial motivations for this text was to help the author understand Jean Taylor's paper, and possibly help some readers too. Part of the reason for this interest was potential applications of the technical lemmas to the regularity of minimal segmentations for the Mumford-Shah functional in dimension 3, but it is not yet certain that the current text will help.

Since this was the triggering reason for this paper, let us define M S-minimal sets first. We give the definition for (n -1)-dimensional sets in R n , but for the theorem we shall restrict to n = 3. Definition 1.1. Let E be a closed set in R n . A M S-competitor for E is a closed set F ⊂ R n such that we can find R > 0 such that

(1.2) F \ B(0, R) = E \ B(0, R)
and

(1.3) F separates y from z whenever y, z ∈ R n \ (B(0, R) ∪ E) are separated by E.

By "F separates y from z", we just mean that y and z lie in different connected components of R n \ F . Note that if (1.2) and (1.3) hold for R, they also hold for every R ′ > R. Here M S stands for Mumford-Shah; indeed the separation condition (1.3) is the same one as in the definition of global minimizers for the Mumford-Shah functional in R n . Definition 1.4. A M S-minimal set in R n is a closed set E ⊂ R n such that

(1.5) H n-1 (E \ F ) ≤ H n-1 (F \ E)
for every M S-competitor F for E.

Here H n-1 denotes the (n -1)-dimensional Hausdorff measure. Notice that if E is M S-minimal, then H n-1 (E ∩ B(0, R)) < +∞ for R > 0, because we can compare with F = [E \ B(0, R)] ∪ ∂B(0, R). We shall also use a weaker notion of minimality, based on a more restrictive notion of competitors. The notion is the same as for Almgren's restricted sets [Al].

Definition 1.6. Let E be a closed set in R n and d ≤ n-1 be an integer. An Al-competitor for E is a closed set F ⊂ R n that can be written as F = ϕ(E), where ϕ : R n → R n is a Lipschitz mapping such that W ϕ = {x ∈ R n ; ϕ(x) = x} is bounded. An Al-minimal set of dimension d in R n is a closed set E ⊂ R n such that H d (E ∩ B(0, R)) < +∞ for every R > 0 and

(1.7)

H d (E \ F ) ≤ H d (F \ E)
for every Al-competitor F for E.

Here it makes sense to define Al-minimal sets of all dimensions d < n; this is not the case for M S-minimal sets because of the special role played by separation. Also, the reader may be surprised because Definition 1.6 seems simpler than if we used the standard definition of Almgren restricted sets; we shall check in Section 4 that the two definitions coincide.

Remark 1.8. Let us check that Al-competitors for E are automatically M S-competitors for E, and hence M S-minimal sets are also Al-minimal. Let F be an Al-competitor for E, let ϕ and W ϕ be as in Definition 1.6, and choose R so that W ϕ ∪ ϕ(W ϕ ) ⊂ B(0, R). Obviously (1.2) holds. For (1.3), let y, z ∈ R n \ (B(0, R) ∪ E) be separated by E. Consider the continuous deformation given by ϕ t (x) = (1t)x + tϕ(x) for x ∈ R n and 0 ≤ t ≤ 1. Notice that none of the sets E t = ϕ t (E) ever contain y or z, because all the modifications occur in B(0, R). Then F = ϕ 1 (E) = E 1 separates y from z, because E = E 0 does. See for instance 4.3 in Chapter XVII of [Du], on page 360. Actually, that result is only stated when E is a compact set of R n . To accommodate this minor difficulty, we should instead assume that F does not separate y from z, pick a path γ ⊂ R n \ F from y to z, choose S > R such that γ ⊂ B(0, S), and notice that E ′ = [E ∩ B(0, S)] ∪ ∂B(0, S) is compact and separates y from z, but its image ϕ 1 (E ′ ) = [F ∩ B(0, S)] ∪ ∂B(0, S) does not (because of γ). From the ensuing contradiction we would deduce that F is a M S-competitor for E, as needed.

Here is a list of M S-minimal sets in R 3 . First, the empty set, and planes, are M Sminimal. Next let P be a plane, pick a point O ∈ P , and let S be the union of three half lines in P that start from O and make 120 • angles with each other at O. The product of S with a line perpendicular to P is M S-minimal. We shall call such a set a cone of type Y; thus cones of type Y are made up of three half planes with a common boundary L, and that meet along the line L with 120 • angles. See Figure 1.1. For the last example, take a regular tetrahedron R, call O its center, and denote by T the cone centered at O over the union of the six vertices of R. See Figure 1.2. Thus T is composed of six angular faces that meet at O, and are bounded by four half-lines that leave from O with equal (and maximal) angles. We shall call such a set a cone of type T. See K. Brakke's home page http://www.susqu.edu/brakke/ for nicer pictures of T and how Y and T arise as tangent objects to minimal sets.

The verification that cones of type Y and T are M S-minimal can be done with a clever integration by parts; see [LM]. It turns out that modulo sets of measure 0, the list is complete.

Theorem 1.9. If E is an M S-minimal set in R 3 , then there is a set E * such that E * ⊂ E, H 2 (E \ E * ) = 0, and E * is the empty set, a plane, or a cone of type Y or T.

The fact that we may have to remove a set of vanishing H 2 -measure is natural: it is easy to see that the union of a minimal set with a closed set of vanishing H 2 -measure is still minimal.

As we shall see, Theorem 1.9 is a rather simple consequence of the proof of Jean Taylor's regularity theorem, but to the author's surprise, it seems that this was not stated in the literature yet. An analogue of this result for size-minimizing currents is stated in Section 4 of [START_REF] Morgan | Size-minimizing rectifiable currents[END_REF]; the proof sketched there has the same sort of ingredients as in the present paper, but does not seem to me to apply directly.

The situation for Al-minimal sets is a little less clear; although it seem quite probable that Theorem 1.9 extends to Al-minimal sets in R 3 , the author does not know how to prove this without an additional assumption, for instance on the separating properties of E. See Proposition 18.29 and Remark 18.43.

As was hinted before, Theorem 1.9 is much easier to deduce from a proof of Jean Taylor's regularity theorem than from its mere statement; we shall take this as a first excuse to give, in full details, a slightly different proof of that regularity result.

Let us be a little more specific about this. Her theorem concerns almost-minimal sets of dimension 2 in R 3 ; we shall give very precise definitions of almost-minimal sets, but for the moment let us just say that they are like the minimal sets of Definition 1.6, except that we may add a small term to the right-hand side of (1.7), and also localize the definition. Recall that J. Taylor proves that if E is a reduced almost-minimal set of dimension 2 in R 3 , with a sufficiently small gauge function (a function that controls the small term that we add to the right-hand side of (1.7)), then every point of E has a neighborhood where E is C 1 -equivalent to a plane or a cone of type Y or T. [See Definition 2.12 for "reduced sets ".] We shall cut this regularity result in two, and first prove a mere biHölder equivalence to a plane, a Y, or a T, under slightly weaker assumptions on the gauge function. See Theorem 16.1 for a precise statement, and observe that in our definition the biHölder equivalence comes from a local biHölder mapping of the ambient space, so the equivalence also concerns the way E is embedded in R n . This is a little simpler to prove than the C 1 equivalence, and in particular does not require epiperimetry estimates. This is also the only part of the argument that is needed for Theorem 1.9, and possibly for other results as well.

The full C 1 equivalence (or even C 1+α if the gauge function is small enough) seems to need a more precise argument, which uses epiperimetry or some comparison with harmonic functions. We intend to give in [D3] a special argument for this, which uses a little bit of the local biHölder equivalence through separation properties.

At this time, only the present paper extends to sets of dimension 2 in R n , and it could be that when n > 3 the existence of epiperimetric inequalities and the local C 1 equivalence of E to a minimal cone near x depends on the type of tangent cones that E has at x. The author does not even know whether E could have an interesting one-parameter family of different tangent cones at x, which would of course exclude the C 1 equivalence if they are not isometric to each other.

Theorem 16.1. also applies to the two-dimensional almost-minimal sets in R n , n > 3. In this case, we don't know the precise list of minimal cones, and by minimal cone of type T we shall then mean a (reduced) minimal cone T which is neither empty, nor a plane, nor a minimal cone of type Y as above. As we shall see, these are also the minimal cones whose density at the origin is larger than 3π/2. See Section 14 for a rapid description of these cones. Observe that we do not exclude the possibility that E may have two significantly different tangent cones of type T at x, and then E will be locally biHölder equivalent to either cone.

Our proof of biHölder equivalence relies on a few constructions by hand (for instance, to get the monotonicity of density), a result of stability of minimal and almost-minimal sets under Hausdorff limits [D1], and an extension of Reifenberg's topological disk theorem to the case when E is a two-dimensional set in R n that stays close to planes or cones of type Y or T at all scales and locations [DDT]. In particular, we never use currents or epiperimetry arguments.

Many of the ingredients in the proofs can be generalized, but the author does not know exactly how far the generalizations may lead. Because of this, we shall try to keep a reasonably large amount of generality in the intermediate statements, especially (but not always) when this is not too costly. For the same sort of reasons, the author decided to go ahead and prove in the first sections that various definitions of almost-and quasiminimality for sets are equivalent, or imply others. Hopefully this will help some readers, and the other ones will not find it too hard to skip.

The plan for the rest of this paper is as follows. Part A is mostly concerned in the definitions, regularity, and limiting properties of almost-minimal sets.

Section 2 contains the definitions and main regularity properties for a slight generalization of Almgren's restricted sets (we shall call them quasiminimal sets); the small amount of new material will only be used to allow us to simplify our notions of almostminimality (in Section 4), and state some results (like Proposition 5.24 and Theorem 16.1) with a little more generality.

In Section 3 we remind the reader that for sequences of reduced quasiminimal sets with uniform constants, limits are also quasiminimal with the same constants, and we have lower-and upper semicontinuity results for the Hausdorff measure. This stays true for the generalized quasiminimal sets of Section 2. See Lemmas 3.3 and 3.12 in particular.

Section 4 contains various definitions of almost-minimal sets, and the equivalence between the two main ones. Just like Section 2, this one is not really needed in view of Theorems 1.9 and 16.1, for instance.

In Part B we prove the almost monotonicity of the density function for almost-minimal sets with a small enough gauge function, and results of approximation by a minimal cone in annuli where the density is almost constant.

The main result of Section 5 is the fact that if E is a minimal set of dimension d, the density r -d H d (E ∩ B(x, r)) is a nondecreasing function of r (Proposition 5.16). The very standard argument is essentially a comparison of E ∩ B(x, r) with the cone over E ∩ ∂B(x, r). Also see Propositions 5.24 and 5.30 for generalizations to almost-minimal sets.

In Section 6 we show that if E is a reduced minimal set of dimension 2 in B(0, b) ⊂ R 3 and the density r -d H d (E ∩ B(0, r)) is constant on (a, b), then E coincides with a reduced minimal cone on the annulus {z ; a < |z| < b}. The proof is surprisingly painful, in the sense that the author did not manage to avoid using the minimality of E again. See Proposition 6.2.

Section 7 contains a variant of Proposition 6.2 for almost-minimal sets and very small variations of density, obtained from it by a compactness argument. See Proposition 7.1 for the initial statement in a cone, Proposition 7.24 for the variant in a ball which is used most of the time, and Proposition 7.31 for the fact that if the density of E at x exists, the tangent objects to E at x are minimal cones with the same density.

Part C contains a structural description of the reduced minimal cones of dimension 2 in R n . We show that for such a cone E, K = E ∩ ∂B(0, 1) is composed of a finite number of arcs of great circles, that may only meet by sets of three at their ends, and with 120 • angles. See Proposition 14.1 for a more precise description (but observe that we shall not give a complete list of cones).

We start Section 8 with the simple fact that if the product set E × R m is a reduced Al-minimal set of dimension d + m in R n+m , then E is a d-dimensional Al-minimal set in R n . See Proposition 8.3. We don't know about, nor need the converse (see Remark 8.22). Proposition 8.3 applies to the blow-up limits of a minimal cone at a point of the unit sphere, and we can use this to prove that if E is a reduced minimal cone of dimension d and ε > 0 is small, then K = E ∩ ∂B(0, 1) is ε-close (in normalized Hausdorff distance) to a minimal set of dimension d -1 in every small enough ball centered on K.

This would lead to some information on E, but since we want something more precise, we start a different approach and try to study the local regularity of K = E ∩∂B(0, 1) with our hands. In section 9 we show that K is a "weakly almost-minimal set" (where "weakly" comes from the fact that we have to pay some price when we start from competitors for K and construct a competitor for E). See Definition 9.1 and Proposition 9.4.

Then we turn more specifically to one-dimensional sets. In Section 10 we show that the nonempty reduced minimal sets of dimension 1 in R n are the lines and the propellers Y . See Theorem 10.1. The proof is also used in Section 11 to show that weakly almost-minimal set are often close (in Hausdorff distance) to lines or propellers (Proposition 11.2).

In Section 12 we state a simpler, one-dimensional version of the Reifenberg-type theorem that will be stated in Section 15. This is enough to show that when K = E ∩ ∂B as above, K is essentially composed of C 1 arcs. See Proposition 12.6 for a first statement, and Proposition 12.27 for the C 1 version.

In Section 13 we compare the cone over an arc of Lipschitz curve in ∂B(0, 1) with the graph of a harmonic function f ; this is a small computation where we use the fact that the main term in the area functional is the energy |∇f | 2 , and whose conclusion is that the curve cannot yield a minimal cone unless it is an arc of great circle.

We apply all this in Section 14 and find the description of the two-dimensional minimal cones of Proposition 14.1.

Part D studies the local regularity of the almost-minimal sets. We start in Section 15 with the description of a result from [DDT], where a slight generalization of Reifenberg's topological disk theorem is proved. This is the result that will be used in Section 16 to produce parameterizations of E when we know that is is well approximated by twodimensional minimal cones in small balls.

Section 16 contains the main local regularity results for almost-minimal sets of dimension 2 in R n . Theorem 16.1 is the main statement (the promised generalization of J. Taylor's theorem), but some intermediate results may have independent interest, such as Lemmas 16.19 and 16.48 (which gives sufficient conditions for E to be biHölder equivalent to a plane in a given ball B), Proposition 16.24 (which proves the existence of a point of density 3π/2 in B), Lemmas 16.25 and 16.51 (sufficient conditions for E to be biHölder equivalent in B to a cone of type Y), and Lemma 16.56 and Proposition 18.1 (for the biHölder equivalence to a cone of type T). Proposition 16.24 is proved in Section 17, with a topological degree argument.

Theorem 1.9 is proved in Section 18, with another topological argument. Also see Proposition 18.29 for a version for Al-minimal sets of dimension 2 in R 3 , but with an extra separation assumption.

In Section 19 we describe a set which looks like cone of type T at large scales, but has no point of type T ; thus there is no trivial way to make the proof of Theorem 1.9 work for Al-minimal sets. In the very short Section 20 we compute the density of the set T of Figure 1.2.

The author wishes to thank Christopher Colin for help with the topological argument in Section 18, T. De Pauw, J.-C. Léger, P. Mattila, and particularly F. Morgan for interesting discussions about the subject of this paper, and gratefully acknowledges partial support from the european network HARP.

A. ALMOST-MINIMAL SETS

In this first part we shall give a few different definitions of almost-minimal sets of dimension d in R n , check that many definitions are equivalent, discuss some of the main properties of these sets (including local Ahlfors-regularity and local uniform rectifiability), and state convergence theorems that will be used in the rest of this paper.

The equivalence between various definitions of almost-minimality is not vital for the rest of the paper, but it will be more convenient to use simple definitions like Definition 1.6, and good to know that they are equivalent to more classical ones, in particular those that come more naturally from Almgren's definitions in [Al]. The equivalence is not particularly subtle or original, but some argument is needed, and the author decided that this is a good opportunity to settle the issue as explicitly as possible.

It appears that for the proof of equivalence, it is useful to introduce a class of sets which is slightly larger than the class of restricted sets introduced by F. Almgren, and then check that the main regularity properties of restricted sets still hold in this class. This is what we do in Section 2.

In Section 3, we use these properties to show that the Hausdorff measure has good semicontinuity properties when we restrict to bounded sequences of generalized Almgren quasiminimal sets. In other words, the Hausdorff measure of these sets goes to the limit well, as in Lemma 3.3 and Lemma 3.12 below, and we shall be able to apply this to our various types of almost-minimal sets.

The various definitions of almost-minimality, and the equivalence of the main ones, are given in Section 4. We also use the results of Section 3 to show that limits of almostminimal sets are almost-minimal; see Lemma 4.7. We try to arrange this text so that the impatient reader can refer to this part only occasionally, for a definition or a believable statement.

Generalized Almgren quasiminimal sets

The main results of this paper apply not only to minimal sets, but to almost-minimal sets of various types. We shall give in Section 4 a few definitions of almost-minimal sets and clarify the relations between the various classes, but before we do this we need to talk about Almgren restricted sets (we'll call them quasiminimal), and even generalize the notion slightly.

The standard properties described in this section will be used in Section 4 to prove the equivalence between different notions of almost-minimality, and then also in later sections, but only applied to more restricted classes of almost-minimal sets.

A good part of this is not needed for the main results of the paper, the main point of it is merely to try to give some statements in the best possible context. It should be possible for the reader to skip Sections 2 and 4 almost entirely if she is just interested in minimal sets or happy to use a stronger form of almost-minimality.

So we want to discuss minor variations over the notion of "restricted set" of Almgren [Al]. For the general setting we are given an open set U ⊂ R n , a quasiminimality constant M ≥ 1, and a diameter δ ∈ (0, +∞]. Now E is a closed subset of U , and we want to assume that (2.1) H d (E ∩ B) < +∞ for every compact ball B ⊂ U.

As in Definition 1.6, we want to compare E with competitors F = ϕ 1 (E), where ϕ t , 0 ≤ t ≤ 1, is a one-parameter family of continuous functions ϕ t : U → U with the following properties:

(2.2) ϕ 0 (x) = x for x ∈ U,

(2.3) the function (t, x) → ϕ t (x), from [0, 1] × U to U , is continuous,

(2.4) ϕ 1 is Lipschitz, and, if we set (2.5) W t = {x ∈ U ; ϕ t (x) = x} and then W = t∈[0,1] W t ∪ ϕ t (W t ) , then (2.6) W is relatively compact in U and diam( W ) < δ.

Definition 2.7. We say that the set E is an (M, δ)-quasiminimal set in U when E is closed in U , (2.1) holds, and

(2.8)

H d (E ∩ W 1 ) ≤ M H d (ϕ 1 (E ∩ W 1 ))
for every ϕ such that (2.2)-(2.6) hold.

Remarks 2.9. a. This definition is almost the same thing as Almgren's definition of a restricted set. The only difference is that Almgren does not require (2.2) or (2.3). That is, he only considers mappings ϕ 1 : U → U and does not care whether they come from deformations or not.

Our definition of quasiminimal sets is slightly less restrictive (because we add constraints on the mappings ϕ 1 ), but this is all right, because the regularity results are the same. In fact, in most regularity arguments for quasiminimal sets, one uses functions ϕ 1 such that W 1 ∪ ϕ 1 (W 1 ) is contained in a compact ball B contained in U and with diameter at most δ. Then ϕ 1 automatically comes from a one-parameter family of deformations as above, namely ϕ t (x) = (1t)x + tϕ 1 (x), for which W is contained in B by convexity. b. We require ϕ 1 to be Lipschitz (in (2.4)) only to make sure that (2.8) makes sense, but we do not require any Lipschitz bound for ϕ 1 . c. It is important that we do not require ϕ 1 to be injective. That is, we want to allow competitors obtained from E by gluing pieces together, or making entire regions collapse to one point. d. We can take δ = +∞ in Definition 2.7. Our definition of Al-minimality in Definition 1.6 corresponds to U = R n , δ = +∞, and M = 1. In addition, the accounting in (1.7) is not the same as in (2.8), but we shall see soon that in this case they are equivalent. In general δ and U give more flexibility in the definition; they may allow us to localize or account for boundary conditions. For instance, catenoids are (1, δ)-quasiminimal sets in R 3 , but only for δ small enough.

The quasiminimal sets of Definition 2.7 are not yet general enough for us, so we introduce an even larger class. This extension will be needed only because we want to work with different notions of almost-minimal sets; the uninterested reader may skip.

Definition 2.10. Let U and M be as above, and let δ 0 ∈ (0, +∞] and h ∈ [0, 1) be given. We shall denote by GAQ(M, δ 0 , U, h) the set of closed subsets E of U such that (2.1) holds and

(2.11)

H d (E ∩ W 1 ) ≤ M H d (ϕ 1 (E ∩ W 1 )) + hδ d
for every δ ≤ δ 0 and every ϕ such that (2.2)-(2.6) hold.

Elements of GAQ(M, δ 0 , U, h) will be called "generalized Almgren quasiminimal sets". The only difference with the quasiminimal sets from Definition 2.7 is that we added the error term hδ d in the right-hand side of (2.11). So h = 0 corresponds to the previous situation. We should always think that h is small, because in all the regularity results below, h has to be small enough, depending on n, d, and M . Also notice that the only situations where we expect a real difference between (2.11) and (2.8) (possibly with a larger M ) is when H d (E ∩ W 1 ) and H d (ϕ 1 (E ∩ W 1 )) are very small, i.e., when ϕ 1 only moves very few points. The main point of the discussion below is that these situations can be avoided when we prove regularity theorems.

We like quasiminimal sets because they have lots of good regularity properties. We want to say that these properties still hold for generalized quasiminimal sets (always assuming that h is small enough), and that the long proofs given elsewhere still work. So, for the rest of this section, we shall try to describe the regularity results that we want to use, and at the same time say why the proof for quasiminimal sets can be generalized rather easily.

Notice that if E is a quasiminimal set and Z is any closed set such that H d (Z) = 0, then E ∪ Z is also a quasiminimal set, with the same constants. For instance, in the situation of (2.8), we have that

H d ((E ∪ Z) ∩ W 1 ) = H d (E ∩ W 1 ) ≤ M H d (ϕ 1 (E ∩ W 1 )) ≤ M H d (ϕ 1 ((E ∪ Z) ∩ W 1 )
). This is why we introduce the (hopefully simpler) set E * below, and the notion of "reduced set". Definition 2.12. For each closed subset E of U , denote by E * = x ∈ E ; H d (E ∩ B(x, r)) > 0 for every r > 0 the closed support (in U ) of the restriction of H d to E. We say that E is reduced when

E * = E.
It is easy to see that (2.13)

H d (E \ E * ) = 0;
indeed, we can cover E \ E * by countably many balls B j such that H d (E ∩ B j ) = 0 (first reduce to E ∩ K, where K is compact and does not meet E * , and then use the definition of E * and the compactness of E ∩ K).

Remark 2.14. If E is quasiminimal, then E * is quasiminimal with the same constants. This is very easy, because when the ϕ t satisfy (2.2)-(2.6), H d (ϕ 1 (E\E * )) = H d (E\E * ) = 0 by (2.13) and because ϕ 1 is Lipschitz, so the conditions (2.8) and (2.11) are the same for E * as for E. Because of this, it is enough to study reduced quasiminimal sets. The same remark will apply to the various types of almost-minimal sets introduced in Section 4.

We start our list of regularity properties with the basic but useful local Ahlforsregularity of E * . Lemma 2.15. For each M ≥ 1 there exist constants h > 0 and C ≥ 1, that depend only on n, d, and M , such that if E ∈ GAQ(M, δ, U, h), x ∈ E * , and 0 < r < δ are also such that B(x, 2r) ⊂ U , then

(2.16) C -1 r d ≤ H d (E ∩ B(x, r)) ≤ Cr d .
See Proposition 4.1 in [DS] for standard quasiminimal sets, but note that this is only a reasonably minor modification of a result in [Al]. Now we should say a few words about why the proof of (2.16) in [DS] extends to generalized quasiminimal sets. For the upper bound, there is no serious problem; the idea of the proof is to suppose that for some cube Q, diam(Q) -d H d (E ∩ Q) is very large, and then compare E with a set obtained from E by a Federer-Fleming projection on d-dimensional skeletons of cubes inside Q. The numbers at stake in the argument are much larger than diam(Q) d , and the additional small error h diam(Q) d does not really change the estimates. More specifically, the first place where quasiminimality is used (in [DS]) is (4.5), and we should add an extra Ch diam(Q) d in the right-hand side there. [Let us point out to the reader that would wish to check things in [DS], that our standard quasiminimal set is called S there, and the quasiminimality constant M is called k.] The computations can stay the same, up to (4.9) and (4.11) where we should also add an extra Ch diam(Q) d in the right-hand side. Even with h = 1, this term is easily eaten up by the other term C2 j(n-d) diam(Q) d in the right-hand side of (4.9) and (4.11) (because j ≥ 1); the rest of the proof is the same.

The proof of the lower bound in (2.16) is probably the place where we should be the most careful, since the mass of E nearby could be very small and hδ d could be the main term in (2.10).

In the argument of [DS] (which begins a little below (4.21) there), we start with a cube Q, assume that diam(Q) -d H d (E ∩ Q) is very small, and then we construct a competitor F = ϕ(E) (with the constraints (2.2)-(2.6) with any δ > diam(Q)). The cube Q divides into a thin annular region A j (Q) near the boundary ∂Q, and the main part H j (Q) = Q \ A j (Q). Here j is an integer parameter, and the width of A j (Q) is roughly 2 -j diam(Q). In the region H j (Q), we do a Federer-Fleming projection on skeletons of cubes of size 2 -j diam(Q), and it turns out that if diam(Q) -d H d (E ∩ Q) is small enough, depending on j, the projection contains no full face of dimension d. This allows us to continue the Federer-Fleming projection one more step, up to (d-1)-dimensional faces. As a result, our final mapping ϕ is such that H d (ϕ(E∩H j (Q))) = 0. In the intermediate region A j (Q), we construct a mapping that matches the desired values on the two sides of ∂A j (Q). We do this so that

H d (ϕ(E ∩ A j (Q))) ≤ CH d (E ∩ A j (Q)). The comparison argument in [DS] yields H d (E ∩ H j (Q)) ≤ CM H d (E ∩ A j (Q))
, where M is the quasiminimality constant. See (4.23)-(4.25) in [DS], and the sentence that follows. Here it should be (2.17)

H d (E ∩ H j (Q)) ≤ CM H d (E ∩ A j (Q)) + Ch diam(Q) d ,
because of the extra term in (2.10). It is important that the constant C does not depend on j, so that we can still take j as large as we wish; the only price to pay is that we have to assume that diam(Q) -d H d (E ∩ Q) is accordingly small at the beginning of the argument, if we want our story about no full face in H j (Q), and then (2.17), to hold.

The way we can deduce the lower bound in (2.16) from this is slightly different from what was done in [DS]. First we observe that it is enough to prove this lower bound for

x ∈ E such that (2.18) lim sup r→0 r -d H d (E ∩ B(x, r)) ≥ c 0 ,
where c 0 is some positive dimensional constant. Indeed, if c 0 is chosen small enough, (2.18) holds for almost-every x ∈ E (see for instance [Ma], Theorem 6.2 on p.89, or even [D2], p.113). Then every point x ∈ E * is a limit of points x k ∈ E such that (2.18) holds, and we easily get the desired lower bound on H d (E ∩ B(x, r)) by using the lower bound that we can get for a smaller ball centered at some x k . Suppose that (2.18) holds, but not the lower bound in (2.16). Then we can find r > 0 such that r < δ, B(x, 2r) ⊂ U , but H d (E ∩ B(x, r)) < αr d , where α is the inverse of C in (2.16), and can be chosen as small as we want. Set η = 2 -d-3 . If α is sufficiently small compared to c 0 , we can pick a new radius r 1 < r such that

(2.19) H d (E ∩ B(x, r 1 )) ≤ αr d 1 but (2.20) H d (E ∩ B(x, ηr 1 )) ≥ α(ηr 1 ) d .
Indeed, we can try successively all the radii η k r, k ≥ 0 ; if none of them works, we can easily prove by induction that

H d (E ∩ B(x, η k r)) < α(η k r) d for every k > 0,
H d (E ∩ A j (Q)) ≤ C2 -j H d (E ∩ B(x, r 1 )) ≤ C2 -j αr d 1 .
Indeed we can easily find C -1 2 j cubes Q such that (2.21) holds, and for which the A j (Q) are disjoint, so (2.22) is just obtained by picking the one for which

H d (E ∩ A j (Q)) is the smallest. On the other hand, (2.23) H d (E ∩ H j (Q)) ≥ H d (B(x, ηr 1 )) ≥ αη d r
H d (E ∩ A j (Q)) ≤ C2 -j H d (E ∩ H j (Q)).
Recall that we can take j as large as we want. If we do so, and then take α small enough (to make sure that (2.17) applies to Q), the term CM H

d (E ∩ A j (Q)) in (2.17) is smaller than 1 2 H d (E ∩ H j (Q))
, and so (2.17) says that

H d (E ∩ H j (Q)) ≤ 2Ch diam(Q) d .
Recall that diam(Q) and r 1 are comparable, by (2.21). So if h is now chosen small enough, we get the desired contradiction with (2.23). This completes our proof of Lemma 2.15.

The next important property of quasiminimal sets is the local uniform rectifiability of E * . Let us not give a precise statement here, and refer to Theorem 2.11 in [DS] instead. We claim that this extends to E ∈ GAQ(M, δ, U, h), provided again that we take h small enough, depending on n, d, and M .

The proof in [DS] is rather long and complicated, but fortunately a lot of it consists in manipulations on Ahlfors-regular sets, and there are only a limited number of places where the quasiminimality of E is used.

In Chapter 5, it is used to get a Lipschitz mapping from E to R d , with big image. Quasiminimality is used in (5.7), but not in (5.8)-(5.10) (that is, only through Ahlforsregularity), so the effect of an additional Ch diam(Q) d in the right-hand side of (5.7) is just an extra Ch diam(Q) d in (5.11). If h is small enough, this is easily eaten up by the C -1 diam(Q) d in the left-hand side, and we can continue the argument with no further modification.

In Chapter 6 we show that if E ⊂ U ⊂ R n is a quasiminimal set and g : U → R l is Lipschitz, the new set S = {(x, y) ; x ∈ S and y = g(x)} ⊂ R n+l is a quasiminimal set too. Quasiminimality is used in (6.10), and our extra Ch diam(Q) d gets multiplied by a constant C (that depend also on the Lipschitz constant for g) in (6.12) and (6.13). Then we add things that do not depend on quasiminimality, and eventually we get an extra Ch diam(Q) d in (6.16). That is, S ∈ GAQ( M , δ, U , Ch), where U = U × R l , M is a constant that depends on M , n, d and the Lipschitz constant for g, and C also depends on these constants.

Chapter 7 only talks about general Ahlfors-regular sets, and Chapter 8 talks about how to deduce the local uniform rectifiability of E * from a main lemma, but only uses results from previous chapters, with no comparison argument. We can also relax through most of Chapter 9, because the first sections concern a deformation lemma that has nothing to do with quasiminimal sets. Even Section 9.2, whose goal is to apply the deformation lemma to quasiminimal sets, starts with a general discussion with dyadic cubes and Ahlforsregularity. Quasiminimality itself appears only on page 75, where a deformation φ is constructed. Thus φ in [DS] plays the role of ϕ 1 in (2.2)-(2.6), and W in [DS] is the same as W 1 here. Recall also that S and k there play the roles of E and M here.

By (9.92), W ∪ φ(W ) ⊂ B ′ , where B ′ is a ball of radius C 3 N + nN + C that is compactly contained in U (by (9.56)). In this argument, N is a very large integer, to be chosen near the end, and we shall see soon that for the present discussion we do not really need to know who the large constant C 3 is, because our small constant h can be chosen last. We can make sure that our set W from (2.5) is also contained in B ′ , because B ′ is convex (see Remark 2.9.a above).

The quasiminimality of E is used in (9.93), and we should add an extra h diam(B ′ ) ≤ h(C + C 3 )N d to the right-hand side, because of the additional term in (2.11). Then there are a few inclusions and identities, and (9.93) is not used until (9.103), where we should also add ChN d to the right-hand side (let us drop the dependence on C 3 and other constants).

Thus (9.103) should be replaced with (2.25)

H d (E ∩ W ) ≤ CM N d-1 + ChN d .
The computations after (9.103) can stay the same; they just use previous results to show that H d (E ∩ W ) ≥ C -1 N d (see (9.105) and two lines above it). The desired contradiction is then derived by choosing N large enough and, in our case, h small. This takes care of local uniform rectifiability.

We are also interested in Chapter 10 of [DS], where one shows that E * also has big projections locally. Quasiminimality shows up in (10.22), which should be replaced with (2.26)

H d (E ∩ W ) ≤ M H d (φ(E ∩ W )) + Chr d 1 .
But since we see (independently) in (10.25) that

H d (E ∩ W ) ≥ C -1 r d 1
, the extra term in (2.26) can be eaten up by the term in the left-hand side, and then one can proceed as in [DS].

This completes our discussion of the extension of results from [DS], and in particular of Theorem 2.11 there, to generalized quasiminimal sets in GAQ(M, δ, U, h). But we are not finished yet, because now we want to extend results from [D1], that concern limits of sequences of quasiminimal sets.

Limits of quasiminimal sets

We want to show that the Hausdorff measure goes to the limit well along bounded sequences of generalized quasiminimal sets; this will be used in the next section to prove the equivalence of various definitions, and also in later sections, although we shall only apply it to almost-minimal sets then.

Let U ⊂ R n , M ≥ 1, δ > 0 and h ≥ 0 be given, and let {E k } be a sequence in GAQ(M, δ, U, h). We shall assume that the E k are reduced, i.e, that E * k = E k (otherwise, we could always replace E k with E * k , which is a generalized quasiminimal set with the same constants by Remark 2.14).

We shall also assume that {E k } converges to a closed set E in U . By this we mean that for each compact set H ⊂ U ,

(3.1) lim k→+∞ d H (E, E k ) = 0,
where the local variant d H of the Hausdorff distance is defined by

(3.2) d H (E, F ) = sup{dist(x, F ) ; x ∈ E ∩ H} + sup{dist(x, E) ; x ∈ F ∩ H}.
We use the convention that sup{dist(x, F ) ; x ∈ E ∩ H} = 0 when E ∩ H is empty. By the way, if we did not assume the E k to be reduced, they could converge to essentially anything, because we could not keep track of the E k \E * k . Indeed it is easy to find sequences of finite sets that converge to any given closed set E.

Lemma 3.3. Let U, M, δ, h, {E k }, and E be as before, and assume that h is small enough, depending on n, d, and M . Then E is reduced,

(3.4) H d (E ∩ V ) ≤ lim inf k→+∞ H d (E k ∩ V ) for every open set V ⊂ U,

and

(3.5) E ∈ GAQ(M, δ, U, h) (i.e., with the same constants).

When h = 0, this is the main result of [D1]. Here we just have to say why the proof of [D1] goes through when h is small. As we shall see soon, E is locally Ahlfors-regular, so it is reduced. For (3.4) (i.e., Theorem 3.4 in [D1]), there is nothing to do: the proof just consists in putting together results from [DS] that we already discussed in Section 2. So we may now discuss the proof of (3.5) (Theorem 4.1 in [D1]).

The verification that E is locally Ahlfors-regular, with big pieces of Lipschitz graphs, does not need to be modified (it is again a mechanical consequence of the results of Section 2). The same comment applies to the remarks about rectifiability and the existence of tangent planes, which leads us near (4.13) in [D1].

Then we come to the verification of quasiminimality itself. Let us remind the reader of how it goes. We are given a deformation f : U → U , which can be written as f = ϕ 1 for a family of mappings ϕ t that satisfy (2.2)-(2.6), where (2.6) holds for some δ ′ ≤ δ, and we want to prove (2.11), i.e., that

(3.6) H d (E ∩ W f ) ≤ M H d (f (E ∩ W f )) + h(δ ′ ) d ,
where we set

W f = {x ∈ U ; f (x) = x}.
We know that the sets E k have the same property, but we cannot take limits directly, because

H d (f (E ∩W f )) could be much smaller than the H d (f (E k ∩ W f )).
A typical unfriendly situation where this may happen would be when f is not injective on E and sends two different pieces of E to the same set, while it is still injective on each of the E k . So we replace f with a different deformation h which does not have the same defect. That is, we try to force h to merge points systematically near the places where we know that it merges points of E. This is easier to do at places where E is close to a tangent plane P and f is close to a linear function A, because there we can compose f with the orthogonal projection onto A(P ). A good part of the argument of [D1] consists in doing this in many places, gluing the different functions that we construct, and checking that the places where we do the gluing or where we cannot control things do not matter too much, because the part of E that lives there has small measure.

Fortunately we can keep the construction of h as it is in [D1]; we just have to say how the final estimates go. Things start being dangerous near (4.94), when the function h is finally defined and we introduce its variant h 1 . Quasiminimality is used in (4.95), with the function h 1 and the set E k . So we add the usual Ch(δ ′ ) d to the right-hand side of (4.95), and get that

(3.7) H d (E k ∩ W h 1 ) ≤ M H d (h 1 (E k ∩ W h 1 )) + h(δ ′ ) d .
Then we quietly follow the computations, up to (4.107) which gives an upper bound on

H d (h 1 (E k ∩ W h 1 )).
We pick the various constants as explained after (4.107), so as to get the equivalent but simpler (4.108), and then prove that

(3.8) H d (h 1 (E k ∩ W h 1 )) ≤ H d (f (E ∩ W f )) + 2η 1 ,
where η 1 is as small as we want. In [D1], (3.8) follows from (4.108) (our upper bound for

H d (h 1 (E k ∩ W h 1 )
)) and a lower bound (4.109) for H d (f (E ∩ W f )). This bound itself is the direct consequence of Lemma 4.111. Unfortunately, we need to say a little more about the proof of Lemma 4.111, because it is very badly written. The author claims at the beginning of the proof that E is a quasiminimal set, and this is a rather stupid thing to say, because this is more or less what we try to prove. So the reader is asked not to pay attention to any of the remarks about quasiminimality, and follow quietly the construction of the sets S 1 (just the intersection of E with a ball B 1 ), then S 2 = {(x, f (x)) ; x ∈ S 1 } ⊂ R 2n , and then S 3 , which is the image of S 2 by some simple affine bijection ϕ of R 2n , whose goal is just to make some tangent plane horizontal. Then we prove (4.113), which says that S 3 stays very close to a horizontal d-plane in some ball. This part only uses the geometry of the construction, so it is all right.

Now replace E with E k ; this gives a new set S 3,k , and this set is a quasiminimal set, by the same argument that was falsely applied to E in [D1]; this involves constructions from [DS], in particular Chapter 5, which also apply to generalized quasiminimal sets, by Section 2. In addition, E k also satisfies the distance estimate (4.113) for k large, because the E k converge to E. So we can apply Lemma 2.18 in [D1] (or Chapter 10 of [DS]) to get that for k large, the projection of S 3,k on some d-plane contains some fixed ball. Then this is also true of S 3 , by compactness, and we can conclude as was claimed in [D1].

So (3.8) holds, and hence (3.9)

H d (E k ∩ W h 1 ) ≤ M H d (f (E ∩ W f )) + 2M η 1 + h(δ ′ ) d
by (3.7). We also know that (3.10)

H d (E ∩ W h 1 ) ≤ H d (E k ∩ W h 1 ) + η 1
for k large, by (3.4). Recall from (4.94) in [D1] and the line below it that W h 1 ⊂ W f , but we may choose the function ϕ in (4.94) so that the difference is as close to ∂W f as we want.

As a result, we can make H d (E ∩ W f \ W h 1 ) (incidentally, another typo in [D1]) as small as we want. When we combine this with (3.10) and (3.9), and then let small constants like η 1 tend to 0, we get that

(3.11) H d (E ∩ W f ) ≤ M H d (f (E ∩ W f )) + h(δ ′ ) d .
This is just the same as (2.11), and this completes our proof of Lemma 3.3.

We shall also need an upper semicontinuity result for H d , which unfortunately the author did not care to write down in [D1].

Lemma 3.12. Let U, M, δ, h, {E k }, and E be as in Lemma 3.3. In particular, assume that h is small enough, depending on d and M . Then

(3.13) lim sup k→+∞ H d (E k ∩ H) ≤ (1 + Ch)M H d (E ∩ H) for every compact set H ⊂ U.
Here C depends only on d and M .

This will be a consequence of the rectifiability of E. Since E is rectifiable, E has a d-dimensional "approximate tangent plane" P x at x for H d -almost every x ∈ E. See for instance [Ma], Theorem 15.19 on page 212. Since in addition E is locally (2.16) holds), P x is a true tangent plane at x. See for instance [D2], Exercise 41.21 on page 277. In addition,

(3.14) lim r→0 r -d H d (E ∩ B(x, r)) = c d for H d -almost every x ∈ E,
B = B(x, r) such that x ∈ E ′ ∩ H, 2r < δ, (3.15) B(x, 2r) ⊂ H,
where H is some compact subset of U that we choose so that H ⊂ int( H),

(3.16) H d (E ∩ ∂B) = 0, (3.17) dist(y, P x ) ≤ εr for y ∈ E ∩ B(x, 2r), and
(3.18)

|r -d H d (E ∩ B(x, r)) -c d | ≤ ε.
Notice that for x ∈ E ′ ∩ H, B(x, r) ∈ B for almost-every small enough r : (3.17) and (3.18) come from the definition of E ′ , and the set of small r such that (3.16) fails is at most countable, because H d (E) is locally finite. Thus the balls B, B ∈ B, form what is called a Vitali covering of E ′ ∩ H. Let us check that there is an at most countable collection {B i } i∈I in B such that the B i , i ∈ I, are disjoint and almost cover E ′ ∩ H in the sense that

(3.19) H d [E ′ ∩ H] \ i∈I B i = 0.
Let us define successive finite collections

{B i } i∈I k . Start with I 0 = ∅. If I k was already defined, set H k = [E ′ ∩ H] \ i∈I k B i .
For each x ∈ H k , there are lots of balls B ∈ B centered on x, and such that 5B does not meet the B i , i ∈ I k . This gives a covering of H k by balls of B; by the standard 5-covering lemma (see the first pages of [St]), we can find an at most countable subcollection

B i , i ∈ J k , such that the B i , i ∈ J k , are disjoint, and the 5B i , i ∈ J k , cover H k . Then (3.20) i∈J k H d (E ∩ B i ) ≥ C -1 i∈J k H d (E ∩ 5B i ) ≥ C -1 H d (H k ),
i∈J k H d (E ∩ B i ) ≤ H d (E ∩ H) < +∞
because the B i are disjoint and contained in H (by (3.15)), H can be covered by a finite collection of compact balls in U , and by (2.1). So we can pick a finite collection

J ′ k ⊂ J k , so that (3.22) i∈J ′ k H d (E ∩ B i ) ≥ 1 2 i∈J k H d (E ∩ B i ) ≥ (2C) -1 H d (H k ),
and then set

I k+1 = I k ∪ J ′ k .
This gives a definition of I k by induction, and then we take

I = ∪ k I k . The B i , i ∈ I, are disjoint by construction. Then (3.23) k i∈J ′ k H d (E ∩ B i ) ≤ H d (E ∩ H) < +∞,
by the proof (3.21). In particular, the series in k converges, and (3.22) says that by (3.16). This proves (3.19).

H d (H k ) tends to 0. But (3.24) H d [E ′ ∩ H] \ i∈I B i ≤ H d [E ′ ∩ H] \ i∈I k B i = H d [E ′ ∩ H] \ i∈I k B i = H d (H k ),
Let us cover [E ∩ H] \ i∈I B i by a collection of open balls B j , j ∈ J, such that diam(B j ) < δ, 2B j ⊂ U for each j, and

(3.25) j∈J r d j ≤ ε,
where r j denotes the radius of B j . Now the balls B i , i ∈ I ∪ J, cover E ∩ H, and by compactness we can find finite sets I ′ ⊂ I and

J ′ ⊂ J such that E ∩ H ⊂ i∈I ′ ∪J ′ B i .
Since the E k converges to E, we also get that by (2.16) and (3.25). Next pick i ∈ I ′ , and write B i = B(x, r). Notice that for k large,

(3.26) E k ∩ H ⊂ i∈I ′ ∪J ′ B i for k large enough, hence (3.27) H d (E k ∩ H) ≤ i∈I ′ ∪J ′ H d (E k ∩ B i ) ≤ i∈I ′ H d (E k ∩ B i ) + Cε,
(3.28) dist(y, P x ) ≤ 2εr for y ∈ E k ∩ B(x, 3r/2), by (3.17) and because the E k converge to E. Also recall that we only have a finite set I ′ of indices i, so for k large, (3.28) holds for all of them at the same time. In addition, (3.28) also holds if we replace P x with any plane P parallel to P x , and close enough to it. We want to use the quasiminimality of E k , so we define a Lipschitz deformation ϕ. Take

(3.29) ϕ(y) = y for y ∈ [R n \ B(x, r)] ∪ y ∈ R n ; dist(y, P ) ≥ 3εr ,
then call π the orthogonal projection onto P , and set (3.30) ϕ(y) = π(y) on the inside region D = y ∈ B(x, (1-ε)r) ; 0 ≤ dist(y, P ) ≤ 2εr .

On the remaining annular region A = y ∈ B(x, r) ; dist(y, P ) < 3εr \ D we choose ϕ so that ϕ(A) ⊂ B(x, r) and ϕ is 10-Lipschitz. For instance, take ϕ(x) = θ(y)y +(1-θ(y))π(y) for y ∈ A, with θ(y) = Min(1, (2εr)

′ ∈ (2r, δ)). That is, (3.31) H d (E k ∩ W 1 ) ≤ M H d (ϕ 1 (E k ∩ W 1 )) + (2r) d h. Set B ′ = B(x, (1 -ε)r) and notice that if y ∈ E k ∩ B ′
, then dist(y, P ) ≤ 2εr by (3.28), so ϕ(y) = π(y) by (3.30). Thus ϕ(y) = y if and only if y ∈ P , and

(3.32) E k ∩ W 1 ∩ B ′ = y ∈ E k ∩ B ′ ; 0 < dist(y, P ) ≤ 2εr = E k ∩ B ′ \ P.
There are at most countably many planes P parallel to P x and for which H d (E k ∩B(x, r)∩ P ) > 0, so we may choose P such that this is not the case. Then

(3.33) H d (E k ∩ B ′ ) = H d (E k ∩ W 1 ∩ B ′ ) ≤ M H d (ϕ(E k ∩ W 1 )) + (2r) d h ≤ M H d (ϕ(E k ∩ B ′ )) + M H d (ϕ(E k ∩ B(x, r) \ B ′ )) + (2r) d h ≤ M H d (ϕ(E k ∩ B ′ )) + 10 d M H d (E k ∩ B(x, r) \ B ′ ) + (2r) d h
by (3.32) and (3.31), because ϕ 1 = ϕ, and because

E k ∩ W 1 ⊂ E k ∩ B(x, r) ⊂ (E k ∩ B ′ ) ∪ (E k ∩ B(x, r) \ B ′ ) and ϕ is 10-Lipschitz. Next let us check that (3.34) H d (E k ∩ B(x, r) \ B ′ ) ≤ Cεr d ,
where C may depend on the quasiminimality constants, but not on ε. We can cover

P x ∩ B(x, r) \ B ′ by Cε -d+1 balls B(z j , εr) centered on P x ∩ B(x, r) \ B ′ . By (3.28), the B(z j , 3εr) cover E k ∩ B(x, r) \ B ′ . Notice also that the B(z j , 10εr) are contained in U , by (3.15). Then H d (E k ∩ B(z j , 3εr)) ≤ Cε d r
H d (E k ∩ B(x, r)) = H d (E k ∩ B ′ ) + H d (E k ∩ B(x, r) \ B ′ ) ≤ M H d (ϕ(E k ∩ B ′ )) + CH d (E k ∩ B(x, r) \ B ′ ) + (2r) d h ≤ M H d (ϕ(E k ∩ B ′ )) + Cεr d + (2r)
(y) = π(y) for y ∈ E k ∩ B ′ , so ϕ(E k ∩ B ′ ) ⊂ P ∩ B(x, r) and (3.36) H d (E k ∩ B(x, r)) ≤ M H d (P ∩ B(x, r)) + Cεr d + (2r) d h ≤ M c d r d + Cεr d + (2r) d h,
where c d still denotes the H d -measure of the unit ball in R d . On the other hand, (3.18) says that (3.37)

H d (E ∩ B(x, r)) ≥ c d r d -εr d .
Incidentally, we could also have obtained something like (3.37) directly from (3.17) and the facts that x ∈ E and E is a reduced quasiminimal sets as in (3.5), by the same proof as in Lemma 10.10 in [DS]. But (3.18) is not that hard to obtain anyway.

Recall that B i = B(x, r); when we put (3.36) and (3.37) together, we get that

(3.38) H d (E k ∩ B i ) ≤ M H d (E ∩ B i ) + Cεr d + (2r) d h ≤ (1 + Cε + Ch)M H d (E ∩ B i ).
Recall that the B i , i ∈ I ′ , are disjoint, so (3.39)

H d (E k ∩ H) ≤ i∈I ′ ∪J ′ H d (E k ∩ B i ) ≤ i∈I ′ H d (E k ∩ B i ) + Cε ≤ (1 + Cε + Ch)M i∈I ′ H d (E ∩ B i ) + Cε ≤ (1 + Cε + Ch)M H d E ∩ (∪ i∈I ′ B i ) +
H d (E k ∩ H) ≤ (1 + Cε + Ch)M H d (E ∩ V ) + Cε
for k large, by (3.39). Since ε is as small as we want,

(3.41) lim sup k→+∞ H d (E k ∩ H) ≤ (1 + Ch)M H d (E ∩ V ). Finally, H d (E∩V ) is as close to H d (E∩H) as we want, because V is any open neighborhood of E ∩ H and H d (E ∩ V ) < +∞ for some V . So (3.42) lim sup k→+∞ H d (E k ∩ H) ≤ (1 + Ch)M H d (E ∩ H),
and this completes our proof of Lemma 3.12.

Almost-minimal sets

For us, almost-minimal sets will be sets that look more and more like minimal sets when you look at them at smaller scales. There are a few different possible definitions, even in the same context as in Sections 2 and 3 where we only consider Almgren competitors. We shall give three of them. The first one (with A + ) is the closest to our definition of quasiminimal sets (Definition 2.7), and it has the advantage that we essentially do not need this section or many details of the previous two to use it. So the reader that would not be interested in subtleties about classes of almost-minimal sets is encouraged to have a look at Definition 4.1 and skip the rest.

The other two notions (with A and A ′ ) are weaker, and turn out to be equivalent to each other (see Proposition 4.10). For A, which is very close to our notion of generalized quasiminimal sets, we have a nice convergence result (Lemma 4.7). The definition for A ′ is a little bit simpler, and this is the definition that we used for Al-minimal sets (in Definition 1.6), so the equivalence between the two will make some of our statements a little simpler.

For all these definitions (with A + , A, and A ′ ), we shall measure closeness to minimality with a nondecreasing function h : (0, +∞) → [0, +∞], such that lim δ→0 h(δ) = 0. We shall call such a function a gauge function.

We start with a first notion which is reasonably close to Definition 2.7.

H d (E ∩ W 1 ) ≤ (1 + h(δ)) H d (ϕ 1 (E ∩ W 1 )),
where W 1 = {x ∈ U ; ϕ 1 (x) = x}, as in (2.5).

In other words, we require that for each δ > 0, E be an (1 + h(δ), δ)-quasiminimal set in U (as in Definition 2.7). The main difference is that M (δ) = 1 + h(δ) tends to 1 when δ tends to 0. Observe that we allowed the possibility that h(δ) = +∞ for δ ≥ δ 0 , to make it plain that in some cases we do not have any information coming from large scales. This is the same as saying that we only consider almost-minimality at scales δ ≤ δ 0 .

Definition 4.1 looks nice, but we shall prefer to use slightly weaker versions where the error term is allowed not to depend on the set W 1 where ϕ 1 (x) = x, but only on δ. The next definition is slightly simpler, but still uses the same sort of accounting method as in Definitions 2.7 and 4.1.

Definition 4.3. We say that the closed set E in U is an A-almost-minimal set in U , with gauge function h, if (2.1) holds and, for each δ > 0 and each family {ϕ t } 0≤t≤1 such that (2.2)-(2.6) hold,

(4.4) H d (E ∩ W 1 ) ≤ H d (ϕ 1 (E ∩ W 1 )) + h(δ)δ d . Again, W 1 = {x ∈ U ; ϕ 1 (x) =
dist(W 1 , R n \ U ) ≥ δ, because W 1 ⊂ U τ (by (2.6))
and we took δ < τ . Then H d (E ∩ W 1 ) ≤ Cδ d , by the local Ahlfors-regularity of E * (see Lemma 2.15, but we are still in the standard quasiminimal case). Here C depends on

M = 1 + h(δ), but this is all right. If H d (E ∩ W 1 ) ≤ H d (ϕ 1 (E ∩ W 1 )), then (4.4) holds stupidly. Otherwise, (4.6) H d (E ∩ W 1 ) ≤ (1 + h(δ))H d (ϕ 1 (E ∩ W 1 )) ≤ H d (ϕ 1 (E ∩ W 1 )) + h(δ)H d (E ∩ W 1 ) ≤ H d (ϕ 1 (E ∩ W 1 )) + Ch(δ)δ d , by (4.
2). This is (4.4), with h(δ) = Ch(δ).

The author does not know for sure that Definition 4.1 is strictly stronger than Definition 4.3, but it also seems difficult to prove that they are equivalent. This is why we spent some time dealing with generalized quasiminimal sets in the previous sections. Notice that E is an A-almost-minimal set in U , with gauge function h, if and only if it lies in the class GAQ(1, δ, U, h(δ)) of Definition 2.10 for every δ > 0 . If δ is small, h(δ) is as small as we want, and then we can apply all the results of Sections 2 and 3. In particular, we have the following consequence of Lemma 3.3. Lemma 4.7. Let {E k } be a sequence of A-almost-minimal sets in U , with the same gauge function h. Assume that each E k is reduced (i.e., that E * k = E k ), and that the sequence {E k } converges to the closed set E in U (i.e., (3.1) holds for each compact H ⊂ U ). Then E is a reduced A-almost-minimal set in U , with the same gauge function h.

To be fair, since Lemma 3.3 needs h to be small enough, it seems that we should replace h(δ) with +∞ for δ large, but in fact we do not even need to do this. We can apply Lemma 3.3 with δ small, and this gives (3.4) and all the good local properties of E and the E k , which we can then use as in the proof of Lemma 3.3 to give the desired result even when h(δ) is large. But anyway it makes no real difference, because we shall not be able to use the conclusion when h(δ) is large.

Let us now give a variant of Definition 4.3 where the accounting is done more simply, as in Definitions 1.4 and 1.6. Definition 4.8. Let E be a closed subset of U such that (2.1) holds. We say that E is an A ′ -almost-minimal set in U , with gauge function h, if for each δ > 0 and each family {ϕ t } 0≤t≤1 such that (2.2)-(2.6) hold, (4.9)

H d (E \ F ) ≤ H d (F \ E) + h(δ)δ d ,
were we set F = ϕ 1 (E).

When E and F coincide out of some compact ball B ⊂ U , (4.9) simply says that

H d (E ∩ B) ≤ H d (F ∩ B) + h(δ)δ d ;
we wrote (4.9) the way we did just to avoid mentioning the existence of such a B. Notice that the Al-minimal sets of Definition 1.6 are the same as A ′ -almost-minimal sets in R n , with gauge function h = 0. Definition 4.8 looks simpler, but is in fact equivalent to Definition 4.3. Proposition 4.10. Let E be a closed subset of U such that (2.1) holds. Then E is an A-almost-minimal set in U with gauge function h if and only if E is an A ′ -almost-minimal set in U with (the same) gauge function h.

First suppose that E is an A ′ -almost-minimal set, and let {ϕ t } satisfy (2.2)-(2.6). We know (4.9) and we want to check (4.4).

Set by (4.11) and (4.9). Observe that

E 1 = E ∩ W 1 , E 2 = E \ E 1 , and F = ϕ 1 (E). Then (4.11) H d (E ∩ W 1 ) = H d (E 1 ) = H d (E 1 ∩ F ) + H d (E 1 \ F ) trivially. Now E 2 ⊂ F because ϕ 1 (x) = x on E 2 , so E 1 \ F = E \ F . Then (4.12) H d (E ∩ W 1 ) = H d (E 1 ∩ F ) + H d (E \ F ) ≤ H d (E 1 ∩ F ) + H d (F \ E) + h(δ)δ d ,
F = ϕ 1 (E) = ϕ 1 (E 1 ) ∪ ϕ 1 (E 2 ) and ϕ 1 (E 2 ) = E 2 ⊂ E because ϕ 1 (x) = x on E 2 . Then E 1 ∩ F = E 1 ∩ [ϕ 1 (E 1 ) ∪ ϕ 1 (E 2 )] = E 1 ∩ ϕ 1 (E 1 ) because E 1 ∩ ϕ 1 (E 2 ) = E 1 ∩ E 2 = ∅. Similarly, F \ E = ϕ 1 (E 1 ) \ E because ϕ 1 (E 2 ) ⊂ E. Thus (4.12) says that (4.13) H d (E ∩ W 1 ) ≤ H d (E 1 ∩ ϕ 1 (E 1 )) + H d (ϕ 1 (E 1 ) \ E) + h(δ)δ d ≤ H d (E 1 ∩ ϕ 1 (E 1 )) + H d (ϕ 1 (E 1 ) \ E 1 ) + h(δ)δ d = H d (ϕ 1 (E 1 )) + h(δ)δ d = H d (ϕ 1 (E ∩ W 1 )) + h(δ)δ d ,
as needed for (4.4). This proves our first inclusion. Now let us assume that E is an A-almost-minimal set, and prove that it is an A ′almost-minimal set with the same gauge function. Again let {ϕ t } satisfy (2.2)-(2.6); we would like to show that (4.9) holds. We cannot deduce this so easily from (4.4), because if we are too unlucky, ϕ 1 (E ∩ W 1 ) may have a big part in E \ W 1 , which is accounted for in (4.4) (and helps it hold) but not in (4.9). This may happen if W 1 is somewhat smaller than its image; if this is the case, we shall need to modify the ϕ t a bit, to make W 1 larger before we apply (4.4).

Set ϕ = ϕ 1 to save notation. We want to replace ϕ with functions ϕ k defined by

(4.14) ϕ k (x) = ϕ(x) + ε k ψ(ϕ(x)) v,
where v is a given unit vector in R n , {ε k } is a sequence that tends to 0, and ψ : U → R is a correctly chosen smooth function with compact support in U . Recall that there is a relatively compact set W ⊂ U , defined by (2.5) and that satisfies (2.6). Let S and S ′ be compact subsets of U such that (4.15) W ⊂ S ⊂ int(S ′ ) ⊂ S ′ ⊂ U and diam(S ′ ) < δ;

other conditions may show up later. We choose ψ so that (4.16) 0 ≤ ψ(x) ≤ 1 everywhere, (4.17) ψ(x) = 1 for x ∈ S, and ψ(x) = 0 for x ∈ U \ S ′ .

Since we want to apply (4.4) to ϕ k , we need to check that it is the endpoint of an acceptable family ϕ k,t , 0 ≤ t ≤ 1. Set

(4.18) ϕ k,t (x) = ϕ t (x) + tε k ψ(ϕ t (x))v.
The conditions (2.2)-(2.4) clearly hold for this family, so we just have to check (2.6). Set W k,t = {x ∈ U ; ϕ k,t (x) = x}. Then W k,t ⊂ S ′ , because if x ∈ S ′ , (4.15) says that ϕ t (x) = x, and then (4.17) and (4.18) say that ϕ k,t (x) = x. Let us also check that

(4.19) ϕ k,t (W k,t ) ⊂ S ′ k =: y ; dist(y, S ′ ) ≤ ε k . Notice that (4.20) |ϕ k,t -ϕ t (x)| ≤ ε k for x ∈ U,
H d (E ∩ W k,1 ) ≤ H d (ϕ k (E ∩ W k,1 )) + h(δ)δ d .
Now we shall cut our sets into pieces to control the overlaps and eventually get something that looks like (4.9). What comes out of the small annular region A = S ′ \ S will be estimated brutally, and later thrown out to the trash, so we shall concentrate on

E 1 = E ∩S and E 2 = E \ S ′ .
First observe that ϕ k (x) = x for x ∈ E 2 , by (4.15) and (4.17). Also, ϕ(E 1 ) ⊂ S because if ϕ(x) = x then x ∈ W 1 and ϕ(x) ∈ S by (4.15). Consequently, ϕ k (E 1 ) ⊂ S ′ for k large enough, by (4.20). In particular,

ϕ k (E 1 ) does not meet ϕ k (E 2 ) = E 2 . Observe also that W k,1 does not meet E 2 = E \ S ′ . Thus (4.22) H d (ϕ k (E ∩ W k,1 )) = H d (ϕ k (E ∩ W k,1 ∩ S ′ )) ≤ H d (ϕ k (E 1 ∩ W k,1 )) + H d (ϕ k (E ∩ A)) = H d (ϕ(E 1 ∩ W k,1 )) + H d (ϕ k (E ∩ A)),
where the last equality comes from the fact that ϕ kϕ is constant on E 1 , by (4.17) and because we just said that ϕ(E 1 ) ⊂ S.

It is not necessarily true that E 1 ⊂ W k,1 , but the difference is small: we claim that (4.23)

E 1 \ W k,1 ⊂ Z k =: x ∈ E 1 ; 0 < |ϕ(x) -x| ≤ ε k .
Indeed let x ∈ E 1 be given, and assume that it does not lie in the set on the right-hand side. If ϕ(x) = x, then ϕ k (x) = x because ϕ(x) ∈ S and (4.17) says that ψ(ϕ(x)) = 1.

H d (E 1 ) = H d (E 1 ∩ W k,1 ) + H d (E 1 \ W k,1 ) ≤ H d (ϕ k (E ∩ W k,1 )) + h(δ)δ d + H d (Z k ) ≤ H d (ϕ(E 1 ∩ W k,1 )) + H d (ϕ k (E ∩ A)) + h(δ)δ d + H d (Z k ) ≤ H d (ϕ(E 1 )) + H d (ϕ k (E ∩ A)) + h(δ)δ d + H d (Z k )
by (4.21) and (4.23), and then (4.22). To get closer to (4.9), observe that (4.25)

H d (E \ ϕ(E)) = H d (E ∩ S ′ \ ϕ(E)) = H d (E ∩ S ′ ) -H d (E ∩ S ′ ∩ ϕ(E)) because E 2 = E \ S ′ is contained in ϕ(E). Similarly, (4.26) H d (ϕ(E) \ E) ≥ H d (ϕ(E) ∩ S ′ \ E) = H d (ϕ(E) ∩ S ′ ) -H d (E ∩ S ′ ∩ ϕ(E)). Set ∆ = H d (E \ F ) -H d (F \ E) = H d (E \ ϕ(E)) -H d (ϕ(E) \ E)
. This is the quantity that we want to majorize, and we just proved that

(4.27) ∆ ≤ H d (E ∩ S ′ ) -H d (ϕ(E) ∩ S ′ ).
Recall that ϕ(E 1 ) ⊂ S (see four lines above (4.22)), so ϕ(E) ∩ S ′ contains ϕ(E 1 ) and (4.27) implies that

(4.28) ∆ ≤ H d (E ∩ S ′ ) -H d (ϕ(E 1 )) = H d (E 1 ) + H d (E ∩ A) -H d (ϕ(E 1 )), because E ∩ S ′ = (E ∩ S) ∪ (E ∩ S ′ \ S) = E 1 ∪ (E ∩ A) by definitions. Then (4.24) yields (4.29) ∆ ≤ H d (E ∩ A) + H d (ϕ k (E ∩ A)) + h(δ)δ d + H d (Z k ).
Notice that ϕ k = h k • ϕ, where h k is Lipschitz, with a Lipschitz constant that stays below 2 when k is large enough. Then for k large,

H d (ϕ k (E ∩ A)) ≤ C ϕ H d (E ∩ A)
, with a constant C ϕ that depends on ϕ but not k or ψ. On the other hand, when ε k tends to 0, the set Z k in (4.23) decreases to the empty set, and so lim k→+∞ H d (Z k ) = 0. Thus (4.29) yields

(4.30) ∆ ≤ (1 + C ϕ )H d (E ∩ A) + h(δ)δ d .
Recall that A = S ′ \ S, where we only required that S and S ′ be compact and satisfy (4.15). We can do this so that H d (E ∩ A) is as small as we want, and so ∆ ≤ h(δ)δ d , as needed for (4.9). Thus E is an A ′ -almost-minimal set, with the same gauge function h as for its A-almost-minimality. This completes our proof of Proposition 4.10.

B. MONOTONICITY OF DENSITY AND APPROXIMATION BY CONES

Density is nearly nondecreasing

In this section we show that if E is a minimal set of dimension d in R n , then, for each choice of x ∈ R n , r -d H d (E ∩ B(x, r)) is a nondecreasing function of r. We also give a local version of this, and generalizations to almost-minimal sets with sufficiently small gauge functions. See Propositions 5.16, 5.24, and 5.30 below. Incidentally, all this is supposed to be very classical, and the general idea will be to compare E with a cone. We do not do this directly because the cone is not exactly a competitor, and to avoid (probably minor) problems of definitions or regularity. Instead we first prove an integrated version.

Lemma 5.1. Let U ⊂ R n be open, and let E be a reduced A-or A ′ -almost-minimal set in U , with gauge function h. Let x ∈ E and 0 < ρ 1 < ρ 2 be such that B(x, ρ 2 ) ⊂ U . Then (5.2)

H d (E∩B(x, ρ 1 )) ≤ h(2ρ 2 )(2ρ 2 ) d + 1 ρ 1 -ρ 2 E∩B(x,ρ 2 )\B(x,ρ 1 ) r(y) cos θ(y) d dH d (y),
where we set r(y) = |y -x|, and θ(y) ∈ [0, π/2] denotes the (smallest) angle between the radius [x, y] and the tangent d-plane to E at y.

See Definition 4.3 or 4.8 for almost-minimal sets. Also recall that E is rectifiable, by Theorem 2.11 in [DS] (generalized as in Section 2). The rectifiability of E gives an approximate tangent plane to E at y for H d -almost every y ∈ E, which is enough to define θ(y). But here E is locally Ahlfors-regular too, so the approximate tangent plane is even a true tangent d-plane (for instance by Exercise 41.21 on page 277 of [D2]).

To prove the lemma, we may assume that x = 0. Let 0 < r 1 < r 2 be such that B(0, r 2 ) ⊂ U ; we want to compare E with the competitor F = φ(E), where φ is the radial function defined by φ(rξ) = ϕ(r) ξ for r ≥ 0 and ξ ∈ ∂B(0, 1), and where ϕ is the piecewise linear function such that (as suggested by Figure 5.1)

(5.3) ϕ(r) = 0 for 0 ≤r ≤ r 1 , ϕ(r) = r 2 (r -r 1 ) r 2 -r 1 for r 1 ≤ r ≤ r 2 ,
and ϕ(r) = r for r ≥ r 2 .

Note that the one-parameter family φ t defined by φ t (y) = (1t)y + tφ(y) satisfies (2.2)-(2.6) with any δ > 2r 2 . Here W 1 = B(0, r 2 ) \ {0}, and (4.4) says that

(5.4)

H d (E ∩ B(0, r 2 )) ≤ H d (φ(E ∩ B(0, r 2 ))) + h(δ)δ d .
Notice that (4.9) yields exactly the same thing, after adding H d (E ∩ φ(E) ∩ B(0, r 2 )), so we don't need to worry about which definition of almost-minimality we take. We shall remember that (5.4) holds for all δ > 2r 2 .

Next we need to evaluate H d (φ(E ∩ B(0, r 2 )). Set A = B(0, r 2 ) \ B(0, r 1 ). We just need to evaluate H d (φ(E ∩ A)), because E ∩ B(0, r 1 ) is mapped to the origin.

We shall use the rectifiability of E. Let y ∈ E be given, and assume that E has a tangent d-plane at y. Denote by P this d-plane and by P ′ the vector d-plane parallel to P . Also set r(y) = |y| and y = y/r(y).

We want to compute how the differential Dφ(y) acts on vectors. In the radial direction, we just use the fact that for t > 0 such that t y ∈ A, φ(t y) = ϕ(t) y = r 2 (tr 1 ) r 2r 1 y, by (5.3), so Dφ(y) • y = [r 2 /(r 2r 1 )] y. For vectors w orthogonal to y, (5.3) yields Dφ(y) • w = [ϕ(r(y))/r(y)]w. If P is orthogonal to [0, y] at y, Dφ(y) acts on P ′ by multiplication by ϕ(r(y))/r(y). Otherwise, denote by p the orthogonal projection on P of the origin, and set v = (y-p) |y-p| ∈ P ′ . [See Figure 5.2.] By definition, θ(y) is the angle of [0, y] with v, so v = cos θ(y) y + sin θ(y) w for some unit vector w orthogonal to y. Then Dφ(y)•v = [r 2 /(r 2 -r 1 )] cos θ(y) y+ [ϕ(r(y))/r(y)] sin θ(y) w and |Dφ(y) • v| = β(y), where we set (5.5) β(y) = [r 2 /(r 2r 1 )] 2 cos 2 θ(y) + [ϕ(r(y))/r(y)] 2 sin 2 θ(y)

1/2
.

We are ready to apply the area formula to compute H d (φ(E ∩ A)) (in other words, we cover E ∩ A with disjoint measurable subsets of C 1 surfaces, plus a set of vanishing H d measure, and then compute the Hausdorff measure of images on each C 1 piece with the derivative of φ; we don't need to worry about overlaps, since φ is bijective on A). We get that

(5.6)

H d (φ(E ∩ B(0, r 2 ))) = H d (φ(E ∩ A)) = E∩A [ϕ(r(y))/r(y)] d-1 β(y)dH d (y),
where the extra [ϕ(r(y))/r(y)] d-1 comes from the directions of P ′ that are orthogonal to v. When P is orthogonal to [0, y] at y, (5.6) also gives the right factor [ϕ(r(y))/r(y)] d , because θ(y) = π/2. We plug this back into (5.4) and get that (5.7)

H d (E ∩ B(0, r 2 )) ≤ h(δ)δ d + E∩A [ϕ(r(y))/r(y)] d-1 β(y)dH d (y).
It is tempting to let r 2 -r 1 tend to 0 now, but we decided to take an average before we do this, and this is what will lead to (5.2). So we fix 0 < ρ 1 < ρ 2 such that B(0, ρ 2 ) ⊂ U , pick a small η > 0, and for each t ∈ (ρ 1 , ρ 2 ), we apply (5.7) with r 2 = t, r 1 = tη, and δ = 2ρ 2 . [Recall that we can take any δ > 2r 2 in (5.4) and (5.7).] Then we average over t and get that (5.8)

H d (E ∩ B(0, ρ 1 -η)) ≤ h(2ρ 2 )(2ρ 2 ) d + 1 ρ 2 -ρ 1 I(η)
where (5.9)

I(η) = t∈(ρ 1 ,ρ 2 ) y∈E ; t-η≤r(y)≤t [ϕ t (r(y))/r(y)] d-1 β t (y)dH d (y)dt,
with obvious notations concerning ϕ t and β t . By Fubini,

(5.10)

I(η) = y∈E ; ρ 1 -η<r(y)<ρ 2 g η (y) dH d (y), with (5.11) g η (y) = t∈(ρ 1 ,ρ 2 )∩[r(y),r(y)+η] [ϕ t (r(y))/r(y)] d-1 β t (y)dt.
First observe that here r 2 /(r 2r 1 ) = t/η. Then (5.5) yields

(5.12) β t (y) = (t/η) 2 cos 2 θ(y) + [ϕ t (r(y))/r(y)] 2 sin 2 θ(y)

1/2 ≤ (t/η) 2 cos 2 θ(y) + 1 1/2 ≤ (t/η) cos θ(y) + 1 because ϕ t (r(y)) ≤ r(y) and a 2 + b 2 ≤ (a + b) 2 . This yields (5.13)

β t (y) ≤ η -1 t cos θ(y) + 1 ≤ η -1 [(r(y) + η) cos θ(y) + η]
on the domain of integration in (5.11). Also

(5.14)

ϕ t (r(y)) r(y) = r 2 r(y) r(y) -r 1 r 2 -r 1 = t r(y) r(y) -t + η η ≤ η -1 r(y) + η r(y) r(y) -t + η ,
by (5.3) and because t ≤ r(y) + η on the domain of integration, so

g η (y) ≤ η -d [(r(y) + η) cos θ(y) + η] r(y) + η r(y) d-1 t∈[r(y),r(y)+η] (r(y) -t + η) d-1 dt = 1 d [(r(y) + η) cos θ(y) + η] r(y) + η r(y) d-1 ≤ 1 d r(y) cos θ(y) + Cη,
where C is allowed to depend on many things, but not on η. We return to (5.10) and get that I(η) ≤ 1 d y∈E ; ρ 1 -η<r(y)<ρ 2 [r(y) cos θ(y) + Cη] dH d (y). Then (5.8) becomes

(5.15)

H d (E∩B(0, ρ 1 -η)) ≤ h(2ρ 2 )(2ρ 2 ) d + 1 ρ 2 -ρ 1 1 d y∈E ; ρ 1 -η<r(y)<ρ 2 [r(y) cos θ(y) + Cη] dH d (y).
Then we let η tend to 0 and get (5.2). This completes our proof of Lemma 5.1.

Remark. The deformation φ that we used in the proof of Lemma 5.1 is not a bijection, but we could have used the homeomorphism φ t (y) = (1t)y + tφ(y), with t < 1 very close to 1, and then go to the limit and get (5.2) anyway. So Lemma 5.1, and Propositions 5.16 and 5.30 below, also hold when E lies in the larger class of almost-minimal sets where the deformations in (2.2)-(2.6) are also required to be homeomorphisms, at least if we assume that E is rectifiable. I do not think that Proposition 5.24 below extends to this context (we shall also use local Ahlfors-regularity there).

We want to use Lemma 5.1 to get monotonicity results for r -d H d (E ∩ B(x, r)). We start with the easier case of minimal sets.

Proposition 5.16. Let E be a minimal set in U . This means that E is almost-minimal in U , with gauge function h = 0 (and Definitions 4.1, 4.3, and 4.8 are all equivalent in this case). Let x ∈ U and R > 0 be such that

B(x, R) ⊂ U . Then θ(x, r) = r -d H d (E ∩B(x, r)) is a nondecreasing function of r on (0, R].
Let B(x, R) be as in the statement, and define a locally finite Borel measure µ on [0, R) by (5.17)

µ(A) = H d (E ∩ π -1 (A)) for every Borel set A ⊂ [0, R),
where π is the radial projection defined by π(y) = |x -y|. Then decompose µ into its absolutely continuous part µ a and its singular part µ s , and denote by a(r) the density of µ a with respect to the Lebesgue measure. The Lebesgue density theorem allows us to compute a(r) at almost-every point r ∈ [0, R) as a limit of densities, i.e.,

(5.18) a(r) = lim

t→0 + 1 t µ([r -t, r)) = lim t→0 + 1 t H d E ∩ B(x, r) \ B(x, r -t) .
Similarly a(r

) = lim t→0 + 1 t H d (E ∩ B(x, r + t) \ B(x, r)) for almost-every r. Consequently, if we set (5.19) l(r) = H d (E ∩ B(x, r)) for 0 ≤ r < R,
then l(r) is differentiable almost-everywhere on (0, R), and l ′ (r) = a(r) almost-everywhere.

Let us only assume for the moment that E is almost-minimal, and apply Lemma 5.1 with ρ 2 = r ∈ (0, R) and ρ 1 = rt. We can even drop cos θ(y) and get that

(5.20) l(r -t) ≤ (2r) d h(2r) + r td H d (E ∩ B(x, r) \ B(x, r -t)).
Then we let t tend to 0, use (5.18), and get that (5.21) l(r) ≤ (2r

) d h(2r) + r d a(r) = (2r) d h(2r) + r d l ′ (r) for almost-every r ∈ (0, R).
Here E is even a minimal set, so (5.21) says that l(r) ≤ r d l ′ (r). Recall that l(r) = [0,r) dµ; we claim that for 0 < a < b < R,

(5.22) θ(x, b) -θ(x, a) = [a,b) t -d dµ(t) -d [a,b) t -d-1 l(t)dt.
Set g(t) = t -d ; thus θ(x, r) = g(r)l(r), and we just want to show that the derivative of the product gl is gdµ + g ′ l. Let 0 < a < b < R be given, and apply Fubini's theorem to

I = a≤s≤t<b g ′ (s)dsdµ(t). One computation gives I = [a,b) [g(t) -g(a)
]dµ(t), and the other one yields

I = [a,b) g ′ (s)[l(b) -l(s)]ds. Then [a,b) g(t)dµ(t) + [a,b) g ′ (s)l(s)ds = I + g(a)
[a,b)

dµ(t) + -I + l(b) [a,b) g ′ (s)ds = g(a)(l(b) -l(a)) + l(b)(g(b) -g(a)) = g(b)l(b) -g(a)l(a). (5.23)
This is just (5.22); note that (5.23) would work with any smooth function g.

We may now conclude. Recall that t

-d dµ(t) ≥ t -d l ′ (t)dt ≥ d t -d-1 l(t)dt because dµ(t) ≥ dµ a (t) = l ′ (t)
dt and by (5.21) with h(2r) = 0, so (5.22) says that θ(x, •) is nondecreasing, and Proposition 5.16 follows.

Proposition 5.24. There is a constant λ > 1, that depends only on n and d, such that the following holds. Suppose that E is an A-or A ′ -almost-minimal set in U , with gauge function h. Let x ∈ E * ∩ U and R > 0 be such that B(x, R) ⊂ U , and also assume that h(R) is small enough (depending only on n and d) and that R 0 h(2r) dr r < +∞. Then

(5.25) θ(x, r) e λA(r) is a nondecreasing function of r ∈ (0, R),

where we set

A(r) = r 0 h(2t) dt r for 0 < r < R.
Notice that here we had to take x ∈ E * ∩ U to get the conclusion (Recall from Definition 2.12 that E * is the closed support of H d |E ). Since lim r→0 A(r) = 0, e λA(r) tends to 1 and can be seen as a mild penalty that measures a potential defect in monotonicity.

We start the proof as in Proposition 5.16, but we multiply l(t) with g(t) = t -d e λA(t) . By (5.23), the distribution derivative of g(t)l(t) is g(t)dµ(t) + g ′ (t)l(t)dt. Also, dµ(t) ≥ dµ a (t) = a(t)dt = l ′ (t)dt, so

(t) ≥ d t [l(t) -(2t) d h(2t)]
, so it is enough to check that (2t) d ≤ λl(t). This comes from Lemma 2.15; we just have to take λ ≥ 2 d C, where C is as in (2.16). This is the reason why we have to take x ∈ E * and assume that h(R) is small enough.

The case of A + -almost-minimal sets is a little easier, because we may replace the error term h(δ)δ d in (5.4) with h(δ)H d (φ(E ∩ B(0, r 2 )). [We use (4.2) instead of (4.4).] When we follow the computations, we see that we can replace (5.2) with (5.28)

H d (E ∩ B(x, ρ 1 )) ≤ 1 + h(2ρ 2 ) ρ 1 -ρ 2 E∩B(x,ρ 2 )\B(x,ρ 1 ) r(y) cos θ(y) d dH d (y).
Then (5.21) may be replaced with

(5.29) l(r) ≤ [1 + h(2r)] r d l ′ (r) for almost-every r ∈ (0, R),
and (5.27) allows us to show that θ(x, r)e λA(r) is nondecreasing as soon as [1 +h(2t)] -1 d t -

d t + λ h(2t) t ≥ 0, i.e., when λ ≥ d {1 -[1 + h(2t)] -1 }.
This proves the following intermediate statement for A + -almost-minimal sets, where the interest is that we no longer need to take a center in E * . Proposition 5.30. For each λ > 0 we can find τ > 0, that depends only on n, d, and λ, with the following property. Let E be an A + -almost-minimal set in U , with gauge function h. Let x ∈ U and R > 0 be such that B(x, R) ⊂ U , and suppose that h(2R) ≤ τ and R 0 h(2r) dr r < +∞. Set A(r) = r 0 h(2r) dt t for 0 < r < R. Then

(5.31) θ(x, r)e λA(r) is a nondecreasing function of r ∈ (0, R).

Remark 5.32. It is tempting to try to prove Propositions 5.16, 5.24, and 5.30 without using the rectifiability of E. It would be enough to prove that (5.33)

H d (E ∩ B(x, ρ 1 )) ≤ h(2ρ 2 )(2ρ 2 ) d + 1 ρ 1 -ρ 2 E∩B(x,ρ 2 )\B(x,ρ 1 ) r(y) d dH d (y)
(or the corresponding variant in the case of Proposition 5.30) instead of (5.2) (i.e., without the cosine), and then continue as before. To do so, we would need to change the argument near (5.5), when we compute the effect of Dφ on the tangent plane. We observed that Dφ(y).v = [r 2 /(r 2 -r 1 )]v in the radial direction and Dφ(y).v = [ϕ(r(y))/r(y)]v in the other direction, so we expect that φ will at most multiply the Hausdorff measure of E by [r 2 /(r 2r 1 )][ϕ(r(y))/r(y)] d-1 at y, with the effect of replacing β(y) in (5.5) with [r 2 /(r 2r 1 )] and eventually leading to (5.33). But the author is not aware of a generalization to unrectifiable sets of the obvious estimate on Hausdorff measure of affine images (of rectifiable sets) that would allow to conclude here. We could also try to replace the Hausdorff measure with a more friendly variant. Anyway, this may not be so useful, because the cosine that shows up in Lemma 5.1 will really be used in the next section.

Minimal sets with constant density

In addition to the monotonicity of density proved in the last section, it will be useful to know that when E is a minimal set and (6.1)

θ(r) = r -2 H d (B(0, r))
is constant on some interval (a, b), then E * coincides on B(0, b) \ B(0, a) with a minimal cone (i.e., a minimal set which is a cone) centered at the origin.

(6.3) θ(r) = θ 0 for a < r < b.
Then there is a reduced minimal cone C centered at the origin such that

(6.4) E ∩ [B(0, b) \ B(0, a)] = C ∩ [B(0, b) \ B(0, a)].
Recall that our two definitions of minimality are equivalent. We shall not need to know this, because it will be just as simple to prove the theorem for both notions. Of course Theorem 6.2 also applies if E is minimal in a domain that contains B(0, b).

Observe also that if we did not assume that E is reduced, we would still get that

E * ∩ [B(0, b) \ B(0, a)] = C ∩ [B(0, b) \ B(0, a)
] (instead of (6.4)); this is easy, because E * is a reduced minimal set, with the same density, by Remark 2.14 and (2.13), so we can apply Theorem 6.2 to it.

Let E be as in the theorem. We first look at Proposition 5.16 to see if we can derive extra information because (6.3) holds. Set A = B(0, b) \ B(0, a) = π -1 ((a, b)), where π denotes the radial projection defined by π(y) = |y|. We want to prove that (6.5) cos θ(y) = 1 for

H d -almost every y ∈ E ∩ A,
where cos θ(y) is as in Lemma 5.1. Let ν be the finite Borel measure on (a, b) defined by (6.6) ν(S) =

E∩π -1 (S) cos θ(y)dH 2 (y)

for S ⊂ (a, b). Notice that ν ≤ µ, where µ is as in (5.17) and corresponds to replacing cos θ(y) with 1. By (6.3), µ([α, β)) = θ 0 (βα) for a < α < β < b, so dµ = θ 0 dx on (a, b). Then ν is absolutely continuous too. Let f denote its density (with respect to the Lebesgue measure). Notice that f (r) = lim t→0 + 1 t ν([rt, r)) almost-everywhere, as in (5.18). We apply Lemma 5.1 with ρ 2 = r ∈ (a, b) and ρ 1 = rt, as we did for (5.20), and we get that

(6.7) l(r -t) ≤ r dt H d (E ∩ B(x, r) \ B(x, r -t))
instead of (5.20) (just because h = 0 here). Then we let t tend to 0 and get that l(r) ≤ r d f (r) (instead of (5.21)). We also have that f (r) ≤ l ′ (r), because l ′ (r) is the density of µ and ν ≤ µ. In addition, l(r) = r d θ(r) = r d θ 0 by (6.1) and (6.3), so l ′ (r) = dr d-1 θ 0 = dl(r)/r. Altogether (6.8)

l(r) ≤ r d f (r) ≤ r d l ′ (r) = l(r) almost-everywhere on (a, b),
and in particular f (r) = l ′ (r) almost-everywhere on (a, b). Recall that f is the density of the absolutely continuous measure ν, so (6.9)

ν((a, b)) = b a f (r)dr = b a l ′ (r)dr = µ((a, b)),
and (6.5) follows by comparison. In geometrical terms, (6.5) just says that for H d -almost-every point y ∈ E ∩ A, the tangent plane to E at y goes through the origin. This looks like pretty strong evidence that E coincides with a cone in A, but apparently we need some nontrivial amount of additional work to get the result. In particular, we shall use the minimality of E again.

Denote by π S the radial projection onto the unit sphere. Thus π S (y) = y/|y|. We claim that (6.10)

H d (π S (E ∩ A)) = 0.
This is an easy consequence of (6.5). Indeed recall from Section 2 that E is both rectifiable and locally Ahlfors-regular, so by Exercise 41.21 on page 277 of [D2]) it has a true tangent d-plane at almost every point. [We don't really need to know this here, but this will make the discussion simpler.] By rectifiability, we can cover E ∩ A, up to a negligible set N such that H d (N ) = 0, by a countable collection of d-dimensional surfaces Γ i of class C 1 . For almost-every point y of E ∩ A ∩ Γ i , the tangent d-plane to E at y is the same as the tangent d-plane to Γ i at y, because we may assume that y is a point of positive upper density of E ∩ Γ i ), and then (6.5) says that this d-plane goes through the origin. So y is a critical point of π S on Γ i , and Sard's theorem (applied to π S on Γ i ) says that H d (π S (E ∩ A ∩ Γ i )) = 0. Since H d (π S (N )) = 0 because π S is Lipschitz on A, we just need to sum over i to get (6.10). Proposition 6.11. Let y ∈ E ∩ A be such that E has a tangent d-plane P at y which goes through 0. Then E contains the radial line segment L = A ∩ π -1 S (π S (y)) through y.

Let us first check that Proposition 6.11 implies that (6.12) E coincides on A with a cone;

we shall only see later that this cone is a reduced minimal set. Call E 1 the set of points y ∈ E ∩ A that satisfy the conditions of Proposition 6.11; by the discussion above (the proof of (6.10)), almost-every point of

E ∩ A lies in E 1 . In particular, E 1 is dense in E ∩ A. Call C the cone over E 1 ; then E ∩ A ⊂ E 1 ⊂ C.
Conversely, C ∩ A ⊂ E, by Proposition 6.11 and because E is closed. So (6.12) follows from Proposition 6.11.

The proof of Proposition 6.11 will take some time. Let y ∈ E * ∩ A, P , and L be as in the proposition, and suppose in addition that L is not contained in E. We want to construct a deformation of E and eventually get a contradiction with its minimality.

We need lots of notation before we start. Since L is not contained in E, we can pick x ∈ L \ E. Let ε 0 be small, to be chosen near the end. Then pick a small radius r y such that (6.13) dist(z,

P ) ≤ ε 0 |z -y| for z ∈ E ∩ 2B y ,
where we set B y = B(y, r y ) and 2B y = B(y, 2r y ). Notice that (6.13) holds for r y small enough, because P is tangent to E at y. We choose r y < b/100 and so small that 2B y is contained in A and does not contain x. Now let r x be a small radius, smaller than r y and such that if B x = B(x, r x ), then

(6.14) 2B x ⊂ A \ [E ∪ 2B y ].
We shall also need to take r x small enough, depending on r y , for the final estimate to work. Without loss of generality, we may assume that coordinates were chosen so that L is contained in the positive first axis, and P is the horizontal d-plane where x d+1 = • • • = x n = 0. We know that L ⊂ P , by (6.5), so this is coherent. We shall identify R n with L ′ × P ′ × Q, where L ′ is the line that contains L and the first axis, P ′ is the orthogonal complement of L ′ in P (so P ′ corresponds to

x 1 = x d+1 = • • • = x n = 0), and Q is the orthogonal of P in R n . Thus Q is (n -d)-dimensional.
We shall often use this identification to denote points of R n by z = (z 1 , z 2 , z 3 ), with z 1 ∈ L ′ , z 2 ∈ P ′ , and z 3 ∈ Q. Also, we identify L ′ with R, so there is an order on L ′ , and we assume that 0 < x < y in L ′ ; the other case would be treaded similarly.

So we want to construct a deformation f and use it to reach a contradiction. We start with a first deformation f 1 that tries to send to P points that lie near L and roughly between x and y. The effect of this map will be to free from E \ P a small tunnel near L. This map f 1 will be composed with other maps f j , j ≥ 2, that will be defined later; the main goal is to make most of E ∩ B y disappear, by retracting it on a boundary; the tunnel will be used as a clean path that leads from x to y. Lemma 6.15. We can find ξ ∈ P ′ ∩ B(x, r x /100) and ρ ∈ (r x /2, 2r x /3) such that E ∩ A does not meet the cone over S ξ,ρ =

x + ξ + [Q ∩ ∂B(0, ρ)].
Thus S ξ,ρ is a (nd -1)-dimensional sphere centered at x + ξ. When d = n -1, S ξ,ρ is just composed of two points that lie on different sides of P . See Figure 6.1 for two special cases when n = 3.

To prove Lemma 6.15, pick any small constant τ > 0 and set Z τ = (z 1 , z 2 , z 3 ) ∈ π S (A ∩ E) ; z 1 ≥ τ , where π S still denotes the radial projection on the unit sphere. We know from (6.10) that H d (Z τ ) = 0. Then define ψ τ : Z τ → P ′ × R, by ψ τ (z 1 , z 2 , z 3 ) = (xz 2 /z 1 , x|z 3 |/z 1 ). Notice that ψ τ is Lipschitz (with a bound that depends on τ , but this does not matter) so H d (ψ τ (Z τ )) = 0.

Recall that P ′ × R is d-dimensional, so we can pick ξ ∈ P ′ ∩ B(x, r x /100) and ρ ∈ (r x /2, 2r x /3) such that (ξ, ρ) lies in no ψ τ (Z τ ). Let us check that the cone over S ξ,ρ does not meet E ∩ A. Otherwise, we can find z = (z 1 , z 2 , z 3 ) ∈ E ∩ A and λ ≥ 0 such that λz ∈ S ξ,ρ . This last forces λz 1 = x > 0 (we identify x with its first component), so z 1 > 0 and π S (z) lies in Z τ for τ small enough. Next

ψ τ (π S (z)) = ψ τ (z/|z|) = (xz 2 /z 1 , x|z 3 |/z 1 ) = (λz 2 , λ|z 3 |) because λz 1 = x. But λz ∈ S ξ,ρ , so λz 2 = ξ and λ|z 3 | = ρ, hence ψ τ (π S (z)) = (ξ, ρ)
, which is impossible by choice of ξ and ρ. So the cone over S ξ,ρ does not meet E ∩ A; this proves Lemma 6.15. Choose a pair (ξ, ρ) as in Lemma 6.15. For a ≤ s ≤ t ≤ b, set (6.16)

V (s, t) = z ∈ R n ; s ≤ z 1 ≤ t .
With this notation, S ξ,ρ is contained in the vertical hyperplane V (x, x). Pick a 1 and b 1 on the first axis, with a < a 1 < b 1 < b, so that (6.17)

B x ∪ B y ⊂ V (a 1 , b 1 ). Set S ε ξ,ρ = z ∈ V (x, x) ; dist(z, S ξ,ρ ≤ 2ε for ε > 0. Also call C ε the cone over S ε ξ,ρ . If r x and ε are small enough, C ε ∩ V (a 1 , b 1 ) is a compact subset of A.
Then, by compactness and because the cone over S ξ,ρ does not meet E ∩ A, we can choose ε so small that (6.18)

C ε ∩ V (a 1 , b 1 ) does not meet E.
Of course ε may be very small, so we shall need to be careful and avoid letting our final estimates depend on ε. Also choose ε smaller than r x /100 to avoid possible geometric complications.

Our first map f 1 will act inside vertical hyperplanes V (s, s), and we first want to describe how it works in V (x, x). The reader may refer to Figure 6.2, which is a picture in V (x, x). The dimensions of the picture correspond to n = 3 and d = 2; in general, V (x, x) is n -1 dimensional, P ′ is (d -1)-dimensional, and Q is (nd)-dimensional, but the picture is simple in P ′ , and things will be invariant under rotations in x + Q, so Figure 6.2 still gives a fair idea of what happens.

Set J = P ′ ∩ B(x + ξ, r x /10) and S ′ = x + ξ + Q ∩ ∂B(0, ρ + ε) (a sphere concentric with S ξ,ρ , but just a bit larger), and then denote by T the convex hull of J ∪ S ′ . Thus (6.19)

T = (x, z 2 , z 3 ) ; |z 2 -ξ| ≤ (ρ + ε -|z 3 |) r x /(10ρ + 10ε) . Set f (z) = z in V (x, x)\T . For z ∈ T \C ε , write z = (x, z 2 , z 3 ), call z * = (x, ξ, ρz 3 /|z 3 |) the closest point of S ξ,ρ
, and then let f (z) be the radial projection of z onto P ′ ∪ ∂T , with origin z * . That is, f (z) lies on P ′ ∪ ∂T , z lies on the segment [z * , f(z)], and the open segment does not meet P ′ ∪ ∂T . Obviously f is continuous across ∂T .

The reader may be worried about the Lipschitz-continuity of f inside T \ C ε , so let us say a few words about that. Let us first check that (6.20)

|z 3 | ≤ ρ -ε when (x, z 2 , z 3 ) ∈ T \ C ε , which is already almost obvious from Figure 6.2. Otherwise, |z 2 -ξ| ≤ (ρ+ε-|z 3 |)r x /(10ρ+ 10ε) ≤ 2εr x /10ρ ≤ 4ε/10, by (6.19) and because ρ ≥ r x /2. Recall that z * = (x, ξ, ρz 3 /|z 3 |), so |z -z * | 2 = |z 2 -ξ| 2 + (ρ -|z 3 |) 2 ≤ 16ε 2 /100 + (ρ -|z 3 |) 2 ≤ 16ε 2 /100 + ε 2 < 4ε 2 . This is impossible because z / ∈ C ε ; (6.20) follows. Figure 6.2. A picture in V (x, x)
We may now compute the radial projection of z onto P ′ , again with origin z * . This point should be of the form

z ′ = z * + λ(z -z * ), with λ ≥ 1. Since the last coordinate z * 3 + λ(z 3 -z * 3 ) vanishes and z * 3 = ρz 3 /|z 3 |, we get that λ = ρ/(ρ -|z 3 |)
, which is indeed defined by (6.20) and larger than 1. In addition, λ and z ′ are Lipschitz functions of z ∈ T \ C ε . Now the radial projection of z onto ∂T , again with origin z * , is also Lipschitz

on T \ C ε . And so is f (z), because the infimum of two Lipschitz functions λ is Lipschitz. Since f is continuous across ∂T , f is Lipschitz on V (x, x) \ C ε .
Extend f to the whole V (x, x), so that it is Lipschitz and f (T ) ⊂ T . [If this last condition is not satisfied, we can always compose the restriction of f to T with a Lipschitz retraction onto T .] Finally extend f to V (a 1 , b 1 ), by homogeneity: set

(6.21) f (z 1 , z 2 , z 3 ) = z 1 x f (x, z 2 x/z 1 , z 3 x/z 1 ).
Notice that f moves the points inside vertical planes V (s, s).

We are now ready to define f 1 . Set (6.22)

x 0 = x - r x 5 and y 0 = y + r y 5 ,
and then let ψ be the continuous cut-off function such that (6.23)

ψ(t) = 0 out of (x 0 , y 0 ), ψ(t) = 1 on [x, y],
and ψ is affine on the two remaining intervals.

Finally set (6.24)

f 1 (z) = ψ(z 1 ) f (z) + (1 -ψ(z 1 )) z for z ∈ R n ,
where z 1 still denotes the first coordinate of z. Notice that z ∈ V (a 1 , b 1 ) when ψ(z 1 ) = 0, by (6.17).

Let us record a few properties of f 1 before we compose with a second deformation. As promised, f 1 only moves the points inside vertical planes V (s, s). Notice also that (6.25)

f 1 (z) = z out of T ∩ V (x 0 , y 0 ),
where T denotes the cone over T , and that (6.26)

f 1 T ∩ V (x 0 , y 0 ) ⊂ T ∩ V (x 0 , y 0 ).
We also want to see where the set

E 1 = f 1 (E) lies. Let us check that (6.27) E 1 ∩ T ∩ V (x, y) ⊂ ∂ T ∪ J ∩ V (x, y),
where J is the cone over J (also notice that ∂ T is the cone over the boundary of T in V (x, x)). Indeed, let w ∈ E 1 ∩ T ∩ V (x, y) be given, and let z ∈ E be such that f (z) = w. By (6.25), z ∈ T . Also, z ∈ V (x, y) because f 1 moves points vertically. By (6.23) and (6.24), w = f (z). By (6.21), f (z) = z 1 x f (xz/z 1 ). By (6.18), z and xz/z 1 lie out of C ε , and then f (xz/z 1 ) is the projection on ∂T ∪ J of some point z * ; (6.27) follows.

Near x, things are also easy. Indeed T ∩ V (x 0 , x) ⊂ B x by construction, and B x does not meet E by (6.14), so (6.28)

E 1 ∩ T ∩ V (x 0 , x) = ∅
(again recall that f 1 only moves points on planes V (s, s), and inside T ).

Near y, we want to use the fact that (6.29) dist(z, P ) ≤ ε 0 |z -y| for z ∈ E ∩ 2B y (by (6.13)) to show that

(6.30) dist(w, P ) ≤ 2ε 0 |w -y| for w ∈ E 1 ∩ B y Let w ∈ E 1 ∩ B y be given, and let z ∈ E be such that w = f 1 (z). Notice that z ∈ V (y -r y , y + r y ) because f 1 moves points vertically. If f 1 (z) = z, then dist(w, P ) = dist(z, P ) ≤ ε 0 |w -y| by (6.29).
Otherwise, z ∈ T by (6.25). If r x is small enough (depending on x, y, and r y , but this is all right), T ∩ V (yr y , y + r y ) ⊂ 2B y , so z ∈ 2B y and (6.29) holds.

Set z ′ = xz/z 1 , and let us first worry about f (z ′ ). Recall from (6.18) that z and z ′ lie out of C ε ; then z ′ ∈ T \ C ε , and (6.20) says that |z ′ 3 | ≤ ρε. The point z * used to compute f (z ′ ) lies "above" z ′ , and so

(6.31) dist( f (z ′ ), P ) = | f (z ′ ) 3 | ≤ |z ′ 3 | = dist(z ′ , P ).
Let us also check that

(6.32) | f (z ′ ) -z ′ | ≤ 6 dist(z ′ , P ).
If dist(z ′ , P ) ≥ ρ/2, then (6.32) holds because the diameter of T is less than 3ρ (recall that ρ ≥ r x /2 ≥ 5 diam(J)/2). Otherwise, set z * = (x, ξ, ρz ′ 3 /|z ′ 3 |), as in the definition of f (z ′ ), and observe that f (z ′ )z * = λ(z ′z * ), for some λ ≥ 1. In addition, λ ≤ ρ/(ρ -|z ′ 3 |), which corresponds to the radial projection of z ′ onto P ; see the computation above (6.21). Then So (6.32) holds in all cases. By homogeneity, (6.32

(6.33) | f (z ′ ) -z ′ | ≤ (λ -1)|z * -z ′ | ≤ |z ′ 3 | ρ -|z ′ 3 | |z * -z ′ | ≤ 6|z ′ 3 | = 6 dist(z ′ , P ) because ρ -|z ′ 3 | ≥ ρ/2 and |z * -z ′ | ≤ diam(T ) ≤ 3ρ.
) yields | f (z) -z| ≤ 6 dist(z, P ). Then (6.34) |f 1 (z) -z| ≤ 6 dist(z, P ) ≤ 6ε 0 |z -y|, because f 1 (z) lies on the segment [z, f (z)],
and then because (6.29) holds for z. Similarly, dist( f (z), P ) ≤ dist(z, P ) by (6.31) and homogeneity, and then

(6.35) dist(f 1 (z), P ) ≤ dist(z, P ) because f 1 (z) lies on [z, f (z)]. Next |f 1 (z) -y| ≥ |z -y| -|f 1 (z) -z| ≥ (1 -6ε 0 )|z -y|, by (6.34). Therefore (6.36) dist(f 1 (z), P ) ≤ dist(z, P ) ≤ ε 0 |z -y| ≤ 2ε 0 |f 1 (z) -y|
by (6.35) and (6.29). Recall that w = f 1 (z); hence (6.36) proves (6.30).

To summarize, E 1 ∩ T ∩ V (x, y) is contained in the union of the flat floor J and the triangular tunnel ∂ T , by (6.27). In addition, there is no piece of E 1 at the end of T near x (by (6.28)), and near y we know that E 1 stays very close to P (by (6.30)), so the end of the tunnel near y is not completely blocked by E 1 either.

We are now ready to construct a second deformation f 2 , to be composed with f 1 . Its goal is to project E 1 ∩ B y onto P . We take (6.37)

f 2 (z) = z for z ∈ R n \ B y ,
and inside B y we want f 2 to look like a radial projection onto P ∪ ∂B y . It will act a little like f 1 above, but we won't need to move points in vertical planes. On the other hand, we want to keep the invariance under rotations in Q.

Set S y = (y/x)S ξ,ρ = y + (y/x)ξ + Q ∩ ∂B(0, ρ 1 ), where we set ρ 1 = yρ/x. The precise choice of sphere will be useful later. Also set S y,+ = z ∈ R n ; dist(z, S y ) ≤ ρ 1 /100 . Notice that S y and S y,+ both lie well inside B y , because ρ ≤ r x and we take r x much smaller than r y . See Figure 6.3.

Figure 6.3 For z ∈ B y \S y,+ , denote by z ♯ the closest point of S y . Thus z ♯ = (y, (y/x)ξ, ρ 1 z 3 /|z 3 |).
Then let f 2 (z) be the radial projection of z onto ∂B y ∪ P , centered at z ♯ . Let us check that f 2 is Lipschitz on B y \ S y,+ . Let τ > 0 be very small. In the region where |z 3 | ≤ ρ 1τ , the radial projection of z onto P is well defined, and Lipschitz (we even checked below (6.20) that it is equal to z

♯ + λ(z -z ♯ ), with λ = ρ 1 (ρ 1 -|z 3 |) -1 .
Then f 2 is Lipschitz on this region. In the region where |z 3 | > ρ 1τ , and if τ is small enough, the projection of z onto P is either not defined, or else lies out of B y . Then f 2 (z) is the projection onto ∂B y , and it is Lipschitz on the region. Notice also that f 2 (z) = z on ∂B y , so it is continuous across ∂B y and Lipschitz on R n \ S y,+ .

We now extend f 2 to the whole R n , so that it is Lipschitz (probably with a very bad constant) and maps B y to itself (otherwise, compose its restriction to B y with a Lipschitz retraction onto B y ).

We need to know where the set

E 2 = f 2 (E 1 ) lies. Notice that (6.38) E 2 \ B y = E 1 \ B y and E 2 ∩ B y = f 2 (E 1 ∩ B y )
by ( 6.37) and because f 2 (B y ) ⊂ B y . Set (6.39)

H = z ∈ R n ; dist(z, P ) ≤ 2ε 0 |z -y| ;
thus (6.30) says that E 1 ∩ B y is contained in H. It will be good to know that (6.40)

E 1 ∩ B y ⊂ B y \ S y,+ ,
and for this it is enough to check that S y,+ does not meet

H. Let z = (z 1 , z 2 , z 3 ) ∈ S y,+ be given; then ||z 3 | -ρ 1 | ≤ dist(z, S y ) ≤ ρ 1 /100, so dist(z, P ) = |z 3 | ≥ 99ρ 1 /100. On the other hand, |z -y| ≤ dist(z, S y ) + ρ 1 + |(y/x)ξ -y| ≤ ρ 1 /100 + ρ 1 + yr x /100x ≤ ρ 1 /100 + ρ 1 + yρ/50x ≤ 2ρ 1 (
E 2 ∩ B y ⊂ P ∪ [∂B y ∩ H].
Let w ∈ E 2 ∩ B y be given, and let z ∈ E 1 be such that w = f 2 (z). By (6.37), z ∈ B y . By (6.40), z lies out of S y,+ , so w is the radial projection of z onto P ∩ ∂B y , with origin z ♯ as above. In particular, w ∈ P ∪ ∂B y , and we just have to check that w ∈ H if w / ∈ P . First assume that z, z ♯ , and y are not collinear, and let V denote the plane that contains them. Also assume that P ∩ V is a line; then H ∩ V is a thin double cone centered at y that contains P ∩ V . Now z ♯ lies out of H (because z ♯ ∈ S y ⊂ S y,+ ), and z ∈ H by (6.30). So the half line [z ♯ , z) starts out of H, then gets in H before it hits z (or at the same time), and it does not leave H before it meets P . This last may not happen, but if it happens, it happens after [z ♯ , z) hits z (the open line segment (z ♯ , z) does not meet P by definition of z ♯ ). In both cases, w = f 2 (z) lies in H.

If P ∩ V is not a line, it is reduced to {y} (because z ♯ lies out ou P ). Then the line through z and z ♯ does not meet P (recall that z, z ♯ , and y are not collinear). By definition of z ♯ = (y, (y/x)ξ,

ρ 1 z 3 /|z 3 |), this can only happen if |z 3 | = ρ 1 . In this case, dist(w, P ) = dist(z, P ) ≤ 2ε 0 |z -y| ≤ 2ε 0 r y , because z ∈ H ∩ B y .
In addition, w ∈ ∂B y (because it does not lie in P ), hence |w -y| = r y and w ∈ H.

Finally suppose that z, z ♯ , and y are collinear. In this case z lies between z ♯ and y (because otherwise z would not lie in H; already z ♯ does not), and w = y (because y ∈ P and [z ♯ , y] does not meet ∂B y ). This completes our proof of (6.41).

Let us also check that E 2 does not block our access to the tube T , i.e., that (6.42)

E 2 ∩ B y ∩ V (x, y) does not meet int( T ) \ P.
Let w ∈ E 2 ∩ B y ∩ V (x, y) be given, and suppose it lies in int( T ) \ P . Let z ∈ E 1 be such that f 2 (z) = w. Then z ∈ B y , because otherwise w = z by (6.37), and (6.27) says that this is impossible. In particular, w is obtained from z by the usual radial projection process with z ♯ . Then z / ∈ P , because otherwise w = z, and we assumed that w / ∈ P . Since w ∈ V (x, y) and z ♯ ∈ V (y, y), z lies in V (x, y). Then (6.27) says that z lies out of int( T ). Call w ′ , z ′ , and z * the radial projections of w, z,, and z ♯ on V (x, x). Thus w ′ ∈ int(T ), z ′ / ∈ int(T ), and

z * = (x, ξ, ρz ′ 3 /|z ′ 3 |) lies in int(T )
. This is impossible, because T is convex (see near (6.19)) and z ′ lies on [z * , w]. So (6.42) holds.

Let us summarize the situation now. Set (6.43)

Z = [ T ∩ V (x 0 , y)] ∪ B y .
Recall from (6.25), the fact that T ∩ V (y, y 0 ) ⊂ B y if r x is small enough (see (6.22)), and (6.37) that (6.44)

f 2 (z) = f 1 (z) = z for z ∈ R n \ Z.
Let us also check that (6.45) by (6.37). Next suppose that z ∈ T ∩ V (y, y 0 ); then f 1 (z) ∈ T ∩ V (y, y 0 ) ⊂ B y by (6.26), because f 1 moves points vertically, and because r x is very small. Then

f 2 • f 1 (Z) ⊂ Z. First take z ∈ T ∩ V (x 0 , y), f 1 (z) ∈ T ∩ V (x 0 , y), by (6.26) and because f 1 moves points vertically. If f 1 (z) lies in B y , f 2 • f 1 (z) ∈ B y ⊂ Z because f 2 (B y ) ⊂ B y . Otherwise, f 2 • f 1 (z) = f 1 (z) ∈ T ∩ V (x 0 , y) ⊂ Z,
f 2 • f 1 (z) ∈ B y ⊂ Z, as before. Finally, if z ∈ B y \ [ T ∩ V (x 0 , y 0 )],
f 1 (z) = z by (6.25), and

f 2 • f 1 (z) = f 2 (z) ∈ B y ⊂ Z; (6.45) follows.
We now verify that (6.46)

E 2 ∩ Z ⊂ P ∪ [∂ T ∩ V (x, y) \ B y ] ∪ [∂B y ∩ H \ (int( T ) ∩ V (x, y))].
Indeed there is no part in V (x 0 , x) because of (6.28) and (6.38). The part of E 2 ∩ Z \ P that lies in V (x, y) \ B y lies in ∂ T , by (6.27) and (6.38). The rest lies in B y , hence in ∂B y ∩ H by (6.41), and out of int( T ) ∩ V (x, y) by (6.42). This is somewhat better than before; we now have a nice floor P ∩ Z, and we managed to surround its interior with a tunnel that leads from x to y and a big cavity B y \ P .

Our third mapping f 3 will contract most of the floor P ∩ Z onto its boundary; this will be good, because it will make a big part of the floor vanish. More precisely, set F = P ∩ Z ∩ V (x, b). See Figure 6.4. Thus F is the union of J ∩ V (x, y) and P ∩ B y , so it looks like a long d-dimensional cylinder attached to a much larger d-ball. [Recall that J is the cone over the (d -1)-ball J = P ′ ∩ B(x + ξ, r x /10); see above (6.19).] There is a bilipschitz mapping ψ from F to the cylinder C = [x, y] × J × {0}, and which fixes the points of the end F ∩ V (x, x) = {x} × J × {0}. (open at one side) with center (x 0 , ξ, 0). [We don't care about the precise value; the point is that x 0 < x, so ϕ is Lipschitz. The contraction that we use on

F is ψ -1 • ϕ • ψ. It is Lipschitz on F ,
even though with a very large constant. We want to set (6.47)

f 3 (z) = ψ -1 • ϕ • ψ(z) for z ∈ F and (6.48) f 3 (z) = z for z ∈ R n \ Z, so let us check that this defines a Lipschitz function f 3 on F ∪ (R n \ Z). The main point is that f 3 (z) = z on the part of F that touches R n \ Z; the rest of the verification is straightforward. Indeed, we want to check that |f 3 (z 2 ) -f 3 (z 1 )| ≤ C|z 2 -z 1 | when z 1 and z 2 lie in F ∪ (R n \ Z)
. This is trivial if z 1 and z 2 both lie in F or in R n \ Z, so we may assume that z 1 ∈ F and z 2 ∈ R n \ Z. But, due to the simple geometry of Z, we can find

z 3 ∈ F ∩ ∂Z such that |z 3 -z 1 | ≤ C|z 2 -z 1 |, and then |f 3 (z 2 ) -f 3 (z 1 )| ≤ |f 3 (z 2 ) -f 3 (z 3 )| + |f 3 (z 3 ) -f 3 (z 1 )| ≤ |z 3 -z 2 | + C|z 3 -z 1 | ≤ |z 3 -z 1 | + |z 1 -z 2 | + C|z 3 -z 1 | ≤ C ′ |z 2 -z 1 |, as needed.
Extend f 3 to the rest of Z, so that (6.49)

f 3 is Lipschitz on R n and f 3 (Z) ⊂ Z
(the existence of a Lipschitz extension is a well-known general fact, and we can arrange that f 3 (Z) ⊂ Z by composing the restriction of f 3 to Z with a Lipschitz retraction onto Z (such a retraction exists because there is a bilipschitz mapping of R n that maps Z to a cylinder).

Set E 3 = f 3 (E 2 ), and let us check that (6.50)

E 3 ∩ Z ⊂ Z 1 ∪ Z 2 ,
where (6.51) by (6.43) and (6.28).

Z 1 = V (x, y) ∩ ∂ T \ B y and Z 2 = ∂B y ∩ H. Let w ∈ E 3 ∩Z and let z ∈ E 2 be such that w = f 3 (z). Notice that z ∈ Z because f (z) = z for z / ∈ Z. Next, z lies out of V (x 0 , x), because E 2 ∩ V (x 0 , x) = E 1 ∩ V (x 0 , x) by (6.38), and E 1 ∩ Z ∩ V (x 0 , x) ⊂ E 1 ∩ T ∩ V (x 0 , x) = ∅
First suppose that z ∈ P . Then z ∈ F = P ∩ Z ∩ V (x, b), and (6.47) shows that

w = f 3 (z) ∈ ψ -1 [x, y] × ∂J × {0} ∪ {y} × J × {0}
. This last set is the closure of ∂F \ V (x, x) (where ∂F denotes the boundary of F in P ), and it is contained in Z 1 ∪ Z 2 . Now suppose that z / ∈ P ; then (6.46) says that z ∈ ∂Z, so w = f 3 (z) = z by (6.48), and w lies in the set in Z 1 ∪ Z 2 by (6.46) again. This proves (6.50).

Let us also record the fact that (6.52)

f 3 • f 2 • f 1 (z) = z for z / ∈ Z, and f 3 • f 2 • f 1 (Z) ⊂ Z. When d = n-1, this is enough to conclude because H n-1 (E 3 ∩Z) ≤ H n-1 (Z 1 ∪Z 2 ) ≤ Cε 0 r d y +C x,y r d-1
x (maybe see the proof of (6.80) if you are worried about the second term), and for this case the reader may set f 4 (z) = f 5 (z) = z for all z, Z * = Z, and go directly to the line below (6.89) for the final accounting that leads to a contradiction.

For the general case, we still need to compose f 3 • f 2 • f 1 with two last mappings f 4 and f 5 , whose effect will be to project the potentially large sets E 3 ∩ Z 1 and E 3 ∩ Z 2 onto d-dimensional skeletons of cubes. In both cases, we shall use a standard Federer-Fleming argument.

Some amount of notation concerning dyadic cubes will be needed. Here dyadic cubes are closed cubes obtained from [0, 2 k ] n for some k ∈ Z by a translation in 2 k Z n . For such a dyadic cube Q, we shall denote by S d (Q) the d-dimensional skeleton of Q, i.e., the union of the d-dimensional faces of Q.

Also let j ∈ N be given; in this paper j will be chosen once and for all, sufficiently large for some geometric conditions explained below to hold, but for all practical purposes it is a geometric constant that depends only on n. We denote by ∆ j (Q) the set of dyadic cubes of sidelength 2 k-j that are contained in Q, and then by S d,j (Q) the union of all the ddimensional faces of dyadic cubes R ∈ ∆ j (Q). In other words, S d,j (Q) = R∈∆ j (Q) S d (R).

We shall use the following lemma, which is part of Proposition 3.1 on page 13 of [DS]. We also refer to [DS] for additional background on dyadic cubes and skeletons. Lemma 6.53. Let Q be a dyadic cube in R n , and let

F ⊂ Q be closed, with H d (F ) < +∞. Then there is a Lipschitz mapping ϕ Q : R n → R n such that (6.54) ϕ Q (z) = z for z ∈ R n \ Q. (6.55) ϕ Q (F ∩ Q) ⊂ S d,j (Q) ∪ ∂Q, (6.56) ϕ Q (R) ⊂ R for R ∈ ∆ j (Q), and
(6.57) H d (ϕ Q (F ∩ R)) ≤ CH d (F ∩ R) for R ∈ ∆ j (Q).
The constant C depends only on n and d, not on Q, j, or F .

We want to start with the function f 4 that will take care of

E 3 ∩ Z 1 . Let k ∈ Z be such that (6.58) r x /32 ≤ 2 k √ n ≤ r x /16. Let D k denote the set of dyadic (closed) cubes Q of R n with sidelength 2 k such that dist(Q, Z 1 ) ≤ 2 k+1 √ n. For each Q ∈ D k ,
we apply Lemma 6.53 to Q and the set F = E 3 ∩ Q, and get a mapping ϕ Q .

We compose all the mappings ϕ Q , Q ∈ D k , (in some randomly decided order) and get a mapping φ 0 . Notice that for a given z ∈ R n , at most one ϕ Q may move z, the one (if it exists) for which z lies in the interior of Q. Even then, ϕ Q (z) ∈ Q, and by (6.54), the other

ϕ Q ′ , Q ′ = Q, do not move z or ϕ Q (z). Because of this, φ 0 does not change if we compose the ϕ Q , Q ∈ D k , in any other order. Set E 3,0 = φ 0 (E 3 ).
It should be a little better than E 3 near Z 1 because we were able to send lots of points on the S d,j (Q), but there might still be some part of φ 0 (E 3 ) left on the boundaries ∂Q. To fix this, we shall compose with variants of φ 0 for which the boundaries ∂Q lie in different places. Denote by v 0 , v 1 , • • • v m the vectors of R n whose coordinates are either 0 at 2 k-1 . Thus m = 2 n -1. To make the notation coherent, do this so that v 0 = 0.

We now construct φ 1 . For each Q ∈ D k , we apply Lemma 6.53 to Q + v 1 and the set

F = E 3,0 ∩ [Q + v 1 ].
The fact that Q is not a dyadic cube itself, but just a translation of such a cube by v 1 , does not matter, but we shall use the fact that the cubes R ∈ ∆ j (Q+v 1 ) are still dyadic cubes of sidelength 2 k-j .

Call ϕ Q,1 the map given by Lemma 6.53. Then compose the new ϕ Q,1 in any order, and get a new mapping φ 1 . Finally set E 3,1 = φ 1 (E 3,0 ).

We iterate this construction in the obvious way: once we have φ l-1 and E 3,l-1 for some l ≤ m, we apply Lemma 6.53 to the

Q + v l , Q ∈ D k , with F = E 3,l-1 ∩ [Q + v l ]
. This gives a mapping ϕ Q,l . We compose all these mappings to get φ l , and set E 3,l = φ l (E 3,l-1 ).

We end with φ m , and set (6.59)

f 4 = φ m • • • • φ 0 and E 4 = f 4 (E 3 ) = E 3,m .
Our next goal is to control E 4 away from Z 2 , as follows.

Lemma 6.60. Let R be a dyadic cube of sidelength 2 k-j such that

(6.61) dist(R, Z 1 ∪ Z 2 ) ≤ r x /2 and dist(R, Z 2 ) ≥ 2 k . Then (6.62) H d (R ∩ E 4 ) ≤ C2 kd .
Let us first see how

H d (E 3,l ∩ R) evolves with time. Set E 3,-1 = E 3 and ϕ Q,0 = ϕ Q to make the notation uniform. Recall from (6.57) with F = E 3,l-1 ∩ [Q + v l ] that (6.63) H d (ϕ Q,l (R ∩ E 3,l-1 )) ≤ CH d (R ∩ E 3,l-1 ) for 0 ≤ l ≤ m and Q ∈ D k . This is only useful when Q is the cube such that Q + v l contains R, because for the other cubes Q ′ , (6.54) says that ϕ Q ′ ,l (z) = z on R. Thus (6.64) H d (φ l (R ∩ E 3,l-1 )) ≤ CH d (R ∩ E 3,l-1 ).
Observe also that φ l (R ′ ) ⊂ R ′ for every dyadic cube R ′ of sidelength 2 k-j , because (6.54) and (6.56) say that this is true for every

ϕ Q,l . Thus, if z ∈ R ′ and φ l (z) lies in R, R ′ meets R. That is, (6.65) R ∩ E 3,l = R ∩ φ l (E 3,l-1 ) ⊂ R ′ ∈N (R) φ l (R ′ ∩ E 3,l-1 ),
where we denote by N (R) the set of dyadic cubes R ′ of sidelength 2 k-j that touch R.

Consequently, (6.66)

H d (R ∩ E 3,l ) ≤ R ′ ∈N (R) H d (φ l (R ′ ∩ E 3,l-1 )) ≤ C R ′ ∈N (R) H d (R ′ ∩ E 3,l-1 ),
by (6.64). Iterations of this yield (6.67)

H d (R ∩ E 4 ) = H d (R ∩ E 3,m ) ≤ C R ′ ∈N m+1 (R) H d (R ′ ∩ E 3 ),
where the set N l (R) of neighbors of order l of R is just defined by induction by

N 1 (R) = N (R) and N l+1 (R) = R ′ ∈N l (R) N (R ′
). An obvious induction yields

(6.68) dist(R ′ , R) ≤ (l -1)2 k-j √ n for R ′ ∈ N l (R), so, if z 0 is any point of R, (6.69) R ′ ∈N m+1 (R) R ′ ⊂ B z 0 , (m + 2)2 k-j √ n = B z 0 , (2 n + 1) 2 k-j √ n ⊂ B(z 0 , 2 k-2 )
because m = 2 n -1, and if j is chosen large enough. When R lies far enough from Z 1 ∩ Z 2 , all this will lead to (6.62) rather directly. Let us first check that for a dyadic cube S of sidelength 2 k-j , (6.70)

H d (E 3 ∩ S) ≤ C2 kd when dist(S, Z 1 ∪ Z 2 ) ≤ r x and S does not meet Z 1 ∪ Z 2 .
Indeed S does not meet E 3 ∩ Z by (6.50), and then S ∩ E 3 ⊂ S ∩ E by (6.52). So (6.70) follows from the local Ahlfors-regularity of E (Lemma 2.15 above, but only the easier upper bound and in the easier special case of locally minimal sets); this is why we assumed that dist(S, Z 1 ∪ Z 2 ) ≤ r x , to make sure that S ⊂ B for some small ball B such that 2B ⊂ A ⊂ B(0, b). See in particular (6.14), the definition of A above (6.5), and the statement of Theorem 6.2. Now we can already check (6.62) when (6.71)

2 k ≤ dist(R, Z 1 ∪ Z 2 ) ≤ r x /2.
Indeed, by (6.69) and (6.58), all the cubes S ∈ N m+1 (R) satisfy the condition in (6.70), and (6.62) follows from (6.67). We are thus left with the case of cubes R such that dist(R, Z 1 ∪ Z 2 ) < 2 k and hence, by (6.61)

(6.72) dist(R, Z 1 ) ≤ 2 k .
Pick z 0 ∈ R, denote its coordinates by z 0,i , 1 ≤ i ≤ n, and then choose

ε i ∈ {0, 2 k-1 } so that z 0,i -ε i lies at distance at least 2 k-2 from 2 k Z. Recall that there is a v l such that v l = (ε 1 , • • • , ε n ), and then every coordinate of z 0 -v l lies at distance at least 2 k-2 from 2 k Z. Now (6.73) dist(z 0 , v l + ∂Q) ≥ 2 k-2 for Q ∈ D k ,
because the points in the faces that compose ∂Q have at least one coordinate in 2 k Z. Now let S denote any cube of N m+1 (R), and let

Q be such that Q + v l contains S. Observe that (6.74) dist(Q, Z 1 ) ≤ |v l | + dist(S, Z 1 ) ≤ |v l | + 2 k-2 + dist(z 0 , Z 1 ) ≤ |v l | + 2 k-2 + diam(R) + dist(R, Z 1 ) ≤ 2 k-1 √ n + 2 k-2 + 2 k-j √ n + dist(R, Z 1 ) ≤ 2 k+1 √ n
by (6.69) and (6.72) in particular. So Q ∈ D k (see the definition below (6.58)), and (6.55) says that

(6.75) ϕ Q,l (E 3,l-1 ∩ S) ⊂ ϕ Q,l (E 3,l-1 ∩ [v l + Q]) ⊂ v l + [S d,j (Q) ∪ ∂Q]. Now S ∈ N m+1 (R), so S ⊂ B(z 0 , 2 k-2
) by (6.69), and (6.73) says that S does not meet v l + ∂Q. Recall from (6.56

) that ϕ Q,l (S) ⊂ S ; then ϕ Q,l (E 3,l-1 ∩ S) ⊂ v l + S d,j (Q), and
H d (ϕ Q,l (E 3,l-1 ∩ S)) ≤ C2 kd . Now (6.76) H d (φ l (E 3,l-1 ∩ S)) ≤ C2 kd
too, because by (6.54) all the other ϕ Q ′ ,l that compose φ l leave the points of S alone. We shall remember that (6.76) holds for all S ∈ N m+1 (R).

We may now conclude as before, with repeated uses of (6.64). That is, (6.77)

H d (R ∩ E 4 ) = H d (R ∩ E 3,m ) = H d (R ∩ φ m (E 3,m-1 )) ≤ S∈N 1 (R) H d (φ m (S ∩ E 3,m-1 )) ≤ C S∈N 1 (R) H d (S ∩ E 3,m-1 ) = C S∈N 1 (R) H d (S ∩ φ m-1 (E 3,m-2 )) ≤ C S∈N 2 (R) H d (φ m-1 (S ∩ E 3,m-2 )) ≤ C 2 S∈N 2 (R) H d (S ∩ E 3,m-2 ) • • • ≤ C S∈N m-l+1 (R) H d (φ l (S ∩ E 3,l-1 )) ≤ C2 kd
by (6.76). This completes our proof of Lemma 6.60.

Let us record a few facts about f 4 before we move to f 5 . Recall that if R is a dyadic cube of sidelength 2 k-j , then all the mappings that compose

f 4 send R to R, so f 4 (z) ∈ R for z ∈ R. Consequently, (6.78) |f 4 (z) -z| ≤ 2 k-j √ n for z ∈ R n . Also set Z * 1 = z ∈ R n ; dist(z, Z 1 ) ≤ 2 k+2 √ n . Notice that Z * 1 contains all the cubes Q + v l , Q ∈ D k . [See below (6.58) for the definition of D k .] Then (6.79) f 4 (z) = z for z / ∈ Z * 1 and f 4 (Z * 1 ) ⊂ Z * 1
because the φ l only act nontrivially on the cubes Q + v l , and preserve them. Let us also check that (6.80)

H d z ∈ E 4 ∩ Z * 1 ; dist(z, Z 2 ) ≥ 2 k+1 ≤ C x,y r d-1 x ,
where C x,y depends on x, y, but not on r x and k.

Call Y the set on the left-hand side of (6.80). Cover Y with dyadic cubes R of sidelength 2 k-j . Notice that (6.61) holds when T meets Y , so

H d (E 4 ∩ R) ≤ C2 kd by Lemma 6.60. Recall from (6.51) that Z 1 = V (x, y) ∩ ∂ T \ B y is contained in a thin tube of width Cr x around the segment [x, y],
where C may depend on x and y (typically, through the ratio |y|/|x|). Then Z 1 , and also Y , can be covered by less than C|x -y|/r x cubes R as above (recall from (6.58) that r x and 2 k are comparable); each cube contributes at most C2 kd ≤ Cr d

x by (6.62), so (6.80) follows. The reader should not be shocked by the crude estimate above; we shall compensate by taking r x really small. Lemma 6.60 only controls E 4 = f 4 (E 3 ) far from Z 2 , and indeed E 4 could still contain a big piece of Z 2 = ∂B y ∩ H. So we shall compose with a very last mapping f 5 . The strategy is the same as for Z 1 , but this time we shall work at the scale 2 t , where t ∈ Z is such that

(6.81) 2 t-1 ≤ ε 0 r y ≤ 2 t ,
instead of 2 k . Notice that we shall choose r x very small, depending on r y and ε 0 , so we can assume that k is much smaller than t.

Denote by D t the set of dyadic cubes Q of sidelength 2 t such that

(6.82) dist(Q, Z 2 ) ≤ 2 t+1 √ n
We shall repeat exactly the same construction as before. Denote by w 0 , • • • w m the 2 n vectors with coordinates in {0, 2 t-1 }. We inductively build successive mappings φ l , 0 ≤ l ≤ m, and each one is obtained by composing Lipschitz functions ϕ l,Q , Q ∈ D t . We start with E 4,-1 = E 4 , and later set E 4,l = φ l (E 4,l-1 ). Each ϕ l,Q is obtained by applying Lemma 6.53 to Q +w l and the set

F = E 4,l-1 ∩[Q +w l ].
All this defines the mappings ϕ l,Q and φ l by induction. At the end, we set

f 5 = φ m • • • • φ 0 and E 5 = f 5 (E 4 ) = φ m (E 4,m-1 ).
Here is the analogue of Lemma 6.60. Lemma 6.83. Let R be a dyadic cube of sidelength 2 t-j such that dist(R, Z 2 ) ≤ r y /2. Then (6.84)

H d (R ∩ E 5 ) ≤ C2 td .
We shall start with the easier cases where we can prove that (6.85)

H d (E 4 ∩ S) ≤ C2 td for S ∈ N m+1 (R),
where N m+1 (R) still denotes the set of dyadic cubes of sidelength 2 t-j that are neighbors of (n + 1) st order of R.

Indeed, once we have (6.85), we can obtain (6.84) by the same argument as before: the analogue of (6.67) holds and yields (6.84).

Let us first take care of the contribution of E in (6.85). Choose z 0 ∈ R and let S ∈ N m+1 (R) be given; the proof of (6.69) says that S ⊂ B(z 0 , 2 t-2 ). Since we want to apply the local Ahlfors-regularity of E, we should check that B(z

0 , 2 t-1 ) ⊂ B(0, b) (because E is minimal in B(0, b)). And indeed B(z 0 , 2 t-1 ) ⊂ 2B y because dist(R, B y ) ≤ dist(R, Z 2 ) ≤ r y /2
and both 2 t and diam (R) are much smaller than r y , and 2B y ⊂ A ⊂ B(0, b) by definition of r y (below (6.13)). The (easier) upper bound in Lemma 2.15 says that (6.86)

H d (E ∩ S) ≤ H d (E ∩ B(z 0 , 2 t-2 )) ≤ C2 td ;
so it will be enough to control S ∩ E 4 \ E.

We start with the case when dist(R,

Z 1 ∪ Z 2 ) > 2 t+1 . Recall that S ⊂ B(z 0 , 2 t-2 ), so dist(S, Z 1 ∪ Z 2 ) ≥ 2 t .
In particular S does not meet the set Z * 1 of (6.79), and (6.79) says that E 4 ∩ S = E 3 ∩ S. In addition, E 3 ∩ S does not meet Z 1 ∪ Z 2 , so (6.50) says that it does not meet Z. By (6.52), E 3 ∩ S = E 3 ∩ S \ Z ⊂ E. Altogether, E 4 ∩ S ⊂ E, and (6.85) follows from (6.86).

We may now assume that dist(R, Z 1 ∪ Z 2 ) ≤ 2 t+1 , and our second case will be when

dist(R, Z 2 ) ≥ 2 t+1 . Let z ∈ E 4 ∩ S be given. Then dist(z, Z 2 ) ≥ dist(z 0 , Z 2 ) -2 t-2 ≥ dist(R, Z 2 ) -diam(R) -2 t-2 ≥ 2 t because S ⊂ B(z 0 , 2 t-2 ). If z ∈ Z *
1 , it lies in the set of (6.80); therefore, (6.87)

E 4 ∩ S ∩ Z * 1 ≤ C x,y r d-1 x ,
and we can assume that z /

∈ Z * 1 . Then z ∈ E 3 , by (6.79). If z ∈ Z, (6.50) says that z ∈ Z 1 ∪ Z 2 . This is impossible, because Z 1 ⊂ Z * 1 and dist(z, Z 2 ) ≥ 2 t . So z / ∈ Z. Let w ∈ E be such that z = f 3 • f 2 • f 1 (w)
. By (6.52), w ∈ Z, and then z = w. So z ∈ E, and (6.86) already takes care of it. This proves (6.85) and (6.84) in our second case.

We are left with the case when dist(R, Z 2 ) ≤ 2 t+1 . In this case we cannot prove (6.85), but instead we proceed as in the last case of Lemma 6.60 (when (6.72) holds). The point is that R is so close to Z 2 that it lies in the middle of the grid D t , and there is an l ≤ m such that none of the S ∈ N m+1 (R) meets a face w l + ∂Q, Q ∈ D t . Then the corresponding images φ l (E 4,l-1 ∩ S) are contained in d-dimensional grids S d,j (Q), whose H d -measures are controlled. The proof is the same as before, and completes our proof of Lemma 6.83.

Set Z * 2 = z ∈ R n ; dist(z, Z 2 ) ≤ 2 t+2 √ n . Notice that Z * 2 contains all the cubes Q + w l , Q ∈ D t ;
the argument here and below is just the same as for Z * 1 , near (6.79). Then (6.88)

f 5 (z) = z for z / ∈ Z * 2 and f 5 (Z * 2 ) ⊂ Z * 2
because the φ l only act nontrivially on the cubes Q + w l , and preserve them.

Recall from (6.51) that Z 2 = ∂B y ∩H, where H is a small neighborhood of the d-plane P , as in (6.39); then we can cover Z 2 by less than Cε -d+1 0 balls of radius ε 0 r y . By the same argument, and because (6.81) says that 2 t and ε 0 r y are comparable, we need less than

Cε -d+1 0 dyadic cubes of sidelength 2 t-j to cover Z * 2 . All these cubes R, if they touch Z * 2 , are such that dist(R, Z 2 ) ≤ r y /2, so Lemma 6.83 says that H d (R ∩ E 5 ) ≤ C2 td ≤ Cε d 0 r d y . Altogether, (6.89) H d (E 5 ∩ Z * 2 ) ≤ Cε 0 r d y .
We are ready for the final comparison. Set

f = f 5 • f 4 • f 3 • f 2 • f 1 , and then ϕ t (z) = tf (z) + (1 -t)z for 0 ≤ t ≤ 1 and z ∈ R 3 . The maps ϕ t satisfy the properties (2.2)-(2.4). Set Z * = Z ∪ Z * 1 ∪ Z * 2 .
Notice that (6.90) f (z) = z for z / ∈ Z * , and f (Z * ) ⊂ Z * by (6.52), (6.79), and (6.88). So, if we define the W t and W as in (2.5), we get that W is contained in the convex hull of Z * , and (2.6) holds (with

U = B(0, b)) because Z * ⊂ B(0, b).
We can apply (4.9) or (4.4) (depending on which definition of minimality we use), with h(δ) = 0. Let us first see how to derive the final contradiction if we use (4.9). We get that

H d (E \ E 5 ) ≤ H d (E 5 \ E) (because here F = ϕ 1 (E) = f (E) = E 5 ), or equivalently (6.91) H d (E ∩ Z * ) ≤ H d (E 5 ∩ Z * ), because E and E 5 coincide out of Z * . Let us check that (6.92) H d (E 5 ∩ Z * ) ≤ Cε 0 r d y + C x,y r d-1 x . Let z ∈ E 5 ∩ Z * be given. The case when z ∈ Z * 2 is taken care of by (6.89). If z ∈ Z * 1 \ Z * 2 , (6.88) says that z ∈ E 4 ; then dist(z, Z 2 ) ≥ 2 t+2 √ n because z / ∈ Z * 2
, so z lies in the set of (6.80); (6.80) takes care of this case. So we may assume that z ∈ Z * \ Z * 1 ∪ Z * 2 . By (6.88) and (6.79), z ∈ E 3 . Also, z ∈ Z, so (6.50) says that z ∈ Z 1 ∪ Z 2 . This case is impossible, because Z i ⊂ Z * i ; (6.92) follows.

On the other hand, (6.93)

H 2 (E ∩ Z * ) ≥ H 2 (E ∩ B y ) ≥ C -1 r d y ,
by Lemma 2.15, because y ∈ E, E is reduced, and 2B y ⊂ B(0, b).

Incidentally, (6.93) would be easier to prove here because we have (6.13). Or we could have used the fact that the upper density of E at almost every point is larger than some positive constant, restricted to such an y in Proposition 6.11, and then chosen r y specifically so that (6.93) holds.

We get the desired contradiction with (6.91) or (6.92) by choosing ε 0 small, and then r x very small (depending on x, y, and r y ).

The reader may be worried about the case when d = 1; this case is sufficiently simple to be treated directly, but let us say a few words about it for the sake of completeness. In this case, P ′ = {0} and lemma 6.15 says that there is a ρ ∈ (r x /2, 2r x /3) such that the cone S ρ over the (n -2)-dimensional sphere

S ρ = x + [Q ∩ ∂B(0, ρ)] in the vertical hyperplane x + Q, does not meet A ∩ E.
At the same time, (6.13) says that the (n -1)-dimensional cone Σ centered at y and defined by dist(z, P ) = 2ε 0 |z -y| does not meet E ∩ B y \ {y}. We choose r x so small that S ρ meets Σ well inside B y . Denote by Ω the region of V (x, y) bounded by S ρ ∪ Σ; thus Ω is the intersection of V (x, y) with a thin double-cone centered at 0 and y. Notice that Ω ∩ V (x, x) does not meet E, by (6.14). It is easy to contract Ω onto the line segment [x, y] by radial vertical motion, and since E does not meet ∂Ω except at y, we can make sure that the contraction f is Lipschitz and does not move the points of E \ Ω. In addition, f (E ∩ Ω) does not meet B x (again by (6.14)), so we can contract the whole piece f (E ∩ Ω) ⊂ [x, y] to the point y along [x, y]. The final contradiction is now obtained as in the general case.

Return to the case when d > 1. If we use (4.4) (instead of (4.9)), we get that (6.94)

H d (E ∩ W ) ≤ H d (f (E ∩ W )), with W = {w ∈ R 3 ; f (w) = w}. Recall that f (z) = z out of Z * (by (6.90)), so W ⊂ Z * and (6.95) H d (f (E ∩ W )) ≤ H d (f (E ∩ Z * )) ≤ H d (E 5 ∩ Z * )
(by (6.90) again). Notice that (6.96)

|f 5 (z) -z| ≤ 2 t-j √ n for z ∈ R n ,
by the proof of (6.78). Then E 5 = f 5 • f 4 (E 3 ) does not meet B(y, r y /2), by (6.78), (6.96), and because E 3 does not meet B y (by (6.50) and the definition (6.43)). Hence

E ∩ B(y, r y /2) ⊂ E ∩ W \ E 5 , just because f (z) = z if z ∈ E \ E 5 , and (6.97) H d (E ∩ B(y, r y /2)) ≤ H d (E ∩ W ) ≤ H d (f (E ∩ W )) ≤ H d (E 5 ∩ Z * ) ≤ Cε 0 r 2 y + C x,y r d-1
x , by (6.94), (6.95) and (6.92). On the other hand, H d (E∩B(y, r y /2)) ≥ C -1 r d y by Lemma 2.15, so we also get a contradiction when we use (4.4) in the accounting.

This completes our proof of Proposition 6.11. As was explained before, (6.12) follows and so E coincides with a cone inside A.

We return to the proof of Theorem 6.2. So far we proved that there is a cone C centered at the origin, that satisfies (6.4). We still need to check that C is a reduced minimal set. Notice that it is reduced, by (6.4) and because E is reduced. [See Definition 2.12 for the definition.] Suppose it is not minimal. Then we can find a competitor F = ϕ 1 (C) where {ϕ t } satisfies (2.2)-(2.6) with U = R 3 , and such that (6.98)

H 2 (F \ C) < H 2 (C \ F ).
Observe that we use Definition 4.8 for minimality, but Proposition 4.10 tells us that we could also have used Definition 4.3 (or even Definition 4.1, since here h(r) = 0). Recall from (2.6) that the set W of (2.5) is bounded. Since C is a cone, we may always dilate everything by a small constant (this does not change (6.98)), so we may assume that W ⊂ B(0, a).

We want to use F and the ϕ t to construct a deformation of E and contradict its minimality. First observe that

C ∩ B(0, b) is a deformation of E in B(0, b). Indeed, pick any b ′ ∈ (a, b), and set g(x) = 0 for x ∈ B(0, a), g(x) = b ′ (|x|-a) b ′ -a x |x| for x ∈ B(0, b ′ )\B(0, a), and g(x) = x for x ∈ B(0, b) \ B(0, b ′ ). Then set f t (x) = tg(x) + (1 -t) x for 0 ≤ t ≤ 1. It is easy to see that the g t satisfy (2.2)-(2.6) with U = B(0, b), and that g 1 (E) = C ∩ B(0, b)
(by (6.4)). We now put our two deformations together. That is, we set ψ t (x) = g 2t (x) for 0 ≤ t ≤ 1/2 and ψ t (x) = ϕ 2t-1 (g 1 (x)) for 1/2 ≤ t ≤ 1, and get a new family that satisfies (2.2)-(2.6) with U = B(0, b). By minimality of E, (6.99)

H 2 (ψ 1 (E) \ E) ≥ H 2 (E \ ψ 1 (E)). But ψ 1 (E) = F ∩ B(0, b), so (6.99) just says that H 2 (E) ≤ H 2 (F ∩ B(0, b)). Recall from (6.3) that H 2 (E ∩ B(0, r)) = θ 0 r 2 for a < r < b. Also, H 2 (C ∩ A) = H 2 (E ∩ A) by (6.4), so H 2 (C ∩ B(0, r)) = θ 0 r 2 for r > 0 and H 2 (C ∩ B(0, b)) = H 2 (E). Altogether, H 2 (C ∩ B(0, b)) = H 2 (E) ≤ H 2 (F ∩ B(0, b))
, which contradicts (6.98). This completes the proof of Theorem 6.2.

Almost constant density

In this section we use Theorem 6.2 to show that if E is a reduced almost-minimal set of dimension d in B(0, b) ⊂ R n , with sufficiently small gauge function, and if θ(r) = r -d H d (E ∩ B(0, r)) varies sufficiently little between a and b, then E is very close to a reduced minimal cone in A = B(0, b) \ B(0, a).

We also have a (simpler) result when a = 0 (Proposition 7.24), and we shall see in Proposition 7.31 that every blow-up limit of a reduced almost-minimal set of dimension d in U ⊂ R n is a reduced minimal cone.

Proposition 7.1. Let n, d (the dimensions), a, and b be given, with 0 < a < b < +∞. For each τ > 0, we can find ε > 0 so that the following holds. Let E be a reduced almost-minimal set of dimension d in B(0, b) ⊂ R n , with gauge function h, and assume that h(2b) ≤ ε. Also assume that θ(b) ≤ θ(a) + ε. Then there is a reduced minimal cone C centered at the origin such that

(7.2) dist(x, C) ≤ τ for x ∈ E ∩ B(0, b -τ ) \ B(0, a + τ ), (7.3) dist(x, E) ≤ τ for x ∈ C ∩ B(0, b -τ ) \ B(0, a + τ ), (7.4) |H d (E ∩ B(x, r)) -H d (C ∩ B(x, r))| ≤ τ for x ∈ R n and r > 0 such that B(x, r) ⊂ B(0, b -τ ) \ B(0, a + τ ), and
(7.5) |θ(r) -r -d H d (C ∩ B(0, r))| ≤ τ for a + τ ≤ r ≤ b -τ.
See Definition 2.12 for the notion of reduction, and observe that if E is not reduced, Remark 2.14 allows us to apply Proposition 7.1 to E * . Our statement allows C = ∅, but only if θ(r) is very small (by (7.5)). Notice that (7.5) forces θ(r) to be very close to the constant density for C. This is a constraint on θ(r), but also can be a way to determine C in terms of θ(r).

We shall prove the proposition by compactness. Fix n, d, a, and b, and consider a sequence {E n } of reduced almost-minimal sets on B(0, b), with gauge functions h n such that ε n = h n (2b) tends to 0. It is a standard fact about the Hausdorff distance that we can extract a subsequence {n k } so that the sets E n k converge to a limit F ⊂ B(0, b), as in the beginning of Section 3 (see for instance [D2], Proposition 34.6 on page 214).

Set h l = sup n≥l h n for l ≥ 0. Notice that h l is nondecreasing (just like the h n ), and lim r→0 h l (r) = 0, because each h n tends to 0 and h n (r) ≤ ε n for r ≤ 2b. Then Lemma 4.7 (applied to the E n k ) says that for each l, F is almost-minimal in B(0, b), with gauge function h l . But since we can take l as large as we want and h l (2b) tends to 0, we get that

(7.6) F is a minimal set in B(0, b).
We also get that

(7.7) H d (F ∩ V ) ≤ lim inf k→+∞ H d (E n k ∩ V ) for every open set V ⊂ B(0, b), by (3.4) in Lemma 3.3 (applied with U = B(0, b), M = 1, δ = 2b
, and h = h l (2b) for some large l). We also need an inequality in the other direction, so we apply Lemma 3.12 to E n k , with the same choice of U , M , δ, and h, and then let l tend to +∞; this yields

(7.8) lim sup k→+∞ H d (E n k ∩ H) ≤ H d (F ∩ H) for every compact set H ⊂ B(0, b).

52

Notice that H d (F ∩ ∂B(x, r)) = 0 for almost-every r ∈ (a, b); (7.7) and (7.8) yield (7.9)

lim sup k→+∞ H d (E n k ∩ B(x, r)) ≤ H d (F ∩ B(x, r)) = H d (F ∩ B(x, r)) ≤ lim inf k→+∞ H d (E n k ∩ B(x, r))
for such r, so that (7.10)

H d (F ∩ B(0, r)) = lim k→+∞ H d (E n k ∩ B(0, r)) for almost-every r ∈ (a, b). Set θ k (r) = r -d H d (E n k ∩B(0, r)) for r ≤ b, and suppose in addition that θ k (b) ≤ θ k (a)+ε k .
Pick a ′ and b ′ , with a < a ′ < b < b ′ , such (7.10) holds for a ′ and b ′ . Then (7.11)

H d (F ∩ B(0, b ′ )) = lim k→+∞ H d (E n k ∩ B(0, b ′ )) ≤ lim inf k→+∞ H d (E n k ∩ B(0, b)) = b d lim inf k→+∞ θ k (b) ≤ b d lim inf k→+∞ [θ k (a) + ε k ] = b d lim inf k→+∞ θ k (a) = b d a -d lim inf k→+∞ H d (E n k ∩ B(0, a)) ≤ b d a -d lim k→+∞ H d (E n k ∩ B(0, a ′ )) = b d a -d H d (F ∩ B(0, a ′ )).
We let b ′ tend to b and get that (7.12)

θ(b) =: b -d H d (F ∩ B(0, b)) ≤ a -d H d (F ∩ B(0, a ′ )) = a -d (a ′ ) d θ(a ′ ).
Recall from Proposition 5.16 and (7.6) that θ is nondecreasing on (a, b), so θ(a + ) = lim a ′ →a + θ(a ′ ) exists, and (7.12) says that θ(b) ≤ θ(a + ). But θ(b) ≥ θ(a + ) by Proposition 5.16, so θ(b) = θ(a + ) and θ is constant on (a, b). By Theorem 6.2, there is a reduced minimal cone C centered at the origin such that (7.13)

F ∩ A = C ∩ A.
In particular, (7.2) and ( 7.3) hold for E n k and k large enough (and with any τ > 0 given in advance).

It will be good to know that

(7.14) H d (C ∩ ∂B) = 0 for every ball B,
but since this is easy to believe, we shall only check it in Lemma 7.34, at the end of the section. Now let a ball B be given, with B ⊂ B(0, bτ /2) \ B(0, a + τ /2). By (7.14), (7.13), (7.7), (7.8), and the proof of (7.10),

(7.15) lim k→+∞ H d (E n k ∩ B) = H d (F ∩ B) = H d (C ∩ B),
so the analogue of (7.4) for E n k and B holds for k large.

We are now ready to prove Proposition 7.1. Let a, b, and τ be given, and suppose that we cannot find ε so that the proposition holds. Then we can find E n ⊂ B(0, b) that satisfies the hypotheses with ε n = 2 -n , but not the conclusion. We extract a convergent subsequence, as above, and get a set F such that (7.6)-(7.13) hold. Let us show that (7.2)-(7.5) hold for k large; this will prove the desired contradiction.

We already know that (7.2) and (7.3) hold for k large. For (7.4) we only know that it holds for k large when x and r are fixed, so we need a little bit of uniformity. By (7.14), H d (C ∩ B(x, r)) is a continuous function of x and r, so we can find a finite collection of balls B i , i ∈ I, such that B i ⊂ B(0, bτ /2) \ B(0, a + τ /2) for i ∈ I, and such that whenever B(x, r) is a ball such that B(x, r) ⊂ B(0, bτ ) \ B(0, a + τ ) (as in (7.4)), we can find i, j ∈ I such that (7.16)

B i ⊂ B(x, r) ⊂ B j and H d (C ∩ B j \ B i ) ≤ τ /5.
We also know from (7.15) that for k large enough, (7.17)

|H d (C ∩ B i ) -H d (E n k ∩ B i )| ≤ τ /5 for i ∈ I.
Now let B(x, r) be as in (7.4), and let i, j ∈ I be such that (7.16) holds. Then (7.18)

H d (C ∩ B i ) ≤ H d (C ∩ B(x, r)) ≤ H d (C ∩ B j ) ≤ H d (C ∩ B i ) + τ /5
and, for k large (depending on I, but not on B(x, r)), We are left with (7.5) to prove. Call θ 0 the constant density of C; that is,

(7.19) H d (C ∩ B i ) -τ /5 ≤ H d (E n k ∩ B i ) ≤ H d (E n k ∩ B(x, r)) ≤ H d (E n k ∩ B j ) ≤ H d (C ∩ B j ) + τ /5 ≤ H d (C ∩ B i ) + 2τ /
θ 0 = r -d H d (C ∩ B(0, r)) for r > 0. Notice that H d (C ∩ ∂B(0, r)) = 0 for a < r < b, so (7.20) lim k→+∞ H d (E n k ∩ B(0, r)) = H d (F ∩ B(0, r)) = H d (C ∩ B(0, r)) = θ 0 r d ,
by (7.7), (7.8), and (7.13). As before, we need some uniformity, so we pick a finite (but large) collection {r i } i∈I in (a, b). We know that for k large enough,

(7.21) |r -d i H d (E n k ∩ B(0, r i )) -θ 0 | ≤ τ /2 for i ∈ I.
Let r ∈ [a + τ, bτ ] be given. If our collection {r i } is dense enough we can find i, j ∈ I such that r i ≤ r ≤ r j and r j /r i is as close to 1 as we want. Then for k large

(7.22) θ k (r) = r -d H d (E n k ∩B(0, r)) ≤ r -d H d (E n k ∩B(0, r j )) ≤ r -d r d j (θ 0 +τ /2) ≤ θ 0 +τ, by (7.21). Similarly, (7.23) θ k (r) ≥ r -d H d (E n k ∩ B(0, r i )) ≥ r -d r d i (θ 0 -τ /2) ≤ θ 0 -τ,
and (7.5) holds for k large. This completes the proof of Proposition 7.1 by contradiction.

Proposition 7.1 easily extends to the case when a = 0. Let us give a slightly more invariant statement for this to simplify later uses.

Proposition 7.24. For each τ > 0, we can find ε > 0 so that the following holds. Let E be a reduced almost-minimal set in the open set U ⊂ R n , with gauge function h. Let x ∈ E and r > 0 be such that B(x, r) ⊂ U , h(2r) ≤ ε, and

(7.25) θ(x, r) ≤ inf 0<t<r/100 θ(x, t) + ε, where we set θ(x, t) = t -d H d (E∩B(x, t)).
Then there is a reduced minimal cone C centered at x, such that

(7.26) dist(y, C) ≤ τ r for y ∈ E ∩ B(x, (1 -τ )r), (7.27) dist(y, E) ≤ τ r for y ∈ C ∩ B(x, (1 -τ )r), and
(7.28) |H d (E ∩ B(y, t)) -H d (C ∩ B(y, t))| ≤ τ r d for y ∈ R n and t > 0 such that B(y, t) ⊂ B(x, (1 -τ )r).
By scale invariance, is enough to prove this when x = 0, r = 1, and E is almostminimal in (an open set U that contains) B(0, 1). We can proceed as in the proof of Proposition 7.1 (with a = 0 and b = 1), up to (7.10) included. Then pick a 1 ∈ (0, 1/100) and b ′ ∈ (a 1 , 1) such that (7.10) holds for a 1 and b ′ . By (7.25), θ k (1) ≤ θ k (a 1 ) + ε k , so (7.29)

H d (F ∩ B(0, b ′ )) = lim k→+∞ H d (E n k ∩ B(0, b ′ )) ≤ lim inf k→+∞ H d (E n k ∩ B(0, 1)) = lim inf k→+∞ θ k (1) ≤ lim inf k→+∞ [θ k (a 1 ) + ε k ] = lim inf k→+∞ θ k (a 1 ) = a -d 1 lim inf k→+∞ H d (E n k ∩ B(0, a 1 )) = a -d 1 H d (F ∩ B(0, a 1 )),
by (7.10). We let b ′ tend to 1 and get that (7.30)

H d (F ∩ B(0, 1)) ≤ a -d 1 H d (F ∩ B(0, a 1 )).
Now F is minimal in B(0, 1), by (7.6), so Proposition 5.16 says that θ(r) = r -d H d (F ∩ B(0, r)) is nondecreasing in (0, 1). Then θ(1) = lim r→1 θ(r) ≥ θ(a 1 ). But (7.30) says that θ(1) ≤ θ(a 1 ), so θ(r) is constant on [a 1 , 1], hence also on (0, 1] because we can take a 1 as close to the origin as we want. Theorem 6.2 says that F is a reduced minimal cone, and we can continue the argument as in Proposition 7.1.

Let us also give a baby version of Proposition 7.24.

Proposition 7.31. Let E be a reduced almost-minimal set in the open set U , and let

x ∈ E be such that

(7.32) θ(x) = lim r→0 r -d H d (E ∩ B(x, r)) exists.
Then every blow-up limit of E at x is a reduced minimal cone F centered at the origin, and

H d (F ∩ B(0, 1)) = θ(x).
See Definition 2.12 for "reduced", and Definition 4.8 or 4.3 for almost-minimality. We do not care about the precise values of the gauge function h(r) here, we just need it to tend to 0 when r tends to 0, but of course (7.32) is easier to obtain when h is small, by Proposition 5.24. A blow-up limit of E at x is any closed set in R n that can be obtained as the limit (as in (3.1)) of a sequence {r -1 k (Ex)}, with lim k→+∞ r k = 0. To prove the proposition, assume that the sequence {r -1 k (Ex)} converges to some limit F . Notice that for every small a > 0,

F k = r -1 k (E -x)
is, for k large enough, a reduced almost-minimal set with gauge function h(t) = h(at). Then Lemma 4.7 says that F is a reduced almost-minimal set, with gauge function h. But since we can take a as small as we want, and lim a→0 h(t) = 0 for each t, F is a reduced minimal set in R n .

We also get that

(7.33) H d (F ∩ B(0, r)) = lim k→+∞ H d (F k ∩ B(0, r)) for almost every r > 0,
by the proof of (7.10). But

H d (F k ∩ B(0, r)) = r -d k H d (E ∩ B(x, r k r))
, which tends to r d θ(x) by (7.32). Thus H d (F ∩ B(0, r)) = r d θ(x) for almost every r > 0. Theorem 6.2 says that F is a cone, and Proposition 7.31 follows.

Propositions 7.1 and 7.24 will help us find many places where E looks a lot like a minimal cone. We shall then use a variant of Reifenberg's topological disk theorem to get a local parametric description of E.

We end this section with the proof of (7.14).

Lemma 7.34. Let C be a closed cone in R n (centered at the origin), and assume that

H d (C ∩ B(0, 1)) < +∞. Then H d (C ∩ ∂B) = 0 for every ball B.
Let 0 < a < b < +∞ be given, and set

A = x ∈ R n ; a ≤ |x| ≤ b ; it is enough to show that H d (C ∩ A ∩ ∂B) = 0.
Denote by S = ∂B(0, 1) the unit sphere, and set

ϕ(r, θ) = rθ for r ∈ [a, b] and θ ∈ S. Notice that ϕ is a bilipschitz mapping from [a, b] × S to A, and ϕ -1 (C ∩ A) = [a, b] × T , where T = S ∩ C. Then (7.35) H d ([a, b] × T ) = H d (ϕ -1 (C ∩ A)) ≤ CH d (C ∩ A) < +∞.
Let us first check that (7.36)

H d-1 (T ) ≤ CH d ([a, b] × T ) < +∞,
where C depends also on a and b, but we don't care. Let ε > 0 be given. By (7.35), there is a covering of [a, b] × T by balls

B i = B(x i , r i ), i ∈ I, with r i ≤ ε for i ∈ I and (7.37) i∈I r d i ≤ CH d ([a, b] × T ) + ε.
By compactness, we may assume that I is finite. For t ∈ [a, b], set P t = {t} × S. Set B i,t = B i ∩ P t , and denote by I t the set of i ∈ I such that B i,t = ∅. Notice that the balls B i,t = B i ∩ P t , i ∈ I t form a covering of T by balls of radius at most ε, so

(7.38) H d-1 ε (T ) ≤ c d-1 i∈I t diam(B i,t ) d-1 ≤ C i∈I t H d-1 (B i,t ) = C i∈I H d-1 (B i,t )
where

H d-1 ε (T )
is the usual infimum over coverings that is used to define H d-1 (T ). We average this over [a, b] and get that (7.39) by (7.37). Then we let ε tend to 0 and we get (7.36). By (7.36), we can apply Theorem 2.10.45 in [Fe], which says that there are positive constants C 1 and C 2 such that (7.40)

H d-1 ε (T ) ≤ C [a,b] i∈I H d-1 (B i,t )dt = C i∈I [a,b] H d-1 (B i,t )dt ≤ C i∈I r d i ≤ CH d ([a, b] × T ) + Cε
C 1 H d (U × V ) ≤ H 1 (U )H d-1 (V ) ≤ C 2 H d (U × V ) for U ⊂ [a,
v ∈ T , ϕ -1 (C ∩ ∂B ∩ A)) does not meet [a, b] × {v} more than twice. Then (7.41) H d (C ∩ ∂B ∩ A) ≤ CH d (ϕ -1 (C ∩ ∂B ∩ A)) ≤ C v∈T H 1 [a, b] × {v} ϕ -1 (C ∩ ∂B ∩ A) dH d-1 (v) = 0
because ϕ is bilipschitz and by Fubini. Lemma 7.34 follows.

C. MINIMAL CONES OF DIMENSION 2

In this part we study the reduced minimal cones E of dimension 2 in R n . The goal is not to obtain a finite list as when n = 3, but a fairly simple description of K = E ∩∂B(0, 1) as a finite collection of geodesic arcs of circles in ∂B(0, 1), with lengths bounded from below, and that meet only by sets of three with 120 • angles. See Proposition 14.11.

We start with a description of the tangent cones to E, where we show that if x lies in K = E ∩ ∂B(0, 1), every blow-up limit of K at x is a one-dimensional reduced minimal set. See Theorem 8.23 for a more precise statement where K is approximated by minimal sets in small balls centered at x. This description is based on the fact that if

E × R m is a reduced (m + d)-dimensional minimal set in R n × R m , then E is a d-dimensional minimal set in R n . See Proposition 8.3.
With this and the simple version of Reifenberg's topological disk theorem which we describe in Section 12, we could easily obtain information on the bi-Höder structure of E, but since it is preferable to use C 1 information, we will show that K satisfies a weak form of almost-minimization (see Definition 9.1 and Proposition 9.4), and then follow some of the general theory to prove that these weak almost-minimal sets of dimension 1 have some form of C 1 regularity. This is done in Sections 11-13, and for the local description of K by curves we use a simple version of Reifenberg's Topological Disk Theorem (Proposition 12.6).

In the mean time we show in Section 10 that the reduced minimal sets of dimension 1 are cones. See Theorem 10.1.

We state and prove the description of K = E ∩ ∂B(0, 1) when E is a reduced minimal cone of dimension 2 in Section 14. The fact that K is composed of C 1 curves comes from the description of the weak almost-minimal sets given in Sections 11-13, and then a construction of harmonic competitors for a cone, which is done in Section 13, allows us to check that all the little arcs that compose K are arcs of great circles of ∂B(0, 1).

Minimality of a product and tangent sets

We start our study of minimal cones with what would be the most natural attempt to get a nice description of the d-dimensional minimal sets by induction on d. It will be easy to control the minimal sets of dimension 1. Then assume that we have a reasonable description of the minimal sets of dimension d; we can try to use this to get information on minimal cones of dimension d + 1, because we shall see in this section that blow-up limits of these cones are products of a minimal set of dimension d by a line. In turn a good control on the minimal cones may lead to valuable information on minimal sets of the same dimension, by the results of Section 7. We shall only be able to make this plan work for 2-dimensional minimal sets, but we shall write down the first stage in full generality.

In this section, we first show that if the product E × R m is a reduced minimal set of dimension d + m, then E is a reduced dimensional minimal set of dimension d. We shall then use this to show that if the cone over K is a reduced minimal set of dimension d + 1, then the blow-up limits of K are minimal sets of dimension d, and even that K is close to minimal sets of dimension d in small balls.

Let us recall what we mean by minimal sets. This is the same as almost-minimal sets as in Definition 4.3 or 4.8, with h(δ) = 0 for all δ, but since the definition simplifies slightly, let us state it again for future reference.

Definition 8.1. The closed set E ⊂ R n is a minimal set of dimension d if, whenever ϕ : R n → R n is a Lipschitz function such that ϕ(x) = x out of some compact set of R n , then (8.2) H d (E \ ϕ(E)) ≤ H d (ϕ(E) \ E).
We say that E is reduced when H d (E ∩ B(x, r)) > 0 for x ∈ E and r > 0.

See Definition 2.12 for the original definition of reduced sets. Concerning our new definition of minimality, it is easily seen to be equivalent to Definition 4.3 with the gauge function h = 0. The point is that if ϕ is as in Definition 8.1, we can set ϕ t (x) = tϕ(x) + (1t)x for 0 ≤ t ≤ 1; the family {ϕ t } satisfies (2.2)-(2.6) for some δ > 0, and then (8.2) is the same as (4.9) with h(δ) = 0. Let us also recall that Definitions 4.3 and 4.8 are equivalent, by Proposition 4.10.

The first result of this section is a simple observation on the structure of some minimal product sets.

Proposition 8.3. Let m, n, and d be integers, and let E × R m be a reduced

(d + m)- dimensional minimal set in R n+m . Then E is a reduced d-dimensional minimal set in R n .
Let ϕ : R n → R n be a lipschitz function such that ϕ(x) = x out of some compact set; we want to show that (8.2) holds. Our plan is to use ϕ to construct a Lipschitz function f : R n+m → R n+m and apply the minimality of E × R m .

Let R > 0 be large, to be chosen near the end, and choose a smooth radial cut-off function ψ on R m , with (8.4)

ψ(y) = 1 for |y| ≤ R, 0 ≤ ψ(y) ≤ 1 for R < |y| < R + 1, ψ(y) = 0 for |y| ≥ R + 1,
and |∇ψ(y)| ≤ 2 everywhere. Then define g :

R n × R m → R n by (8.5) g(x, y) = ψ(y)ϕ(x) + (1 -ψ(y)) x for x ∈ R n and y ∈ R m
and set f (x, y) = (g(x, y), y). Notice that g(x, y) = x + ψ(y)(ϕ(x)x), and since ϕ(x)x is bounded, (8.6) g and f are Lipschitz, with bounds that do not depend on R.

Set W ♯ = {(x, y) ∈ R n × R m ; f (x, y) = (x, y)}. If (x, y) ∈ W ♯ , then ψ(y) = 0 and ϕ(x) = x. That is, W ♯ ⊂ W × B(0, R + 1), where W = {x ∈ R n ; ϕ(x) = x}. In addition, if B is a ball that contains W ∪ ϕ(W ), then (8.7) W ♯ ∪ f (W ♯ ) ⊂ Σ, where we set Σ = B × B(0, R + 1).
By assumption, E ♯ = E × R m is minimal, and (8.2) says that (8.8)

H d+m (E ♯ \ f (E ♯ )) ≤ H d+m (f (E ♯ ) \ E ♯ ).
By (8.7), both sets are contained in Σ; we add H d+m (E ♯ ∩ f (E ♯ ) ∩ Σ) to both sides and get that (8.9)

H d+m (E ♯ ∩ Σ) ≤ H d+m (f (E ♯ ∩ Σ)).
We shall need to be a little careful when we compute Hausdorff measures of product spaces, because the Hausdorff measure of a product is not always exactly the product of the Hausdorff measures. Fortunately, E ♯ is rectifiable (with dimension d+m), by Section 2 (or directly [Al]), so it will be possible to use the following variant of the coarea formula, which we borrow from [Fe], Theorem 3.2.22. We change the notation a little, and reduce slightly the generality.

Lemma 8.10. Let d, m, n be nonnegative integers, let Z ⊂ R n be a rectifiable set of dimension m + d, with H m+d (E) < +∞, and let π : Z → R m be Lipschitz. Then

(8.11) π -1 (y) is a d-dimensional rectifiable set for H m -almost every y ∈ R m ,
and

(8.12) Z∩A apJ m π(z) dH d+m (z) = y∈R m z∈π -1 (y) 1 A∩Z (z) dH d (z) dH m (y)
for every Borel set A ⊂ R n , where the approximate Jacobian apJ m π of dimension m is defined (almost-everywhere) by apJ m π(z) = || ∧ m Dπ(z)||; see 1.3.1 in [Fe].

We apply Lemma 8.10 to Z = E ♯ ∩ Σ, and where π is the restriction to Z of the second projection π : R n+m → R m . Here π is even linear, and since π is 1-Lipschitz and the tangent plane to E ♯ always contains the m-plane {0} × R m where Dπ is the identity, the approximate Jacobian apJ m π is equal to 1.

Also, π -1 (y) = E ∩ B for y ∈ B(0, R + 1) ⊂ R m , and ∅ otherwise, so (8.11) says that (8.13) E ∩ B is rectifiable, and (8.12) with A = Σ says that (8.14)

H d+m (E ♯ ∩ Σ) = H d (E ∩ B) H m (R m ∩ B(0, R + 1)).
Similarly set Σ 1 = B × B(0, R), and observe that f (x, y) = (ϕ(x), y) for (x, y) ∈ Σ 1 , by (8.5) and because ψ(y) = 1 for y ∈ B(0, R).

Then (8.15) f (E ♯ ∩ Σ 1 ) = f (E ∩ B) × B(0, R) = ϕ(E ∩ B) × B(0, R) = [ϕ(E) ∩ B] × B(0, R) since ϕ(E ∩ B) = ϕ(E) ∩ B because by definition of B, ϕ(B) ⊂ B and ϕ(x) = x out of B. Notice that f (E ♯ ) is rectifiable, because E ♯ is rectifiable and f is Lipschitz. Apply Lemma 8.10 to the product set f (E ♯ ∩ Σ 1 ) = [ϕ(E) ∩ B] × B(0, R) (which is rectifiable because it is contained in f (E ♯ ))
, and with the same projection π as before. The proof of (8.14) yields (8.16)

H d+m (f (E ♯ ∩ Σ 1 )) = H d+m [ϕ(E) ∩ B] × B(0, R) = H d (ϕ(E) ∩ B) H m (B(0, R)). Set Ξ = E ♯ ∩ (Σ \ Σ 1 ) = [E ∩ B] × [B(0, R + 1) \ B(0, R)]. Now (8.17) H d+m (f (Ξ)) ≤ CH d+m (Ξ) = CH d (E ∩ B) H m (B(0, R + 1) \ B(0, R)) ≤ CR m-1
because f is Lipschitz and by Lemma 8.10 (applied to the rectifiable product set Ξ ; the proof is again the same as for (8.14)). Here C depends on the Lipschitz constant for ϕ through |∇f |, and on E ∩B. This is unusual, but the main point is that C does not depend on R, by (8.6).

Return to the right-hand side of (8.9). Let z lie in f (E ♯ ∩ Σ), and let (x, y) R), and (8.9) yields (8.18)

∈ E ♯ ∩ Σ be such that z = f (x, y). If y ∈ B(0, R + 1) \ B(0, R), then (x, y) ∈ Ξ and z ∈ f (Ξ). Otherwise, ψ(y) = 1, g(x, y) = ϕ(x), and z = (ϕ(x), y) ∈ [ϕ(E) ∩ B] × B(0, R). Thus f (E ♯ ∩ Σ) ⊂ f (Ξ) ∪ [ϕ(E) ∩ B] × B(0,
H d+m (E ♯ ∩ Σ) ≤ H d+m (f (E ♯ ∩ Σ)) ≤ H d+m (f (Ξ)) + H d+m ([ϕ(E) ∩ B] × B(0, R)) ≤ CR m-1 + H d+m (f (E ♯ ∩ Σ 1 )) ≤ CR m-1 + H d (ϕ(E) ∩ B) H m (B(0, R))
by (8.17), (8.15), and (8.16). Now apply (8.14), divide by H m (B(0, R + 1)), and use (8.18) to get that

(8.19) H d (E ∩ B) = H m (B(0, R + 1)) -1 H d+m (E ♯ ∩ Σ) ≤ H d (ϕ(E) ∩ B) + CR -1 .
Then let R tend to +∞, remove from both sides H d (E ∩ ϕ(E) ∩ B), and get (8.2).

We still need to show that E is reduced. So let x ∈ E and r > 0 be given. The same computations as above show that (8.20)

H d+m (E ♯ ∩ [B(x, r) × B(0, 1)]) = H d (E ∩ B(x, r)) H m (B(0, 1) ∩ R m ).
The left-hand side is positive because E ♯ is reduced, so H d (E ∩ B(x, r)) > 0, as needed. This completes our proof of Proposition 8.3.

Remark 8.21. The proof of Proposition 8.3 relies on the fact that we work with minimal sets, so that we were able to use the Lipschitz norm of ϕ in the argument above, compensate by taking R enormous, and get away with it. There will be an argument of the same type in the next section, but we will then be working with almost-minimal sets, and have to pay by including the Lipschitz properties of ϕ in the definition of weak almost-minimal sets.

Remark 8.22. Probably the proof of Proposition 8.3 allows us to show that if a product

E ♯ = E × F of a d 1 -dimensional closed set E ⊂ R n 1 with a d 2 -dimensional closed set F ⊂ R n 2 is a (d 1 + d 2 )-dimensional minimal set in R n 1 +n 2
, and if some mild assumptions on the behavior of

H d i (E i ∩ B(0, R)) for R large are satisfied, then each E i is a minimal set of dimension d i in R n i .
I don't know whether the converse is true, i.e., whether products of minimal sets are necessarily minimal. There is an argument in [D2] for the product of a minimal set with R n 2 (see page 530 and Exercise 76.16 in [D2]), but in the different context of Mumford-Shah minimal sets in codimension 1; the argument does not apply here, even for

E ♯ = E × R n 2 , because the sections ϕ 1 (E) ∩ [R n 1 × {y}], y ∈ R n 2 , are not always competitors for E × {y} in R n 1 × {y}.
Theorem 8.23. Let the integers 0 < d < n be given. For each ε > 0, we can find r 0 ∈ (0, 1) such if E is a reduced minimal cone of dimension d + 1 in R n , x is a point of K = E ∩ ∂B(0, 1), 0 < r ≤ r 0 , and P denotes the tangent hyperplane to ∂B(0, 1) at x, then there is a reduced minimal set L of dimension d in P such that (8.24) dist(y, L) ≤ εr for y ∈ K ∩ B(x, r) and dist(z, K) ≤ εr for z ∈ L ∩ B(x, r), and

(8.25) H d (L ∩ B(y, t -εr)) -εr d ≤ H d (K ∩ B(y, t)) ≤ H d (L ∩ B(y, t + εr)) + εr d
for every ball B(y, t) contained in B(x, r).

We start with a few comments before the proof. The definition of minimal sets was recalled (and slightly simplified) in Definition 8.1. A minimal cone is simply a minimal set which happens to be a cone centered at the origin. The two possible definitions of reduced minimal sets in P (where we see L as a subset of P , or directly as a subset of R n ) are equivalent. Indeed it is obvious that if L is minimal in R n , then it is minimal in P because every deformation in P is a deformation in R n . Conversely, suppose that L is not minimal in R n . Let ϕ : R n → R n be Lipschitz, with ϕ(x) = x out of a compact set, and such that

H d (E \ ϕ(E)) > H d (ϕ(E) \ E). Set ϕ = π • ϕ,
where π denotes the orthogonal projection on P . Let B be a large ball centered on P such that ϕ(x) = x out of B and ϕ(B) ⊂ B. Thus E, ϕ(E), and ϕ(E) coincide out of B. In addition, (8.26)

H d ( ϕ(E) ∩ B) = H d (π(ϕ(E)) ∩ B) = H d (π(ϕ(E) ∩ B)) ≤ H d (ϕ(E) ∩ B) < H d (E ∩ B),
so E is not minimal in P either.

Observe also that Theorem 8.23 contains the slightly easier observation that any blowup limit of K = E ∩ ∂B(0, 1) at a point x ∈ K is a reduced minimal set of dimension d in the tangent plane to ∂B(0, 1) at x. For a direct proof of this observation, simply follow the proof of the theorem up to (8.30).

Our estimate (8.25) looks a little ugly, compared for instance to its analogue (7.28); the point is that we don't know that H d (L ∩ ∂B(y, t)) = 0 (as happened with Lemma 7.34). Incidentally, if t ≤ εr, simply understand B(y, tεr) as the empty set (and the first half of (8.25) is trivial). We do not really need (8.25) for the rest of this text (we use it for (14.2), but we could also use (8.24) and a compactness argument using Lemma 3.12 there, and the argument would be a tiny bit simpler), but we decided to include (8.25) for the sake of completeness and to avoid potential unpleasant surprises later. We shall start with the simpler proof of (8.24) anyway.

We now prove the theorem, by contradiction and compactness. Suppose that we can find ε such that no r 0 works. Then for each integer k ≥ 0, the statement fails with r 0 = 2 -k and there is a reduced minimal cone E k , a point x k ∈ K k = E k ∩ ∂B(0, 1), and a radius r k ∈ (0, 2 -k ) such that no minimal set L k satisfies (8.24) and (8.25). We want to derive a contradiction.

Set A = (1, 0, • • • , 0). By rotation invariance, we can assume that x k = A for every k, and then P is always the same vertical plane through A.

Set F k = r -1
k (E k -A), so that now {F k } is a sequence of reduced minimal sets through the origin. We can extract a subsequence, so that after extraction {F k } converges to some closed set F ⊂ R n . See (3.1) for the definition, and recall that the existence of converging subsequences is a standard consequence of the fact that the Hausdorff distance on the set of compact subsets of a compact metric space makes this set compact too. Then Lemma 4.7 (with the gauge function h = 0) says that (8.27)

F is a reduced minimal set of dimension d + 1.

Next denote by P 0 the hyperplane parallel to P through the origin and by D the first axis (the vector line perpendicular to P 0 ). Let us check that (8.28)

F = D × F ♯ , where F ♯ = F ∩ P 0 .
Let y ∈ F be given, and let us verify that y + D ⊂ F . Since the F k tend to F , we can find y k ∈ F k so that {y k } tends to y. Set z k = r k y k + A; notice that z k ∈ E k by definition of F k , and that z k tends to A because r k tends to 0. Fix λ ∈ R, and set

v k = (1 + λr k )z k . Then v k ∈ E k for k large, because E k is a cone, and so w k = r -1 k (v k -A) lies in F k . But (8.29) w k = r -1 k ((1 + λr k )z k -A) = r -1 k ((1 + λr k )(r k y k + A) -A) = r -1 k (r k y k + λr 2 k y k + λr k A) = y k + λA + λr k y k ,
so w k tends to y + λA (because y k tends to y and r k tends to 0), and y + λA ∈ F because F is the limit of the F k . Thus y + D ⊂ F when y ∈ F ; this proves (8.28). We deduce from (8.28), (8.27), and Proposition 8.3 (with m = 1) that (8.30) F ♯ is a reduced minimal set of dimension d in P 0 .

[This is where a proof of our description of blow-up limits of K would end.]

Set L k = r k F ♯ + A; this is a reduced minimal set of dimension d in P . Let us check that for k large, L k satisfies (8.24) with respect to K k = E k ∩ ∂B(0, 1).

First let y ∈ L k ∩ B(A, r k ) be given, and set z = r -1 k (y -A). Thus z ∈ F ♯ ∩ B(0, 1) ⊂ F ∩ B(0, 1). Let v ∈ F k minimize the distance to z ; thus |v -z| ≤ ε k , where ε k measures the distance from F k to F in B(0, 2), say, and the sequence {ε k } converges to 0. Set

w = r k v + A. Then w ∈ E k , and |w -y| = r k |v -z| ≤ r k ε k . Also, y ∈ P ∩ B(A, r k ), so |y -1| ≤ Cr 2 k (because P is tangent to ∂B(0, 1) at A), and |w -1| ≤ |w -y| + |y -1| ≤ r k ε k + Cr 2 k . Now w |w| ∈ K k , and | w |w| -y| ≤ |w -1| + |w -y| ≤ 2r k ε k + Cr 2 k
, which is less than εr k for k large. This proves the second half of (8.24). Now consider y ∈ K k ∩ B(A, r k ), set z = r -1 k (y -A), and observe that z ∈ F k ∩ B(0, 1) because y ∈ E k . We can find v ∈ F such that |v -z| ≤ ε k , with ε k as above. Observe that dist(y, P ) ≤ Cr 2 k because P is tangent to ∂B(0, 1) at A, so dist(z, P 0 ) ≤ Cr k and dist(v, P 0 ) ≤ Cr k + ε k . Recall from (8.28) that F = D × F ♯ , so the point v ′ at the intersection of v + D and P 0 lies in F ♯ , and of course |v

′ -v| = dist(v, P 0 ) ≤ Cr k + ε k . Now w = r k v ′ + A lies in L k , and |w -y| = r k |v ′ -z| ≤ r k (Cr k + 2ε k ) < εr k for k large.
Thus (8.24) holds for k large. Our proof of Theorem 8.23 would end here with a contradiction if we did not include (8.25) in the statement.

By assumption, there is a ball B k = B(y k , t k ) ⊂ B(A, r k ) such that (8.25) fails, i.e., (8.31)

H d (L k ∩ B(y k , t k -εr k )) > H d (K k ∩ B k ) + εr d k or (8.32) H d (L k ∩ B(y k , t k + εr k )) < H d (K k ∩ B k ) -εr d k .
We want to produce a contradiction. Set (8.33)

z k = r -1 k (y k -A) , ρ k = r -1 k t k , and B ′ k = B(z k , ρ k ) = r -1 k (B k -A).
Notice that B ′ k ⊂ B(0, 1) because B k ⊂ B(A, r k ). Since we may replace {E k } with a subsequence, we may assume that {B ′ k } converges to some ball B(z, ρ) ⊂ B(0, 1). That is, (8.34) {z k } converges to z and {ρ k } converges to ρ ≥ 0.

We want to use the results of Section 4 to estimate the measure of some sets, but we need a small amount of space to move around. So let τ > 0 be very small (to be chosen later). We shall start with upper bounds for H d (E k ∩ B k ). Identify D with the real line, set I = [0, 1] ⊂ D, and apply Lemma 3.12 to the sequence {F k } of (d + 1)-minimal sets, in the compact set (8.35)

D 1 = I × B(z, ρ + τ ).
We can take M = 1 and h = 0 in (3.13), and we get that

(8.36) lim sup k→+∞ H d+1 (F k ∩ D 1 ) ≤ H d+1 (F ∩ D 1 ).
By (8.27), F is rectifiable, and we can apply Lemma 8.10 (or Theorem 3.2.22 in [Fe]) to the product

F ∩ D 1 = [I × F ♯ ∩ B(z, ρ + τ )]
, in the same conditions as above. This yields (8.37)

H d+1 (F ∩ D 1 ) = H 1 (I) H d (F ♯ ∩ B(z, ρ + τ )) = H d (F ♯ ∩ B(z, ρ + τ )).
Let us check that for k large, (8.38)

H d (F ♯ ∩ B(z, ρ + τ )) ≤ r -d k H d (L k ∩ B(y k , t k + εr k )).
Indeed B(z, ρ + τ ) ⊂ B(z k , ρ k + ε) for k large, by (8.34) and if we choose τ < ε. Set ψ(x) = A + r k x for a moment; we know that L k = ψ(F ♯ ), and

ψ(B(z k , ρ k + ε)) = B(y k , t k + εr k ) by (8.33). So (8.39) ψ(F ♯ ∩ B(z, ρ + τ )) ⊂ ψ(F ♯ ∩ B(z k , ρ k + ε)) = L k ∩ B(y k , t k + εr k ),
and (8.38) follows. We deduce from (8.36), (8.37), and (8.38) that for k large, (8.40)

H d+1 (F k ∩ D 1 ) ≤ τ + r -d k H d (L k ∩ B(y k , t k + εr k )).
We also need to compare

H d+1 (F k ∩ D 1 ) with H d (K k ∩ B k ). Set (8.41) E ′ k = λv ; 1 + τ r k ≤ λ ≤ 1 + r k -τ r k and v ∈ K k ∩ B k .
and

E ′′ k = r -1 k (E ′ k -A).
We claim that for k large, (8.42)

E ′′ k ⊂ F k ∩ D 1 .
Indeed let λ and v be as in the definition of E ′ k ; we want to show that w = r -1 k (λv-A) lies in

F k ∩ D 1 . First, λv ∈ E k because v ∈ K k ⊂ E k and E k is a cone, so w ∈ F k = r -1 k (E k -A). We still need to check that w ∈ D 1 . Write v = A + (y k -A) + (v -y k )
, and also denote by π the orthogonal projection onto P 0 . Observe that

(8.43) |π(y k -A) -(y k -A)| ≤ Cr k |y k -A| ≤ Cr 2 k because y k ∈ B(A, r k
) and P is tangent to ∂B(0, 1) at A, and similarly

(8.44) |π(v -y k ) -(v -y k )| ≤ Cr k |v -y k | ≤ Cr k t k because v ∈ B k = B(y k , t k
) and both points lie in K k ⊂ ∂B(0, 1). Then (8.45)

w = r -1 k (λv -A) = r -1 k [(λ -1)v + (v -A)] = r -1 k [(λ -1)v + (y k -A) + (v -y k )] = z k + r -1 k [(λ -1)v + (v -y k )], because z k = r -1 k (y k -A) by (8.33). Then (8.46) w -z k -r -1 k [(λ -1)A -π(v -y k )] ≤ r -1 k (λ -1)|v -A| + |π(v -y k ) -(v -y k )| ≤ |v -A| + Ct k ≤ r k + Ct k because 1 ≤ λ ≤ 1 + r k , by (8.44), and because v ∈ B k ⊂ B(A, r k ) by definition of B k = B(y k , t k ). Set w ′ = z + r -1 k (λ -1)A + r -1 k π(v -y k ). Then (8.47) |w -w ′ | ≤ |z -z k | + r k + Ct k .
Observe that z ∈ P 0 , for instance because it is the limit of the z k = r -1 k (y k -A) and by (8.43). Thus the projection of

w ′ on D is r -1 k (λ -1)A ∈ [τ, 1 -τ ],
and by (8.47) the first coordinate of w lies in I for k large.

Similarly, π(w

′ ) = z + r -1 k π(v -y k ), and r -1 k |v -y k | < r -1 k t k = ρ k because v ∈ B k
and by (8.33), so π(w ′ ) ∈ B(z, ρ k ). Since ρ k tends to ρ, we deduce from (8.47) that π(w) ∈ B(z, ρ + τ ) for k large. Altogether, w ∈ D 1 = I × B(z, ρ + τ ) (compare with the definition (8.35)), and we completed our proof of (8.42).

Next we evaluate

H d+1 (E ′′ k ). Recall that E ′′ k = r -1 k (E ′ k -A), so (8.48) H d+1 (E ′′ k ) = r -d-1 k H d+1 (E ′ k ).
Notice that E ′ k ⊂ E k , which is rectifiable, so we may apply Lemma 8.10 with m = 1, Z = E ′ k and to the Lipschitz function π r (x) = |x|. Here (8.12) can be written as (8.49)

E ′ k apJ 1 π r (x) dH d+1 (x) = λ |x|=λ 1 E ′ k (x) dH d (x) dH 1 (λ).
The approximate Jacobian is apJ 1 π r (x) = 1, because π r is 1-Lipschitz, and the tangent plane to E k at x always contains the radial direction where the differential Dπ r has a norm exactly equal to 1. Thus (8.49) and (8.41) say that (8.50)

H d+1 (E ′ k ) = 1+τ r k ≤λ≤1+r k -τ r k H d (λ(K k ∩ B k )) dH 1 (λ) = 1+τ r k ≤λ≤1+r k -τ r k λ d H d (K k ∩ B k ) dH 1 (λ) ≥ (r k -2τ r k )H d (K k ∩ B k ).
Altogether,

(8.51) (r k -2τ r k )H d (K k ∩ B k ) ≤ H d+1 (E ′ k ) = r d+1 k H d+1 (E ′′ k ) ≤ r d+1 k H d+1 (F k ∩ D 1 ) ≤ r d+1 k τ + r k H d (L k ∩ B(y k , t k + εr k ))
for k large, by (8.50), (8.48), (8.42), and (8.40). Thus (8.52)

H d (K k ∩ B k ) ≤ (1 -2τ ) -1 [r d k τ + H d (L k ∩ B(y k , t k + εr k ))] ≤ H d (L k ∩ B(y k , t k + εr k )) + εr d k , because H d (L k ∩ B(y k , t k + εr k )) ≤ Cr d (recall from (8.30) that L k = A + r k F ♯ is a d- dimensional minimal set
, and use Lemma 2.15), and if τ is small enough. We just proved that (8.32) fails for k large.

Now we want to show that (8.31) fails too. The argument will be quite similar to the one above. This time we set I = (0, 1) ⊂ D and D 2 = I × B(z, ρτ ). [If ρ < τ , then ρ k < ε and hence t k < εr k for k large, and (8.31) fails trivially.] Apply Lemma 3.3 to the open set D 2 and the sequence {F k } of (d + 1)-minimal sets. We get that (8.53)

H d+1 (F ∩ D 2 ) ≤ lim inf k→+∞ H d+1 (F k ∩ D 2 ).
As for (8.37), F is rectifiable and we can apply Lemma 8.10 to the product

F ∩ D 2 = I × [F ♯ ∩ B(z, ρ -τ )]. We get that (8.54) H d+1 (F ∩ D 2 ) = H d (F ♯ ∩ B(z, ρ -τ )).
Next we check that for k large, (8.55)

H d (F ♯ ∩ B(z, ρ -τ )) ≥ r -d k H d (L k ∩ B(y k , t k -εr k )).
Observe that for k large,

B(z, ρ -τ ) contains B(z k , ρ k -ε), by (8.34). Again set ψ(x) = A + r k x for x ∈ R n ; thus L k = ψ(F ♯ ) and ψ(B(z k , ρ k -ε)) = B(y k , t k -εr k ) by (8.33). So (8.56) ψ(F ♯ ∩ B(z, ρ -τ )) ⊃ ψ(F ♯ ∩ B(z k , ρ k -ε)) = L k ∩ B(y k , t k -εr k ),
and (8.55) follows. Then (8.57)

H d+1 (F k ∩ D 2 ) ≥ H d+1 (F ∩ D 2 ) -τ ≥ r -d k H d (L k ∩ B(y k , t k -εr k )) -τ
for k large, by (8.53), (8.54), and (8.55).

We again compare

H d+1 (F k ∩ D 2 ) with H d (K k ∩ B k ). We still set E ′′ k = r -1 k (E ′ k -A
), but this time with the larger (8.58)

E ′ k = λv ; 1 -τ r k ≤ λ ≤ 1 + r k + τ r k and v ∈ K k ∩ B k .
We claim that then (8.59)

F k ∩ D 2 ⊂ E ′′ k for k large. Let x ∈ F k ∩ D 2 be given; we need show that w = A + r k x lies in E ′ k . We already know that w ∈ E k , so w/|w| ∈ K k , and it is enough to show that (8.60) w/|w| ∈ B k and 1 -τ r k ≤ |w| ≤ 1 + r k + τ r k Write x = αA + z + v,
where αA is the projection of x on D. Thus 0 < α < 1, and also

v ∈ P 0 ∩B(0, ρ-τ ) (because x ∈ D 2 and z ∈ P 0 ). Now w = A+r k x = (1+αr k )A+r k (z+v) so (8.61) |w| = 1 + αr k + O(r 2 k )
by Pythagoras (recall that z + v ∈ P 0 ). Then (8.62)

w/|w| = (1 -αr k ) w + O(r 2 k ) = A + r k (z + v) + O(r 2 k ) and (8.63) w/|w| -y k ≤ r k |z -z k | + r k |v| + O(r 2 k ) ≤ r k |z -z k | + r k (ρ -τ ) + O(r 2 k ) because y k = A + r k z k and v ∈ B(0, ρ -τ ).
In these estimates, the error terms O(r 2 k ) are less than Cr 2 k , with C independent of x. If k is large enough, (8.61) implies that 1τ r k ≤ |w| ≤ 1 + r k + τ r k (because 0 ≤ α ≤ 1), and (8.63) says that w/|w|y k < t k = r k ρ k (because z k tends to z and ρ k tends to ρ, by (8.34)). This proves (8.60) and (8.59).

We still have that

H d+1 (E ′′ k ) = r -d-1 k H d+1 (E ′ k )
as in (8.48), and Lemma 8.10 now leads to (8.64)

H d+1 (E ′ k ) = 1-τ r k ≤λ≤1+r k +τ r k H d (λ(K k ∩ B k )) dH 1 (λ) = 1-τ r k ≤λ≤1+r k +τ r k λ d H d (K k ∩ B k ) dH 1 (λ) ≤ (r k + 2τ r k )(1 + r k + τ r k ) d H d (K k ∩ B k ).
as in (8.49) and (8.50). Thus (8.65)

H d (K k ∩ B k ) ≥ (1 -Cr k -Cτ ) r -1 k H d+1 (E ′ k ) = (1 -Cr k -Cτ ) r d k H d+1 (E ′′ k ) ≥ (1 -Cr k -Cτ ) r d k H d+1 (F k ∩ D 2 ) ≥ (1 -Cr k -Cτ )H d (L k ∩ B(y k , t k -εr k )) -τ r d k ≥ H d (L k ∩ B(y k , t k -εr k )) -εr d k by (8.59) and (8.57), because H d (L k ∩ B(y k , t k -εr k )) ≤ Cr d
k by Ahlfors regularity of the minimal set L k (see the proof of (8.52)), if τ is small enough, and for k large. That is, (8.31) fails for k large.

We finally managed to show that (8.31) and (8.32) fail for k large; this contradiction with the definition of our sequence {K k } completes the proof of Theorem 8.13. Theorem 8.13 gives some information on the local behavior of K = E ∩ ∂B(0, 1) when E is a reduced minimal cone of dimension d + 1. For instance, if we have a good control on the minimal sets of dimension d, we can try to use a generalization of Reifenberg's topological disk theorem to get local bi-Hölder parameterizations of K. This may not be enough to control K precisely, but it could give some useful topological control.

We shall be interested in the case when d = 1; then we'll see in Section 10 that minimal sets of dimension 1 are lines and sets Y , and a simple case of the generalized Reifenberg theorem shows that K is locally bi-Hölder equivalent to a line or a Y . See Section 15 for the existence of generalized Reifenberg parameterizations, Section 12 for the simpler special case when d = 1, and Section 16 for similar arguments in dimension 2.

But we also want to know more about K when d = 1. In particular, we want to check that it is composed of a finite collection of arcs of great circles, and for this it will be simpler to have a Lipschitz control, rather than merely bi-Hölder.

So we shall start with another approach, where we show that K satisfies a weak almostminimality property, and first prove reasonable properties for weak almost-minimal sets. This will give slightly more precise results than Theorem 8.23, at least when d = 1, and maybe the method can be used in slightly different contexts. But to be fair, one could also start from Theorem 8.23, use Proposition 12.6 to show that K is locally bi-Hölder equivalent to a line or a Y , and then find more direct ways to show that K is composed of arcs of great circles in the regions where it looks a lot like a line. Some arguments of this type will be used in the next paper [D3], to deduce the C 1 -regularity of almost-minimal sets of dimension 2 in R 3 from a local biHölder description, and they would be simpler for one-dimensional sets.

Weak almost-minimality of K

= E ∩ ∂B(0, 1)
Let E be a reduced minimal cone of dimension d + 1 in R n , and set K = E ∩ ∂B(0, 1). We want to show that E is weakly almost-minimal, as in the next definition. Later on we shall use this property to study E when d = 1.

The following definition is a little awkward, because it is a compromise between what we can easily prove for K and what will be needed in later arguments. We state the definition for subsets of ∂B(0, 1), but we could also work directly in R n , or in some affine subspace, with only minor modifications. Definition 9.1. Let K be a closed subset of ∂B = ∂B(0, 1) ⊂ R n . We say that K is a weak almost-minimal set of dimension d, with gauge function h if, for each choice of x ∈ ∂B, r ∈ (0, 1), and a Lipschitz function f : ∂B → ∂B such that

(9.2) W f = y ∈ ∂B ; f (y) = y ⊂ B(x, r) and f (W f ) ⊂ B(x, r),
we have that

(9.3) H d (K ∩ B(x, r)) ≤ H d (f (K) ∩ B(x, r)) + (1 + M d ) h(r)r d ,
whenever there is a partition of K ∩B into a finite number of pieces where f is M -Lipschitz.

As before, our convention is that a gauge function is a nonnegative nondecreasing function h, with lim r→0 h(r) = 0. There are two differences with the definitions of Section 4. The small one is that we work on ∂B instead of an open set in R n . This is just because our examples will live on the sphere, but the same definition would work in R n or in some other smooth manifolds.

The main difference is that here we pay an additional price that depends on the Lipschitz properties of f . The story about the partition of K ∩ B is a mostly a way to avoid discussing local Lipschitz constants or gradients. Note that M may be significantly smaller than the Lipschitz norm of f , or even of its restriction to K ∩ B; we shall use this advantage in some proofs.

We would have a slightly weaker notion of weak almost-minimal set if we replaced M d in (9.3) with some larger functions of M ; I don't think that this would make much difference in the properties of weak almost-minimal set that will be proved below. Also M d+1 would be a little easier to get in the proof of Proposition 9.4. On the other hand M d looks more natural, so we shall keep it in the definition.

Proposition 9.4. Let E be a reduced minimal cone of dimension d + 1 in R n , and set K = E ∩ ∂B(0, 1). Then K is a reduced weak almost-minimal set of dimension d, with gauge function h(r) = Cr.

Let us first check that K is reduced. Recall from Definition 2.12 that this means that (9.5)

H d (K ∩ B(x, r)) > 0
for x ∈ K and r ∈ (0, 1). Let us check this brutally. We apply Lemma 8.10 (i.e., Theorem 3.2.22 in [Fe]) with Z = E ∩ B(x, r/2), m = 1, A = Z, and for π the radial projection defined by π(x) = |x|. The rectifiability of E was discussed in Section 2, in a more general context. So (8.12) yields (9.6)

Z apJ 1 π(x)dH d+1 (x) = ρ Z∩π -1 (ρ)
dH d dρ.

The approximate Jacobian apJ 1 π(x) is easily computed to be 1, because π is 1-Lipschitz and the tangent plane to E always contains the radial direction where the Lipschitz constant for π is exactly 1. Thus (9.6) says that (9.7)

H d+1 (E ∩ B(x, r/2)) = H d+1 (Z) = ρ H d (Z ∩ π -1 (ρ)) dρ = ρ H d (E ∩ B(x, r/2) ∩ ∂B(0, ρ)) dρ.
In addition, the only contribution in the right-hand side comes from 1r/2 < ρ < 1 + r/2, for which

E ∩ B(x, r/2) ∩ ∂B(0, ρ) is contained in ρ • (K ∩ B(x, r)) because E is a cone and K = E ∩ ∂B(0, 1). Thus (9.8) H d+1 (E ∩ B(x, r/2)) ≤ 1-r/2<ρ<1+r/2 H d (K ∩ B(x, r)) ρ d dρ.
The left-hand side of (9.8) is positive because E is reduced, hence H d (K ∩ B(x, r)) > 0, (9.5) holds, and K is reduced.

We now prove that K is weakly almost-minimal. Let x ∈ ∂B, r ∈ (0, 1), and f : ∂B → ∂B be as in Definition 9.1, and let us use f to define a competitor for E.

Let ψ : [0, +∞) → [0, 1] be such that (9.9)

ψ(t) = 0 for 0 ≤ t ≤ 1 2 , ψ(t) = 1 for 1 2 + r 10 ≤ t ≤ 1 - r 10 , ψ(t) = 0 for t ≥ 1,
ψ is continuous, and ψ is linear in the two missing interval. Write every point of R n as z = ρy, with ρ = |z| and y = y/|y| (the choice of y when z = 0 does not matter), and set (9.10)

ϕ(z) = ψ(ρ)ρf (y) + (1 -ψ(ρ))ρy for z ∈ R n .
Since E is minimal and ϕ is Lipschitz, Definition 8.1 yields

(9.11) H d+1 (E \ ϕ(E)) ≤ H d+1 (ϕ(E) \ E).
Let us compute what this means in terms of K and f (K). Observe that (9.12) ϕ(z) = z for z ∈ R n \ B(0, 1) and ϕ(B(0, 1)) ⊂ B(0, 1), so ϕ(E) coincides with E out of B(0, 1) and there will be no contribution to (9.11) coming from R n \ B(0, 1). The most interesting contributions will come from the annulus (9.13)

A = z ∈ R n ; 1 2 + r 10 ≤ |z| ≤ 1 - r 10 .
Let us again apply Lemma 8.10 to the mapping π : z → |z|, from Z = E ∩ A to I = [ 1 2 + r 10 , 1 -r 10 ]. We get (9.6) as before, again with apJ 1 π(x) = 1, so

(9.14) H d+1 (E ∩ A) = H d+1 (Z) = I H d (Z ∩ π -1 (t)) dt = I H d (E ∩ ∂B(0, t)) dt = I H d (t • K) dt = I t d H d (K) dt = αH d (K), because π -1 (t) = ∂B(0, ρ) ⊂ A for t ∈ I, and E is a cone. The precise value of α = I t d dt won't matter much. Observe that ψ(|z|) = 1 when z ∈ A, so ϕ(z) = ρf (y). Thus (9.15) ϕ(E ∩ A) ∩ ∂B(0, ρ) = ρf (K) for ρ ∈ I,
and it is empty otherwise. The same computation as for E∩A, but with the other rectifiable set ϕ(E ∩ A), yields (9.16)

H d+1 (ϕ(E ∩ A)) = I H d (ϕ(E ∩ A) ∩ ∂B(0, t)) dt = αH d (f (K)).
71

We shall also need to estimate the contributions of the two annuli (9.17)

A 1 = z ∈ R n ; 1 2 ≤ |z| ≤ 1 2 + r 10 and A 2 = z ∈ R n ; 1 - r 10 ≤ |z| ≤ 1 . Set Z i = z ∈ E ∩ A i ; ϕ(z) = z .
We want to estimate the Lipschitz properties of ϕ on Z i , and these will depend on the Lipschitz properties of f on K ∩ B(x, r). Let us assume, as in Definition 9.1, that we have a partition K ∩ B(x, r) = ∪ j K j of K ∩ B(x, r) into a finitely number of sets K j where f is M -Lipschitz. Note that we may assume that the K j are Borel sets, because f is automatically M -Lipschitz on the K j , which enables us to first replace K j with K j , and then build a Borel partition out of the covering. If z ∈ Z i , y = z/|z| lies in K, because z ∈ E and E is a cone, and f (y) = y because ϕ(z) = z (see (9.10)). Then y ∈ K ∩ B(x, r), by (9.2). Thus the sets (9.18)

Z i,j = z ∈ Z i ; y = z/|z| ∈ K j form a partition of Z i . Let us check that (9.19) ϕ is (25 + 2M )-Lipschitz on each Z i,j .
It will be enough to control g(z) = ϕ(z)z = ρψ(ρ)(f (y)y). We write that (9.20)

g(z 1 ) -g(z 2 ) = ρ 1 ψ(ρ 1 )(f (y 1 ) -y 1 ) -ρ 2 ψ(ρ 2 )(f (y 2 ) -y 2 ) = ∆ 1 -∆ 2 , with ∆ 1 = [ρ 1 ψ(ρ 1 ) -ρ 2 ψ(ρ 2 )](f (y 1 ) -y 1 ) and ∆ 2 = ρ 2 ψ(ρ 2 )[(f (y 2 ) -y 2 ) -(f (y 1 ) -y 1 )]. Since |∇(ρψ(ρ))| ≤ ψ(ρ) + ρ|ψ ′ (ρ)| ≤ 1 + 10 r , we get that (9.21) |∆ 1 | ≤ 11 r |ρ 2 -ρ 1 ||f (y 1 ) -y 1 | ≤ 22|ρ 2 -ρ 1 |,
because y 1 and f (y 1 ) lie in B(x, r), by (9.2). Also, z 1 and z 2 both lie in Z i,j , so y 1 and y 2 lie in K j . Then

(9.22) |∆ 2 | ≤ |(f (y 2 ) -y 2 ) -(f (y 1 ) -y 1 )| ≤ |y 2 -y 1 | + |f (y 2 ) -f (y 1 )| ≤ (1 + M )|y 2 -y 1 |,
by definition of M and the K j . Altogether

(9.23) |g(z 1 ) -g(z 2 )| ≤ |∆ 1 | + |∆ 2 | ≤ 22|ρ 2 -ρ 1 | + (1 + M )|y 2 -y 1 | ≤ (24 + 2M )|z 2 -z 1 | and (9.19) follows because ϕ(z) = z + g(z). An immediate consequence of (9.19) is that (9.24) H d+1 (ϕ(Z i,j )) ≤ C(1 + M ) d+1 H d+1 (Z i,j ).
This would be enough to prove (9.3) with M d replaced by M d+1 , but since we decided to go for the more restrictive definition, we shall need to improve (9.24) a little bit. Observe that Z i,j is rectifiable, because it is contained in E and by the rectifiability results of Section 2, so we can use the area formula (see for instance Corollary 3.2.20 in [Fe]). That is, (9.25)

H d+1 (ϕ(Z i,j )) ≤ Z i,j J ϕ (z) dH d+1 (z).
Our improvement on (9.24) will come from a better control of the (d + 1)-dimensional Jacobian J ϕ due to a better control on the derivative of ϕ in the radial direction.

It may comfort the reader to notice that since E is Ahlfors-regular and rectifiable, it even has a true tangent (d + 1)-plane at z for H d+1 -almost every z. This plane z + P (z) contains the radial direction because E is a cone, and if z + Q(z) denotes the d-plane in P (z) perpendicular to the radial direction, it is easy to see that K has a tangent plane at y = z/|z|, which is y + Q(z). Note that we already knew from the first part of Lemma 8.10 (applied as before to the radial projection π) that K is rectifiable.

The derivative of g(z) = ρψ(ρ)(f (y)y) in the radial direction is bounded by

(9.26) [ψ(ρ) + ρψ ′ (ρ)]|f (y) -y| ≤ Cr -1 |f (y) -y| ≤ C, by (9.2).
The derivative in all the tangential directions is bounded by 25 + 2M , because we restrict our attention to Z i,j and by (9.19). So J ϕ (z) ≤ C(1 + M ) d . Indeed, after some manipulations, we can compute J ϕ (z) as a determinant of Dϕ in some set of bases, and if we start the basis of P (z) with a radial vector, the first row of the corresponding matrix has bounded terms, while the terms in the other rows are bounded by C(1 + M ). Thus (9.25) yields (9.27)

H d+1 (ϕ(Z i,j )) ≤ C(1 + M ) d H d+1 (Z i,j ).
Incidentally, we have been trying to avoid talking about approximate differentials of f and ϕ on K and E, but the natural way to deal with (9.25) would use these. Our discussion leads to J ϕ (z) ≤ C|D E ϕ(z)| d , where |D E ϕ(z)| is the norm of the approximate differential of the restriction of ϕ to E (which exits H d+1 -almost everywhere on E). Then the proof of (9.19) yields

|D E ϕ(z)| ≤ C + |D K f (y)|, where |D K f (y)|
is the norm of the approximate differential of the restriction of f to K. Thus a more precise version of (9.27) is that (9.28)

H d+1 (ϕ(Z i )) ≤ C Z i (1 + |D K ϕ(y)| d ) dH d+1 (z).
Notice that we don't need to cut K ∩ B(x, r) or the Z i into pieces in this context, because

|D K ϕ(y)| is local.
We return to (9.27) and estimate H d+1 (Z i,j ). Recall from (9.18) that Z i,j is contained in the cone over K j ⊂ K ∩ B(x, r). It is also contained in the thin annulus A i of (9.17), so ∪ j Z i,j is contained in a ball B i of radius 2r. Then (9.29)

j H d+1 (Z i,j ) = H d+1 ∪ j Z i,j ≤ H d+1 (E ∩ B i ) ≤ Cr d+1
because the Z i,j are disjoint and contained in E, and by the easiest part of Lemma 2.15 (the Ahlfors-regularity of the minimal set E).

We are almost ready to return to (9.11). Set (9.30)

S = E ∩ B(0, 1/2) ∪ ∪ i∈{1,2} E ∩ A i \ Z i ) ,
and check that ϕ(z) = z on S. When z ∈ B(0, 1/2), this comes directly from (9.10) and the fact that ψ(ρ) = 0 by (9.9). On E ∩ A i \ Z i , this is the definition of Z i (just below (9.17)). Next (9.31)

ϕ(E) ∩ B(0, 1) = ϕ(E ∩ B(0, 1)) ⊂ ϕ(S) ∪ ϕ(E ∩ A) ∪ ∪ i ϕ(Z i ) = S ∪ ϕ(E ∩ A) ∪ ∪ i ϕ(Z i ) by (9.12), because E ∩ B(0, 1) = S ∪ (E ∩ A) ∪ [∪ i Z i ],
and because ϕ(z) = z on S. So (9.32)

H d+1 (ϕ(E) ∩ B(0, 1)) ≤ H d+1 (S) + H d+1 (ϕ(E ∩ A)) + C(1 + M ) d r d+1
by (9.27) and (9.29). Similarly, (9.33)

H d+1 (E ∩ B(0, 1)) ≥ H d+1 (S) + H d+1 (E ∩ A)
because these sets are disjoint. On the other hand, H d+1 (E \ ϕ(E)) ≤ H d+1 (ϕ(E) \ E) by (9.11), and since ϕ(E) coincides with E out of B(0, 1) by ( 9.12), we get that (9.34)

H d+1 (E ∩ B(0, 1)) ≤ H d+1 (ϕ(E) ∩ B(0, 1))
by adding H d+1 (E ∩ ϕ(E) ∩ B(0, 1)) to both sides. Altogether, (9.35)

H d+1 (E ∩ A) ≤ H d+1 (E ∩ B(0, 1)) -H d+1 (S) ≤ H d+1 (ϕ(E) ∩ B(0, 1)) -H d+1 (S) ≤ H d+1 (ϕ(E ∩ A)) + C(1 + M ) d r d+1
by (9.33), (9.34), and (9.32). By (9.14) and (9.16), this implies that (9.36)

αH d (K) ≤ αH d (f (K)) + C(1 + M ) d r d+1 , with α = I t d dt ≥ 9/10 6/10 t d dt (recall that I = [ 1 2 + r 10 , 1 -r 10
], see above (9.14)). This is just the same thing as (9.3), with h(r) = Cr. Proposition 9.4 follows.

Remark 9.37. We can improve slightly our estimate of H d+1 (ϕ(Z i )), by using (9.28) instead of (9.27). The same computations as above then lead to (9.38)

H d (K) ≤ H d (f (K)) + C Z 1 ∪Z 2 (1 + |D K ϕ(y)| d ) dH d+1 (z),
where we still use the convention that y = z/|z|. Recall from just above (9.18) that if z ∈ Z 1 ∪ Z 2 , then f (y) = y. In particular, y ∈ K ∩ B(x, r), by (9.2). Also, z ∈ A 1 ∪ A 2 . We can apply Lemma 8.10 to Z = Z 1 ∪ Z 2 (which is rectifiable because it is contained in E) and the radial projection

π from Z to J = [ 1 2 , 1 2 + r 10 ] ∪ [1 -r 10 , 1]
. This time we use sets A ⊂ Z to show that both sides of (8.12) define equal measures, and we get that (9.39)

Z (1 + |D K ϕ(y)| d ) dH d+1 (z) = ρ∈J z∈Z∩π -1 (ρ) (1 + |D K ϕ(y)| d ) dH d (z) dρ,
where we omitted the approximate Jacobian from the left-hand side because it is equal to 1, as for (9.6) and (9.14). Recall that π -1 (ρ) = ∂B(0, ρ) and that Z is contained in the cone over W f = {y ∈ ∂B(0, 1) ; f (y) = y} ⊂ K ∩ B(x, r). Thus (9.40)

z∈Z∩π -1 (ρ) (1 + |D K ϕ(y)| d ) dH d (z) ≤ ρW f (1 + |D K ϕ(y)| d ) dH d (z) = ρ d W f (1 + |D K ϕ(y)| d ) dH d (y).
We integrate this over J and get that (9.41)

H d (K) ≤ H d (f (K)) + Cr W f (1 + |D K ϕ(y)| d ) dH d (y),
by (9.38) and (9.39). This is slightly more precise and elegant than (9.36), but will not be used here.

One-dimensional minimal sets in R n

We want to describe here the reduced minimal sets of dimension 1 in R n . There are two simple examples, the lines and the Y -sets. In this section, a Y -set will be a union of three half lines contained in an affine plane, with the same endpoint, and that make angles of 120 • at that point.

It is fairly easy to see that lines and Y -sets are indeed minimal. See [D2], Lemma 59.1 on page 392, but here is a rapid sketch of the argument for the typical case when Y is a Y -set centered at the origin. Let Z = ϕ(Y ) be a deformation of Y , choose R so large that Y = Z out of B(0, R), and denote by a 1 , a 2 , a 3 the three points of Y ∩ ∂B(0, R). The images under ϕ of the three segments of Y ∩ B(0, R) are three arcs that connect ϕ(0) to the a j . Then (if H 1 (Z ∩ B(0, R)) < +∞) we can find a point z ∈ Z ∩ B(0, R) and three simple arcs in B(0, R) ∩ Z that go from z to the a j , and whose interiors are disjoint. Replacing these by line segments makes Z shorter, and a variational argument shows that the optimal case is when z = 0 and Z = Y .

To the mild surprise of the author, there is no other one-dimensional reduced minimal set, even in large ambient dimensions.

Theorem 10.1. Every nonempty reduced minimal set of dimension 1 in R n is a line or a Y -set.

The rest of this section is devoted to the proof. Even though a direct approach would be possible, it will be easier for us to apply some general arguments, and in particular use blow-down limits. We start with the simpler case of minimal cones.

Lemma 10.2. Every nonempty reduced minimal cone of dimension 1 is a line or a Y -set.

Let E be a reduced minimal cone of dimension 1 (centered at the origin), and set K = E ∩ ∂B(0, 1). Notice that K is finite, because otherwise H 1 (E ∩ B(0, 1)) is infinite. Write K = {a 1 , • • • , a m }, and denote by L j the half line through a j . Let us first show that

(10.3) Angle(L i , L j ) ≥ 2π 3 for i = j.
Choose i = j so that θ = Angle(L i , L j ) is minimal, and suppose that θ < 2π 3 . We want to get a contradiction by replacing [L i ∪ L j ] ∩ B(0, 1) with a shorter fork F t defined as follows. Set z t = t(a i + a j ) for 0 < t < 10 -1 , say, and

F t = [0, z t ] ∪ [z t , a i ] ∪ [z t , a j ]. A simple differentiation of ℓ(t) = H 1 (F t ) at t = 0 shows that H 1 (F t ) < H 1 ([L i ∪L j ]∩B(0, 1)) for t small (because θ < 2π
3 ). We choose t so that this holds, and set (10.4)

E = E \ [L i ∪ L j ] ∩ B(0, 1) ∪ F t .
Notice that H 1 ( E ∩ B(0, 1)) < H 1 (E ∩ B(0, 1)) and the two sets coincide out of B(0, 1), so

H 1 ( E \ E) < H 1 (E \ E).
We want to construct a Lipschitz mapping ϕ : R n → R n such that ϕ(x) = x out of B(0, 1) and ϕ(E) ⊂ E. Since E is minimal, Definition 8.1 will allow us to apply (8.2) to ϕ, and get that

H 1 (E \ ϕ(E)) ≤ H 1 (ϕ(E) \ E), hence H 1 (E \ E) ≤ H 1 ( E \ E)
. This is just the opposite of what we have, so we shall get the desired contradiction as soon as we have ϕ.

Let L denote the half line through a i + a j , and set C α = y ∈ R n ; dist(y, L) ≤ α|y| , with α = dist(a i , L). That is, C α is the smallest revolution cone around L that contains L i and L j . Notice that since |a ia j | is minimal, dist(a l , L i ) ≥ dist(a j , L i ) for l = i. Also, a i and a j are furthest possible in C α ∩ ∂B(0, 1), so the only point of C α ∩ ∂B(0, 1) such that dist(y, L i ) ≥ dist(a j , L i ) is y = a j . Put in other words, a l / ∈ C α unless l = j. Let us choose α ′ > α so close to α that no a l , l = i, j, lies in C α ′ either.

Set ϕ(y) = y for y ∈ R n \[C α ′ ∩B(0, 1)]. Thus we still need to define ϕ on C α ′ ∩B(0, 1), and we start with (L i ∪ L j ) ∩ B(0, 1). Define ϕ on L i ∩ B(0, 1) so that it is Lipschitz and maps L i ∩ B(0, 1) into F t ∩ B(0, 1), with ϕ(0) = 0 and ϕ(x) = x near a i . Proceed similarly with L j . Because there are small security cones around L i and L j that are contained in

C α ′ , our map ϕ is Lipschitz on R n \ [C α ′ ∩ B(0, 1)] ∪ L i ∪ L j .
We simply use the Whitney extension Theorem to extend it to the rest of R n , so that it stays Lipschitz.

Notice that ϕ does not move the half lines L l , l = i, j; then ϕ(E) ⊂ E, where E is as in (10.4). This completes our construction of ϕ, and we have noted earlier that the ensuing contradiction proves (10.3).

Next we claim that if j, k, l are distinct, the three lines L j , L k , L l lie in a plane, and L j ∪ L k ∪ L l is a Y -set. Denote by P the plane that contains L j and L k , and identify P with C. By symmetry, we may assume that a j = e iθ and a k = e -iθ , with 0 < θ ≤ π/2. By (10.3), θ ≥ π/3. Denote by ξ the orthogonal projection of a l on P ; then (10.3) says that ξ, a j = a l , a j ≤ cos(2π/3) = -1/2, and similarly ξ, a j ≤ -1/2. Since ξ lies in the closed unit ball, this forces the half line through ξ to make angles at least 2π/3 with L i and L j . The only possibility is that θ = π/3, ξ = -1 and a l = -1. Our claim follows.

We are now ready to prove the lemma. If E ∩∂B(0, 1) is empty, then E is empty. Next E ∩ ∂B(0, 1) cannot be reduced to one point, because a half line is not minimal (we can contract it along its end to make it shorter). If it has exactly two points, it is easy to check that E is a line (otherwise, deform E ∩ B(0, 1) into a segment with the same endpoints, and you'll save some length). If it has more than two points, choose two of them (say, a i and a j ); the argument above shows that all the other points a k lie just opposite to a i + a j , so there is only one of them, and E is a Y -set. This proves Lemma 10.2. Now we want to deal with the general case of Theorem 10.1. We shall use results of the previous parts to save time, but more direct proofs would be possible (and also the theorems that we use are easier for one-dimensional sets).

So let E be a nonempty reduced minimal set of dimension 1. Let us assume for simplicity that the origin lies in E, and set θ(r) = r -1 H 1 (E ∩ B(0, r)). By Lemma 2.15, θ is bounded. By Proposition 5.16, θ is nondecreasing, so θ ∞ = lim r→+∞ θ(r) exists and is finite.

Let τ > 0 be small, and let ε > 0 be as in Proposition 7.1, with a = 1 and b = 2. Then let R ≥ 1 be such that θ(r) ≥ θ ∞ε for r ≥ R. For r ≥ R, apply Proposition 7.1 to r -1 E; the density assumption is satisfied because θ ∞ε ≤ θ(r) ≤ θ(2r) ≤ θ ∞ . The conclusion is that there is a minimal cone C centered at the origin, that is close to E ′ = r -1 E in various respects, but in particular such that (10.5)

|ρ -1 H 1 (E ′ ∩ B(0, ρ)) -ρ -1 H 1 (C ∩ B(0, ρ))| ≤ τ for 1 + τ ≤ ρ ≤ 2 -τ,
as in (7.5). Take ρ = 3/2; we get that

(10.6) θ(3r/2) = ρ -1 H 1 (E ′ ∩ B(0, ρ)) ≤ ρ -1 H 1 (C ∩ B(0, ρ)) + τ ≤ 3 + τ
by Lemma 10.2. Denote by N (t) the number of points in E∩∂B(0, t). By [D3], Lemma 26.1 on page 160, or [Fe], Theorem 3.2.22, (10.7)

3r/2 0 N (x)dx ≤ H 1 (E ∩ B(0, 3r/2)) = (3r/2)θ(3r/2) ≤ 3(3 + τ )r/2
so we can find ρ ∈ (r/10, 3r/2) such that N (ρ) ≤ 10 14r 3(3+τ )r 2

< 4 (if τ is small enough). That is, (10.8) E ∩ ∂B(0, ρ) has at most three points.

Lemma 10.9. If E is a reduced minimal set that contains the origin, and ρ > 0 is such that (10.8) holds, then E coincides with a line or a Y -set in B(0, ρ).

Theorem 10.1 will follow as soon as we prove this lemma. Indeed, we found arbitrarily large radii ρ such that (10.8) holds, hence such that E coincides with a line or a Y -set in B(0, ρ). It is easy to see that E is then a line or a Y -set.

Our proof of Lemma 10.9 will take some time, in particular because we want to be able to use the same construction with weak almost-minimal sets.

Before we really start, let us construct retractions h F onto some simple sets F . Our first case is when F = Y ∩ B, where Y is a Y -set and B is a closed ball that contains the center of Y . Let P denote the plane that contains Y . We first construct a retraction h 1 : P → Y . Denote by L 1 , L 2 , L 3 the three half lines that compose Y , and by V i the connected component of P \ Y opposite to L i . For y ∈ V i , h 1 (y) is the projection of y onto L j ∪ L k along the direction of L i , where L j and L k are the two other half lines. In other words, h 1 (y) is the intersection of L j ∪ L k with the line through y parallel to L i . It is easy to see that h 1 is 2-Lipschitz on V i . For instance, observe that we can split V i into two equal sectors where h 1 is the oblique projection onto the line L j or L k , along a direction that makes an angle π/3 with that line. Then h 1 is also 2-Lipschitz on P , because the various definitions coincide with the identity on Y .

We compose h 1 with the retraction h 2 from Y to F , which maps y ∈ F to itself, and maps y ∈ Y \ F to the point of Y ∩ ∂B that lies on the same branch of Y as y. Obviously h 2 is 1-Lipschitz. Finally set (10.10)

h F = h 2 • h 1 • π P on R n
, where π P denotes the orthogonal projection onto P . Then (10.11) h F is a 2-Lipschitz retraction of R n onto F .

We also need to construct h F when F = B ∩ (L 1 ∪ L 2 ), where L 1 and L 2 are half lines with the same endpoint and that make an angle at least 2π/3 at that point, and B is a closed ball that contains that point. Let us first define a retraction h 1 on the plane P that contains Z (if L 1 ∪ L 2 is a line L, choose any plane that contains L; anyway h 1 • π P will end up being the orthogonal projection onto L). Call D the line in P that goes through the common endpoint of L 1 and L 2 and lies at equal distance from L 1 and L 2 and, for y ∈ P , let h 1 (y) be the intersection of L 1 ∪ L 2 with the line through y parallel to D. Now h 1 is still 2-Lipschitz on P , because D makes an angle of π/3 or more with the lines that contain L 1 and L 2 . We then define h F as before, by (10.10), and (10.11) holds for the same reasons.

Let E be as in Lemma 10.9. We want to treat a few cases at the same time, so some notation will be useful. Set B = B(0, ρ), and denote by a j , 1 ≤ j ≤ N , the points of E ∩ ∂B. Let us still keep the number N of points in E ∩ ∂B open, except that (10.8) says that N ≤ 3.

Then denote by E j the connected component of E ∩ B that contains a j . Call M the number of components that we get, and renumber the a j so that the E j , 1 ≤ j ≤ M , are disjoint, and each a k , k > M , lies in some E j such that j ≤ M . Notice that the E j are compact, so (10.12)

dist(E i , E j ) > 0 for 1 ≤ i, j ≤ M, i = j.

When M > 1, we shall also need to dispatch the rest of E ∩ B among the E j , as follows.

Lemma 10.13. We can find a partition of E ∩ B into sets E ′ j , 1 ≤ j ≤ M , such that E j ⊂ E ′ j for j ≤ M and

(10.14) dist(E ′ i , E ′ j ) > 0 for i, j ≤ M, i = j.
We leave the proof of Lemma 10.13 for the end of the section, because it is a pure fact about connected components, and it does not interfere with the rest of the proof.

Let us return to the proof of Lemma 10.9. For 1 ≤ j ≤ M , denote by F j the shortest connected set that contains all the points of E j ∩ ∂B. We claim that if E j ∩ ∂B has three points, i.e., if E ∩ ∂B has three points which lie in the same component of E ∩ B, then F 1 is either the intersection of B with a Y -set whose center lies in B, or else we can rename the three points a j so that F = [a 1 , a 2 ] ∪ [a 2 , a 3 ], and these segments make an angle at least 2π/3 at a 2 .

To prove the claim, we consider any connected set G with finite length that contains the three a j . By Proposition 10.37 in [D2] (for instance), we can find a simple arc γ in G from a 1 to a 2 , and a simple arc γ 1 in G from a 3 to a 1 . We denote by z the first point of γ 1 ∩ γ when we start from a 3 and run along γ 1 . Possibly z is one of the a i , but this is all right. Call g 1 the arc of γ from a 1 to z, g 2 the arc of γ from z to a 2 , and g 3 the arc of γ 1 from a 3 to z. These three arcs are disjoint, except for z, and they are contained in G. Then H 1 (G) ≥ H 1 (∪ j g j ) = j H 1 (g j ) ≥ H 1 (∪ i [a i , z]), because the g i are disjoint. So it is enough to minimize length among sets of the form ∪ i [a i , z]. By differentiating in z, it is easy to see that H 1 (∪ j g j ) is minimal only when ∪ j g j is contained in a Y -set centered at z, or else z is one of the a j . In this last case, if for instance z = a 2 , it is easy to show directly that the segments [a 1 , z] and [z, a 3 ] make an angle at least 2π/3 at z (otherwise, a Y -set would be shorter). This proves our claim. Notice that in both cases, we constructed a Lipschitz retraction h F j onto F j .

When E j contains exactly two points a i , F j is the line segment between these points, and we also constructed h F j above (this is the case when L 2 lies opposite to L 1 ). Finally, when E j ∩ ∂B is reduced to a j , F j = {a j } and we simply take h F j constant equal to a j .

We are now ready to define a competitor ϕ(E). First set (10.15) ϕ(y) = y for y ∈ R n \ B(0, ρ).

We would have liked to set ϕ(y) = h F j on E ′ j , but then ϕ would not necessarily be Lipschitz, because E ∩ B(0, ρ) may approach the a j tangentially, and the definitions of ϕ from inside and outside B would not match well. So we shall need a small cut-off function ψ. For this reason we introduce a small positive number ε, set d(y) = dist(y, E ∩ ∂B), and then (10.16)

ψ(y) = 0 for 0 ≤ d(y) ≤ ε, ψ(y) = ε -1 d(y) -1 for ε ≤ d(y) ≤ 2ε, ψ(y) = 1 for d(y) ≥ 2ε.
Next set (10.17) ϕ j (y) = y + ψ(y)[h F j (y)y] for y ∈ R n , and (10.18) ϕ(y) = ϕ j (y) for y ∈ E ′ j .

Notice that ϕ j (y) = y on F j , which contains all the points of E ′ j ∩ ∂B, because F j contains all the points of E j ∩ ∂B, and E ′ j \ E j does no meet ∂B (every point of E ∩ ∂B lies in some E j ). Thus the two definitions of ϕ(y) coincide on E ∩ ∂B. Let us check that for each j, (10.19) ϕ j is 10-Lipschitz.

Fix j, and start the verification in the small balls

B k = B(a k , 2ε) centered at the points of E ∩ ∂B. Notice that |h F j (y) -y| ≤ 6ε for y ∈ B k , because h F j is 2-Lipschitz and h F j (a k ) = a k (see (10.11)). Then (10.20) |ψ(y)[h F j (y) -y] -ψ(y ′ )[h F j (y ′ ) -y ′ ]| ≤ [ψ(y) -ψ(y ′ )][h F j (y) -y] + ψ(y ′ ) [h F j (y) -y] -[h F j (y) -y] ≤ ε -1 |y -y ′ | h F j (y) -y + 3ψ(y ′ )|y -y ′ | ≤ 9|y -y ′ | for y, y ′ ∈ B k , again because h F j is 2-Lipschitz. Thus ψ(y)[h F j (y) -y] is 9-Lipschitz on B k
, and ϕ j is 10-Lipschitz there. Out of the B k , ψ(y) = 1 by (10.16), so ϕ j (y) = h F j (y), which is 2-Lipschitz. This proves (10.19).

Next we check that ϕ is Lipschitz on

[R n \ B(0, ρ)] ∪ [E ∩ B(0, ρ)], i.e., that (10.21) |ϕ(x) -ϕ(y)| ≤ C|x -y|
for some (very large) C. If x and y lie out of B(0, ρ), this is clear because of (10.15); so let us assume that y ∈ B(0, ρ). Notice that y ∈ E ′ j for some j ≤ M , by Lemma 10.13, so ϕ(y) = ϕ j (y) by (10.18), hence ϕ(y) lies in the segment [y, h F j (y)] by (10.17). In particular, we just checked that

ε = E ∩ B \ ∪ k B(a k , ε). Notice that K ε does not meet R n \ B(0, ρ), because E ∩ ∂B reduces to the a k , so |y -x| ≥ d ε = dist(K ε , R n \ B(0, ρ)) > 0, and |ϕ(x) -ϕ(y)| ≤ 4ρ ≤ 4ρd -1
ε |y -x| because ϕ(x), ϕ(y) ∈ B(0, 2ρ), so (10.21) holds. We are left with the case when x, y ∈ E ∩ B(0, ρ). Then x also lies in some E ′ i . If i = j, we just use (10.18) and (10.19). Otherwise, |ϕ(x)ϕ(y)| ≤ 2ρ, and |y -x| ≥ dist(E ′ j , E ′ i ) > 0 by Lemma 10.13, so (10.21) holds in this last case as well. Let us now use the Whitney extension theorem and extend ϕ to the rest of R n , so that it stays Lipschitz. For Lemma 10.9 and Theorem 10.1, we don't really need that (10.23) ϕ(B(0, ρ)) ⊂ B because there will be no gauge function to pay for, but let us require this anyway, for later uses of the argument. This is easy to arrange; simply compose the restriction of ϕ to B(0, ρ) with the radial contraction on B defined by f (z) = z for z ∈ B, and f (z) = z/|z| otherwise. This does not change the values of ϕ(y) when y ∈ E ∩ B(0, ρ), by (10.22), and ϕ stays Lipschitz on R n because ϕ(y) = y = f (y) on ∂B.

Our argument so far would work with minor modifications for weak almost-minimal sets. We shall return to this soon, but for the moment let us do the accounting in the situation of Lemma 10.9 and Theorem 10.1, where E is a reduced minimal set. Since ϕ is Lipschitz and ϕ(x) = x out of B, we can apply (8.2), which says that

H 1 (E \ ϕ(E)) ≤ H 1 (ϕ(E) \ E)
. Since E and ϕ(E) coincide out of B (by (10.15) and (10.22)), we get that (10.24)

H 1 (E ∩ B) ≤ H 1 (ϕ(E) ∩ B) = H 1 (ϕ(E ∩ B))
(by (10.15) and (10.22) again). Let y ∈ E ∩ B be given. Thus y lies in some E ′ j . If y lies out of the B(a k , 2ε), ψ(y) = 1, and ϕ(y) = ϕ j (y) = h F j (y) ∈ F j (by (10.11)). Hence (10.25)

H 1 (ϕ(E ∩ B)) ≤ j≤M H 1 (F j ) + j k H 1 (ϕ j (E ′ j ∩ B ∩ B(a k , 2ε))) ≤ j≤M H 1 (F j ) + 10 j k H 1 (E ′ j ∩ B ∩ B(a k , 2ε)) = j≤M H 1 (F j ) + 10 k H 1 (E ∩ B ∩ B(a k , 2ε))
by ( 10.18) and (10.19). Hence (10.26)

H 1 (E ∩ B) ≤ j≤M H 1 (F j ) + 10 k H 1 (E ∩ B ∩ B(a k , 2ε))
by (10.24). We may now let ε go to 0, and we get that (10.27)

H 1 (E ∩ B) ≤ j≤M H 1 (F j )
because E ∩ ∂B is finite. Now recall that F j is the connected set that contains the points of E j ∩ ∂B for which H 1 (F j ) is minimal (the description above gives the uniqueness). In particular, H 1 (F j ) ≤ H 1 (E j ) because E j is connected. The E j are distinct and contained in E ∩ B by construction, so (10.27) says that H 1 (E j ) = H 1 (F j ), whence E j = F j . In addition, H 1 (E ′ j \ E j ) = 0. Suppose for a moment that E ′ j \ E j is not empty. Then we can find y ∈ E ′ j \ E j and r > 0 so small that B(y, r) does not meet the closed sets E j = F j , nor R n \ B(0, ρ) (because y ∈ E ∩ ∂B is impossible). Then E ∩ B(y, r) is contained in the union of the E ′ j \ E j , so H 1 (E ∩ B(y, r)) = 0, which contradicts our assumption that E is reduced. So E ′ j \ E j is empty, and E ∩ B is the union of the F j .

If F 1 ∩ ∂B contains three points, then there is no other F j (by (10.8)), and E ∩ B = F 1 ∩ B. In addition, the origin lies in F 1 because we assumed that it lies in E.

If F j does not come from a Y -set, it is of the form [a 1 , a 2 ] ∪ [a 2 , a 3 ]. Assume for instance that 0 ∈ [a 1 , a 2 ]. Then [a 1 , a 2 ]
is a diameter of B, and then its angle with [a 2 , a 3 ] it less than π/2. We saw that this is not possible (by minimality of F j ). So E coincides with a Y -set in B, and we are happy.

If F 1 ∩ ∂B contains two points a 1 and a 2 , F 2 either does not exist (if N =2) or is reduced to one point, and E ∩ B(0, ρ) = F 1 ∩ B(0, ρ) = (a 1 , a 2 ). This case is fine too. The same thing happens if some other F j ∩ ∂B has exactly two points.

The other cases where no E j contains more than one point of ∂B are impossible, because then the F j are reduced to one point, and the origin cannot lie in E.

This completes our proof of Lemma 10.9 and Theorem 10.1, modulo Lemma 10.13 which we prove now.

Proof of Lemma 10.13. For each small η > 0, denote by G j,η the set of points y ∈ E ∩B that can be connected to E j by an η-string in E ∩ B, i.e., a finite sequence {y k } in E ∩ B, with y 0 = y, |y ky k-1 | ≤ η, and the last y k lies in E j . Let i ≤ M be given, with i = j, and let us check that for η small enough, G j,η does not meet E i .

Suppose not. We can find a sequence {η l } that tends to 0, and for each l an η l -string {y k } in E ∩ B that goes from some z l ∈ E i to some w l ∈ E j . Set Γ l = ∪ k≥1 [y k-1 , y k ]; thus Γ l is compact, connected, and contained in B. Replace {η l } with a subsequence for which {z l } converges to some z, {w l } converges to some w, and {Γ l } converges to some compact set Γ ⊂ B.

Notice that y ∈ E i (because E i is closed), z ∈ E j , and also Γ ⊂ E, because each point of Γ l lies within η l of E, and E is closed. In addition, we claim that Γ is connected.

Again, suppose this is not the case. Split Γ as Γ = G 1 ∪ G 2 , where the G i are disjoint nonempty closed subsets. Set d = dist(G 1 , G 2 ) > 0, and take l so large that every point of Γ l lies within d/3 of Γ, and every point of Γ lies within d/3 of Γ l . For m = 1, 2, set G l,m = y ∈ Γ l ; dist(y, G m ) ≤ d/3 . First, G l,m is closed. It is not empty, because G m contains some y, then there is a y ′ ∈ Γ l such that |y ′ -y| ≤ d/3, and obviously y ′ ∈ G l,m . Also, dist(G l,1 , G l,2 ) ≥ d/3 by definitions, and finally Γ l = G l,1 ∪ G l,2 because every point of Γ l lies within d/3 of Γ = G 1 ∪ G 2 . All this contradicts the connectedness of Γ l , and our claim that Γ is connected follows.

We managed to find a connected set Γ ⊂ E ∩ B that contains z ∈ E i and w ∈ E j , but this contradicts our assumption that E i and E j are different connected components of E ∩ B. Thus G j,η does not meet E i for η small. Then dist(G j,η , G i,η ) ≥ η, because otherwise we could put two η-strings together to go from E i to E j . Now choose η so small that all the G j,η lie at distance ≥ η from each other, and set E ′ j = G j,η for j > 1 and E ′ 1 = E ∩ B \ ∪ j>1 G j,η . These sets are disjoint because the G j,η don't meet. It is obvious that E ′ j ⊃ E j for j > 1, and for j = 1 this is true because G 1,η does not meet the other G j,η . Their union is E ∩ B by definition of E ′ 1 . We still need to check (10.14). Let us even check that dist(E ′ i , E ′ j ) ≥ η for i = j. When i and j are larger than 1, this comes directly from our choice of η. If i = 1, we simply observe that if y ∈ E ′ 1 , then dist(y, E ′ j ) = dist(y, G j,η ) ≥ η because otherwise y would lie in G j,η . This completes our proof of Lemma 10.13. Lemma 10.9 and Theorem 10.1 follows as well.

Local regularity for 1-dimensional weak almost-minimal sets

We still want to describe the two-dimensional minimal cones in R n , and to this effect we shall need information on the reduced weak almost-minimal sets of dimension 1. In this section we let E ⊂ ∂B(0, 1) ⊂ R n be a closed set, and we assume that (11.1) E is a reduced weak almost-minimal set of dimension 1, with gauge function h [see Definition 9.1 for weak almost-minimal sets, and Definition 2.12 (or (9.5)), for the notion of reduction], and we try to derive some amount of regularity from mild assumptions on E ∩ ∂B(x, ρ). The main result of this section is the following.

Proposition 11.2. Let E ⊂ ∂B(0, 1) ⊂ R n satisfy (11.1), and suppose that E ∩ ∂B(x, ρ) has at most three points. Also suppose that r + h(r) is small enough (depending only on n). Then for y ∈ E ∩ B(x, ρ/10) and 0 < r ≤ ρ/10, we can find a one-dimensional minimal cone Z that contains y (i.e., a line or a Y -set through y) such that

(11.3) dist(z,Z) ≤ C[r + h(10r)] 1/2 r for z ∈ E ∩ B(y, r)
and dist(z, E) ≤ C[r + h(10r)] 1/2 r for z ∈ Z ∩ B(y, r).

Observe that we do not require Z to be centered at y when it is a Y , and the type of Z may depend on y and r (but in fact not wildly). If E, x, and ρ are as in the proposition and h(ρ) is small enough, it follows from Proposition 12.6 (the one-dimensional case of the generalization of Reifenberg' theorem that will be stated in Section 15) that E is biHölder-equivalent to a line or a Y -set in a ball that contains B(y, ρ/20).

In the case when E = K ∪ ∂B(0, 1) for a reduced minimal set K of dimension 2, we could also get the biHölder equivalence from Theorem 8.23 and Theorem 10.1, but in this case Proposition 9.4 says that (11.1) and then (11.3) hold with h(r) ≤ Cr; we shall see in Proposition 12.27 that (11.3) then gives a C 1 control of E near y. This better control will make it easier to prove that E is composed of arcs of great circles which only meet with 120 • angles; see Section 14.

Let us return to the general case where (11.1) holds, forget about Proposition 11.2 for the moment, and see what the proof of Lemma 10.9 yields in this context. We let x ∈ ∂B(0, 1) and ρ ∈ (0, 1/2) be given, and start with the only additional assumption that (11.4) E ∩ ∂B(x, ρ) is finite.

As before, we denote by a j , j ≤ N , the points of E ∩ ∂B(x, ρ) (if E ∩ ∂B(x, ρ) = ∅), and by E j the connected component of a j in E ∩ B(x, ρ). We number the points a j so that the first E j , j ≤ M , are distinct, while each E k , k > M , coincides with some previous E j . And we denote by F j the shortest connected set in R n that contains all the points of E j ∩ ∂B(x, ρ).

Lemma 11.5. There is a constant C ≥ 1, that depends only on n, such that

(11.6) H 1 (E ∩ B(x, ρ)) ≤ j≤M H 1 F j ) + C [h(ρ) + ρ] ρ.
Let us repeat the construction of the function ϕ as in Lemma 10.9, except that we replace the origin with x, and we only define ϕ on ∂B(0, 1). We cannot use ϕ directly, because of the constraint that competitors lie in ∂B(0, 1), so we shall need to project back onto ∂B(0, 1).

Denote by ξ the point of R n such that the cone centered at ξ over ∂B(x, ρ) ∩ ∂B(0, 1) is tangent to B along ∂B(x, ρ) ∩ ∂B(0, 1). That is, ξ lies on the line through 0 and x and zξ is orthogonal to zx for z ∈ ∂B(x, ρ) ∩ ∂B(0, 1). Notice that z lies on the half line [0, -x) because we may assume that ρ ≤ 1. Set η(z) = z for z ∈ ∂B(0, 1) \ B(x, ρ) and, for z ∈ B(z, ρ), denote by η(z) the radial projection of z onto ∂B(0, 1) ∩ B(x, ρ) centered at ξ. That is, η(z) is the only point of ∂B(0, 1) ∩ B(x, ρ) that lies on the line through 0 and z. Observe that η is 2-Lipschitz on [∂B(0, 1) \ B(x, ρ)] ∪ B(x, ρ); we shall not need to define it on a larger set.

We set f (y) = η(ϕ(y)) for y ∈ ∂B(0, 1). Observe that we still have that (11.7) f (y) = ϕ(y) = y for y ∈ ∂B(0, 1) \ B(x, ρ) by (10.15), and ϕ(y) ∈ B = B(x, ρ) for y ∈ B(x, ρ), by (10.23), so that in particular f (y) is well defined, and

(11.8) f (B(x, ρ)) ⊂ B(x, ρ) ∩ ∂B(0, 1).
The composition with η does not radically change the Lipschitz properties of ϕ, so f is Lipschitz (by (10.21)), and (11.9) the restriction of f to each E ′ j is 20-Lipschitz, by (10.19) and (10.18). Let us apply the weak almost-minimality of E, as defined in Definition 9.1. The set W f = y ∈ ∂B(0, 1) ; f (y) = y is contained in B(x, ρ) by (11.7), and f (W f ) ⊂ B(x, ρ) by (11.8). In addition, we have a partition of E ∩ B(x, ρ) into sets E ′ j , as in Lemma 10.13, and f is 20-Lipschitz on each E ′ j by (11.9). Thus we can take M = 20 in (9.3), which yields (11.10)

H 1 E ∩ B(x, ρ) ≤ H 1 f (E) ∩ B(x, ρ) + C h(ρ)ρ.
We want to estimate H 1 f (E) ∩ B(x, ρ) . We get no contribution from the complement of B(x, ρ), by (11.7). So it is enough to control the sets f

(E ′ j ) (because E ∩ B(x, ρ) = ∪ j E ′ j ).
Recall that E ∩ ∂B(x, ρ) is composed of a finite number of points a k . Observe that (11.11)

j k H 1 (f (E ′ j ∩ B(a k , 2ε))) ≤ 20 j k H 1 (E ′ j ∩ B(a k , 2ε)) ≤ 20 k H 1 (E ∩ B(a k , 2ε))
by (11.9). When y ∈ E ′ j lies out of the B(a k , 2ε), we know that ϕ(y) = h F j (y) ∈ F j , where F j still denotes the shortest connected set that contains the points of E j ∩ ∂B(x, ρ). Then f (y) ∈ η(F j ). Altogether, (11.12)

H 1 f (E) ∩ B(x, ρ) ≤ j H 1 (η(F j )) + j k H 1 f (E ′ j ∩ B(a k , 2ε) ≤ j H 1 (η(F j ))+ ≤ 20 k H 1 (E ∩ B(a k , 2ε))
by (11.11). We replace in (10.10), let ε tend to 0, and get that (11.13)

H 1 (E ∩ B(x, ρ)) ≤ j H 1 (η(F j )) + C h(ρ)ρ ≤ j H 1 F j ) + C [h(ρ) + ρ] ρ
because η barely deforms F j (recall that the center ξ of the radial projection is far from x, and that F j lies in the convex hull of the points a k of E ∩ ∂B(x, ρ), hence close to the surface ∂B(0, 1)). This proves Lemma 11.5.

Lemma 11.14. There is a constant C 0 ≥ 1, that depends only on n, such that if (11.1) holds, then

(11.15) H 1 (E ∩ B(x 0 , r)) ≥ r/2 for x 0 ∈ E and r > 0 such that r + h(r) ≤ C -1 0 .
Indeed suppose that x 0 ∈ E and r > 0 are such that H 1 (E ∩B(x 0 , r)) < r/2. Pick any x ∈ ∂B(0, 1)∩B(x 0 , r/4), and set δ(y) = |y -x| for y ∈ R n ; then H 1 (δ(E ∩B(x 0 , r))) < r/2 because δ is 1-Lipschitz, so we can find ρ ∈ (r/4, 3r/4) such that ρ / ∈ δ(E ∩B(x 0 , r)). Since B(x, ρ) ⊂ B(x 0 , r), this means that ∂B(x, ρ) does not meet E.

Thus we can apply Lemma 11.5, with no set F j , and get that

(11.16) H 1 (E ∩ B(x, r/4)) ≤ H 1 (E ∩ B(x, ρ)) ≤ C [h(ρ) + ρ] ρ ≤ C [h(r) + r] r < r/8
because h is nondecreasing (it is a gauge function) and if h(r) + r is small enough. We can iterate this argument, this time keeping the same point x, and we get that (11.17)

H 1 (E ∩ B(x, 4 -k r)) < 4 -k-1 r for k ≥ 1
and even, by (11.16) again, (11.18)

H 1 (E ∩ B(x, 4 -k-1 r)) ≤ C[h(4 -k r) + 4 -k r] 4 -k r.
Then, if t is small and k is chosen such that 4 -k-2 ≤ t ≤ 4 -k-1 , we get that

(11.19) t -1 H 1 (E ∩ B(x, t)) ≤ t -1 H 1 (E ∩ B(x, 4 -k-1 )) ≤ Ct -1 [h(4 -k r) + 4 -k r] 4 -k r ≤ 4C[h(16t) + 16t].
We proved that (11.20)

lim t→0 t -1 H 1 (E ∩ B(x, t)) = 0
for every x ∈ ∂B(0, 1)∩B(x 0 , r/4). That is, B(x 0 , r/4) contains no point of positive density for E. But it is a simple and standard result that as soon as H 1 (E) < +∞, H 1 -almost every point of E has positive upper density; see for instance [Ma], Theorem 6.2 on page 89 or Lemma 19.1 in [D2]. Thus H 1 (E ∩ B(x 0 , r/4)) = 0, and x / ∈ E because E is reduced. Lemma 11.14 follows.

We start the local study of E that will lead to Proposition 11.2 with the simpler case when (11.1) holds and E ∩ ∂B(x, ρ) has two points.

Lemma 11.21. There is a constant C 1 ≥ 1 such that the following holds. Let E ⊂ ∂B(0, 1) satisfy (11.1), and let x ∈ ∂B(0, 1) and ρ ∈ (0, 1/2) be such that

E ∩ ∂B(x, ρ) = {a 1 , a 2 }, E ∩ B(x, ρ/2) = ∅, and r + h(r) ≤ C -1 1 . Then (11.22) H 1 (E ∩ B(x, ρ)) ≤ |a 2 -a 1 | + C[ρ + h(ρ)] ρ,
where C is as in (11.6),

(11.23) dist(y, [a 1 , a 2 ]) ≤ C 1 [ρ + h(ρ)] 1/2 ρ for y ∈ E ∩ B(x, 9ρ/10), and
(11.24) dist(z, E ∩ B(x, ρ)) ≤ C 1 [ρ + h(ρ)] 1/2 ρ for z ∈ [a 1 , a 2 ].
Let E, x, and ρ be as in the lemma. First observe that if we take C 1 ≥ C 0 , we can apply Lemma 11.14 to the pair (x 0 , ρ/2), where x 0 is any point of E ∩ B(x, ρ/2), and get that (11.25)

H 1 (E ∩ B(x, ρ)) ≥ H 1 (E ∩ B(x 0 , ρ/2)) ≥ ρ/4.
Denote by E 1 the connected component of E ∩ B(x, ρ) that contains a 1 . First suppose that E 1 does not contain a 2 . Then Lemma 11.5 applies, with and (11.6) by (11.25). So E 1 contains a 2 , we can Lemma 11.5 with F 1 = [a 1 , a 2 ], and we get (11.22).

F 1 = {a 1 } and F 2 = {a 2 },
says that H 1 (E ∩ B(x, ρ)) ≤ C[ρ + h(ρ)] ρ. This is impossible (if ρ + h(ρ) is small enough),
Next let y ∈ E 1 be given, set d = dist(y, [a 1 , a 2 ]), and denote by Z the shortest connected set that contains y, a 1 , and a 2 . Thus (11.26) by (11.22). by (11.26).

H 1 (Z) ≤ H 1 (E 1 ) ≤ |a 2 -a 1 | + C[ρ + h(ρ)] ρ,
If Z = [a 1 , a 2 ] ∪ [a i , y] for some i, we get that d ≤ |y -a i | ≤ C[ρ + h(ρ)] ρ,
If Z = [a 1 , y] ∪ [y, a 2 ], (11.26) and Pythagoras yield d ≤ C[ρ + h(ρ)] 1/2 ρ. The last case is when Z = [y, z] ∪ [a 1 , z] ∪ [a 2 , z] is a piece of Y -set, but then dist(z, [a 1 , a 2 ]) ≤ C[ρ+h(ρ)] 1/2 ρ because H 1 ([a 1 , z]∪[a 2 , z]) ≤ H 1 (Z)
and by (11.26), and

|y -z| = H 1 (Z)- H 1 ([a 1 , z] ∪ [a 2 , z]) ≤ H 1 (Z) -|a 2 -a 1 | ≤ C[ρ + h(ρ)] ρ. In this last case too, d ≤ C[ρ + h(ρ)] 1/2 ρ. So we proved that (11.27) dist(y, [a 1 , a 2 ]) ≤ C 1 [ρ + h(ρ)] 1/2 ρ for y ∈ E 1 .
This proves (11.24), because if z ∈ [a 1 , a 2 ] is given, the hyperplane through z which is orthogonal to [a 1 , a 2 ] meets E 1 (because E 1 connects a 1 to a 2 ), and the intersection lies within C 1 [ρ + h(ρ)] 1/2 ρ of z by (11.27).

We still need to prove (11.23). Let y ∈ E ∩ B(x, 9ρ/10) be given, and set δ = Min(dist(y, E 1 ), ρ/10). Thus B(y, δ) ⊂ B(x, ρ) \ E 1 , and (11.28)

H 1 (E ∩ B(y, δ)) ≤ H 1 (E ∩ B(x, ρ)) -H 1 (E 1 ) ≤ H 1 (E ∩ B(x, ρ)) -|a 2 -a 1 | ≤ C[ρ + h(ρ)] ρ,
because E 1 is connected and contains a 1 and a 2 and by (11.22). On the other hand, Lemma 11.14 says that

H 1 (E ∩ B(y, δ)) ≥ δ/2 (if C 1 ≥ C 0 ), so δ ≤ C[ρ + h(ρ)] ρ.
In particular, δ < ρ/10, so dist(y, (11.27). This proves (11.23) (with a slightly larger C 1 ); Lemma 11.21 follows.

E 1 ) = δ ≤ C[ρ + h(ρ)] ρ and then dist(y, [a 1 , a 2 ]) ≤ C[ρ + h(ρ)] 1/2 ρ by
Lemma 11.21 can be iterated to give more information. The following lemma gives a proof of Proposition 11.2 in the special case when E ∩ B(x, ρ) has at most two points.

Lemma 11.29. Let E ⊂ ∂B(0, 1) ⊂ R n satisfy (11.1), and suppose that E ∩ ∂B(x, ρ) has at most two points. Also suppose that r + h(r) is small enough (depending only on n). Then for y ∈ E ∩ B(x, ρ/10) and 0 < r ≤ ρ/10, we can find a line L that contains y such that

(11.30) dist(z,L) ≤ C[r + h(10r)] 1/2 r for z ∈ E ∩ B(y, r) and dist(z, E) ≤ C[r + h(10r)] 1/2 r for z ∈ L ∩ B(y, r).
Let E, x, ρ, and y be as in the lemma. Then we are in the situation of Lemma 11.21 (because E ∩ B(x, ρ/2) contains y). By (11.22),(11.31) H 1 (E ∩ B(y, 9ρ/10)) ≤ 2ρ + C[ρ + h(ρ)] ρ < 21ρ/10, so we can find r ∈ (ρ/10, 9ρ/10) such that E ∩ ∂B(y, r) has at most two points (otherwise H 1 (E ∩ B(y, 9ρ/10)) \ B(y, ρ/10)) ≥ 24ρ/10 by the same computation as for (10.7)).

In fact, E ∩ ∂B(y, r) has exactly two points, because otherwise Lemma 11.5 says that H 1 (E ∩ B(y, r)) ≤ C[r + h(r)] r, which contradicts Lemma 11.14. So we can apply Lemma 11.21 again to the pair (y, r).

Let us iterate this with the same y. We get a sequence of radii r k , with r 0 = r and r k ∈ (r k-1 /10, 9r k-1 /10) for k ≥ 1, such that Lemma 11.21 applies to the pair (y, r k ). For each k, we get a line segment I k with extremities on ∂B(y, r k ), such that

(11.32) dist(y,I k ) ≤ C 1 [r k + h(r k )] 1/2 r k for y ∈ E ∩ B(y, 9r k /10) and dist(z, E ∩ B(y, r k )) ≤ C 1 [r k + h(r k )] 1/2 r k for z ∈ I k .
The sequence {r k } is dense enough; we get that for 0 < t < 8ρ/10, we can find a line L such that

(11.33) dist(z, L) ≤ C[t + h(10t)] 1/2 t for z ∈ E ∩ B(y, t) and dist(z, E) ≤ C[t + h(10t)] 1/2 t for z ∈ L ∩ B(y, t).
Notice that dist(y, L) ≤ C[t + h(10t)] 1/2 t because y ∈ E, so the line L ′ through y and parallel to L also satisfies (11.33) (with a larger constant). This proves Lemma 11.29.

In the case of Lemma 11.29, the classical topological disk theorem of Reifenberg, in the simple case of sets of dimension 1, says that E is a Hölder curve near x. If in addition h(t) ≤ Ct α for some α > 0, say, then (11.30) implies that E has a tangent everywhere on B(x, ρ/10), and the direction of this tangent is Hölder continuous, so that E is in fact a C 1,α/2 curve near x.

Let us now turn to the case when (11.1) holds and E ∩ ∂B(x, ρ) has three points, and discuss the analogue of Lemmas 11.21 and 11.29. Assume that (1.1) holds, that (11.34) E ∩ B(x, ρ/10) = ∅, and that

(11.35) E ∩ ∂B(x, ρ) = {a 1 , a 2 , a 3 }.
Recall that E j is the component of a j in E ∩ B(x, ρ). If the three E j are different, Lemma 11.5 applies with F j = {a j }, and yields

H 1 (E ∩ B(x, ρ)) ≤ C [h(ρ) + ρ] ρ.
This is impossible, by (11.34) and Lemma 11.14. So we may assume that E 1 contains a 1 and a 2 . If E 1 does not contain a 3 , we can apply the proof of Lemma 11.21, with F 1 = [a 1 , a 2 ] and F 2 = {a 3 }, and get the conclusions (11.22)-(11.24), and then (11.31)-(11.33). In particular, the conclusions of Lemma 11.29 and Proposition 11.2 are valid.

So let us assume that E 1 contains the three a j . Let F 1 be the shortest connected set that contains the three a j . We know from Lemma 11.5 that (11.36)

H 1 (E ∩ B(x, ρ)) ≤ H 1 (F 1 ) + C [h(ρ) + ρ] ρ,
and we claim that this implies that

(11.37) dist(y, F 1 ) ≤ C [h(ρ) + ρ] 1/2 ρ for y ∈ E 1 .
Let G 1 be a connected set that contains y and the three a j , and whose length is minimal. Thus (11.36) says that (11.38)

H 1 (G 1 ) ≤ H 1 (E 1 ) ≤ H 1 F 1 ) + C [h(ρ) + ρ] ρ.
Let us describe G 1 as we did for F j in Lemma 10.13. As before, we can find a simple arc γ 1 ⊂ G 1 from a 1 to a 2 , then a simple arc γ 2 ⊂ G 1 from a 3 to some point z 1 ∈ γ 1 (where z 1 is the first point of γ 1 ∩ γ 2 when we run along γ 2 starting from a 3 ), and then a simple arc γ 3 that starts from y and hits γ 1 ∪ γ 2 at some point z 2 (and not before). We do not exclude the cases when z 1 is one of the a i or z 2 = y. Set γ = ∪ 1≤i≤3 γ i . Since G 1 has minimal length and γ connects y and the three a j , G 1 = γ. In addition, the arcs of γ between two consecutive nodes or endpoints are line segments. That is, G 1 = γ is composed of five or less line segments. The question is how far from F 1 can it get, knowing (11.38).

Set

G 2 = G 1 \ [y, z 2 [.
Observe that G 2 still connects the three a i , and by (11.38). Thus (11.37) will follow as soon as we check that

H 1 (G 2 ) = H 1 (G 1 )-|y-z 2 |. By minimality of F 1 , H 1 (F 1 ) ≤ H 1 (G 2 ), so |y-z 2 | = H 1 (G 1 )-H 1 (G 2 ) ≤ H 1 (G 1 ) -H 1 (F 1 ) ≤ [h(ρ) + ρ] ρ,
(11.39) dist(z 2 , F 1 ) ≤ C [h(ρ) + ρ] 1/2 ρ.
Recall that z 2 lies in γ 1 ∪ γ 2 . First suppose that z 2 ∈ γ 2 . Thus G 2 is composed of the four arc segments [a 1 , z 1 ], [a 2 , z 1 ], [z 1 , z 2 ], and [z 2 , a 3 ], where we do not exclude the possibility that some of these segments are reduced to one point. If we replace [z 1 , z 2 ] ∪ [z 2 , a 3 ] with [z 1 , a 3 ] in G 2 , we get a shorter connected set G 3 that contains the three a j , and so by (11.38). Thus dist(z 2 , [z 1 , a 3 ]) ≤ C [h(ρ) + ρ] 1/2 ρ, and it will be enough to show that

H 1 (G 2 ) -H 1 (G 3 ) ≤ H 1 (G 2 ) -H 1 (F 1 ) ≤ [h(ρ) + ρ] ρ,
(11.40) dist(z, F 1 ) ≤ C [h(ρ) + ρ] 1/2 ρ. for z ∈ G 3 .
In the other case where

z 2 ∈ γ 1 , G 2 is either composed of the four segments [a 1 , z 2 ], [z 2 , z 1 ], [z 1 , a 2 ], and [z 1 , a 3 ] (if z 2 lies between a 1 and z 1 on γ 1 ), or [a 1 , z 1 ], a 3 , z 1 ], [z 1 , z 2 ],
and [z 2 , a 2 ] (if z 2 lies between z 1 and a 2 ). In both cases we set G 3 = ∪ j [a j , z 1 ] as before, and it is still enough to prove (11.40). Notice that (11.41) (11.38). Also, F 1 = ∪ j [a j , z], for some z that may also be equal to one of the a j , and it is enough to show that

H 1 (G 3 ) ≤ H 1 (G 1 ) ≤ H 1 F 1 ) + C [h(ρ) + ρ] ρ by
(11.42) |z -z 1 | ≤ C [h(ρ) + ρ] 1/2 ρ. Suppose that z 1 = z, set v = (z 1 -z)/|z 1 -z|, and set f (t) = j |z + tv -a j | for t ≥ 0.
Let us also assume that z + tv = a j for t = 0, so that we can differentiate f . Let us compute some derivatives. Set r j (t) = |z+tv-a j | and u j (t) = (z+tv-a j )/r j (t). Then r ′ j (t) = v, u j (t) and u ′ j (t) = v/r j (t) -(z + tva j ) r j (t) -2 v, u j (t) , so

(11.43) r ′′ j (t) = v, u ′ j (t) = r j (t) -1 -v, z + tv -a j r j (t) -2 v, u j (t) = r j (t) -1 1 -v, u j (t) 2
(where we used again the definition of u j (t)). Thus f ′ (t) = j v, u j (t) and

f ′′ (t) = j r j (t) -1 1 -v, u j (t) 2 ≥ 0.
Notice that f has a minimum at t = 0. Then the half derivative f ′ (0 + ) of f at 0 is nonnegative. As long as (11.44)

j 1 -v, u j (t) 2 > 10 -2
for 0 ≤ t < t 0 , we get that f ′′ (t) ≥ (200ρ) -1 (because r j ≤ 2ρ, since all our points lie in B(x, ρ)). Then (11.45)

f ′ (t) ≥ (200ρ) -1 t for 0 ≤ t ≤ t 0 .
If (11.44) fails for some t 0 , the three v, u j (t 0 ) 2 are close to 1, which means that the three u j (t 0 ) are close to v or -v. Recall that f ′ (t) = j v, u j (t) , and f ′ (t) ≥ 0 (because f ′ (0 + ) ≥ 0 and f ′′ ≥ 0), so at least two of the u j (t 0 ) are close to v. This is even more so for t > t 0 (we move z + tv away from the corresponding two a j ), so f ′ (t) ≥ 1/2 for t ≥ t 0 , which is even better than (11.45) if t ≤ 2ρ.

We may now integrate (11.45) or the better estimate between 0 and t 1 = |z 1 -z| and get that f and (11.42) follows. This completes our proof of (11.37).

(t 1 ) -f (0) ≥ (400ρ) -1 t 2 1 . But (11.46) f (t 1 ) -f (0) = j |z 1 -a j | - j |z -a j | = H 1 (G 3 ) -H 1 (F 1 ) ≤ C [h(ρ) + ρ] ρ by (11.41), so |z 1 -z| 2 = t 2 1 ≤ C [h(ρ) + ρ] ρ 2 ,
Observe that by (11.36) and because E 1 also connects the three a j , so

H 1 (E ∩ B(x, ρ)) ≤ H 1 (F 1 ) + C [h(ρ) + ρ] ρ ≤ H 1 (E 1 ) + C [h(ρ) + ρ] ρ,
H 1 (E ∩ B(x, ρ) \ E 1 ) ≤ C [h(ρ) + ρ] ρ. Then (11.47) dist(y, F 1 ) ≤ C [h(ρ) + ρ] 1/2 ρ for y ∈ E ∩ B(x,
∩ B(x, ρ)) ≤ dist(y, E 1 ) ≤ C [h(ρ) + ρ] 1/2 ρ for y ∈ F 1 ,
because E 1 connects the three a i and is contained in a small strip around F 1 , by (11.37).

[Cut the strip transversally by a segment of length C [h(ρ) + ρ] 1/2 ρ, and observe that if the segment did not meet E 1 , it would separate some a j from the other ones.] Finally notice that since E meets B(x, ρ/10) by (11.34), F 1 gets very close to B(x, ρ/10). If h(r) is small enough, this actually prevents F 1 from being a simple union of two intervals [a i , a j ], because one of them would be close to a diagonal, and then they would make an angle smaller than 2π/3. So F 1 is the intersection of B(x, ρ) with a Y -set.

We just finished a verification of the analogue of Lemma 11.21, and we are ready to complete the verification of Proposition 11.2 in the present remaining case. Let y ∈ E ∩ B(x, ρ/10) be given; as for (11.31), (11.49) In all cases we can find a sequence of radii r k , with r 0 = r and r k ∈ (r k-1 /10, 9r k-1 /10) for k ≥ 1, so that for each k, we can apply Lemma 11.21 or the discussion above to the pair (y, r k ). By (11.23) and (11.24) or (11.47) and (11.48), there is minimal cone Z k (a line or a Y -set) such that

H 1 (E ∩ B(y, 9ρ/10)) ≤ H 1 (E ∩ B(x, ρ)) ≤ H 1 (F 1 ) + C [h(ρ) + ρ] ρ, ≤ 3ρ + C[ρ + h(ρ)] ρ < 31ρ/
(11.50) dist(z,Z k ) ≤ C 1 [r k + h(r k )] 1/2 r k for z ∈ E ∩ B(y, 9r k /10) and dist(z, E ∩ B(y, r k )) ≤ C 1 [r k + h(r k )] 1/2 r k for z ∈ Z k ∩ B(y, r k ).
If Z k does not go through y, we can replace it with a translation of Z k , without affecting (11.50). We also get a similar result (with a constant C 1 ten times larger) for every r ∈ (0, ρ/10), by the same argument as for (11.33). [Apply (11.50) to r k , where r ≤ r k ≤ 10r.] This completes our proof of Proposition 11.2.

Remark 11.51. It is very likely that we can use Lemma 11.5 to prove a rapid decay estimate on the density of E when E ∩ ∂B(x, ρ) has more than four points. Indeed, if

E ∩ ∂B(x, ρ) = {a 1 , • • • , a N }, with N ≥ 4
, the length of the shortest connected set that contains the a j is at most τ N ρ, where the constant τ < 1 depends on the dimension, and can be taken very small when N is large.

Once we get such a decay, we can probably find an effective radius r 0 > 0 (that depends on the dimension) such that for y ∈ E and 0 < r ≤ r 0 , we can find ρ > r so that E ∩ ∂B(x, ρ) has at most three points. Then there is also an effective radius r 1 > 0 such that for ∈ E, E is biHölder-equivalent to a line or a Y in a ball that contains B(y, r 1 ).

A baby extension of Reifenberg's theorem

If we want to deduce a good description of E ∩ B(x, ρ/10) from the conclusion of Proposition 11.2, the simplest is to use a slight improvement of Reifenberg's topological disk theorem, in the special case of one-dimensional sets, but where we only assume that E is close enough to lines or Y -sets. The statement below is a special case of the generalization of Reifenberg's theorem from [DDT] that will be described in Section 15, but for the convenience of the reader, we shall try to give a short account of how it could be proved.

In this section, E ⊂ R n is a compact set. Denote by Y the set of nonempty onedimensional minimal cones in R n . That is, Y ∈ Y if Y is a line or a Y -set (three half lines in a same plane, with the same endpoint, and that make 120 • angles at that point). We measure the closeness of E to sets of Y with (12.1)

β Y (x, r) = inf Y ∈Y d x,r (E, Y ),
where (12.2)

d x,r (E, Y ) = 1 r sup dist(y, Y ) ; y ∈ E ∩ B(x, r) + 1 r sup dist(y, E) ; y ∈ Y ∩ B(x, r) .
We shall only use β Y (x, r) and d x,r (E, Y ) when E and Y meet B(x, r), so we don't need to worry about the definition of the supremums in (12.2) when some sets are empty. Notice that then β Y (x, r) ≤ 2 (take any set Y through x). We could easily restrict to Y -sets (and forget about lines), because for each line L, there is a Y that coincides with L in B(x, 4r), but it is more natural to keep the lines. If Y was reduced to the set of lines, the β Y (x, r) would be the bilateral P. Jones numbers that are often used to study the uniform rectifiability of sets. We shall assume that the origin lies in E (a normalization), and that

(12.3) β Y (x, r) ≤ ε for x ∈ E ∩ B(0, 2) and 0 < r ≤ 2,
where ε > 0 is a small constant that we can choose later. Let us immediately choose, for

x ∈ E ∩ B(0, 2) and 0 < r ≤ 2, a set Y (x, r) ∈ Y such that (12.4)

d x,r (E, Y (x, r)) ≤ β Y (x, r) ≤ ε.
Our first statement will be a description of E near the origin as a biHölder image of a line or a Y -set. Later in this section we will prove the C 1 equivalence under a suitable decay assumption on the β Y (x, r). Our initial description will be slightly different, depending on whether or not (12.5) Y (0, 2) is a Y -set centered in B(0, 12/10).

Proposition 12.6. For each small constant τ > 0, there is a constant ε > 0, that depends only on n and τ , such that if E ⊂ R n is a compact set such that 0 ∈ E and (12.3) holds, we have the following description. If (12.5) does not hold, there is a line L through the origin and a biHölder mapping f : L ∩ B(0, 11/10) → R n , such that

(12.7) |f (x) -x| ≤ τ for x ∈ L ∩ B(0, 11/10), (12.8) (1 -τ )|x -y| 1+τ ≤ |f (x) -f (y)| ≤ (1 + τ )|x -y| 1-τ for x, y ∈ L ∩ B(0, 11/10), (12.9) E ∩ B(0, 1 -τ ) ⊂ f (L ∩ B(0, 1)) ⊂ E ∩ B(0, 1 + τ ).
If (12.5) holds, there is a Y -set Y centered at some z ∈ E ∩ B(0, 13/10), and a biHölder mapping f : Y ∩ B(0, 14/10) → R n , such that f (z) = z,

(12.10) |f (x) -x| ≤ τ for x ∈ Y ∩ B(0, 14/10), (12.11) (1 -τ )|x -y| 1+τ ≤ |f (x) -f (y)| ≤ (1 + τ )|x -y| 1-τ for x, y ∈ Y ∩ B(0, 13/10), (12.12) E ∩ B(0, 13/10 -τ ) ⊂ f (L ∩ B(0, 13/10)) ⊂ E ∩ B(0, 13/10 + τ ).
Of course the precise values of the explicit constants are not so important. The main point is a local description of E in B(0, 1τ ) as a simple curve or a union of three curves that we control well. We can also extend our mapping f to R n , so that it is still a biHölder homeomorphism of R n that satisfies (12.7) and (12.8) (or (12.10) and (12.11)) for x, y ∈ B(0, 2), but we shall not need this. See Section 15 for a more general statement and [DDT] for a proof of that result. Also see Proposition 12.27 for a C 1 description.

For the proof of Proposition 12.6, we suppose that the reader knows a proof of the standard Reifenberg theorem, and we explain how to deduce the Proposition 12.6 from that proof. We start with the observation that the set Y (x, r) ∈ Y that we chose in (12.4) does not depend too wildly on x and r. Let us check that (12.13)

d y,9t/10 (Y (x, r), Y (y, t)) ≤ 100β Y (x, r) + 2β Y (y, t) ≤ 102ε when (12.14) x, y ∈ E ∩ B(0, 2), 0 ≤ r ≤ 1, B(y, t) ⊂ B(x, r
), and t ≥ r/90.

The point will be that Y (x, r) and Y (y, t) are both very close to E in B(y, t). First let z ∈ Y (y, t) ∩ B(y, 9t/10) be given. By (12.4), we can find z 1 ∈ E such that |z 1 -z| ≤ β Y (y, t)t. Then z 1 ∈ B(y, t) ⊂ B(x, r), and we can find

z 2 ∈ Y (x, r) such that |z 2 -z 1 | ≤ β Y (x, r)r. Altogether, |z 2 -z| ≤ β Y (y, t)t+β Y (x, r)r ≤ β Y (y, t)t+90β Y (x, r)t (by (12.14)). Similarly, if z ∈ Y (x, r) ∩ B(y, 9t/10), we can find z 1 ∈ E such that |z 1 -z| ≤ β Y (x, r)r ≤ 90β Y (x, r)t, so z 1 ∈ B(y, t) and we can find z 2 ∈ Y (y, t) so that |z 2 -z 1 | ≤ β Y (y, t)t and |z 2 -z| ≤ β Y (y, t)t + 90β Y (x, r)t; (12.13) follows.
Denote by c(x, r) the center of Y (x, r) if Y (x, r) is a Y -set, and set c(x, r) = ∞ otherwise. A direct consequence of (12.13) is that c(x, r) does not vary too rapidly, and in particular that

(12.15) |c(x, r) -c(y, t)| ≤ C β Y (x, r) + β Y (y, t) r ≤ Cεr
when (12.14) holds and c(x, r) or c(y, t) lies in B(y, t/2).

Let us first prove the proposition when (12.5) fails, i.e., when c(0, 2) / ∈ B(0, 12/10). Then (12.15) says that c(y, 1/10) / ∈ B(y, 1/20) for y ∈ E ∩ B(0, 11/10). Repeated applications of (12.15) show that c(y, t) / ∈ B(y, t/2) for y ∈ E ∩ B(0, 11/10) and 0 < t ≤ 1/10. In this case we have a good approximation of E by a line in B(y, t/4) for each y ∈ E ∩ B(0, 11/10) and 0 < t ≤ 1/10, and we can conclude with the standard Reifenberg theorem.

So we may now assume that (12.5) holds. Let us first look for the point z ∈ E where the three curves that compose E near B(0, 1) meet. By definition of Y (0, 2), we can find by (12.15). Notice that |c(z 0 , 1)z 0 | ≤ Cε, so c(z 0 , 1) lies in the middle of B(z 0 , 1) and we can find

z 0 ∈ E such that |z 0 -c(0, 2)| ≤ 2β Y (0, 2) ≤ 2ε. Then (12.16) |c(z 0 , 1) -c(0, 2)| ≤ C β Y (0, 2) + β Y (z 0 , 1) ≤ Cε,
z 1 ∈ E such that |z 1 -c(z 0 , 1)| ≤ β Y (z 0 , 1) ≤ Cε.
Then we iterate. By induction, we can find a sequence {z k } in E, so that (12.17)

|z k+1 -c(z k , 2 -k )| ≤ 2 -k β Y (z k , 2 -k ) ≤ Cε2 -k , hence (12.18) |c(z k+1 , 2 -k-1 ) -c(z k , 2 -k )| ≤ Cε2 -k by (12.15), then |c(z k+1 , 2 -k-1 ) -z k+1 | ≤ 2Cε2 -k < 2 -k-2
, we can find z k+2 , and so on. Notice also that

(12.19) |z k+2 -z k+1 | ≤ |z k+2 -c(z k+1 , 2 -k-1 )| + |z k+1 -c(z k , 2 -k )| + |c(z k+1 , 2 -k-1 ) -c(z k , 2 -k )| ≤ Cε2 -k
by (12.17) and (12.18), so the z k converge to some z ∈ E ∩ B(0, 13/10). We shall prove that near B(0, 13/10), E is composed of three Reifenberg-flat curves that meet at z. by (12.19) and (12.17). Denote by

Let k ≥ 0 be given. Observe that |z -c(z k , 2 -k )| ≤ |z -z k+1 | + |z k+1 -c(z k , 2 -k )| ≤ Cε2 -k ,
Y k the translation of Y (z k , 2 -k ) by z-c(z k , 2 -k ).
Thus Y k is a Y -set centered at z, and it also gives a very good approximation of E in B(z, 2 -k-1 ). Indeed, (12.20)

d z,2 -k-1 (Y k , E) ≤ 2d z k ,2 -k (Y (z k , 2 -k ), E) + 2 k+1 |z -c(z k , 2 -k )| ≤ Cε,
by the same proof as for (12.13). We also have a similar control out of B(z, 1/2). Indeed (12.21) d 0,2 (Y 0 , E) ≤ Cε because d 0,2 (Y (0, 2), E) ≤ Cε and Y 0 is so close to Y (0, 2). This last comes from (12.13) which says that d z 0 ,9/10 (Y (0, 2), Y (z 0 , 1)) ≤ Cε, and because |zz 0 | ≤ Cε.

Next we use (12.20) and (12.21) to get a good description of E in the annuli (12.22)

A 0 = B(0, 5/3) \ B(z, 2 -6 ) and A k = B(z, 2 -k-3 ) \ B(z, 2 -k-6 ), k ≥ 1.
We claim that

(12.23) c(x, t) / ∈ B(x, t/2) for x ∈ A k and t ≤ 2 -k-7 .
We start with the case when 2 -k-8 ≤ t ≤ 2 -k-7 . Let us first check that

(12.24) dist(w, Y k ) ≤ Cε2 -k for w ∈ Y (x, t) ∩ B(x, t).
By (12.4), we can find

w 1 ∈ E so that |w 1 -w| ≤ εt ≤ ε2 -k-7 ; hence w 1 ∈ B(z, 2 -k-2 ) (because x ∈ B(z, 2 -k-3
)), and (12.20) says that dist(w 1 , Y k ) ≤ Cε2 -k ; (12.24) follows.

Recall that Y k is a Y -set centered at z and that |z -x| ≥ 2 -k-6 ; then (12.24) prevents Y (x, t) from being a Y -set with a center in B(x, t/2). That is, (12.23) holds for 2

-k-8 ≤ t ≤ 2 -k-7 .
Finally, if (12.23) holds for some t ≤ 2 -k-7 , but c(x, t/2) ∈ B(x, t/4), (12.15) says that |c(x, t/2)-c(x, t)| ≤ Cεt, and c(x, t) ∈ B(x, t/2), a contradiction. This proves (12.23).

Because of (12.23), we can use the standard Reifenberg theorem to control E in each annulus A k . We get three biHölder parameterizations f i,k , defined (for instance) on the three intervals J i,k that compose Y k ∩ A k , with properties similar to (12.7) and (12.8) above, and such that E ∩ A k ⊂ ∪ i f i,k (J i,k ) ⊂ E. The argument even works on A 0 , even though A 0 is not perfectly symmetric.

Recall that the A k have large overlaps, so we can easily put together our descriptions of E in the various A k . We get three biHölder curves γ i , that leave from y and end out of B(0, 3/2), such that E ∩ B(0, 3/2) \ {y} ⊂ ∪ i γ i ⊂ E. These curves are disjoint (they are even far from each other on each A k ), and they can easily be parameterized (with gluing of the f i,k ), so that they have a single parameterization defined on some Y -set centered at z, and that satisfies the constraints (12.10)-(12.12). [Incidentally, we shall not really need (12.11) here.] Proposition 12.6 follows.

Next we want to study the regularity of the branches of E when the numbers β Y (x, r) tend to 0 uniformly and sufficiently fast. Let E satisfy the assumptions of Proposition 12.6, say, with τ = 10 -2 , and assume in addition that there are numbers ε k > 0, k ∈ N, such that (12.25)

k≥0 ε k < +∞ and (12.26) β Y (x, 2 -k ) ≤ ε k for y ∈ E ∩ B(0, 2) and k ≥ 0.
Let L or Y and f be as in Proposition 12.6. Set γ 1 = f (L ∩ B(0, 11/10)) when (12.5) does not hold. When (12.5) holds, denote by ℓ 1 , ℓ 2 , and ℓ 3 , the three components of Y ∩ B(14/10) \ {z}, and set γ j = f (ℓ j ) for 1 ≤ j ≤ 3.

Proposition 12.27. Let E satisfy the assumptions of Proposition 12.6, and also (12.25) and (12.26). Then each γ j is of class C 1 , and the tangent direction τ j (x) of γ j at x ∈ γ j is such that (12.31) for the definition of dist(τ j (x), τ j (y))]. In the case when (12.5) holds, the three γ j have tangent half lines at z that are coplanar and make 120 • angles.

(12.28) dist(τ j (x), τ j (y)) ≤ C k∈N ; 2 -k ≤16|x-y| ε j for x, y ∈ γ j [see
In particular, the curves γ j are of class C 1,α if ε k ≤ C2 -αk for some α ∈ (0, 1). Let us only prove the proposition when (12.5) holds; the other case is merely simpler. Let x ∈ γ j be given, and let k ≥ 0 be such that x ∈ A k . Also let t ≤ 2 -k-7 be given. Recall from (12.23) that c(x, t) / ∈ B(x, t/2). Also observe that Y (x, t) comes very near x, just by (12.4). Then Y (x, t) coincides with a line in B(x, t/2), and we denote this line by L(x, t). Thus (12.29)

d x,t/4 (E, L(x, t)) ≤ 4d x,t (E, Y (x, t)) ≤ 4β Y (x, t)
by definition of Y (x, t). It will be enough to restrict to powers of 2. We just proved that (12.30) by (12.26).

d x,2 -l-2 (E, L(x, 2 -l )) ≤ 4β Y (x, 2 -l ) ≤ 4ε l for l ≥ k + 7,
Denote by τ l (x) the direction of L(x, 2 -l ) for l ≥ k + 7. We see τ l (x) as a vector in the unit sphere S n-1 , determined modulo multiplication by ±1, and we use the natural distance

(12.31) dist(τ, τ ′ ) = Min(|τ -τ ′ |, |τ + τ ′ |)
on the sphere modulo ±1. By (12.30) and its analogue for l + 1, every point of L(x, 2 -l-1 ) ∩ B(x, 2 -l-1 ) lies within ε l+1 2 -l+1 from E, and then within

ε l+1 2 -l+1 + ε l 2 -l+2 from L(x, 2 -l ); hence (12.32) dist(τ l+1 (x), τ l (x)) ≤ Cε l + Cε l+1 .
The triangle inequality yields dist(τ l (x), τ j (x)) ≤ C m≥l ε m < +∞ when j ≥ l, where the finiteness comes from (12.25). Thus τ (x) = lim l→+∞ τ l (x) exists, and

(12.33) dist(τ (x), τ l (x)) ≤ C m≥l ε m for l ≥ k + 7.
Denote by L(x) the line through x with direction τ (x), and let us check that (12.34)

d x,2 -l-3 (E, L(x)) ≤ C m≥l ε m for l ≥ k + 7.
First let w ∈ L(x) ∩ B(x, 2 -l-3 ) be given. Observe that dist(x, L(x, 2 -l )) ≤ 2 -l ε l by (12.30), so dist(w, L(x, 2 -l )) ≤ 2 -l ε l + 2 -l-3 dist(τ (x), τ l (x)) ≤ C2 -l m≥l ε m by (12.33). So we can find w 1 ∈ L(x, 2 -l ) such that |w 1 -w| ≤ C2 -l m≥l ε m . Notice that w 1 ∈ B(x, 2 -l-2 ) (or else some point near x will do even better), so by (12.30) we can find y ∈ E such that |yw 1 | ≤ 2 -l ε l . Altogether, |y -w| ≤ C2 -l m≥l ε m , as needed. Conversely, let y ∈ E ∩ B(x, 2 -l-3 ) be given. By (12.30), we can find w ∈ L(x, 2 -l ) such that |w-y| ≤ 2 -l ε l . Then we can find w 1 ∈ L(x) such that |w 1 -w| ≤ C2 -l m≥l ε m . This completes our verification of (12.34).

The right-hand side of (12.34) tends to 0 when l tends to +∞, hence L(x) is tangent to E at x. That is, our curves γ j have tangents everywhere.

We still need to estimate the oscillation of τ (x). Let y be another point of the same γ j as x, and assume for the moment that y lies in the same annulus A k as x, and that |y -x| ≤ 2 -k-11 . Choose l so that 2 -l-4 ≤ |y -x| < 2 -l-3 ; then l ≥ k + 7, and we can apply (12.30). We also have the analogue of (12.30) for y, i.e., d y,

2 -l-2 (E, L(y, 2 -l )) ≤ 4ε l . So E is very close to L(x, 2 -l ) and L(y, 2 -l ) in B(x, 2 -l-3 ) ⊂ B(x, 2 -l-2 ) ∩ B(y, 2 -l-2 ), and hence (12.35) dist(τ l (x), τ l (y)) ≤ Cε l ,
for instance by the proof of (12.32). Then by (12.35) and (12.33), which is enough for (12.28).

(12.36) dist(τ (x), τ (y)) ≤ dist(τ (x), τ l (x)) + dist(τ (y), τ l (y)) + Cε l ≤ C m≥l ε m ≤ C 2 -m ≤2 -l ε m ≤ C 2 -m ≤16|z-x| ε m ,
Next assume that y lies in the same A k as x, but that |y -x| > 2 -k-11 . By (12.20) or (12.21) and because y lies in the same γ j as x, there is a chain of points

x ℓ , 0 ≤ ℓ ≤ N , in E ∩ A k , with x 0 = x, |x ℓ -x ℓ-1 | ≤ 2 -k-11 for ℓ ≥ 1, x N = y, and N ≤ 2 15 . Then dist(τ k+7 (x ℓ ), τ k+7 (x ℓ-1 )) ≤ Cε k+7
, by the proof of (12.35), and hence

(12.37) dist(τ k+7 (x), τ k+7 (y)) ≤ Cε k+7 . Now (12.38) dist(τ (x), τ (y)) ≤ dist(τ k+7 (x), τ k+7 (y)) + dist(τ k+7 (x), τ (x)) + dist(τ k+7 (y), τ (y)) ≤ Cε k+7 + C m≥k+7 ε m ≤ C 2 -m ≤16|x-y| ε m
by (12.37), (12.33), and because |y -x| > 2 -k-11 . Again this is enough for (12.28).

We are left with the case when y lies in an annulus A l = A k . Without loss of generality, we may assume that l ≥ k. We may also assume that k was chosen as large as possible. Thus, by (12.22)

, x ∈ A k \ A k+1 = B(z, 2 -k-3 ) \ B(z, 2 -k-4 ) if k ≥ 1, and x ∈ B(0, 5/2) \ B(z, 1/16) if k = 0.
We can use the fairly large overlap of the annuli A m to find a chain of points x m , k ≤ m ≤ l, that all lie in the same γ j as x and y, and so that each

x m lies in A m ∩ A m+1 . Observe that for k ≤ m ≤ l -1, (12.39) dist(τ m+7 (x m ), τ m+8 (x m+1 )) ≤ dist(τ m+7 (x m ), τ m+7 (x m+1 )) + dist(τ m+7 (x m+1 ), τ m+8 (x m+1 )) ≤ Cε m+7 + Cε m+8
by (12.32) and (12.37) or the proof of (12.35). We also have similar estimates on dist(τ k+7 (x), τ k+7 (x k )) and dist(τ l+7 (y), τ l+7 (x l )), hence

(12.40) dist(τ k+7 (x), τ l+7 (y)) ≤ C m≥k+7 ε m ,
and then

(12.41) dist(τ (x), τ (y)) ≤ C m≥k+7 ε m , by (12.33). Recall that |z -x| ≥ 2 -k-4 because x ∈ A k \ A k+1 .
At the same time, y ∈ A l for some l > k, so |z -y| ≤ 2 -k-4 . Since in addition z / ∈ A k , we get that |z -y| ≤ 2 -k-6 (compare with (12.22)). Altogether, 16|x -y| ≥ 16|x -z| -16|z -y| ≥ 2 -k-7 , and (12.41) implies (12.28).

We still need to show that the three tangent half lines to the γ j at z are coplanar and make a 120 • angle. Recall from (12.20) that d z,2 -k-1 (Y k , E) ≤ Cε, but the same proof shows that (12.42) lim

k→+∞ d z,2 -k-1 (Y k , E) = 0
(that is, (12.25) allows us to replace ε > 0 with any small number in the proof of (12.20)). Fix a γ j and an annulus A k , and denote by L j,k the branch of Y k that gets close to γ j ∩ A k . Let τ j,k denote the direction of L j,k . Pick any x ∈ A k ∩ γ j , and apply (12.34) with l = k + 7. We get that

(12.43) dist(τ j,k , τ (x)) ≤ Cd z,2 -k-1 (Y k , E) + Cd x,2 -k-10 (E, L(x)) ≤ Cd z,2 -k-1 (Y k , E) + C m≥k+7 ε m .
Now let k tend to +∞. The right-hand side of (12.43) tends to 0, by (12.25) and (12.42). Also, τ (x) has a limit, by (12.28), so Y k has a limit Y . The three branches of Y are tangent half lines for the corresponding γ j at z, by (12.42), and they make 120 • angles by definition. In addition, Y lies in a plane because each Y k lies in a plane. This completes our proof of Proposition 12.27.

Harmonic improvement over a small Lipschitz cone

In this section we consider the cone over a small Lipschitz graph defined on an arc of circle, and show that we almost always reduce its surface when we replace it with the graph of a harmonic function with the same boundary values. This will be useful in the next section, to show that the reduced two-dimensional minimal cones in R n are composed of arcs of great circles.

We start with a Lipschitz function f : [0, T ] → R m , and we assume that

(13.1) 0 < T < π , f (0) = f (T ) = 0, and ||∇f || ∞ ≤ τ
for some small τ > 0. We associate to f a Lipschitz function F 0 on the sector (13.2)

D T = {(ρ cos t, ρ sin t) ; 0 ≤ ρ ≤ 1 and 0 ≤ t ≤ T } ⊂ R 2 by (13.3) F (ρ cos t, ρ sin t) = ρf (t) for 0 ≤ ρ ≤ 1 and 0 ≤ t ≤ T.
Thus the graph of F is a part of the cone over the graph of the lipschitz function (cos t, sin t) → f (t) that lies above D T . We want to compare the surface measure of this graph with what we get with a harmonic function G with the same values on ∂D T . Since f (0) = 0, we can extend f to [-T, T ] so that it is odd, and then f (-T ) = f (T ) = 0 by (13.1). So we may write f as a sum of sines, i.e., (13.4)

f (t) = k≥1 β k sin(πkt/T ),
for some coefficients

β k ∈ R m such that k |kβ k | 2 < +∞, because the periodized extension is Lipschitz. Let us check that (13.5) π 2 2T k≥1 k 2 |β k | 2 = ||f ′ || 2 2 = [0,T ] |f ′ (t)| 2 dt. Indeed f ′ (t) = k≥1 πk T β k cos(πkt/T ) and the cosine functions are orthogonal on [0, T ] because [0,π] cos kt cos lt dt = 1 2 [-π,π] cos kt cos lt dt = 0 for k = l, so (13.6) ||f ′ || 2 2 = T 2 k≥1 πk T 2 |β k | 2 = π 2 2T k≥1 k 2 |β k | 2 .
Our new extension G is defined on D T by (13.7) G(ρ cos t, ρ sin t) = k≥1 β k ρ πk/T sin(πkt/T ).

Notice that it has the same boundary values as F , i.e., that

(13.8) G(z) = F (z) for z ∈ ∂D T .
Indeed G(cos t, sin t) = f (t) by (13.4), and G(ρ cos t, ρ sin t) = 0 when t = 0 or t = T . We do not really need to know that G is harmonic on D T , but this is the case, essentially because (13.7) defines a harmonic function when T = π and D T is a half disk, while in general G T is obtained by composition with the conformal mapping z → z π/T which sends D T to a half-disk. Of course it was reasonable to pick a harmonic function G, because we want its energy to be as small as possible.

Lemma 13.9. The functions F and G lie in W 1,2 (D T ), with (13.10)

D T |∇G| 2 = π 2 k≥1 k|β k | 2 ≤ ||f ′ || 2 2 and (13.11) 1 2 ||f ′ || 2 2 ≤ D T |∇F | 2 = k≥1 k 2 π 2 + T 2 4T |β k | 2 ≤ ||f ′ || 2 2 .
Let us first compute the energy of G. We start with the radial derivative (13.12) ∂G ∂ρ (ρ cos t, ρ sin t) = k≥1 β k πk T ρ ρ kπ/T sin(πkt/T ).

For ρ fixed, the sine functions are orthogonal on [0, T ] (again because [0,π] sin kt sin ltdt = 1 2 [-π,π] sin kt sin lt dt = 0 for k = l by evenness), so (13.13)

T 0 ∂G ∂ρ (ρ cos t, ρ sin t) 2 dt = T 2 k≥1 |β k | 2 πk T ρ 2 ρ 2πk/T = π 2 2T k≥1 k 2 |β k | 2 ρ 2πk/T ρ -2
and then (13.14)

D T ∂G ∂ρ 2 = π 2 2T k≥1 k 2 |β k | 2 1 0 ρ 2πk/T ρ -1 dρ = π 2 2T k≥1 k 2 |β k | 2 T 2πk = π 4 k≥1 k|β k | 2 .
Similarly, the derivative in the tangential direction is

(13.15) 1 ρ ∂G ∂t (ρ cos t, ρ sin t) = 1 ρ k≥1 β k ρ kπ/T πk T cos(πkt/T ),
and the same computation (only with sines replaced by cosines) leads to (13.16)

D T 1 ρ ∂G ∂t 2 = π 4 k≥1 k|β k | 2 .
We add (13.14) and (13.16) and get the first half of (13.10). The fact that the result does not depend on T is not too surprising, because energy integrals are invariant under the conformal mapping z → z π/T . For the second half of (13.10), we simply use (13.5) and the fact that kπ/2 ≤ k 2 π 2 /2T for k ≥ 1, by (13.1).

Let us turn to F . Recall from (13.3) et (13.4) that (13.17)

F (ρ cos t, ρ sin t) = ρf (t) = ρ k≥1 β k sin(πkt/T ).
The radial derivative is k≥1 β k sin(πkt/T ) and the integral of its square is

T 4 k≥1 |β k | 2 . The tangential derivative is 1 ρ ∂F ∂t = k≥1 β k πk T cos(πkt/T ),
D T |∇F | 2 = k≥1 k 2 π 2 + T 2 4T |β k | 2 .
This is the main part of (13.11); for the rest, just use (13.5) and notice that 0 ≤ T 2 ≤ π 2 k 2 . Lemma 13.9 follows.

Observe that when T = π and k = 1, the energy terms in (13.10) and (13.11) are both equal to π 2 |β 1 | 2 , which is not surprising because for f (t) = sin t, the harmonic extension is the same as the homogeneous extension. Here T < π, and we shall see now that all the values of k give an advantage to G. This is not surprising, since the sine function in no longer possible, by the constraints in (13.1) and because a plane that passes through the origin, (1, 0, 0) and (cos T, sin T, 0), has to be the horizontal plane through the origin.

Set λ = T /π < 1 (by (13.1)), and let us check that (13.19)

D T |∇G| 2 ≤ 2λ 1 + λ 2 D T |∇F | 2 and 2λ 1 + λ 2 < 1.
For the first inequality, we just need to compare the coefficients in (13.10) and (13.11), and show that for k ≥ 1, (13.20)

kπ 2 ≤ 2λ 1 + λ 2 k 2 π 2 + T 2 4T = 2λ 1 + λ 2 k 2 π 2 + λ 2 π 2 4πλ . So we just need to check that k ≤ k 2 + λ 2 1 + λ 2 , but k(1 + λ 2 ) ≤ k 2 + λ 2 because k 2 + λ 2 - k(1 + λ 2 ) = k(k -1) -λ 2 (k -1) ≥ 0,
so the first part of (13.19) holds. For the second part, we just need to observe that 2λ 1+λ 2 is an increasing function of λ ∈ [0, 1]. Let Σ F and Σ G denote the respective graphs of F and G (both defined on D T ). We want to use the computations above to compare H 2 (Σ F ) and H 2 (Σ G ). This will be easy because we can assume that ||f ′ || ∞ ≤ τ , which is as small as we want.

Lemma 13.21. We have that

(13.22) H 2 (Σ G ) -H 2 (D T ) ≤ 1 2 D T |∇G| 2 ≤ λ 1 + λ 2 D T |∇F | 2 and (13.23) H 2 (Σ F ) -H 2 (D T ) ≥ 1 2 1 - (1 + T ) 2 τ 2 2 D T |∇F | 2 .
Before we prove the lemma, observe that since we can take τ as small as we want (and

T ≤ π anyway), 1 -(1+T ) 2 τ 2 2
is as close to 1 as we want. Since λ 1+λ 2 ≤ 1 2 by (13.19), for each choice of T < π we can find η ∈ (0, 1) such that (for τ small enough, depending on T through λ), (13.22) and (13.23) imply that (13.24) , by (13.11).

H 2 (Σ G ) -H 2 (D T ) ≤ η H 2 (Σ F ) -H 2 (D T ) . Also, (13.25) H 2 (Σ F ) -H 2 (D T ) ≥ 1 3 D T |∇F | 2 ≥ 1 6 ||f ′ || 2 ∞ if τ is small enough
Now we prove the lemma. Notice that G is smooth in the interior of D T ; the series in (13.12) and (13.15) that give ∇G converge rather brutally because of the term in ρ kπ/T . Thus (13.26)

H 2 (Σ G ) = D T 1 + |∇G| 2 1/2 . Notice that (1 + u) 1/2 ≤ 1 + u/2 for u ≥ 0, so (13.27) H 2 (Σ G ) ≤ D T 1 + 1 2 |∇G| 2 = H 2 (D T ) + 1 2 D T |∇G| 2 ,
which gives the first inequality in (13.22). The second one comes from (13.19). For (13.23), it will be useful to know that F is Lipschitz with a small norm. And indeed F (ρ cos t, ρ sin t) = ρf (t) by (13.3), so (13.28)

|∇F (ρ cos t, ρ sin t)| ≤ |f (t)| + |f ′ (t)| ≤ (1 + T ) ||f ′ || ∞ ≤ (1 + T ) τ,
where we also used the fact that f (0) = 0 (by (13.1)). Now we use the fact that

(13.29) (1 + u) 1/2 ≥ 1 + u 2 - u 2 4 ≥ 1 + u 2 1 - (1 + T ) 2 τ 2 2 for 0 ≤ u ≤ (1 + T ) 2 τ 2 ,
which we apply with u = |∇F | 2 , to obtain that (13.30)

H 2 (Σ G ) -H 2 (D T ) = D T 1 + |∇F | 2 1/2 -1 ≥ 1 2 1 - (1 + T ) 2 τ 2 2 D T |∇F | 2 .
This proves (13.23); Lemma 13.21 follows.

Structure of the two-dimensional minimal cones

We are finally ready to give a description of the reduced two-dimensional minimal cones in R n . Recall that by minimal cone, we simply mean a minimal set (as in Definition 8.1), which is a cone. We say that the closed set E is reduced when E * = E; see Definition 2.12. We shall denote by Z the set of non empty reduced minimal cones of dimension 2 in R n , not necessarily centered at the origin, and by Z 0 the subset of elements of Z that are centered at the origin. Proposition 14.1. Let E ∈ Z be given, and set K = E ∩ ∂B(0, 1). Then K is a finite union of great circles or arcs of great circles C j , j ∈ J. The C j can only meet at their extremities, and each extremity is a common extremity of exactly three C j , which meet with 120 • angles. In addition, the length of each C j is a least l 0 , where l 0 > 0 depends only on n.

See (14.10) and Lemma 14.12 for additional constraints on K. Thus K is an array of arcs of great circles, with some constraints, but starting from the ambient dimension n = 4, we do not know exactly which arrays give rise to minimal cones. When n = 3, the answer was given by E. Lamarle [La], A Heppes [He], and J. Tayor [Ta], and indeed some arrays in S 2 do not give a minimal cone. See [Ta] and Ken Brakke's home page for descriptions. When n ≥ 4, it is expected that the disjoint union of two disjoint great circles, contained in almost orthogonal two-planes, gives a minimal cone, but the precise condition for this to happen is not known. There may even be other, combinatorially more complicated, continuous families of non mutually isometric minimal cones.

With the help of Proposition 14.1, we shall be able to give a local biHölder description of the two-dimensional almost-minimal sets E, but if we want a C 1 description near a point x, as in the original Jean Taylor theorem, it is probable that more information on the blowup limits of E at x will be needed.

Let us prove the proposition. Take ε > 0 small; ε = 10 -2 will be more than enough. Apply Theorem 8.23 with d = 1. We get r 0 > 0, that depends only on n and ε, such that if E ∈ Z, then for x ∈ K = E ∩ ∂B(0, 1) and 0 < r < r 0 , we can find a reduced minimal set L of dimension 1 such that (8.24) and (8.25) hold (that is, K is close to L in B(x, r), both in terms of Hausdorff distance and measure).

Continue with a given choice of x and r. Observe that L is not empty, by (8.24) and because x ∈ E, so Theorem 10.1 says that L is a line or a Y -set (not necessarily centered at x). By (8.25) with y = x and t = r, we get that (14.2)

H 1 (K ∩ B(x, r)) ≤ H 1 (L ∩ B(x, r + εr)) + εr ≤ 34r/10.
Denote by N (t) the number of points in K ∩ ∂B(x, t). By [D3], Lemma 26.1 on page 160, or [Fe], Theorem 3.2.22,

(14.3) r 0 N (t)dt ≤ H 1 (K ∩ B(x, r
)) ≤ 34r/10, so by Chebyshev we can find ρ ∈ (r/10, r) such that N (ρ) < 4.

By Proposition 9.4, K is a reduced weak almost-minimal set, with gauge function h(s) ≤ Cs. If r 0 was chosen small enough, Proposition 11.2 applies and yield the good approximation of K by one-dimensional minimal cones in every B(y, t), y ∈ K ∩B(x, ρ/10) and 0 < t < ρ/10, as in (11.3). This is enough to apply Proposition 12.6, with the unit ball replaced with B(x, ρ/20) ⊃ B(x, r/200) (and again, if r 0 is small enough). We get that there is an open set V , with B(x, r/300) ⊂ V ⊂ B(x, r/100), where K is either composed of a single biHölder curve, or of three biHölder curves that end at a some point. In addition, (11.3) says that (12.25) and (12.26) hold, with

ε k ≤ C[2 -k + h(2 -k )] 1/2 ≤ C2 -k/2
. So Proposition 11.27 says that each of these curve is of class C 1+α . If we prove that every point of these curves C j has a small neighborhood where C j is a great circle, Proposition 14.1 will follow, with l 0 = r 0 /1000, say.

So let x lie in the interior of a curve C j , and choose r > 0 so small that K ∩ B(x, r) = C j ∩ B(x, r) is a single C 1+α curve with two endpoints on C j ∩ ∂B(x, r), and the oscillation of the direction of the tangent line to C j along this curve is very small (how small will be decided soon). Without loss of generality, we can assume that the two points of K ∩ ∂B(x, r) are (1, 0, 0) and (cos T, sin T, 0) for some small positive T < π (and where the last coordinate 0 lies in R n-2 ). If the oscillation of the tangent direction of K in B(x, r) is small enough, K ∩B(x, r) is a small lipschitz graph over the arc of circle J = {(cos t, sin t, 0) ; 0 ≤ t ≤ T }.

Define the Lipschitz function f on [0, T ] by the fact that for 0 ≤ t ≤ T , (cos t, sin t, f (t)) is the point of K ∩ B(x, r) above (cos t, sin t, 0) ∈ J. We can assume that ||f ′ || ∞ is as small as we want, by taking r very small (recall that C j is C 1+α ).

Define F on the cone over J by F (ρ cos t, ρ sin t) = ρf (cos t, sin t) for ρ ≥ 0 and 0 ≤ t ≤ T . Thus F is the homogeneous extension of the function defined in (13.3). Denote by C(x, r) the cone over ∂B(0, 1) ∩ B(x, r) and by Γ the graph of F . Thus (14.4)

E ∩ C(x, r) = Γ,
because E and Γ are cones, and both coincide on the unit sphere with K ∩ B(x, r). We want to use section 13 to construct a competitor for E.

Let G be as in (13.7). Also denote by π the orthogonal projection on the horizontal plane, and by D T = {(ρ cos t, ρ sin t, 0) ; 0 ≤ t ≤ T and 0 ≤ ρ ≤ 1} the same sector as in (13.2). We define a function ϕ on Γ by

(14.5) ϕ(z) = (π(z), G(π(z))) for z ∈ Γ ∩ π -1 (D T ) and ϕ(z) = z for z ∈ Γ \ π -1 (D T ).
Notice that ϕ(z) = z on Γ ∩ π -1 (∂D T ), by (13.8), so ϕ is continuous across Γ ∩ π -1 (∂D T ). We should also check that ϕ is Lipschitz on Γ, or equivalently that G is Lipschitz, but we don't need to have precise bounds. One could easily get rid of this difficulty by replacing G with a lipschitz function G ′ , which is so close to G in the Sobolev W 1,2 -norm that the strict inequalities of Section 13 stay true for G ′ . But we can also observe that G is Lipschitz in the present situation. Indeed, f is C 1+α (by Proposition 11.27), and G is obtained from f by a composition with a dilation (to get a map defined on [0, π]), an extension by antisymmetry (to get a function on [-π, π] whose 2π-periodic extension is still C 1+α ), then a Poisson extension to the disk (which is Lipschitz because f is C 1+α ), which we restrict to the half disk and compose with the conformal mapping z → z π/T to get a function defined on D T , which is still Lipschitz because T < π. Set (14.6)

ϕ(z) = z for z ∈ R n \ C(x, r).
Recall that Γ∩∂C(x, r) is the union of the two half lines through (1, 0, 0) and (cos T, sin T, 0), so (13.8) says that ϕ(z) = z there, and ϕ is continuous across ∂C(x, r). In fact, due to the fact that Γ is transverse to ∂C(x, r), our mapping ϕ, which is now defined on Γ ∪ [R n \ C(x, r)], is Lipschitz there. In addition, ϕ(z) = z for |z| ≥ 2. So we can extend ϕ to R n in a Lipschitz way, and so that ϕ(z) = z for |z| ≥ 2. Now E is a minimal set, and Definition 8.1 says that (14.7) by (14.5) and (14.6).

H 2 (E \ ϕ(E)) ≤ H 2 (ϕ(E) \ E). Set E 1 = E ∩C(x, y)∩π -1 (D T ) = Γ∩π -1 (D T ) (by (14.4)). Observe that ϕ(z) = z for z ∈ E \ E 1 ,
Then E \ ϕ(E) = E 1 \ ϕ(E) and ϕ(E) \ E = ϕ(E 1 ) \ E ⊂ ϕ(E 1 ) \ E 1 , so (14.7) says that (14.8) H 2 (E 1 \ ϕ(E)) = H 2 (E \ ϕ(E)) ≤ H 2 (ϕ(E) \ E) ≤ H 2 (ϕ(E 1 ) \ E 1 ),
and hence (14.9)

H 2 (E 1 ) = H 2 (E 1 \ ϕ(E)) + H 2 (E 1 ∩ ϕ(E)) ≤ H 2 (ϕ(E 1 ) \ E 1 ) + H 2 (E 1 ∩ ϕ(E)) = H 2 (ϕ(E 1 ) \ E 1 ) + H 2 (E 1 ∩ ϕ(E 1 )) = H 2 (ϕ(E 1 )), because ϕ(E) \ ϕ(E 1 ) ⊂ ϕ(E \ E 1 ) does not meet E 1 .
On the other hand, E 1 = Γ ∩ π -1 (D T ) is the graph of F over D T , and by (14.5) ϕ(E 1 ) is the graph of G over D T . With the notation of Section 13, E 1 = Σ F and ϕ(E 1 ) = Σ G (see above Lemma 13.21). If we chose r above small enough, (13.1) holds with a τ so small that by Lemma 13.21, (13.24) holds for some η ∈ (0, 1). That is,

H 2 (Σ G ) -H 2 (D T ) ≤ η[H 2 (Σ F ) -H 2 (D T )]. But (14.9) says that H 2 (Σ F ) = H 2 (E 1 ) ≤ H 2 (ϕ(E 1 )) = H 2 (Σ G ), so we get that H 2 (Σ F ) = H 2 (D T ). Then D T |∇F | 2 =
0, by (13.23) (recall that τ is small), and ||f ′ || 2 = 0, by (13.11). That is, f = 0 (by (13.1)).

We just proved that the intersection of the arc C j with B(x, r) is contained in the horizontal plane. That is, C j ∩ B(x, r) is an arc of great circle, as needed. Proposition 14.1 follows, as was noted before.

We complete this section with a little more information on the minimal cones in Z 0 . Let E ⊂ Z 0 be given, and set K = E ∩ ∂B(0, 1). Recall from Proposition 14.1 that K is composed of some great circles C j , j ∈ J 0 , and some arcs of great circles C j , j ∈ J 1 .

Let H denote the set of extremities of the arcs C j , j ∈ J 1 . For each x ∈ H, there are exactly three C j that leave from x ; thus there are 3N/2 arcs, where N is the number of points in H. In particular, N is even and the number of arcs is a multiple of 3.

Next we claim that there is a constant η 0 > 0, that depends only on n, such that (14.10) if i, j ∈ J 0 ∪ J 1 , i = j, x ∈ C i , and dist(x, C j ) ≤ η 0 , then i, j ∈ J 0 , and C i and C j have a common endpoint in B(x, dist(x, C j )).

We shall take η 0 = r 0 /300, where r 0 is as in the proof of Proposition 14.1. Let i, j, and x ∈ C i be such that dist(x, C j ) ≤ η 0 . As before (see below (14.3)), there is a neighborhood V of B(x, r 0 /300) where E is composed of one simple curve, or of three simple arcs of curves that meet at one point a ∈ B(x, r 0 /100). In addition, these arcs are arcs of great circles, and each one is contained in a C k , k ∈ J 0 ∪ J 1 . One of these arcs is contained in C i (because x ∈ C i ), and our assumption that dist(x, C j ) ≤ η 0 for some other j implies that there are three arcs, and another one is contained in C j . These three arcs meet at a with 120 • angles, so |a -x| ≤ dist(x, C j ); (14.10) follows.

The simplest set E ∈ Z 0 is a plane, and then K is a great circle. In dimension n ≥ 2p, p ≥ 2, we may select p planes P j that are orthogonal (or nearly orthogonal) to each other, and set E = ∪ j P j (or equivalently, K is the union of the great circles P j ∩ ∂B(0, 1)). It seems that the precise conditions under which E ∈ Z 0 is not known. More generally, we may always start from a set E ∈ Z 0 that lies in R n-2 , and add to it the plane {0} × R 2 ; this gives a set E ′ that fits the description above; I don't know whether E ′ is necessarily a minimal cone.

The next simplest example is a set of type Y, the product of a Y -set in a plane by a line perpendicular to that plane. It is thus contained in a space of dimension 3, and is composed of three half planes with a common boundary L, that make 120 • angles along L (see Figure 1.1). Notice that it is easy to see that (14.11) if E ∈ Z 0 , K is connected, and H has 2 points, then E is a set of type Y;

indeed E is composed of exactly three arcs of great circles that connect the two points of H, these arcs are determined by their directions at any of the two points of H, and we know that these directions make 120 • angles.

The simplest set for which H has 4 points is the cone over the union of vertices of a regular tetrahedron in R 3 centered at the origin (see Figure 1.2). Lemma 14.12. There is a constant d T > 3π/2, that depends only on n, such that

(14.13) H 2 (E ∩ B(0, 1)) ≥ d T when E ∈ Z 0 is neither a plane nor a set of type Y.
Let us first check that (14.14)

H 2 (E ∩ B(0, 1)) > 3π/2 for such sets E. If E is composed of more than one full great circles, H 2 (E ∩ B(0, 1)) ≥ 2π.
Otherwise, H is not empty. Let y ∈ H be given, and set θ(y, r) = r -2 H 2 (E ∩ B(y, r)) for r > 0. Proposition 5.16 says that θ(y, r) = r -2 H 2 (E ∩ B(y, r)) is nondecreasing, so the limits θ(y) = lim r→0 θ(y, r) and θ ∞ (y) = lim r→+∞ θ(y, r) exist, and θ ∞ (y) ≥ θ(y). In addition, E does not coincide with a cone centered at y in big annuli, so Theorem 6.2 says that θ ∞ (y) > θ(y). By Proposition 14.1, the sets k -1 (Ey) converge to a minimal cone of type Y, and Proposition 7.31 (or directly Lemmas 3.3 and 3.12) says that θ(y) = 3π/2 (the density of a set of type Y). Hence θ ∞ (y) > 3π/2. Finally, (14.15)

H 2 (E ∩ B(0, 1)) = R -2 H 2 (E ∩ B(0, R)) ≥ R -2 H 2 (E ∩ B(y, R -|y|)) = R -2 (R -|y|) 2 θ(y, R -|y|)
for R large, and the last term tends to θ ∞ (y) > 3π/2, so (14.14) holds. Now suppose that (14.13) fails. Then there is a sequence of sets E k in Z 0 , that are neither planes nor sets of type Y, and for which a k = H 2 (E k ∩ B(0, 1)) tends to 3π/2. We can replace {E k } by a subsequence that converges to some limit E. Obviously, E is a cone, and Lemma 4.7 (with h = 0) says that E ∈ Z 0 . In addition, H 2 (E ∩ B(0, 1)) = 3π/2 by Lemmas 3.3 and 3.12, so E is a set of type Y by (14.14).

Denote by H k the set of extremities of the arcs of great circles in E k . By Proposition 14.1 or (14.10), the points of H k lie at distance at least η 1 = Min(l 0 , η 0 ) from each other. If H k has more than 2 points and k is large enough, then some y k ∈ H k lies at distance at least η 1 /2 from the two singular points of K = E ∩∂B(0, 1), and K is composed of at most an arc of great circle in B(y k , η 1 /3). At the same time, Proposition 14.1 says that

E k ∩ B(y k , η 1 /3) is very close, in Hausdorff distance, to a Y -set. This is impossible, because {E k } converges to E.
So H k has at most two points for k large. If it has exactly two points, E k contains a set of type Y by the proof of (14.11), and since it is not a set of type Y, Proposition 14.1 says that it needs to contain at least another arc of circle of length at least l 0 . Thus H 2 (E k ∩ B(0, 1)) ≥ 3π/2 + l 0 /2, which does not tend to 3π/2. If H k is empty, E k is composed of at least two great circles, and H 2 (E k ∩ B(0, 1)) ≥ 2π. In every case we reach a contradiction, which proves Lemma 4.12.

It is believable that we can take d T = 3Argcos(-1/3) ≈ 1.82 • π in Lemma 4.12, which corresponds to the set T based on a tetrahedron in R 3 (see Section 20), but I did not try to check.

D. LOCAL REGULARITY OF 2-DIMENSIONAL ALMOST-MINIMAL SETS

We want to generalize J. Taylor's theorem and give a local biHölder description of E when E is a reduced almost-minimal set of dimension 2 in R n , with sufficiently small gauge function.

The proof will rely on a generalization of Reifenberg's Topological Disk Theorem stated in the next section, together with the approximation results of Part B and a small argument with a topological degree. Most of the local regularity results are stated and proved in Section 16, except for a topological argument which is left for Section 17. In Section 18 we prove a regularity result for almost-minimal sets of dimension 2 in R 3 , which is valid in balls where E is close to a set of type T, but under an additional separation assumption. This is enough to establish Theorem 1.9 (i.e., the fact that every reduced M Sminimal set of dimension 2 in R 3 is a minimal cone). Section 19 contains the description of a set E ⊂ R 3 that looks like a set of type T at large scales, but not have the same topology. We rapidly compute the density of the set of type T of Figure 1.2 in Section 20.

An extension of Reifenberg's Topological Disk Theorem

Our statement of the desired extension of Reifenberg's theorem will use some amount of notation. We shall consider sets of dimension d ≥ 2. The most important case for us is when d = 2; we could also take d = 2, forget about general sets of type T, and get a statement as in Section 12, but let us not do this. We shall allow d > 2, but we shall keep a collection of cones that are just obtained as products of a 2-dimensional cone with an orthogonal (d -2)-dimensional plane.

Let us describe the collection of cones that will be used to approximate our set E. We start with the collection of d-dimensional (affine) planes, which we call P.

Next we define the collection Y of sets of type Y. We start with propellers in a plane, which we called Y -sets in the previous sections, and which are just unions of three half lines with a same endpoint and that make 120 • angles at that point. In [DDT], we even allow sets Y where the three half lines are only required to make angles at least π/2, but we won't need that generality here. We obtain a first set of type Y as the product Y 0 = Y × V , where Y is a Y -set centered at the origin, and V is a (d -1)-dimensional vector space that is orthogonal to the plane that contains Y . We shall call V the spine of Y 0 . Finally, Y is the collection of sets Y of the form Y = j(Y 0 ), where j is an isometry of R n . Then the spine of Y is the image by j of the spine of Y 0 .

The description of the collection T of sets of type T (the sets of type TG3 in [DDT]) will take a little longer. For our main application to two-dimensional almost-minimal sets, we should take a collection that includes the minimal cones of dimension 2, and unfortunately this will force us to modify slightly the definitions of [DDT]. Our set T will be the collection of sets T = j(T 0 × V ), where T 0 lies in a set T 0 of 2-dimensional cones in R n-d+2 , V is the plane of dimension d -2 orthogonal to R n-d+2 in R n , and j is an isometry of R n .

Each T ∈ T 0 will be the cone over a set K ⊂ ∂B(0, 1), with the following properties. First, K = ∪ j∈J C j is a finite union of great circles, or closed arcs of great circles. Denote by H the collection of extremities of arcs C j , j ⊂ J; each point x of H lies in exactly three C j , x is an endpoint for each such C j , and the three C j make 120 • angles at x. The C j can only meet at their endpoints (and hence the full arcs of circles are disjoint from the rest of K). In addition, there is a constant η 0 > 0 such that (15.1)

H 1 (C j ) ≥ η 0 for j ∈ J, and (15.2) if x ∈ C i , and dist(x, C j ) ≤ η 0 for some other j, then C i and C j have a common extremity in B(x, dist(x, C j )).

In addition, we exclude the case when T is a plane or a set of type Y.

When Y ∈ T 0 , the spine of E is the cone over H or, if K is uniquely composed of great circles, the origin. In general, when T = j(T 0 × V ), the spine of T is j(S × V ) where S is the spine of T 0 .

The definition of sets of type T G3 given in [DDT] was a little different. First, we also allowed angles larger than π/2 instead of exactly 2π/3, but we don't need this here. More unfortunately, we forgot the possibility of adding a collection of full great circles in K (provided that they stay far from the rest of K, as in (15.2)). We also decided to take the full collection of sets K as above, but our statement is a little better if we allow a subcollection (both in the assumptions and the conclusion). Nevertheless, the proof given in [DDT] goes through without changes, and leads to the statement below.

Before we start, let us record the fact that the class T 0 of reduced minimal cones of dimension 2 in R n-d+2 that are neither planes nor sets of type Y satisfies the requirements above, with a constant η 0 that depends only on m. This is a consequence of Proposition 4.1, except for (15.2) which was stated separately in (14.10).

Let P, Y, and some choice of T be as above. Set Z = P ∪ Y ∪ T , For the statement below, we are given a closed set E ⊂ R n , and we assume for simplicity that the origin lies in E. As in Section 12, we measure how close E is to sets of Z with (15.3)

β Z (x, r) = inf Z∈Z d x,r (E, Z),
where d x,r (E, Z) is defined as in (12.2). (Compare with (12.1).) We assume that (15.4) β Z (x, r) ≤ ε for x ∈ E ∩ B(0, 3) and 0 < r ≤ 3, where ε ≥ 0 is small, depending on n, d, and the constant η 0 . This means that for x ∈ E ∩ B(0, 3) and 0 < r ≤ 3, we can choose Z(x, r) ∈ Z such that d x,r (E, Z(x, r)) ≤ ε. We want a bilipschitz parameterization of E by a set of Z near the unit ball.

Theorem 15.5 [DDT]. Let Z be as above, and let τ > 0 be given. Then there is a constant ε > 0 that depends only on n, d, η 0 , and τ , such that if E ⊂ R n is a closed set that contains the origin and if (15.4) holds, we can find Z ∈ Z and a a biHölder mapping f : B(0, 2) → R n , such that

(15.6) |f (x) -x| ≤ τ for x ∈ B(0, 2), (15.7) (1 -τ )|x -y| 1+τ ≤ |f (x) -f (y)| ≤ (1 + τ )|x -y| 1-τ for x, y ∈ B(0, 2), (15.8) B(0, 2 -τ ) ⊂ f (B(0, 2)),

and

(15.9)

E ∩ B(0, 2 -τ ) ⊂ f (Z ∩ B(0, 2)) ⊂ E.
Compared to [DDT], we just changed a little some of the constants (such as 2 and 3), but this does not matter.

It will be useful for us that in addition to giving a parameterization of E by Z near 0, f extends to a local homeomorphism of R n . In particular, we know that the topology of R n \ E near 0 is the same as for R n \ Z.

We can say a little more on the choice of Z. First observe that Z has to be fairly close to Z(0, 3) in Hausdorff distance, because (15.6) and (15.9) say that it is fairly close to E.

If Z(0, 3) is a plane, or coincides with a plane near B(0, 29/10), say, we can show that for x ∈ E ∩ B(0, 27/10) and r ≤ 1/10, say, Z(x, r) is a plane, or at least its spine does not get close to x. Then we are in the standard Reifenberg situation, and we can take Z = Z(0, 3) ∈ P.

Similarly, if Z(0, 3) is a set of type Y with a spine through the origin, we can take Z = Z(0, 3) ∈ Y. Slightly more generally (but we won't need that case in the present paper), if Z(0, 3) ∈ Y or coincides with a set of type Y in B(0, 29/10), we can show that for x ∈ E ∩ B(0, 27/10) and r ≤ 1/10, Z(x, r) is never a set of type T centered near x. Then we can take Z ∈ Y, except perhaps in the situation above where it is more clever to take a plane.

If Z(0, 3) is a set of type T centered at the origin, then again we can take Z = Z(0, 3). More generally, if Z(0, 3) ∈ T , with a center in B(0, 25/10), we can first show (by checking things at smaller and smaller scales) that there is a single point x 0 ∈ B(26/10) such that every Z(x 0 , r) is a set of type T centered in B(x 0 , r/10). Then we use Z = Z(0, 3) (or a translation of it if we want f to map the center of Z to x 0 ).

If our class T (modulo identification of isometric sets) is not discrete, it could be that when r varies, the set Z(x 0 , r) changes. This is all right, it just means that some other sets Z ′ ∈ T could be used instead of Z (to parameterize E), and they are all biHölder equivalent to each other. This is not surprising, since Z(x 0 , r) is only allowed to vary very slowly with r (because Z(x 0 , r) stays so close to E). The situation would be different if we wanted C 1 parameterizations. Then we would probably need bounds on how fast the sets Z(x 0 , r) vary and (in the case of minimal sets), it can be expected that these bounds will be harder to prove than when d = 2 and n = 3.

The remaining case when the center of Z(0, 3) lies out of B(0, 25/10), but Z(0, 3) does not coincide with a set of type Y in B(0, 29/10), could be treated similarly, or just eliminated by replacing 3 with a larger constant. We shall not need it here anyway.

See [DDT] for a proof of Theorem 15.5. The proof is a little ugly because there are more cases to study, but we essentially use Reifenberg's scheme, plus an organization of the construction in layers (first find the center if there is one, then find the image of the spine, then define f on Z, and finally extend to the rest of R n ).

There is no immediate obstruction to having an additional layer in the complexity of our sets, but the situation is already complicated enough, and we were lucky that, just because there is at most one center x 0 , we did not have to understand what happens when the isometry class of Z(x 0 , r) changes with r.

We shall get our local descriptions of the almost-minimal sets by brutal applications of Theorem 15.5 in the case when Z(0, 3) is a cone centered at 0. The following definition will make it more convenient.

Definition 15.10. Let E be a closed set in R n . We way that B(0, 1) is a biHölder ball for E, with constant τ ∈ (0, 1), if we can find a cone Z ∈ Z centered at 0, and f : B(0, 2) → R n , with the properties (15.6)-(15.9). We say that B(0, 1) is of type P, Y, or T, depending on whether Z lies in P, Y, or T . Finally, we say that B(x, r) is a biHölder ball for E (with the same parameters) when B(0, 1) is a biHölder ball for r -1 (Ex).

With this definition, we have the following consequence of Theorem 15.5 and the discussion above. We keep the same assumptions on Z.

Corollary 15.11. For each small τ > 0, we can find ε > 0, that depends only on n, d, η 0 , and τ , such that if E ⊂ R n is a closed set, x ∈ E, and r > 0 are such that (15.12)

for y ∈ E ∩ B(x, 3r) and 0 < t ≤ 3r, we can find Z(y, t) ∈ Z such that d y,t (E, Z(y, t)) ≤ ε, and in addition Z(x, 3r) is a cone centered at x, then B(x, r) is a biHölder ball for E, with constant τ , of the same type as Z(x, 3r), and we can even take Z isometric to Z(x, 3r) in Definition 15.10.

A local regularity theorem for two-dimensional almost-minimal sets in R n

We are now ready to state a generalization to higher dimensions of the biHölder part of Jean Taylor's local regularity theorem [Ta].

Theorem 16.1. Let U be an open set in R n and E a closed subset of U . Suppose that E is a reduced A-almost-minimal set in U , with a gauge function h such that (16.2)

r 1 0 h(r)
dr r < +∞ for some r 1 > 0.

Then for each x 0 ∈ E and every choice of τ ∈ (0, 1), there is an r 0 > 0 such that B(x 0 , r 0 ) is a biHölder ball for E, with constant τ .

See Definition 4.3 for A-almost-minimal sets. Notice that we could equivalently have assumed that E is a reduced A ′ -almost-minimal set with the same gauge function h, as in Definition 4.8, because Proposition 4.10 says that the two definitions are equivalent.

Recall that we say that E is reduced when E * = E, where E * is the closed support of the restriction of H 2 to E; see Definition 2.12.

The notion of biHölder ball is defined in Definition 15.10, and here we take for Z the collection of two-dimensional reduced minimal cones in R n , as described in Section 14. Thus, in short, Theorem 16.1 says that every point of E has a small neighborhood where E is biHölder equivalent to a minimal cone of dimension 2.

Jean Taylor [Ta] also gives the C 1 equivalence when n = 3, under conditions on h that are stronger than (16.2) (but it is not clear that she tried to optimize this). We shall see in [D3] how to deduce this from Theorem 16.1 (or some weaker separation result). It is not clear to the author that this more precise result will generalize almost-minimal sets of dimension 2 in R n , n ≥ 4; the answer may depend on which type of blow-up limits for E at x exist.

Curiously, the condition (16.2) is only used to prove control the density, as in (16.6) at the beginning of the argument; I don't know whether it is sharp.

It should be observed that neither here nor in [Ta] (I think) do we get any precise lower bound for r 0 . The proof will say that for each x ∈ E, the density θ(x, r) eventually gets very close to its limit at r = 0, and then B(x, r) is a biHölder ball for E, but we do not know in advance when this happens. Some of our intermediate results, such as Lemma 16.19 and 16.48,Proposition 16.24,Lemmas 16.25,16.51,and 16.56, and later on Proposition 18.1, are more quantitative, though.

We shall prove Theorem 16.1 now, except for Proposition 16.24, which uses a little bit of degree theory and will be treated in Section 17. We shall systematically assume in this section that (16.3) E is a reduced A-almost-minimal set in U , with a gauge function h such that (16.2) holds.

Set

(16.4) θ(x, r) = r -2 H 2 (E ∩ B(x, r))

for x ∈ E and r > 0. Set (16.5)

A(r) = r 0 h(2t) dt t .
By (16.2), we can apply Proposition 5.24 and get that (16.6) θ(x, r)e λA(r) is a nondecreasing function of r, as long as B(x, r) ⊂ U and h(r) stays small enough, and where λ is a positive constant. Notice that A(r) tends to 0 when r tends to 0 (by (16.2)), so (16.6) allows us to define (16.7) θ(x) = lim r→0 θ(x, r)

for x ∈ E. As we shall see soon, there are restrictions on the values that θ(x) can take.

Proposition 7.31 says that every blow-up limit of E at x is a reduced minimal cone, whose density is θ(x).

If θ(x) = π, all the blow up limits are planes, and we shall call x a P -point. If θ(x) = 3π/2, all the blow up limits are minimal cones of type Y, and we shall call x a Y -point. Otherwise, we shall say that x is a T -point. In this case do not know whether the various blow-up limits of E at x are all isometric to each other, but they all have the same density θ(x). Lemma 14.12 then says that their density is at least d T (where d T > 3π/2 is a constant that may depend on n), so (16.8) θ(x) ≥ d T when x is a T -point.

When n = 3, there is only one type of minimal cone of type T (i.e., which is not a plane or a set of type Y), which is the one introduced in Remark 1.8 and Theorem 1.9. Then θ(y) = 3Argcos(-1/3) ≈ 1.82•π when x is a T -point. See the computation in Section 20. In higher dimensions, there may be lots of minimal cones with different densities ≥ d T . But this will not matter much here, because the T -points are isolated.

We start with a simple consequence of (16.6) that will be used repeatedly. We claim that (16.9) θ(x) ≤ e λA(r) θ(x, r)

for x ∈ E and r > 0 such that B(x, r) ⊂ U and h(r) is small enough. Indeed notice that for 0 < s ≤ r, (16.10) θ(x, s) ≤ e λA(s) θ(x, s) ≤ e λA(r) θ(x, r);

we let s tend to 0 and get the result.

The following consequence of Proposition 7.24 will also be used quite often.

Lemma 16.11. For each τ > 0, we can find ε(τ ) > 0 such that if x ∈ E and r > 0 are such that

(16.12) B(x, r) ⊂ U, h(2r) ≤ ε(τ ), 2r 0 h(r) dr r < ε(τ ), and
(16.13) θ(x, r) ≤ θ(x) + ε(τ ),
then for every ρ ∈ (0, 9r/10] there is a reduced minimal cone Z(x, ρ) centered at x, such that for y ∈ R n and t > 0 such that B(y, t) ⊂ B(x, ρ).

(
If θ(x) = π, Z(x, ρ) is a plane; if θ(x) = 3π/2, Z(x, ρ) is a set of type Y, and if θ(x) > 3π/2, Z(x, ρ) is a set of type T.

Let ρ ∈ (0, 9r/10] be given, and let us try to apply Proposition 7.24, with the constant τ ′ = τ /2, to the ball B(x, 10ρ/9). The main assumption is (7.25), which holds because (16.16) θ(x, 10ρ/9) ≤ e λA(10ρ/9) θ(x, 10ρ/9) ≤ θ(x, r)e λA(r) ≤ θ(x, r)e λε(τ )

≤ [θ(x) + ε(τ )] e λε(τ ) ≤ θ(x) + Cε(τ ) ≤ e λA(t) θ(x, t) + Cε(τ ) ≤ e λε(τ ) θ(x, t) + Cε(τ ) ≤ θ(x, t) + Cε(τ ) ≤ θ(x, t) + τ ′
for t < ρ, by (16.6), (16.5), (16.12), (16.13), (16.9), and (16.12), and if ε(τ ) is small enough.

The other assumptions are satisfied because of (16.12), so Proposition 7.24 applies and gives a reduced minimal cone C with the properties (7.26)-(7.28). We take Z(x, ρ) = C, (16.14) follows from (7.26) and (7.27), and (16.15) follows from (7.28), and because we may assume that (1τ ′ ) > 9/10.

We still need to show that the type of Z(x, ρ) is determined by θ(x). Denote by D = H 2 (Z(x, ρ) ∩ B(x, 1)) the density of Z(x, ρ). Then (16.17)

D -θ(x, ρ) = ρ -2 H 2 (C ∩ B(x, ρ)) -H 2 (E ∩ B(x, ρ)) ≤ τ
by (16.15), applied with B(y, t) = B(ρ, r). But θ(x, ρ) ≤ θ(x) + Cε(τ ) by the first two lines of (16.16) (with 10ρ/9 replaced with ρ) and θ(x) ≤ e λA(ρ) θ(x, ρ) ≤ θ(x, ρ) + Cε(τ ) by (16.9) and (16.2), we get that |θ(x, ρ)θ(x)| ≤ Cε(τ ), and then

(16.18) |D -θ(x)| ≤ |D -θ(x, ρ)| + |θ(x, r) -θ(x)| ≤ τ + Cε(τ ).
So |Dθ(x)| is as small as we want. But Z(x, ρ) is a plane as soon as D < π; by Lemma 14.12, it is a set of type Y if π < D < d T , and it is of type T if D > 3π/2; Lemma 16.11 follows.

The following lemma will allow us to deal with P -points in Theorem 16.1

Lemma 16.19. For each choice of τ ∈ (0, 1) there is a constant ε 1 > 0 such that if x 0 ∈ E and r 0 > 0 are such that B(x 0 , 9r 0 ) ⊂ U , h(18r 0 ) ≤ ε 1 , (16.20)

18r 0 0 h(t) dt t ≤ ε 1 , and θ(x 0 , 9r 0 ) ≤ π + ε 1 , then B(x, r
) is a biHölder ball of type P for E, with constant τ , for every choice of x ∈ E ∩ B(x 0 , r 0 ) and 0 < r ≤ r 0 .

Most of the time we shall use this with x = x 0 , but we shall need the additional uniformity provided by the present statement in Section 17.

Let x 0 , r 0 , x, and r be as in the statement. We want to apply Corollary 15.11 to the pair (x, r). Notice however that here all the sets Z(y, t) in (15.12) will be planes, so Corollary 15.11 comes directly from Reifenberg's original theorem.

So we need to show that (15.12) holds for the pair (x, r). That is, for each y ∈ E ∩ B(x, 3r) and 0 < t ≤ 3r, we need to find a plane Z(y, t) such that d y,t (E, Z(y, t)) ≤ ε, where ε comes from Corollary 15.11 and depends on τ . For the rest of the argument, we shall only need to know that y ∈ E ∩ B(x 0 , 4r 0 ) and 0 < t ≤ 3r 0 (which is true because x ∈ E ∩ B(x 0 , r 0 ) and 0 < r ≤ r 0 ).

Let τ 1 be small (much smaller than τ , and to be chosen soon), and apply Lemma 16.11 to B(x 0 , 9r 0 ). The assumptions are satisfied if ε 1 ≤ ε(τ 1 ), and we get that for 0 < ρ ≤ 8r 0 , there is a plane Z(x 0 , ρ) through x 0 , with the properties (16.14) and (16.15). We are only interested in P = Z(x 0 , 8r 0 ), and (16.15) which says that (16.21)

H 2 (E ∩ B(y, t)) -H 2 (P ∩ B(y, t)) ≤ 64τ 1 r 2 0 for y ∈ R n and t > 0 such that B(y, t) ⊂ B(x 0 , 8r 0 ).

Take y ∈ E ∩ B(x 0 , 4r 0 ) as above, and t = 4r 0 . We get that (16.22) θ(y, 4r 0 ) = (4r 0 ) -2 H 2 (E ∩ B(y, 4r 0 ))

≤ (4r 0 ) -2 H 2 (P ∩ B(y, 4r 0 )) + 64τ 1 r 2 0 ≤ π + 4τ 1 by (16.21) and because P is a plane. Notice that θ(y) ≥ π because y ∈ E and E is reduced (see the discussion below (16.7)). On the other hand, (16.23) θ(y) ≤ e λA(4r 0 ) θ(y, 4r 0 ) ≤ e λε 1 θ(y, 4r 0 ) ≤ e λε 1 (π + 4τ 1 ) ≤ π + 4τ 1 + Cε 1 < 3π/2 by (16.9) and (16.20), (16.22), and if τ 1 and ε 1 are small enough. Hence θ(y) = π.

Apply Lemma 16.11 to the ball B(y, 4r 0 ) and with τ = ε, where ε comes from Corollary 15.11 as above. The hypotheses are satisfied, by (16.19) and (16.22) (and if ε 1 and τ 1 are small enough). We obtain that for 0 < t < 36r 0 /10, there is a plane Z(y, t) through y such that (16.15) holds. This means that d y,t (E, Z(y, t)) ≤ ε, and this is exactly what was needed to apply Corollary 15.11. This completes our proof of Lemma 16.19. Lemma 16.19 gives the local regularity of E near every P -point (and for this we may even take x = x 0 ). Next we want to take care of the Y -points. The proof above does not work directly in this case, because we used once or twice the fact that the current density was close to π, and π is the smallest possible density at a point of E. Thus it will be good to know that when E is close to a set of type Y, we have Y -points around to which we shall be able to apply Lemma 16.11 because their density is not too small. This is the point of the next proposition.

Proposition 16.24. There is constant ε 2 > 0, that depends only on n, such that if (16.3) holds, B(x, r) ⊂ U , h(2r) ≤ ε 2 , 2r 0 h(t)dt/t ≤ ε 2 , and there is a reduced minimal cone Y of type Y centered at x and such that d x,r (E, Y ) ≤ ε 2 , then E ∩ B(x, r/100) contains (at least) a Y -point.

The proof of Proposition 16.24 relies on Lemma 16.11 (which we use to get the local regularity of E if there is no Y -point), and a topological argument that counts the points of intersection of E with some (n -2)-dimensional spheres. We shall do the proof in Section 17, and in the mean time we proceed with the regularity of E near a Y -point.

Lemma 16.25. For each choice of τ ∈ (0, 1), there is a constant

ε 3 > 0 such that if x 0 ∈ E is a Y -point, r 0 > 0, (16.26) B(x 0 , 50r 0 ) ⊂ U, h(100r 0 ) ≤ ε 3 , 100r 0 0 h(t) dt t ≤ ε 3 , and
(16.27) θ(x 0 , 50r 0 ) ≤ 3π/2 + ε 3 , then B(x 0 , r 0 ) is a biHölder ball of type Y for E, with constant τ .
It is enough to prove the lemma when x 0 = 0 and r 0 = 1, because it is invariant under rotations and dilations. Let τ 2 be small, to be chosen later, and apply Lemma 16.11 to the ball B(0, 50); the assumptions (16.12) and (16.13) are satisfied if ε 3 is small enough. We get that for 0 < ρ ≤ 45, there is a reduced minimal cone Y (ρ) = Z(0, ρ) of type Y, centered at 0, such that (16.14) We shall need some control on (16.34)

E Y = x ∈ E ; x is a Y -point of E = x ∈ E ; θ(x) = 3π 2 .
Let τ 3 be small, to be chosen soon, and let us apply Lemma 16.11 with τ = τ 3 , and to the ball B(y, 10), where we now assume that y ∈ E Y ∩ B(0, 35). If ε 3 and τ 2 are small enough, the assumption (16.12) follows from (16.26), and (16.13)

H d (E ∩ B(x, r)) ≤ H d (F ∩ B(x, (1 + δ)r)) + δr d .
Suppose that the lemma fails, and choose for each k ≥ 0 sets E k and F k that provide a counterexample with ε = 2 -k . By homogeneity, we can assume that x = 0 and r = 1 for all k. Thus E k and F k are reduced almost minimal sets in B(0, 10/9). It is a standard fact about the Hausdorff metric that we can find a subsequence (which we shall still denote by {E k } to save notation) such that {E k } converges to a limit E locally in B(0, 10/9). See for instance Proposition 34.6 on page 214 of [D2]. Since d 0,10/9 (E k , F k ) ≤ 2 -k by (16.44), {F k } also converges to E locally in B(0, 10/9) (and we don't even need to extract a second subsequence).

Our assumption is that E k and F k are almost minimal in a domain that contains B(0, 10/9), and with a gauge function h k such that h k (20/9) ≤ 2 -k . The sequence {F k } satisfies the assumptions of Lemma 3.3 (and even of Theorem 3.4 in [D1]; we do not need generalized quasiminimal sets here), so (16.46)

H d (E ∩ B(0, 1 + δ)) ≤ lim inf k→+∞ H d (F k ∩ B(0, 1 + δ))
by (3.3), applied with V = B(0, 1). Similarly, {F k } satisfies the assumptions of Lemma 3.12, with M = 1 and the number h as small as we want. This is because (the end of) Remark 4.5 says that F k ∈ GAQ(M, δ, U, h) with M = 1, δ = 20/9, U = B(0, 10/9), and h = h k (20/9). Then (3.13) with H = B(0, 1) says that (16.47) lim sup k→+∞

H d (E k ∩ B(0, 1)) ≤ H d (E ∩ B(0, 1)).
Altogether, lim sup k→+∞ H d (E k ∩B(0, 1)) ≤ lim inf k→+∞ H d (F k ∩B(0, 1+δ)), and (16.45) holds for k large. This contradiction completes our proof of Lemma 16.43.

Here is a variant of Lemma 16.19.

Lemma 16.48. For each choice of τ ∈ (0, 1), there is a constant ε > 0 such that if x 0 ∈ E, r 0 > 0, (16.49) B(x 0 , 10r 0 ) ⊂ U, h(20r 0 ) ≤ ε,

18r 0 0 h(t) dt t ≤ ε,
and there is a plane P such that (16.50) d x 0 ,10r 0 (E, P ) ≤ ε , then for every choice of x ∈ B(x 0 , r 0 ) and 0 < r ≤ r 0 , B(x, r) is a biHölder ball of type P for E, with constant τ .

Indeed, let ε 1 be as in Lemma 16.19. If ε is chosen small enough we can apply Lemma 16.43 to B(x 0 , 9r 0 ), δ = ε 1 /10, and F = P , and get that θ(x 0 , 9r 0 ) ≤ π + ε 1 . Then the conclusion follows from Lemma 16.19 (if in addition ε ≤ ε 1 ). Now we give a variant of Lemma 16.25.

Lemma 16.51. For each choice of τ ∈ (0, 1), there is a constant ε 4 > 0 such that if

x 0 ∈ E, r 0 > 0, (16.52) B(x 0 , 60r 0 ) ⊂ U, h(100r 0 ) ≤ ε 4 , 100r 0 0 h(t) dt t ≤ ε 4 ,
and there is a reduced minimal cone Y of type Y, centered at x 0 , and such that (16.53)

d x 0 ,60r 0 (E, Y ) ≤ ε 4 ,
then B(x 0 , r 0 ) is a biHölder ball of type Y for E, with constant τ .

Let x 0 and r 0 be as in the statement. Let ε 5 be small, to be chosen soon, and apply Proposition 16.24 to the ball B(x 0 , ε 5 r 0 ). Notice that (16.54)

d x 0 ,ε 5 r 0 (E, Y ) ≤ 60ε -1 5 d x 0 ,60r 0 (E, Y ) ≤ 60ε -1 5 ε 4 < ε 2
if ε 4 is small enough, and the other assumptions of Proposition 16.24 are satisfied more trivially, so there is a point x 1 ∈ E ∩ B(x 0 , ε 5 r 0 ) which is of type Y. If ε 4 is small enough, (16.53) and (16.52) allow us to apply Lemma 16.43 to B(x 1 , 500r 0 /9) and get that (16.55) θ(x 1 , 50r 0 ) ≤ 3π/2 + ε 3 , where ε 3 is as in Lemma 16.25.

The pair (x 1 , 50r 0 ) satisfies the hypotheses of Lemma 16.25, so B(x 1 , r 0 ) is a bihölder ball of type Y. Since this is not exactly the conclusion of Lemma 16.43, let us say a little more. The proof of Lemma 16.25 gives that for x ∈ E ∩ B(x 1 , 3r 0 ) and 0 < r ≤ 3r 0 , there is a reduced minimal cone Z(x, r) such that d x,r (E, Z(x, r)) ≤ ε (see (16.41) and below), and then we use Corollary 15.11 to conclude. But if ε 5 is small enough (and ε 5 = 10 -2 would be more than enough), we also get Z(x, r) for x ∈ E ∩B(x 0 , 3r 0 ) ⊂ B(x 1 , 3r 0 +ε 5 r 0 ). In addition, (16.44) allows us to use Y itself (if ε 4 is small enough) as the set Z(x 0 , r 0 ), so the last condition in Corollary 15.11 is also fulfilled. Finally Corollary 15.11 says that B(x 0 , r 0 ) also is a bihölder ball of type Y, and Lemma 16.51 follows.

Return to the slow proof of Theorem 16.1. Lemmas 16.19 and 16.25 show that E has some biHölder regularity near P -points and Y -points. We deal with remaining case of T -points now.

Lemma 16.56. For each choice of τ ∈ (0, 1), there is a constant ε 6 > 0 such that if

x 0 ∈ E is a T -point, r 0 > 0, (16.57) B(x 0 , 10r 0 ) ⊂ U, h(20r 0 ) ≤ ε 6 , 20r 0 0 h(t) dt t ≤ ε 6 ,

and

(16.58) θ(x 0 , 10r 0 ) ≤ θ(x 0 ) + ε 6 then B(x 0 , r 0 ) is a biHölder ball of type T for E, with constant τ .

The proof will go as for Lemma 16.25. By rotation and dilation invariance, we may assume that x 0 = 0 and r 0 = 1. Let τ 4 be small, to be chosen later, and apply Lemma 16.11 with τ = τ 4 and to B(0, 10). If ε 6 ≤ ε(τ 4 ), we get that for 0 < ρ ≤ 9, there is a reduced minimal cone T (ρ) of type T, centered at 0, and such that (16.59)

d 0,ρ (E, T (ρ)) ≤ τ 4 and (16.60) H 2 (E ∩ B(y, t)) -H 2 (T (ρ) ∩ B(y, t)) ≤ τ 4 ρ 2
for y ∈ R n and t > 0 such that B(y, t) ⊂ B(0, ρ).

as in (16.14) and (16.15). We want to control the density of T (ρ) in small balls. Set η = 10 -1 Min(η 0 , l 0 ), where η 0 is as in (14.10) and l 0 comes from Proposition 14.1. It will be useful to know that for ρ > 0 and x = 0, (16.61) T (ρ) coincides with a cone of type Y in B(x, η|x|).

Indeed suppose that B(x, η|x|) meets two of the faces that compose T (ρ). Call F 1 and F 2 these faces, and pick y j in F j ∩ B(x, η|x|). Set z j = y j /|y j |; thus z j lies in the arc of great circle C j = F j ∩ ∂B(0, 1). Observe that |z 1z 2 | < 3η < η 0 , so (14.10) says that the C j have a common endpoint a ∈ B(z 1 , 3η). If B(x, η|x|) meets some other face F 3 of T (ρ), the corresponding arc C 3 also has a common endpoint with F 1 in B(z 1 , 3η). This endpoint is a, because the length of C 1 is at least l 0 . In other words, all the faces that meet B(x, η|x|) are contained in the set of type Y which coincides with T (ρ) in a small neighborhood of a; (16.61) follows.

Next let x ∈ E ∩B(0, 8)\{0} be given, and set ρ(x) = 9|x|/8. Thus T (ρ(x)) is defined, and Next assume that our x ∈ E ∩ B(0, 8) is a Y -point. Let τ 5 be small, and apply Lemma 16.11 with the constant τ 5 , and to the ball B(x, η|x|). The assumption (16.12) follows from (16.57), and (16.13) follows from (16.62) because θ(x) = 3π/2 (if ε 6 and τ 4 are small enough, depending on τ 5 ). We get that for 0 < r ≤ 9η|x|/10, there is a reduced minimal cone Y (x, r) of type Y, centered at x, and such that (16.65) d x,r (E, Y (x, r)) ≤ τ 5 and (16.66)

(16.62) θ(x, η|x|) = (η|x|) -2 H 2 (E ∩ B(x, η|x|)) ≤ (η|x|) -2 H 2 (T (ρ(x)) ∩ B(x, η|x|)) + (η|x|) -2 τ 4 ρ(x) 2 ≤ 3π/2 + 2η
H 2 (E ∩ B(y, t)) -H 2 (Y (x, r) ∩ B(y, t)) ≤ τ 5 r 2
for y ∈ R n and t > 0 such that B(y, t) ⊂ B(x, r).

We also need to approximate E near the points of type P . So let z ∈ E ∩ B(0, 3) be a P -point, and set d(z) = dist(z, {0} ∪ E Y ). Let us show that for 0 < r < 9ηd(z)/10, (16.67) there is a plane P (z, r) through z such that d z,r (E, P (z, r)) ≤ ε.

Let us first assume that (16.68) d(z) ≥ η|z|/2.

Set ρ = 2|z| ≤ 6, and T = T (ρ). We shall use T to compute the density of E near z. Denote by L the spine of T (the cone over the the union of the extremities of the arcs of circles that compose T ∩ ∂B(0, 1)

d ξ,t (E, Y ) = d ξ,t (E, T ) ≤ t -1 ρ d 0,ρ (E, T ) ≤ t -1 ρ τ 4 = 8η -1 d(z) -1 |z| τ 4 ≤ 16η
z)) ≤ (ηd(z)) -2 H 2 (T ∩ B(z, ηd(z))) + (ηd(z)) -2 τ 4 ρ 2 ≤ π + (ηd(z)) -2 τ 4 ρ 2 ≤ π + 4η -4 |z| -2 τ 4 ρ 2 ≤ π + 16η -4 τ 4 by (16.68).
Recall that z is a P -point; if τ 4 is small enough, (16.76) allows us to apply Lemma 16.11 to B(z, ηd(z)), where this time we take τ = ε, with ε coming from Corollary 15.11. We get that for 0 < r ≤ 9ηd(z)/10, there is a plane P (z, r) through z such that d z,r (E, P (z, r)) ≤ ε. In other words, (16.67) holds.

We are left with the case when z ∈ E ∩ B(0, 3) is a P -point as before, but (16.68) fails, i.e., when d(z) ≤ η|z|/2.

Let x ∈ E Y be such that |x -z| ≤ 11d(z)/10 ≤ η|z|/20. Thus x ∈ B(0, 4). Also set r = 3d(z)/2. Then r ≤ 3η|z|/4 < 9η|x|/10, and we have a set Y = Y (x, r) of type Y that 123 satisfies (16.65) and (16.66). Let L denote the spine of Y ; then dist(z, L) ≥ d(z)/4, by the proof of (16.39), and so Y coincides with a plane in B(z, d(z)/4). Now (16.77)

θ(z, d(z)/4) ≤ (d(z)/4) -2 H 2 (Y ∩ B(z, d(z)/4)) + (d(z)/4) -2 τ 5 r 2 ≤ π + (d(z)/4) -2 τ 5 (3d(z)/2) 2 = π + 36τ 5 because B(z, d(z)/4) ⊂ B(x, r) = B(x, 3d (
z)/2), and by (16.66). By Lemma 16.11, we can find planes P (z, t), 0 < t ≤ d(z)/5, such that d z,t (E, P (z, t)) ≤ ε. That is, we have (16.67) for a slightly larger range of radii.

We are now ready to use our various cones to check the assumptions of Corollary 15.11. We need to define a minimal cone Z(x, r) for x ∈ E ∩ B(0, 3) and 0 < r ≤ 3, so that (16.78) d x,r (E, Z(x, r)) ≤ ε as in (15.12). When x = 0, we take Z(x, t) = T (ρ), and (16.78) follows from (16.59). Notice, by the way, that Z(0, 3) is a set of type T centered at 0, as required by Corollary 15.11 if we want a biHölder ball of type T.

Now suppose that x ∈ E Y . If 0 < r < 9η|x|/10, we can use the set Y (x, r) constructed near (16.65), and (16.78) follows from (16.65). Otherwise, we take Z(x, r) = T (|x| + r), which is defined because |x| + r ≤ 6. Notice that B(x, r) ⊂ B(0, |x| + r), so (16.79)

d x,r (E, Z(x, r)) ≤ r -1 (|x| + r) d 0,|x|+r (E, T (|x| + r)) ≤ r -1 (|x| + r) τ 4 ≤ 2η -1 τ 4
by (16.59) and because |x| ≤ 10η -1 r/9. Again (16.78) holds if τ 4 is small enough. We are left with the case when y is a P -point. If 0 < r < 9ηd(x)/10, we can use the plane P (x, r) of (16.67), and (16.78) follows directly from (16.67). So we may assume that r ≥ 9ηd(x)/10.

If r + 2d(x) ≤ η|x|/3, we choose y ∈ E Y such that |x -y| < 2d(x), and try to take Z(x, r) = Y (y, r + 2d(x)). [If d(x) = 0, apply the argument below with d(x) replaced with a very small number (compared to r and |x|).] Let us check that Z(x, r) is defined. First, |x -y| ≤ 2d(x) ≤ η|x|/3, so y ∈ B(0, 4) (because x ∈ B(0, 3)), and also |y| ≥ 2|x|/3. Then r + 2d(x) ≤ η|x|/3 ≤ η|y|/2, which is enough to define Y (y, r + 2d(x)) (see above (16.65)). Now (16.80)

d x,r (E, Z(x, r)) ≤ r -1 (r + 2d(x)) d y,r+2d(x) (E, Z(x, r)) = r -1 (r + 2d(x)) d y,r+2d(x) (E, Y (y, r + 2d(x))) ≤ r -1 (r + 2d(x)) τ 5 ≤ 3η -1 τ 5 because B(x, r) ⊂ B(y, r + 2d(x))
, by (16.65), and because r ≥ 9ηd(x)/10. Thus (16.78) holds if τ 5 is small enough. Finally assume that r + 2d(x) > η|x|/3. Take Z(x, r) = T (r + |x|), which is defined because r ≤ 3 and x ∈ B(0, 3). Then (16.81)

d x,r (E, Z(x, r)) ≤ r -1 (r + |x|) d 0,r+|x| (E, T (r + |x|)) ≤ 2r -1 (r + |x|) τ 4
by (16.59). Since |x| ≤ 3η -1 (r + 2d(x)) ≤ 3η -1 (r + 3η -1 r) because r ≥ 9ηd(x)/10, (16.78) follows from (16.81).

We completed the verification of (16.78) in all cases, Corollary 15.11 applies, and Lemma 16.56 follows.

Let us now see how to deduce Theorem 16.1 from our various lemmas. Let U and E be as in the statement, and let x 0 ∈ E be given. If x 0 is a P -point, we can choose r 0 so small that the assumptions of Lemma 16.19 are satisfied, and then B(x 0 , r 0 ) is a biHölder ball of type P. If x 0 is a Y -point, we use Lemma 16.25 instead. Otherwise, x 0 is a T -point, and we use Lemma 16.56. This completes our proof of Theorem 16.1, except for the fact that the two last lemmas depend on Proposition 16.24, which will be proved in the next section.

Remark 16.82. We do not know whether the obvious analogue of Lemma 16.43 for Tpoints holds. That is, suppose that the locally minimal set E is very close to a set of type T in a ball; is there necessarily a point of type T near the center? However, the answer is yes if n = 3 and we also assume that E separates locally R 3 into at least three big connected components. See Proposition 18.1.

A topological argument for the existence of Y -points

This section is devoted to the proof of Proposition 16.24. So we are given a reduced almost-minimal set E of dimension 2 in the domain U ⊂ R n , with a gauge function h such that (16.2) holds, and a pair (x, r) such that (17.1) B(x, r) ⊂ U , h(2r) ≤ ε 2 , and

2r 0 h(t)dt/t ≤ ε 2 .
We also assume that there is a reduced minimal cone Y of type Y, centered at x and such that (17.2)

d x,r (E, Y ) ≤ ε 2 ,
where ε 2 is s small constant that depends on n, and we want to prove that E ∩ B(x, r/100) contains (at least) a Y -point. By rotation and dilation invariance, we may assume that x = 0 and r = 200. Thus We shall prove the proposition by contradiction. So we assume that B(0, 2) contains no Y -point, and try to prove a contradiction.

Let us check that for any given ε > 0, we have that

(17.4) H 2 (E ∩ B(y, t)) ≤ H 2 (Y ∩ B(y, t)) + ε for y ∈ B(0, 2) and 10 -1 ≤ t ≤ 1 if ε 2 is small enough.
Let ε ′ denote the small constant that we get when we apply Lemma 16.43 with δ = ε/10, and let us check that (17.4) holds as soon as ε 2 ≤ ε ′ /1800. Let y ∈ B(0, 2) and 10 -1 ≤ t ≤ 1 be given; observe that (17.5) Next choose a small constant τ > 0; we claim that we can find η ∈ (0, 1) such that (17.9) B(x, r) is a biHölder ball of type P for E, with constant τ , for every x ∈ B(0, 3/2) and 0 < r ≤ 10η.

d y,10t/9 (E, Y ) ≤ 9 • 200 10t d 0,200 (E, Y ) ≤ 9ε • 200 < ε ′ because B(10t/9) ⊂ B(0,
Indeed, set K = E ∩ B(0, 3/2) and, for x 0 ∈ K, choose r 0 = r(x 0 ) so small that the assumptions of Lemma 16.19 are satisfied. This is possible, because θ(x 0 ) = π by (17.7). Then cover K by a finite number of balls B(x 0 , r(x 0 )). Finally take η smaller than the tenth of the smallest radius r(x 0 ). If x ∈ K, we can find x 0 as above such that x ∈ B(x 0 , r(x 0 )), and then Lemma 16.19 says that B(x, r) is a biHölder ball of type P for E, with constant τ , for 0 < r ≤ r 0 . This proves the claim (17.9).

Let us first give a vague idea of the plan that we would like to follow to get the desired contradiction. First, we would find a sphere S of codimension 2 in B(0, 3/2) that meets E exactly three times. We would deform this sphere through a family S t of spheres that eventually collapse to a point. Then we would use (17.9) to show that along this deformation, the number of points of E ∩ S t can only jump by multiples of two; this is impossible when we start with 3 intersections and end with none.

Things will be a little more complicated because we need to put our objects in general position before we compute the number of intersections; we shall also need to relate our numbers to degrees of mappings on spheres to get some stability.

To compute the number of points in E ∩ S t (maybe after we modify S t slightly), one way would be to define a one-parameter family of functions f t on S, so that S t = f t (S), give equations of E locally, and compute how many points x ∈ S satisfy the equations. We shall find it more convenient to proceed in the dual way, i.e., give equations of the sets S t , and count the number of solutions of these equations in E. It seems a little more convenient because it is apparently easier to localize in E.

I wish to thank Christopher Colin for his help in my initial struggle with degrees, and in particular the suggestion that I use equations.

Let us fix the setting and define a sphere S. Write R n = R 3 × R d , with d = n -3. By rotation invariance, we may assume that Y = Y 0 × {0}, where Y 0 ⊂ R 3 is a vertical propeller (or Y -set). That is, we select three points a 1 , a 2 , and a 3 in the unit circle of R 2 , so that if L j denotes the half line that starts from 0 and goes trough a j , the three L j make 120 • angles at the origin, and then we set

Y 0 = [L 1 ∪ L 2 ∪ L 3 ] × R ⊂ R 3 .
Denote by P the horizontal hyperplane P = R 2 × {0} × R d , and set S = ∂B(0, 1) ∩ P . Observe that S ∩ Y = {a 1 , a 2 , a 3 }; S is the sphere of codimension 2 that we would like to use in the argument suggested above, but we need to modify it slightly near the a j if we want it to meet E exactly three times.

Fix j ≤ 3 and use (17.3) to choose z j ∈ E such that |z ja j | ≤ 300ε 2 . Choose ε in (17.4) smaller than ε 1 /4, where ε 1 is as in Lemma 16.19 (and where, as before, the small constant τ just needs to be reasonably small, depending on the geometry). Then θ(z j , 1/2) ≤ 4H 2 (Y ∩ z j , 1/2) + 4ε ≤ π + 4ε if ε 2 is small enough, by (17.4) and because B(z j , 1/2) only meets one face of Y . Then we can apply Lemma 16.19 to the pair (z j , 1/18), and we get that (17.10) B(z j , r) is a biHölder ball of type P for E for 0 < r ≤ 1/18.

We start with a rough model, which will be used as a pattern, where the S t is (d + 1)dimensional sphere with radius (1t), and centered at b t = t(a 1 + a 2 )/100, say. We choose a final center b 1 = (a 1 + a 2 )/100 purposefully away from Y , to be sure that S t does not meet E for t close to 1. A formula for the pair of equations of S t is f t (x) = 0, where (17.11)

f t (x) = (x 2 , |x -b t | 2 -(1 -t) 2 ).
Let us record the fact that (17.12) by (17.3). We shall not use the one-parameter family {f t } to go from f 0 to f 1 , and instead we shall use a finite number of intermediate functions, so that each time the modifications occur on a small biHölder ball for E. So we shall use a partition of unity. Denote by {x j }, 1 ≤ j ≤ l, a maximal collection of points in E ∩ B(0, 3/2), with the property that |x ix j | ≥ η when i = j. Let ϕ j be a smooth bump function supported in B(x j , 2η) and such that ϕ j (x) = 1 for x ∈ B(x j , η) and 0 ≤ ϕ j (x) ≤ 1 everywhere. Notice that j ϕ j (x) ≥ 1 for x ∈ E ∩ B(0, 3/2), because x lies in some B(x j , η) by maximality of the family {x j }. Also let ϕ 0 be a smooth function on R n such that ϕ 0 (x) = 0 when |x| ≤ 3/2η, ϕ 0 (x) = 1 when |x| ≥ 3/2, and 0 ≤ ϕ 0 (x) ≤ 1 otherwise. Now l j=0 ϕ j (x) ≥ 1 on E, and we set (17.13) ϕ j (x) = ϕ j (x) l j=0 ϕ j (x)

|f 1 (x)| ≥ |x -b 1 | 2 ≥ dist(b 1 , E) 2 ≥ 10 -5 for x ∈ E,
-1

for x ∈ E and 0 ≤ j ≤ l.

Thus the ϕ j keep the same support properties as the the ϕ j , and in particular (17.14) ϕ j is supported in B(x j , 2η) for j ≥ 1, and in addition (17.15) l j=0 ϕ j (x) = 1 for x ∈ E , and

l j=1 ϕ j (x) = 1 for x ∈ E ∩ B(0, 3/2 -η).
(because ϕ 0 (x) = 0 there). Our first approximation of a discrete path from f 0 to f 1 is given by the maps (17.16) by (17.15). Notice that for k ≥ 1, (17.18) by (17.14). Recall that we would like to compute the number of solutions in E ∩ B(0, 3/2) of the systems g k (x) = 0. This will be easier to do with the small modifications of f 0 and the g k that we shall construct next, so that they only have a finite number of simple roots. We start with a modification of f 0 Lemma 17.19. There is a continuous function h 0 on E, such that (17.20) |h 0 (x)f 0 (x)| ≤ 10 -6 for x ∈ E, h 0 has exactly three zeroes ξ 1 , ξ 2 , and ξ 3 in E, one in each B(a i , 1/60), and in addition (17.21) each ξ j is a non-degenerate simple zero of h 0 .

g k = f 0 + 0<j≤k ϕ j (f 1 -f 0 ), with 0 ≤ k ≤ l. Thus g 0 = f 0 and (17.17) g l (x) = f 1 (x) for x ∈ E ∩ B(0, 3/2 -η),
g k (x) -g k-1 (x) = ϕ k (x)(f 1 (x) -f 0 (x)) is supported in B(x k , 2η),
Let us first define (17.21). We shall say that ξ ∈ E is a non-degenerate simple zero of some continuous function h on E if h(ξ) = 0 and there is a ball B(ξ, ρ) and a bi-Hölder map γ from E ∩ B(ξ, ρ) to an open set V of a plane, such that h

• γ -1 is of class C 1 on V and the differential of h • γ -1 at γ(ξ) is of rank 2.
Notice that when this holds, there is a small neighborhood W of γ(ξ) such that h•γ -1 is a homeomorphism from W to its image in R 2 , so if we replace ρ with a sufficiently small radius t > 0, we get that (17.22) the restriction of

h to E ∩ B(ξ, t) is a homeomorphism from E ∩ B(ξ, t) to the open subset h(E ∩ B(ξ, t)) of R 2 .
We now prove the lemma. We shall only modify f 0 near the three points a j of S ∩ Y . Recall from (17.10) that B(z j , 1/18) is a biHölder ball of type P. Thus there is a plane P j centered at z j and a mapping ψ j : B(z j , 1/9) → R n , with properties like (15.6)-(15.9) (see Definition 15.10). In particular, |ψ j (x) -x| ≤ τ /18 for x ∈ B(z j , 1/9) and (17.23) E ∩ B(z j , 10 -1 ) ⊂ ψ j (P j ∩ B(z j , 1/9)) ⊂ E.

We keep h 0 = f 0 out of the three B(z j , 1/30). In B(z j , 1/60), we replace f 0 with h 0 = f 0 • ψ -1 j . In the intermediate regions, we interpolate, i.e., we set (17.24)

h 0 (x) = α(x)f 0 (x) + (1 -α(x))f 0 • ψ -1 j (x), with α(x) = 60|x -z j | -1. Notice that |h 0 (x) -f 0 (x)| ≤ |f 0 (x) -f 0 • ψ -1 j (x)| ≤ Cτ for x ∈ B(z j , 1/30), because |ψ j (x) -x| ≤ τ /
18 and the derivative of f 0 is bounded there; thus (17.20) holds.

Recall from (17.11) that f 0 (x) = (x 2 , |x| 2 -1); then |f 0 (x)| ≥ 1/500 for x ∈ Y \ ∪ j B(x j , 10 -2 ) (essentially, because S ∩ Y = {a 1 , a 2 , a 3 } and |z ja j | ≤ 300ε 2 (see above (17.10)). This implies that (17.25) |h 0 (x)| ≥ 10 -3 for x ∈ E ∩ B(0, 2) \ ∪ j (B(z j , 1/60)), because f 0 varies slowly, and by (17.3) and (17.20). Thus h 0 only has zeroes in the B(z j , 1/60). We still need to check that h 0 has exactly one zero in each B(z j , 1/60), and that (17.21) holds. Set γ j (x) = ψ -1 j (x) for x ∈ B(z j , 1/60). By (17.23) and because ψ j is biHölder, γ j is a homeomorphism from E ∩ B(z j , 1/60) to its image, which is an open subset of P j .

Recall that h

0 = f 0 • ψ -1 j = f 0 • γ j in B(z j , 1/60). Thus, for ξ ∈ E ∩ B(z j , 1/60), ξ is a zero of h 0 if and only if γ j (ξ) is a zero of f 0 (x) = (x 2 , |x| 2 -1) in P j . So we study f 0 in P j .
By (17.3), we can find z ′ j ∈ E such that |z ′ j -(1 + 10 -2 ) a j | ≤ 300ε 3 , and z ′′ j ∈ E such that |z ′′ ja j -10 -2 e 2 | ≤ 300ε 3 , where e 2 = (0, 1, 0, • • • , 0). Set w = γ j (z j ), w ′ = γ j (z ′ j ), and w ′′ = γ j (z ′′ j ). These three points lie in P j , so P j is the plane that contains w and the two vectors v 1 = 100(w ′w) and v 2 = 100(w ′′w).

Notice that |wz j | ≤ τ /18, by (15.6), and similarly for w ′ and w ′′ . Then |f 0 (w)| ≤ 10|wa j | ≤ Cε 3 + Cτ , because f 0 (a j ) = 0.

Since |v 1a j | ≤ Cε 3 + Cτ and |v 2e 2 | ≤ Cε 3 + Cτ , when we can compute the differential of f 0 in P j , we can replace v 1 with a j and v 2 with e 2 , and make errors smaller than Cε 3 + Cτ . But D a j f 0 (x) = (0, 2 x, a j ) and D e 2 f 0 (x) = (1, 2x 2 ). These stay within 10 -1 of (0, 2) and (1, 0) when x ∈ P j ∩ B(a j , 1/50), so, if ε 3 and τ are small enough, the differential of f 0 in P j ∩ B(a j , 1/50) is invertible and almost constant, and f 0 has a unique zero ζ j in P j ∩ B(a j , 1/50), which even lies in B(a j , 10 -3 ) (recall that |f 0 (w)| ≤ Cε 3 + Cτ ).

Thus h 0 = f 0 • ψ -1 j has a unique zero in B(z j , 1/60), namely, ξ j = ψ j (ζ j ). In addition, (17.21) holds, precisely because Thus we keep the same formula as for g k in (17.18), but we add small perturbations θ k to obtain (17.30).

f 0 = h 0 • ψ j = h 0 • γ -1 j is C 1 on P j ,
h k = h k-1 + ϕ k (f 1 -f 0 ) + θ k for 1 ≤ k ≤ l, then (
We shall construct the h k by induction. Let k ≥ 1 be given, and suppose that we already constructed h k-1 so that (17.30) holds. Notice that this is the case when k = 1, by Lemma 17.19.

Notice that h k-1 + ϕ k (f 1f 0 ) coincides with h k-1 out of B(x k , 2η), by (17.14). We select a very thin annulus (17.31)

A = B(x k , ρ 2 ) \ B(x k , ρ 1 ), with 2η < ρ 1 < ρ 2 < 3η,
that does not meet the finite set of zeroes of h k-1 . Recall from (17.9) that B(x k , 10η) is a biHölder ball of type P, so there is a biHölder mapping ψ k : B(x k , 20η) → R n and a plane P k through x k such that |ψ k (x) -x| ≤ 10ητ for x ∈ B(x k , 20η), as in (15.6), and

(17.32) E ∩ B(x k , 19η) ⊂ ψ k (P k ∩ B(x k , 20η)) ⊂ E,
as in (15.9).

We shall take θ k supported in B(x k , ρ 2 ), so (17.27) will hold, and also the desired control on the zeroes of h k out of B(x k , ρ 2 ) will come from the induction assumption on h

k-1 , since h k = h k-1 out of B(x k , ρ 2 ).
We shall also take ||θ k || ∞ smaller than 2 -k 10 -6 , so that (17.28) holds, and smaller than inf x∈A |h k-1 (x)| > 0, so that (17.29) (with the fact that 0 ≤ ϕ k (x) ≤ 1 everywhere) implies that h k has no zero in A.

We still need to control h k in B(x k , ρ 1 ). Set γ(x) = ψ -1 k (x) for x ∈ E ∩ B(x k , ρ 1 ). By (17.32) and because ψ k is biHölder-continuous on B(x k , 20η), γ is a biHölder homeomorphism from E ∩ B(x j , ρ 1 ) to some open subset V of the plane P k , and its inverse is the restriction of ψ k to V .

By density of the C 1 functions (in the space of bounded continuous functions on V , with the "sup" norm), we can choose θ k with the constraints above, and such that (17.33)

h k • ψ k is of class C 1 on V .
We can even keep the option of adding a very small constant w ∈ R

Z y = z ∈ V ; h k • ψ k (z) = y for y ∈ R 2
. By (17.33), we can apply the coarea theorem to h k • ψ k on V . We get that (17.36)

V J(z) dH 2 (z) = y∈R 2 H 0 (Z y ) dH 2 (y),
where J is a bounded Jacobian that we don't need to compute (apply (8.12) with m = 2 and d = 0). Then Z y is finite for almost every y ∈ R 2 . If we choose w such that Z w is finite and add -w to θ k in E ∩ B(x j , ρ 1 ), then the new Z 0 will be finite, i.e., (17.34) will hold.

We also care about the rank of the differential. Sard's theorem says that the set of critical values of h k • ψ k has vanishing Lebesgue measure in R 2 ; if we choose w ∈ R 2 such that w is not a critical value, and add -w to θ k in E ∩ B(x j , ρ 1 ), then the differential of the new h k • ψ k at every zero of h k • ψ k has rank 2.

We select w very small, with the properties above, and add -w t o θ k in E ∩B(x j , ρ 1 ); then h k has a finite number of zeros in E ∩B(x j , ρ 1 ), and they are all non-degenerate simple zeroes of h k (see the definition below Lemma 17.19 and recall that h k • γ -1 = h k • ψ k ). So (17.30) holds, and Lemma 17.26 follows.

Denote by N (k) the number of zeroes of h k in E ∩ B(0, 2). By Lemma 17.9, N (0) = 3. Let us check that for the last index l, N (l) = 0. First observe that (17.37)

h l -h 0 = 1≤k≤l (h k -h k-1 ) = 1≤k≤l ϕ k (f 1 -f 0 ) + 1≤k≤l θ k by (17.29). If x ∈ E ∩ B(0, 4/3), (17.15) says that 1≤k≤l ϕ k (x) = 1, so (17.38) h l (x) = h 0 (x) + f 1 (x) -f 0 (x) + 1≤k≤l θ k (x) and (17.39) |h l (x)| ≥ |f 1 (x)|-|h 0 (x)-f 0 (x)|- 1≤k≤l |θ k (x)| ≥ 10 -5 -10 -6 - 1≤k≤l 2 -k 10 -6 >
(x) = 1 -ϕ 0 (x), so (17.40) h l (x) = h 0 (x) + [1 -ϕ 0 (x)][f 1 (x) -f 0 (x)] + 1≤k≤l θ k (x), and
(17.41) |h l (x)-f 0 (x)-[1-ϕ 0 (x)][f 1 (x)-f 0 (x)]| ≤ |h 0 (x)-f 0 (x)|+ 1≤k≤l |θ k (x)| ≤ 2•10 -6
by (17.20) and (17.28). But the second coordinate of f 0

(x) + [1 -ϕ 0 (x)][f 1 (x) -f 0 (x)] is (17.42) |x| 2 -1+[1-ϕ 0 (x)][|x-b 1 | 2 -|x| 2 +1] = ϕ 0 (x)[|x| 2 -1]+[1-ϕ 0 (x)]|x-b 1 | 2 ≥ 1/3
by (17.11) and because |x| ≥ 4/3. So h l (x) = 0 in this case as well, h l has no zero in E ∩ B(0, 2), and N (l) = 0. We shall reach the desired contradiction as soon as we prove that (17.43)

N (k) -N (k -1) is even for 1 ≤ k ≤ l.
This is the moment where we shall use some degree theory. Since it is easier to study the degree for mappings between spheres, we first replace h k and h k-1 with mappings between 2-spheres.

Recall that h k-1 does not vanish on A (by definition of A), and that we chose ||θ k || ∞ so small that this stays true for h k (see (17.29) and recall that ϕ k = 0 on A, by (17.31) and (17.14)). Set (17.44)

m t (x) = h k-1 (x) + t[h k (x) + h k-1 (x)] for x ∈ E ∩ B(x k , ρ 2 ) and 0 ≤ t ≤ 1. Thus m 0 = h k-1 and m 1 = h k on E ∩ B(x k , ρ 2 ). Notice that m t (x) = h k-1 (x) + tθ k (x) for x ∈ E ∩ A and 0 ≤ t ≤ 1, by (17.29), so m t (x) = 0 if we took ||θ k || ∞ small enough. Choose β k > 0 such that |m t (x)| ≥ β k for x ∈ E∩A and 0 ≤ t ≤ 1. Set S ∞ = R 2 ∪{∞}
(x) = ∞ for x ∈ E ∩ A and 0 ≤ t ≤ 1.
It will also be more convenient to replace the domain E ∩ B(x k , ρ 2 ) with a subset of the plane. Let us use the same mapping ψ k as above (near (17.32)), and its inverse γ, which is a biHölder homeomorphism from E ∩ B(x k , ρ 2 ) to some open subset V ′ of P k (for the same reason as before). For 0 ≤ t ≤ 1, set (17.48)

q t (x) = p t (ψ k (x)) for x ∈ V ′ and q t (x) = ∞ for x ∈ P k \ V ′ .
Let us check that q t is continuous on by (17.32) and because ψ k is biHölder), so some neighborhood of ψ k (x) in B(x k , ρ 2 ) is contained in A, hence p t (w) = ∞ there (by (17.47)), and q t (y) = ∞ near x (and for all t). Extend q t to the sphere S ′ = P k ∪ ∞ by setting q t (∞) = ∞; obviously q t is still continuous on S ′ × [0, 1]. Now we have two mappings q 0 and q 1 from the 2-sphere S ′ to the 2-sphere S ∞ , so we can compute their degrees. See for instance in [Do], to which we shall systematically refer because it contains the needed information rather explicitly. First observe that since q 0 and q 1 are homotopic, they have the same degree (see Proposition 4.2 on page 63 of [Do]).

P k × [0, 1]. It is continuous on V ′ × [0, 1], because p t is continuous on E ∩ B(x k , ρ 2 ) × [0, 1], and also on [P k \ V ′ ] × [0, 1], because it is constant there. We are left with ∂V ′ × [0, 1]. But for x ∈ ∂V ′ , ψ k (x) lies in E ∩ ∂B(x k , ρ 2 ) (again
Fix t ∈ {0, 1}, and try to compute the degree of q t . First observe that the degree (which roughly speaking counts the total number of inverse images of any point of the target space S ∞ , counted with orientation and multiplicity) can be computed locally near any point Q of the target space; see Corollary 5.6 on page 67 of [Do]. Here we shall use (17.48), so p t (ψ k (x)) = 0, and then m t (ψ k (x)) = 0 (by (17.46) and (17.45)). Now (17.49) follows from (17.44). By (17.30), both h k-1 and h k have a finite number of zeroes, so the same thing holds for q t . Then the degree of q t (computed at 0) can be computed locally: Proposition 5.8 of [Do] says that if we cover S ′ by finitely many open sets Y l , so that each Y l contains exactly one zero y l of q t , then the degree of q t is the sum of the degrees of the restrictions of q t to Y l . In addition, Proposition 5.5 of [Do] allows us to replace each Y l with any smaller open neighborhood W l of y l . Let us check that if W l is small enough, (17.50) q t is a homeomorphism from W l to q t (W l ).

Q = 0. Let us check that for x ∈ S ′ , (17.49) if q t (x) = 0, then ψ k (x) is a zero of h k-1 if t = 0, and of h k if t = 1. Indeed, if q t (x) = 0, then x ∈ V ′ (otherwise q t (x) = ∞), hence q t (x) = p t (ψ k (x)) (by
and that there is a minimal cone T of type T, centered at x, such that

(18.3) d x,r (E, T ) ≤ ε 7 .
Denote by a 1 , a 2 , a 3 , and a 4 the points of ∂B(x, r/2) whose distance to T is largest, and by V (a j ) the connected component of a j in B(x, r) \ E. Suppose in addition that at least three of the V (a j ) are distinct. Then E ∩ B(x, 3r/4) contains at least a T -point.

See Definition 4.3 for A-almost-minimal sets, and recall that Definition 4.8 gives an equivalent notion, by Proposition 4.10. See Definition 2.12 for the "reduced sets". Also recall that in R 3 , there is only one sort of set of type T: the cones based on a tetrahedron that are described near Figure 1.2. A T -point is therefore a point x ∈ E such that every blow-up limit of E at x is of this type (see the definition and discussion near (16.8)).

The author doesn't know whether the separation assumption can be removed. At least the most obvious attempt fails: see Section 19.

By scale invariance, it is enough to prove the proposition when x = 0 and r = 2. Our plan is to proceed by contradiction, so we shall assume that there is no point of type T in E ∩ B(0, 3/2) and try to reach a contradiction. This assumption will make it easier for us to study E Y ∩ B(0, 1), where E Y still denotes the set of Y -points of E.

Let α > 0 be small, to be chosen later. Let us first prove that if ε 7 is small enough,

(18.4) dist(y, L) ≤ 2α for y ∈ E Y ∩ B(0, 3/2),
where L denotes the spine of T (a union of four half lines). First observe that if ε 7 is small enough, (18.5) H 2 (E ∩ B(y, α)) ≤ H 2 (T ∩ B(y, α)) + 10 -1 α 2 for y ∈ E ∩ B(0, 3/2), because we can apply Lemma 16.43 to B(y, α) with δ > 0 as small as we want. The details are the same as for (17.4). If y ∈ E ∩B(0, 3/2) and dist(y, L) ≥ 2α, then H 2 (T ∩B(y, α)) ≤ πα 2 , and (18.5) says that θ(y, α) ≤ 11π/10. Then (18.6) θ(y) ≤ e λA(α) θ(y, α) = e λ α 0 h(2t) dt t θ(y, α) ≤ e λε 7 θ(y, α) < 3π/2 by (16.9), (16.5), and (18.2), so y is a P -point. So (18.4) holds.

Next we want to show that every point of E Y ∩ B(0, 4/3) has a neighborhood where E Y is a nice curve. So let y ∈ E Y ∩ B(0, 4/3) be given. Lemma 16.25 says that for r small, B(y, r) is a biHölder ball of type Y for E (with τ = 10 -4 , say) Let r = r y be such a radius, denote by f y : B(y, 2r) → R 3 the corresponding biHölder mapping, and let Y y be the minimal cone of type Y centered at y that goes with it. Here (15.6)-(15.9) say that We start with the first inclusion. Let z ∈ E Y ∩B(y, r/2) be given, and set w = f -1 y (z). Clearly w ∈ B(y, r) (by (18.7)); let us assume that w ∈ L y and find a contradiction. The idea will be that this implies that R 3 \ E has only two local connected components near w, while the fact that z ∈ E Y implies that there are three. But we want to check this more carefully. Let us first check that (18.13) for each ρ 2 > 0, we can find ρ > 0 such that for z 1 , z 2 , z 3 ∈ B(z, ρ) \ E, we can find a path γ ⊂ B(z, ρ 2 ) \ E, that connects two of the z j .

So let z 1 , z 2 , and z 3 lie in B(z, ρ) \ E. Set w j = f -1 y (z j ) (these are well defined if ρ ≤ r, by (18.9)). By the first half of (18.8), (1τ )(|w j -w|/r) 1+τ ≤ |z j -z|/r ≤ ρ/r, or equivalently (18.14) |w j -w| ≤ ρ 1 , with ρ 1 = r [(1τ ) -1 ρ/r] 1/(1+τ ) .

If ρ is small enough, 4ρ 1 ≤ dist(w, L y ), and Y y coincides with a plane in B(w, 2ρ 1 ). Notice that the w j lie out of Y y , because they lie in B(y, 2r) and otherwise z j = f y (w j ) would lie in E by the second part of (18.10). So one of the segments [w j , w k ] does not meet Y y (pick j and k, j = k, so that w j and w k lie on the same side of the plane that coincides with Y y in B(w, 2ρ 1 )).

Set γ = f y ([w j , w k ]); this is (the support of) a path from z j to z k . Recall that z = f y (w). By (18.14) and the second half of (18.8), γ ⊂ B(z, ρ 2 ), where (18.15) ρ 2 = (1 + τ ) r (ρ 1 /r) 1-τ ≤ (1 + τ ) r [(1τ ) -1 ρ/r] (1-τ )/(1+τ ) .

It is a little unpleasant that ρ 2 is not equivalent to ρ (because f y is merely biHöldercontinuous, and not quasisymmetric), but at least it is as small as we want, which will be enough for (18.13). Suppose that γ meets E at some point x. Since γ ⊂ B(y, 19r/10), the first part of (18.10) says that x = f y (ξ) for some ξ ∈ Y y ∩ B(y, 2r). But f y is injective, so ξ ∈ [w j , w k ] and this is impossible because [w j , w k ] does not meet Y y . So γ does not meet E. This proves (18.13).

Next we use the fact that z ∈ E Y , so Lemma 16.25 says that for t small, B(z, t) is a biHölder ball of type Y for E. Choose ρ 2 so small that this is the case for t ≤ 2ρ 2 . Denote Now we consider w ∈ L y ∩ B(y, r) and check that z = f y (w) ∈ E Y (it is clear that f y (w) ∈ B(y, 3r/2), by (18.7)). By (18.10), E is locally biHölder-equivalent to a set of type Y near z, and in particular R 3 \ E has three local components near z. More precisely, choose three points ζ 1 , ζ 2 and ζ 3 in B(y, r) \ Y y , that lie in different connected components of R 3 \ Y y . For each small ρ and each j ≤ 3, choose a point ζ j,ρ ∈ B(w, ρ) \ Y y , that lies in the same component of R 3 \ Y y as ζ j . The segment l j,ρ = [ζ j , ζ j,ρ ] does not meet Y y and is contained in B(y, r), so (18.10) says that γ j,ρ = f y (l j,ρ ) does not meet E. In addition, (18.8) says that the extremities z j,ρ = f y (ζ j,ρ ) lie in B(z, ρ 1 ), where ρ 1 = r(1 + τ )|ρ/r| 1-τ . Notice that ρ 1 tends to 0 when ρ tends to 0.

If z is a P -point, Lemma 16.19 says that for ρ 1 small enough, B(z, ρ 1 ) is a biHölder ball of type P for E. Denote by f : B(z, 2ρ 1 ) → R 3 the corresponding mapping, and P the corresponding plane through z. The three f -1 (z j,ρ ) lie in B(z, 11ρ 1 /10) \ P , by the analogues of (18.7) and (18.10) and because z j,ρ / ∈ E, and at least two of them lie on the same side of P . Choose j and k = j such that Γ = [f -1 (z j,ρ ), f -1 (z k,ρ )] does not meet P . Then f (Γ) ⊂ B(z, 3ρ 1 /2) \ E (by (18.7) and (18.10)). Then γ = γ j,ρ ∪ f (Γ) ∪ γ k,ρ ⊂ B(y, r) \ E, so its inverse image f -1 y (γ) does not meet Y y . But γ contains ζ j and ζ k , which lie in different components of R 3 \ Y y . This contradiction proves that z is not a P -point.

We assumed that there is no T -point in B(0, 3/2), and y ∈ E ∩ B(0, 4/3), so it is a Y -point. [We could also exclude the case when y is a T -point, by an argument similar to the proof of the first inclusion in (18.12).] This proves the second half of (18.12), and Lemma 18.11 follows.

We shall also need to know that near ∂B(0, 1), E Y is composed of three nice arcs of curves. Let T be the minimal cone of (18.3), and denote by L the spine of T . Thus L is composed of four half-line L j , and we choose the names so that the point a j in the statement of Proposition 18.1 lies exactly opposite to L j . Call z j the point of L j ∩∂B(0, 1); thus z j = -a j . Observe that T coincides with a set of type Y in B(z j , 10 -1 ). Choose y j ∈ E such that (18.17) |y jz j | ≤ 2d 0,2 (E, T ) ≤ 2ε 7

(by (18.3)). If ε 7 < 10 -4 ε 4 , where ε 4 comes from Lemma 16.51 (applied with τ = 10 -4 , say), we can apply Lemma 16.51 to B(y j , 10 -3 ), where we use a translation by y jz j of the set of type Y that coincides with T near z j , to make sure that the spine goes through y j . We get that B(y j , 10 -3 ) is a biHölder ball of type Y for E, with constant τ . Denote by f j : B(y j , 2 • 10 -3 ) → R 3 the corresponding biHölder mapping, by Y j the associated minimal cone of type Y and centered at y j , and by ℓ j the spine of Y j . The proof of Lemma 18.11 yields (18.18) E Y ∩ B(y j , 10 -3 /2) ⊂ f j (ℓ j ∩ B(y j , 10 -3 )) ⊂ E Y ∩ B(y j , 3 • 10 -3 /2).

Set g j = ℓ j ∩ B(y j , 10 -3 ) and γ j = f j (g j ); these are nice little arcs, and (18.18) says that γ j ⊂ E Y . Notice that γ j crosses the annulus A = B(0, 1 -10 -4 ) \ B(0, 1 + 10 -4 ), because |f j (x) -x| ≤ τ 10 -3 for x ∈ g j and the extremities of g j lie on both sides of A (and reasonably far from it). Let us also check that

(18.19) E Y ∩ A ⊂ 4 j=1 γ j .
Indeed, if y ∈ E Y ∩ A, (18.4) (with α = 10 -4 /2) says that dist(y, L) ≤ 10 -4 , so y lies in one of the B(y j , 10 -3 /2) (because it lies in A). Hence (18.19) follows from (18.18).

The situation in B(0, 1) will also be easy to control. First observe that (18.20) E Y ∩ B(0, 1) is compact.

Indeed, let {y k } be a sequence in E Y ∩ B(0, 1), and assume that it converges to y. Then y ∈ E, because E is closed in U . If y is a P -point, Lemma 16.19 says that every point x ∈ E near y is a P -point again. [Incidentally, we do not need the whole proof; we could deduce the same result from Lemma 16.11 (and in particular (16.15) with the plane Z(x, ρ)), plus (16.9) to control θ(x).] This is impossible, because y k ∈ E Y , so y is not a P -point. It cannot be a T -point either, because we assumed that there is no T -point in B(0, 3/2). So y ∈ E Y , as needed for (18.20).

Select a small biHölder ball B(y, r y ) for each y ∈ E Y ∩ B(0, 1), as we did for (18.7)-(18.10), and then cover E Y ∩ B(0, 1) by a finite collection of B(y, r y /10), y ∈ J. Thus J is finite and contained in E Y ∩ B(0, 1). Call η the minimum of 10 -5 and the smallest of the r y /10, y ∈ J. Now pick one of the four γ j . Recall that γ j crosses A. Let us continue the branch of γ j that goes inside B(0, 1), by little steps of diameter comparable to η, by a curve in E Y . For each extension that we get, and if the endpoint ξ of the extension lies in B(0, 1), we can find y ∈ J such that ξ ∈ B(y, r y /10). Then by Lemma 18.11 there is a neighborhood of ξ, that contains B(ξ, η), where E Y coincides with the nice simple curve f y (L y ∩ B(y, r y /2)). We continue our extension of γ j by an arc of f y (L y ∩ B(y, r y /2)) of diameter η, in the direction that was not covered (if it exists). That is, since the curve comes from the outside boundary of A, it has to cover at least one side of f y (L y ∩ B(y, r y /2)). We continue the path in the other direction.

We continue the construction like this, as long as we do not leave B(0, 1). Observe that when we hit a point x, our curve really crosses a neighborhood of x in E Y . As a result, our curve is simple: otherwise, consider the first time where we hit a point x that was already on the curve; we cannot come from any of the two branches of E Y that leave from x, because they were covered some time ago and x is the first point of return, but then there is no access left, so we cannot have returned to x. In fact, this argument shows that as soon as our curve meets x, our next extension leaves B(x, η), and then the curve never returns to B(x, η) again.

As a result of this (and because B(0, 1) does not contain too many disjoint balls of radius η/2), our extension process has to stop after a finite number of steps. This means that our extension of γ j eventually hits ∂B(0, 1) again. By (18.19), this happens in one of the γ k . Since our curve is simple, k = j. So we proved that for each j, (18.21) there is a path in E Y ∩ B(0, 1) that connects γ j to some other γ k .

Next we want to study the connected components of the complement of E near E Y . For y ∈ E Y ∩ B(0, 4/3), denote by A(y) the collection of connected components V of B(0, 2) \ E such that y ∈ V . Recall that each y ∈ E Y ∩ B(0, 4/3) has a small neighborhood where E Y is biHölder-equivalent to a set of type Y as in (18.7)-(18.10). Consequently, A(y) has at most 3 elements, and is a locally constant function of y ∈ E Y ∩ B(0, 4/3).

We can control A(y) near the four y j . For 1 ≤ j ≤ 4, pick w j ∈ γ j ∩ ∂B(0, 1).

Lemma 18.22. For each j, (18.23) A(w j ) = V (a i ) ; i = j , where the V (a i ), 1 ≤ i ≤ 4, are as in the statement of Proposition 18.1.

We do not say here that the three components mentioned in (18.23) are different; we shall take care of this later.

Recall that for 1 ≤ i ≤ 4, a i = -z i , where z i is the point of the spine of T that was used to define y i ; see a little above (18.17). And V (a i ) is the connected component of a i in B(0, 2) \ E.

Fix j and i = j. Denote by b i the point of [a i , z j ] ∩ ∂B(z j , 10 -5 ). Notice that

(18.24) dist([a i , b i ], E) ≥ dist([a i , b i ], T ) -2ε 7 ≥ 10 -6
by simple geometry and (18.3), so b i ∈ [a i , b i ] ⊂ V (a i ).

To relate V (b i ) = V (a i ) to A(w j ), we shall use the same biHölder ball B(y j , 10 -3 ) as for (18.18), and the corresponding mapping f j : B(y j , 2 • 10 -3 ) → R 3 . Recall that (18.25) |f j (x) -x| ≤ τ 10 -3 for x ∈ B(y j , 2 • 10 -3 ), (18.26) B(y j , (2τ )10 -3 ) ⊂ f j (B(y j , 2 • 10 -3 )), and (18.27) E ∩ B(y j , (2τ )10 -3 ) ⊂ f j (Y j ∩ B(y j , 2 • 10 -3 ))) ⊂ E.

by (15.6), (15.8) , and (15.9).

Recall that b i ∈ ∂B(z j , 10 -5 ) \ E (by (18.24)), and that |y jz j | ≤ 2ε 7 (by (18.17)). Then c i = f -1 j (b i ) is well defined (by (18.26)), lies in B(y j , 10 -4 ) (by (18.25)), and out of Y j (by (18.27)). Denote by W i the connected component of B(y j , 10 -4 ) \ Y j that contains c i , and let us check that (18.28) the three W i , i = j, are different.

Indeed Y j is quite close to T near y j : every point ξ ∈ Y j ∩B(y j , 2•10 -3 ) lies within τ 10 -3 of f j (ξ) ∈ E, by (18.27), hence dist(ξ, T ) ≤ τ 10 -3 + 2ε 7 , by (18.3). Now the b i lie reasonably

Then E is a reduced minimal cone, i.e., E is empty, a plane, or a cone of type Y or T.

Since we deal with two dimensional sets in R 3 , the reduced minimal cones of type T are the cones based on the edges of a regular tetrahedron, as defined near Figure 1.2.

Recall that an Al-minimal set in R 3 is also the same as an almost-minimal set in R 3 , with gauge function h = 0, as in Definition 4.8 (or even Definitions 4.1 and 4.3, see Proposition 4.10), so we may apply to E all the results from the previous sections.

By Proposition 5.16, θ(x, r) = r -2 H 2 (E ∩ B(x, r)) is a nondecreasing function of r for each x ∈ R 3 . It is also bounded, by the easy part of (2.16), so it has finite limits (18.32) θ(x) = lim r→0 θ(x, r) and θ ∞ = lim r→+∞ θ(x, r).

It is easy to see that θ ∞ does not depend on x ; indeed B(0, r-|x|) ⊂ B(x, r) ⊂ B(0, r+|x|) for r large, so (r -|x|) 2 θ(0, r -|x|) ≤ r 2 θ(x, r) ≤ (r + |x|) 2 θ(0, r + |x|), which yields lim r→+∞ θ(x, r) = lim r→+∞ θ(0, r) after dividing by r 2 and taking limits. We know from the discussion below (16.7) that θ(x) can only take the values 0, π, 3π/2, and d + = Argcos(-1/3), corresponding to the various types of blow-up limits that E may have at x. When x ∈ E, θ(x) = 0 is excluded by (2.16).

But we also have that θ ∞ ∈ {0, π, 3π/2, d + }. Indeed consider a blow-in sequence {E k }, where E k = t -1 k E for some sequence {t k } that tends to +∞. We can extract a subsequence (which we shall still denote by {E k }), which converges to a limit F (in Hausdorff distance on compact subsets of R 3 , as in Section 3). Recall that each E k is a minimal set, by homogeneity. By Lemma 4.7, F is a reduced minimal set in R 3 . By (3.4), (18.33)

H 2 (F ∩ B(0, r)) ≤ lim inf k→+∞ H 2 (E k ∩ B(0, r)) = lim inf k→+∞ t -2 k H 2 (E ∩ B(0, t k r))
= lim inf k→+∞ r 2 θ(0, t k r) = r 2 θ ∞ .

Similarly, Lemma 3.12 says that (18.34) H 2 (F ∩ B(0, r)) ≥ lim sup k→+∞ H 2 (E k ∩ B(0, r))

= lim sup k→+∞ t -2 k H 2 (E ∩ B(0, t k r)) = r 2 θ ∞ , so H 2 (F ∩ B(0, r)) = r 2 θ ∞ for r > 0. By Theorem 6.2, F is a reduced minimal cone. By [Ta], F is the empty set, a plane, or a cone of type Y or T. Then θ ∞ ∈ {0, π, 3π/2, d + }, as announced.

Let us distinguish between cases, depending on the value of θ ∞ . If θ ∞ = 0, θ(0, r) = 0 for all r, E = ∅, and we don't need to worry. If θ ∞ = π, θ(0, r) > 0 for some r, and E is not empty. Pick x ∈ E. We know that θ(x) > 0, by (2.16), so θ(x) ≥ π = θ ∞ . Then θ(x, •) is constant and Proposition 6.2 says that E is a cone (in fact, a plane). Now suppose that θ ∞ = 3π/2. As before, we can find a sequence {t k } that tends to +∞, so that the sets E k = t -1 k E converge to a limit F , and then F is a reduced minimal cone. In fact, (18.33) and (18.34) tell us that the density of F is 3π/2, so F is a cone of type Y centered at the origin.

Each t -1 k E is a reduced minimal set, and the sequence converges to F . For k large, it satisfies the assumptions of Proposition 16.24 (in the unit ball), so there is a Y -point in t -1 k E, hence also in E. For this point x, θ(x) = 3π/2 = θ ∞ . Again, θ(x, •) is constant and Proposition 6.2 says that E is a minimal cone (here a set of type Y).

We are left with the case when θ ∞ = d + . The same argument as before gives a blow-in sequence {E k } = {t -1 k E} that converges to a reduced minimal cone T of type T centered at the origin. This time, we do not have a good enough analogue of Proposition 16.24 that would allow us to conclude, and this is why we added an extra hypothesis.

The sequence {t k } satisfies the assumption (18.30), so we have (18.31). This is exactly what we need to apply Proposition 18.1 to the ball B(0, t k l ) (for l large enough, so that (18.3) holds). We get the existence of a point of type T in E ∩ B(0, t k l ). Then θ(x) = d + = θ ∞ at this point, θ(x, •) is constant, and E is a minimal cone (of type T), as before. This completes the proof of Proposition 18.29.

Proof of Theorem 1.9.

Let us prove Theorem 1.9. Let E be a M S-minimal set in R 3 (see Definition 1.4). By Remark 1.8, E is Al-minimal. By Remark 2.14, E * (the closed support of the restriction to E of H 2 , as in Definition 2.12) is a reduced Al-minimal set. We just need to show that E * is a reduced minimal cone (see (2.13)).

We want to apply Proposition 18.29 to E * , so we need to check the separation assumption. So we assume that {t k } is a sequence such that (18.30) holds for E * , and we want to prove (18.31) (for the same sequence, but k large; we won't need to take a real subsequence).

Let us show the dependence in k in our notation. Call a j,k , 1 ≤ j ≤ 4, the four points of ∂B(0, t k /2) that lie furthest from T , and denote by V j,k the connected component of a j,k in B(0, t k ) \ E * . We want to show that for k large, the V j,k are all different.

Suppose instead that two of the V j,k are the same, for instance V 1,k and V 2,k . We want to show that this is impossible, and for this we intend to construct a better competitor F , obtained by removing a big piece of the wall near a k = 1 2 (a 1,k + a 2,k ). The comparison will be easier once we know a little more about E near a k .

Set E k = t -1 k E * , and a = t -1 k a k (this is a fixed point). By (18.30), the E k converge to T in B(0, 1), so Lemma 3.3 says that (18.35) lim sup k→+∞ H 2 (E k ∩ B(a, 10 -1 )) ≤ H 2 (T ∩ B(a, 10 -1 )) ≤ 10 -2 π,

where the last inequality holds because T coincides with a plane in B(a, 10 -1 ). Pick x k ∈ E k as close to a as possible, and apply Lemma 16.19 to E k in B(x k , r k ), where we choose r k such that 9r k = 10 -1 -|x k -a| and hence B(x k , 9r k ) ⊂ B(a, 10 -1 ).

The assumption (16.20)

) \ E k ⊂ 2 -k V 1,k ;
this is believable because we assumed that the components V 1,k and V 2,k coincide, but let us check the details. Let ζ 1 and ζ 2 be the two points of ∂B(x k , 10 -2 ) \ P k that lie at distance 10 -2 from P k (so they are on opposite sides of P k ). If x ∈ B(x k , 10 -2 ) \ E k , f -1 k (x) is defined (by (15.8)), lies in B(x k , 11 1000 ) (by (15.6)) and away from P k (by (15.9)), so one of the two line segments I = [f -1 k (x), ζ i ] does not meet P k . Then f k (I) is an arc in B(x k , 12 1000 ) \ E (by (15.6) and (15.9)) that goes from x to z i = f (ζ i ). Thus (18.37) will follow if we show that both z i lie in 2 -k V 1,k . Now dist(ζ i , P k ) = 10 -2 , so for k large, because the E k converge to T in B(0, 1) (by (18.30)). Then z i lies in the same component as a 1 or a 2 in B(0, 1) \ E k , or equivalently 2 k z i ∈ V 1,k ∪ V 2,k . But we assumed that V 1,k = V 2,k , so z i ∈ 2 -k V 1,k , as needed for (18.37).

Set F k = E \ B(r k x k , 10 -3 r k ). This set may be a little larger than r k (E k \ B(x k , 10 -3 ) if E is not reduced, and we need to pay some attention to the difference because did not verify that E * is a M S-minimal set. Let us check that (18.41) F k is a M S-competitor for E, as in Definition 1.1. Let us take R = r k in that definition; (1.2) is clear, and we just need to prove (1.3). Let z 1 , z 2 lie in R 3 \ [E ∪ B(0, r k )]. We need to show that if E separates z 1 from z 2 , F k separates them too. Suppose it does not, and pick a path γ from z 1 to z 2 , that does not meet F k . Since γ meets E and the only difference is E ∩ B(r k x k , 10 -3 r k ), γ has to go through ∂B, where we set B = B(r k x k , 2 • 10 -3 r k ). Call z ′ 1 the first point of γ ∩ ∂B when we go from z 1 to z 2 along γ, and z ′ 2 the last point of γ ∩ ∂B. The part of γ between z 1 and z ′ 1 does not meet E, because it does not meet F k (and F k coincides with E out of B). Similarly, the part of γ between z 2 and z ′ 2 does not meet E either. So we just need to show that E does not separate z ′ 1 from z ′ 2 and we shall reach the desired contradiction with our assumption that E separates z 1 from z 2 . 144 Notice that the z ′ i lie on ∂B \ E, hence also on ∂B \ E * . Then the r -1 k z ′ i lie in ∂B(x k , 2 • 10 -3 ) \ E k (recall that E k = r -1 k E * ). By (18.37), the z ′ i lie in V 1,k , which means that there is a path γ 1 ⊂ B(0, r k ) \ E * that goes from z ′ 1 to z ′ 2 . We can even make γ 1 piecewise linear. Now γ 1 may meet ∆ = E \ E * , but observe that H 2 (∆) = 0 by (2.13). Let us modify γ 1 slightly, so that it does not meet E. First we move the endpoints of the segments that compose γ 1 very slightly (to make sure that the new path will not meet E * ), so that the endpoints lie out of ∆ too. This is very easy to arrange, because the complement of ∆ is dense. Note that we can leave z ′ 1 and z ′ 2 alone, since they already lie out of E.

For each of the segments that compose our new path, the orthogonal projection of ∆ on a plane perpendicular to that segment has vanishing Lebesgue measure because H 2 (∆) = 0, so we may translate the segment a tiny bit so that it no longer meets ∆. We do this for each segment, and then connect the new endpoints together by very short line segments to make a new path γ 2 from z ′ 1 to z ′ 2 . Now γ 2 does not meet E. So z ′ 1 and z ′ 2 lie in the same component of R 3 \ E, we reached the desired contradiction, and (18.41) holds.

Since E is M S-minimal, Definition 1.4 says that H 2 (E \ F k ) ≤ H 2 (F k \ E) = 0. But this is not true:

(18.42) H 2 (E \ F k ) = H 2 (E ∩ B(r k x k , 10 -3 r k ) = r 2 k H 2 (E k ∩ B(x k , 10 -3 ) > 0
by (2.13) and (2.16), for instance. This last contradiction completes our proof of (18.31); Proposition 18.29 applies, and Theorem 1.9 follows.

Remark 18.43. So we got a complete description of M S-minimal sets in R 3 , but unfortunately not of Al-minimal sets. It is tempting to try to deduce the separation assumption in Proposition 18.29 from the closeness of E to a cone of type T and its local description of E in Theorem 16.1. But we shall see in Section 19 that it is not too hard to construct sets E ⊂ R 3 that are very close to a cone of type T in the unit ball, satisfy the conclusions of Theorem 16.1, but do not have any T -point and only separate R 3 locally into two connected components only. So a proof of the analogue of Theorem 1.9 for Al-minimal sets would need to use Al-minimality one more time. There is a little more of E that we could construct without thinking. We know that a neighborhood of each Γ i will be composed of three thin stripes that touch Γ i . We already know one of these stripes, which comes from F i , and we could brutally construct the two other ones: we would just need to draw two little curves vaguely parallel to Γ i , and make sure that the two new stripes meet F i neatly along Γ i , with 120 • angles. This is what is suggested by Figure 19.1. We shall not really need this, but maybe knowing this will help the reader follow the description of F below.

The remaining set F = E \ [F 1 ∪ F 4 ] will be a little harder to describe. The contact with F 1 and F 4 is along the Γ i , and through the four stripes alluded to above. Now F is actually a single connected face (we shall see that it is even oriented). Let us use the symmetry of E and F with respect to P 1 to simplify our description. First, the intersection F ∩ P 1 is just a vaguely round closed curve γ, with a singularity at the point of Γ 4 ∩ P 1 (let us denote it by z 0 ), which is actually excluded (because it lies in F 4 ) and where γ has two half tangents that make 120 • angles. See Figure 19.2, but do not pay attention to the shaded area yet. The face F crosses γ perpendicularly to P 1 ; it also approaches Γ 1 from both sides of P 1 , with 60 • angles.

Call F + the part of F that lies on the same side of P 1 as x 3 . Out of B(0, 1), F + coincides with the union of two faces of T (namely the face T 1,3 bounded by L 1 and L 3 , and the face T 2,3 ). Let us look at F + ∩ B(0, 1). Topologically, it is a disk, and its boundary is composed as follows. We may start with the curve Γ 1 ∩ B(0, 1), say from x 1 to x 2 . Then we follow the arc ∂B(0, 1) ∩ T 2,3 , up to x 3 . Then we follow Γ 4 from x 3 to z 0 , then we turn around γ, and return along Γ 4 from z 0 to x 3 . After this we run along the arc ∂B(0, 1)∩T 1,3 up to x 1 , and we are back where we started. Notice that if we include the part of Γ 4 from x 3 to z 0 , F + ∩ B(0, 1) is a topological tube (because of the gluing along Γ 4 ). Similarly, the whole F + is a half-infinite tube, with a gluing along half of Γ 4 and only one "circular" boundary along γ. The reader may try to get a mental picture of F + by looking at the shaded part of Figure 19.2 that lives above Γ 1 and outside of γ, and imagining that it is lifted in the third direction. In particular, imagine that F 4 ∩ P 1 is lifted until it becomes the upper half of Γ 4 . Also recall that F + leaves from Γ 1 and both sides of Γ 4 with 60 • angles, and vertically from γ.

Let F -denote the part of F that lies on the other side of P 1 . Since F -is obtained from F + by symmetry with respect to P 1 , it is also a tube, bounded by γ and glued along the other half of Γ 4 . Finally F is an infinite tube, obtained from the two pieces by gluing along γ. The fact that F + and F -live on different sides of P 1 makes it easy to believe that nothing weird (like self-intersections) happens when we do the gluing. It is easy to see that F can be oriented too.

Hopefully the properties above give a precise idea of the set F , and of E = F 1 ∪F 4 ∪F . The fact that E is locally equivalent to a plane or a Y is easy to check, and E coincides with T out of B(0, 1).

Let us say a few words about the connected components of R 3 \ E. It should be expected from the proof of Theorem 1.9 that E does not separate R 3 into three or four components, and indeed R 3 \ E has only two components. More precisely, call y j = -x j the point of ∂B(0, 1) which is opposite to L j , and denote by U j the connected component of R 3 \ E that contains y j . It is easy to see that U 3 = U 4 , because there is a path from y 3 to y 4 that does not meet E and crosses P 1 inside γ. Similarly, U 1 = U 2 ; rather than trying to use the fact that P 1 and P 4 can be interchanged in the construction, the reader should just look at Figure 19.2 and imagine a curve that runs in P 1 along Γ 1 , at a short distance from but just opposite to F 1 . This curve does not meet E, and it goes from U 3 to U 4 . Of course the two ways to see that U 1 = U 2 are similar, since the analogue of γ here is a loop in P 4 \ F 1 that ends at the point of P 4 ∩ Γ 1 and passes through the closest point of γ on the other side.

It is also true that U 1 = U 3 ; this is not easy to see without a paper model, and not so important, so we skip the proof. Notice that once we know that U 1 = U 2 and U 3 = U 4 , there is no contradiction with the fact that the list A(y) of components that touch y is locally constant along the curves Γ i ; in both cases, A(y) = {U 1 , U 3 } (with one of the components counted twice).

This completes our description of E. Unless we are extremely lucky, our example E will not be minimal. We can try to make it minimal inside B(0, 1) under the constraint that E coincides with T out of B(0, 1); the author does not know whether this is really possible. If we can make H 2 (E ∩ B(0, 1)) smaller than H 2 (T ∩ B(0, 1)), then we can probably minimize H 2 (E ∩ B(0, 1)), but then E is not globally minimal in R 3 , because a dilation would do better. If H 2 (E ∩ B(0, 1)) > H 2 (T ∩ B(0, 1)) for E as above, the only minimum is T (obtained as limit of contractions of sets E). But there is still the unlikely possibility that the minimal value of H 2 (E ∩ B(0, 1)) is precisely H 2 (T ∩ B(0, 1)), and then we could have an exotic minimal set E. The case when E is obtained with curves Γ i that are linked in a complicated way could even be more fun.

Simple computations of density

For the convenience of the reader, we do the simple computations of density for the minimal cones of types Y and T.

Let us first remind the reader that H 2 (Y ∩ B(0, r)) = 3πr 2 /2 when Y is a set of type Y centered at the origin, simply because Y ∩ B(0, r) is composed of three half disks of radius r.

Next let T be a set of type T centered at the origin. We want to check that (20.1) H 2 (T ∩ B(0, r)) = d + r 2 , with d + = 3Argcos(-1/3). Now T is composed of six equal angular sectors, so if α ∈ (π/2, π) denotes the angle of these sectors, H 2 (T ∩ B(0, r)) = 3αr 2 . We just need to show that cos α = -1/3.

Recall that T is the cone over the edges of a regular tetrahedron centered at the origin; we may obtain such a tetrahedron by starting from the cube [-1, 1] 3 , and selecting four vertices (out of the eight) that are not mutually contiguous (see Figure 20.1). The distance between two vertices is 2 √ 2, and the distance form a vertex to the origin is √ 3. Thus (see Figure 20.2) sin(α/2) = √ 2 √ 3 and so cos α = 1 -2 sin 2 (α/2) = 1 -4/3 = -1/3, as needed. This proves (20.1).

Figure 1

 1 Figure 1.1. A cone of type Y Figure 1.2. A set T

Figure

  Figure 5.1.Figure5.2.

  b)e λA(b)θ(x, a)e λA(a) ≥ b a

Figure 6

 6 Figure 6.1.a. n = 3, d = 2 Figure 6.1.b. n = 3, d = 1

Figure 6

 6 Figure 6.4. A picture in P To contract C to a piece of its boundary, we use the mapping ϕ which to z ∈ C associates the radial projection of z onto the box [x, y] × ∂J × {0} ∪ {y} × J × {0}(open at one side) with center (x 0 , ξ, 0). [We don't care about the precise value; the point is that x 0 < x, so ϕ is Lipschitz. The contraction that we use on F is ψ -1 • ϕ • ψ. It is Lipschitz on F , even though with a very large constant. We want to set

(

  10.22) ϕ(y) ∈ B = B(0, ρ) for y ∈ E ∩ B(0, ρ) because F j ⊂ B. If x lies out of B(0, 2ρ), (10.21) holds because |ϕ(x)ϕ(y)| = |xϕ(y)| ≤ |x -y| + |yϕ(y)| ≤ |x -y| + 2ρ ≤ 3|x -y|. Next assume that x ∈ B(0, 2ρ) \ B(0, ρ). If y lies in one of the B(a k , ε), then ψ(y) = 0, ϕ(y) = y, and (10.21) is trivial. Otherwise, y lies in the compact set K

  dt/t ≤ ε 2 and d 0,200 (E, Y ) ≤ ε 2 , by (17.1) and (17.2).

(

  18.7) |f y (x) -x| ≤ 2τ r for x ∈ B(y, 2r), (18.8) (1-τ )|(x-z)/r| 1+τ ≤ |(f (x)-f (z))/r| ≤ (1+τ )|(x-z)/r| 1-τ for x, z ∈ B(y, 2r)E ∩ B(y, 19r/10) ⊂ f (Y y ∩ B(y, 2r)) ⊂ E.Lemma 18.11. Denote by L y the spine of Y y . Then (18.12) E Y ∩ B(y, r/2) ⊂ f y (L y ∩ B(y, r)) ⊂ E Y ∩ B(y, 3r/2).

  (18.38) dist(z i , E k ∩B(x k , (2τ )10 -2 )) ≥ dist(z i , f (P k ∩ B(x k , 2 • 10 -2 ))) ≥ dist(ζ i , P k ∩ B(x k , 2 • 10 -2 )) -2τ 10 -2 ≥ (2 -2τ )10 -2by (15.9), (15.6), and the definition ofζ i . Since (18.39) dist(z i , E k \ B(x k , (2τ )10 -2 )) ≥ (2τ )10 -2 -|z ix k | ≥ (1 -2τ )10 -2by (15.6) and because ζ i ∈ ∂B(x k , 10 -2 ), we get that(18.40) dist(z i , T ) ≥ dist(z i , E k )d 0,1 (E k , T ) ≥ 10 -2 /2

  Remark 18.44. It is possible to define a notion of M S-almost-minimal sets in U ⊂ R 3 , as we did for A ′ -almost-minimal sets in Definition 4.8. That is, if E is a closed subset of U ⊂ R 3 such that H 2 (E ∩ B) < +∞ for every compact ball B ⊂ U , we say that E is a M S-almost-minimal set in U , with gauge function h, if (18.45) H 2 (E \ F ) ≤ H 2 (F \ E) + h(r)r 2 whenever B(x, r) ⊂ U and F is another closed set in U such that F \ B(x, r) = E \ B(x, r),and(18.46) F separates y from z whenever y, z ∈ U \ (B(x, r) ∪ E) are separated by E.When h = 0 and U = R 3 , we get the minimal sets of Definition 1.4.

Figure

  Figure 19.1 Figure 19.2. Intersection with P 1

Figure 20

 20 Figure 20.1 A tetrahedron in its box Figure 20.2. A face

  where c d is the H d -measure of the unit ball in R d . See again[Ma], Theorem 16.2 on page 222. Call E ′ the set of points x ∈ E such that E has a tangent plane P x at x and (3.14) holds; thus almost-every point of E lies in E ′ .Let ε > 0 be small, and denote by B the set of balls

  -1 dist(y, D)). [We added a factor 2 to make sure that ϕ(A) ⊂ B(x, r) if P is close enough to P x , even if it does not exactly go through x.]Set ϕ B i . Recall that diam(B i ) = 2r < δ by definition of B. So we may apply Definition 2.10, and (2.11) holds (with any δ

t (y) = tϕ(y) + (1t)y for 0 ≤ t ≤ 1; this family obviously satisfies (2.2)-(2.6), with W ⊂

  Remark 4.5. So we just replaced the error term h(δ)H d (ϕ 1 (E ∩ W 1 )) in (4.2) with the simpler h(δ)δ d . Let us check that A + -almost-minimal set are automatically A-almostminimal sets, at least in any smaller open set U τ = {x ∈ U ; B(x, τ ) ⊂ U }.Choose δ < τ such that h(δ) < +∞, and let {ϕ t } satisfy (2.2)-(2.6). Observe that diam(W 1 ) ≤ δ, by (2.6), and that

x} as in (2.5).

  by (4.18) and (4.16). If x ∈ W t , then ϕ t (x) ∈ ϕ t (W t ) ⊂ S by (4.15), and soϕ t,k (x) ∈ S ′ k by (4.20). If x ∈ W k,t \ W t , then ϕ t (x) = x ∈ W k,t ⊂ S ′ ,and then ϕ k,t (x) ∈ S ′

	again. This proves (4.19).	k , by (4.20)
	We are now ready to prove (2.6) for the ϕ k,t . Set W k = t [W k,t ∪ ϕ k,t (W k,t )]. We just proved that W k ⊂ S ′ k , and it is obvious from (4.15) and the compactness of S ′ that for k large, S ′ k is a compact subset of U with diam(S ′ k ) < δ. This proves (2.6). So we can apply (4.4), and we get that
	(4.21)

  Cτ 2 for y ∈ E ∩ B(0, 35).Set Y = Y (45), and observe that H 2 (Y ∩ B(y, t)) ≤ 3πr 2 /2 for all y and t because the density is largest for balls centered on the spine of Y , for instance by Proposition 5.16. ≤ e λA(10) θ(y, 10) ≤ e λε 3 θ(y, 10) ≤ θ(y, 10) + Cε 3 by (16.5) and (16.26), so (16.31) says that θ(y) ≤ 3π/2 + Cτ 2 + Cε 3 < d T if τ 2 and ε 3 are small enough. By (16.8), y cannot be a T -point. So we just checked that

	as needed for (16.30). In addition, (16.9) says that
	(16.32) θ(y) (16.33)	there is no T -point in E ∩ B(0, 35).
			and (16.15) hold. That is,
	(16.28)		d 0,ρ (E, Y (ρ)) ≤ τ 2
	and
	(16.29)	H 2 (E ∩ B(y, t)) -H 2 (Y (ρ) ∩ B(y, t)) ≤ τ 2 ρ 2 for y ∈ R n and t > 0 such that B(y, t) ⊂ B(0, ρ).
	Let us check that
	(16.30) θ(y, 10) ≤ 3π/2 + Then apply (16.29) with ρ = 45 and t = 10 (so that B(y, t) ⊂ B(0, 45)). We get that
	(16.31)		θ(y, 10) = 10 -2 H 2 (E ∩ B(y, t)) ≤ 10 -2 H 2 (Y ∩ B(y, t)) + 10 -2 τ 2 ρ 2 ≤ 3π/2 + Cτ 2 ,

  follows from (16.31) because θ(y) = 3π/2 when y ∈ E Y . We get that for Notice that(16.36) gives the sort of control on E ∩ B(y, ρ) that we need in order to apply Corollary 15.11, but only in balls B(y, ρ) that are centered on E Y . We also need to control the balls that are centered on P -points. ≤ 3, because the origin lies in E Y by assumption. Also, d(x) > 0, because the proof of Lemma 16.19 gives a neighborhood of x which contains no Y -point (see below (16.23)). We do not even need to know this, because if d(x) = 0 we can use the cones Y (y, ρ) to control E in the balls centered at x, so we do not need the construction below. See the small argument above (16.42).Pick y ∈ E Y such that |y -x| ≤ 11d(x)/10, and set ρ = 2d(x). Then y ∈ E Y ∩B(0, 10) and ρ ≤ 6, so (16.35) holds and we can set Y = Y (y, ρ), where Y (y, ρ) is as in (16.36) and (16.37). Denote by L the spine of Y ; that is, L is the common boundary of the three half planes that compose Y . It is a line through y. Let us check that Lemma 16.43. For each small δ > 0, there is a constant ε > 0 (that depends only on n and d) such that if E and F are reduced almost minimal sets of dimension d in the open set U ⊂ R n , with gauge function h, and if

	in Lemmas 16.19 and 16.25 with assumptions of good approximation of E by a minimal
	cones.
	(16.44)	B(x, 10r/9) ⊂ U, h(20r/9) ≤ ε, and d x,10r/9 (E, F ) ≤ ε,
	then
	(16.45)
	(16.35)		y ∈ E Y ∩ B(0, 35) and 0 < ρ ≤ 9,
	there is a reduced minimal cone Y (y, ρ) of type Y, centered at y, such that
	(16.36)		d y,ρ (E, Y (y, ρ)) ≤ τ 3
	and
	(16.37)	H 2 (E ∩ B(x, t)) -H 2 (Y (y, ρ) ∩ B(x, t)) ≤ τ 3 ρ 2 for x ∈ R n and t > 0 such that B(x, t) ⊂ B(y, ρ).
	Let x ∈ E ∩ B(0, 3) \ E Y be given, and set
	(16.38)		d(x) = dist(x, E Y ).
	Observe that d(x) (16.39)	dist(x, L) ≥ 3d(x)/4.

  ≤ e λA(η|x|) θ(x, η|x|) ≤ e λε 6 [3π/2 + 2η -2 τ 4 ] ≤ 3π/2 + Cε 6 + 2η -2 τ 4 by (16.9), (16.57), and (16.62). In particular, θ(x) < d T if ε 6 and τ 4 are small enough, and

	by (16.60), because B(x, η|x|) ⊂ B(0, ρ(x)), and by (16.61). Then
	(16.63) θ(x) (16.64)	0 is the only T -point of B(0, 8).
		-2 τ 4

  Suppose instead that we can find ξ ∈ L, with |ξ -z| = dist(z, L) < d(z)/2. Set t = ηd(z)/4; we want to apply Proposition 16.24 to B(ξ, t), so we need a set of type Y. By (16.61), there is a cone Y of type Y that coincides with T in B(ξ, η|ξ|). Observe that Y is centered at ξ, because ξ lies in the spine ofT . Next |ξ| ≥ |z| -|ξ -z| ≥ |z|d(z)/2 ≥ d(z)/2 (because d(z) ≤ |z|), so t = ηd(z)/4 ≤ η|ξ|/2 and Y coincides with T in B(ξ, 2t). Thus d ξ,t (E, Y ) = d ξ,t (E, T ).

		). Let us show that
	(16.69)	dist(z, L) ≥ d(z)/2.
	In addition, |ξ| ≤ |z|+|ξ -z| ≤ |z|+d(z)/2 ≤ 3|z|/2 and ρ = 2|z|, so B(ξ, t) ⊂ B(0, ρ). Hence
	(16.70)

  -2 τ 4 by(16.59) and (16.68).If τ 4 is small enough, (16.70) allows us to apply Proposition 16.24 to B(ξ, t), and get a Y -point y in B(ξ, t/100). Recall that |ξ -z| < d(z)/2; then |y -z| ≤ d(z)/2 + t < d(z), which contradicts the definition of d(z). So (16.69) holds.Otherwise, B(z, ηd(z)) meets two faces of T . Let z 1 , z 2 ∈ B(z, ηd(z)) lie in different faces F 1 and F 2 , and setw i = z i /|z i |.w 2 | ≤ 2|z 1z 2 |/|z| ≤ 4ηd(z)/|z| because the radial projection onto ∂B(0, 1) is (2/|z|)-Lipschitz on R n \ B(0, |z|/2).Next, w 1 and w 2 lie in different arcs of circles C i = F i ∩ ∂B(0, 1) in the description of

	Next we check that
	(16.71)	T coincides with a plane in B(z, ηd(z)).
		First observe that
	(16.72)	|z i -z| ≤ ηd(z) ≤ η|z|,
	so |z i | ≥ |z|/2, and then
	(16.73) |w 1 T ∩ ∂B(0, 1), and
	(16.74)	dist(w

1 , C 2 ) ≤ |w 1w 2 | ≤ 4ηd(z)/|z| ≤ 4η < η 0 because η ≤ η 0 /10 (see below (16.60)). So (14.10) says that C 1 and C 2 have a common endpoint a such that |aw 1 | ≤ |w 1w 2 | ≤ 4ηd(z)/|z|. Of course |z 1 |a lies in the spine L, so (16.75) dist(z 1 , L) ≤ z 1 -|z 1 |a ≤ |z 1 ||w 1 -a| ≤ 4η|z 1 |d(z)/|z| ≤ 8ηd(z).

This is impossible, because z 1 ∈ B(z, ηd(z)) and dist(z, L) ≥ d(z)/2 (by (16.69)). So (16.71) holds, and H 2 (T ∩ B(z, ηd(z))) ≤ (ηd(z)) 2 π. Recall that B(z, ηd(z)) ⊂ B(0, ρ) because ρ = 2|z| and d(z) ≤ |z|, so (16.60) says that (16.76) θ(z, ηd(

  with a differential of full rank at ζ j . This completes proof of Lemma 17.19.Lemma 17.26. We can find continuous functions θ k , 1 ≤ k ≤ l, such that

	(17.27)	θ k is supported in B(x k , 3η),
	(17.28)	||θ k || ∞ ≤ 2 -k 10 -6 ,
	and, if we set
	(17.29)

  2 to θ k in E ∩ B(x j , ρ 1 ) (and interpolating continuously on A). Let us first check that for almost every choice of w,

	(17.34)	h

k has a finite number of zeros in E ∩ B(x j , ρ 1 ) for the modified h k . Set

(17.35)

  , which we see as a 2-dimensional sphere, and define π : R 2 → S ∞ by (x) = π(m t (x)) for x ∈ E ∩ B(x k , ρ 2 ) and 0 ≤ t ≤ 1.Notice that p t (x) is a continuous function of x and t, with values in S ∞ . By definition of

	Then set
	(17.46) p t β k ,
	(17.47)	p t
	(17.45)	π(x) = ∞ if |x| ≥ β k and π(x) =	x β k -|x|	otherwise.

  holds for k large, because (E k ∩ B(x k , 9r k )) ≤ lim sup k→+∞ H 2 (E k ∩ B(a, 10 -1 )) ≤ 10 -2 πby (18.35), and 9r k tends to 10 -1 . So Lemma 16.19 applies, and says that B(x k , 10 -2 ) is a biHölder ball of type P for E k . Let f k : B(x k , 2/100) → R 3 be the corresponding mapping, and P k the associated plane. Let us use f k to show that (18.37) B(x k , 10 -2

	(18.36)	lim sup	H 2
		k→+∞

Suppose that (16.39) fails. Choose z ∈ L so that |z -x| ≤ 3d(x)/4, and apply Proposition 16.24 to the ball B(z, d(x)/10). We need to check the assumption that d z,d(x)/10 (E, Y ) ≤ ε 2 . Indeed, B(z, d(x)/10) ⊂ B(y, ρ) because |z -y| ≤ |z -x| + |x -y| ≤ 3d(x)/4 + 11d(x)/10 and ρ = 2d(x). Then d z,d(x)/10 (E, Y ) ≤ 10d(x) -1 ρd y,ρ (E, Y ) ≤ 20τ 3 < ε 2 if τ 3 is small enough, as needed.

So Proposition 16.24 applies, and there is a point of ξ ∈ E Y inside B(z, 10 -3 d(x)). But then |ξ -x| ≤ 10 -3 d(x) + |z -x| ≤ [10 -3 + 3/4] d(x) < d(x). This is incompatible with the definition of d(x), and so (16.39) holds.

Notice that B(x, d(x)/2) is contained in B(y, ρ) (because |y -x| ≤ 11d(x)/10 and ρ = 2d(x)), so (16.37) says that (16.40) θ(x, d(x)/2) = (d(x)/2) -2 H 2 (E ∩ B(x, d(x)/2) because (16.39) says that Y coincides with a plane in B(x, d(x)/2). Recall that θ(x) = π because we assumed that x ∈ E \E Y (and by (16.33)). Hence, if τ 3 is small enough, (16.40) allows us to apply Lemma 16.11 with τ = ε, where ε comes from Corollary 15.11. We get that for 0 < r < d(x)/3, there is a plane Z(x, r) through x such that d x,r (E, Z(x, r)) ≤ ε, as in (16.14).

We are now ready to deduce Lemma 16.25 from Corollary 15.11. We want to get the constant τ in the conclusion of Lemma 16.25, so we need to prove (15.12), with a constant ε that depends on τ . For each choice of x ∈ E ∩ B(0, 3) and 0 ≤ r ≤ 3, we need to find a minimal cone Z(x, r) such that (16.41) d x,r (E, Z(x, r)) ≤ ε.

When x ∈ E Y , the pair (x, r) satisfies (16.35), so we can use Z(x, r) = Y (x, r), and (16.41) follows from (16.36) if τ 3 < ε. Note that Z(0, 3) is a set of type Y centered at the origin, as required by the last assumption of Corollary 15.11.

When x ∈ E ∩ B(0, 3) \ E Y and r < d(x)/3, we can use the set Z(x, r) that was found a few lines ago, below (16.40).

We are left with the case when x ∈ E ∩ B(0, 3) \ E Y and d(x)/3 ≤ r ≤ 3. Our argument also applies when d(x) = 0, if we pretend that we don't know that

, and r ≤ 3, and also ρ ≤ 9, so (16.35) holds, and we can set

by (16.36) and if τ 3 is small enough. This completes our proof of Lemma 16.25.

The next lemmas are a little bit of a digression, and are not needed for the proof of Theorem 16.1. We shall use a compactness argument to replace the density assumptions Indeed, set ξ l = ψ k (y l ); by (17.49), ξ l is a zero of h = h k-1 or h k , and (17.30) says that it is a non-degenerate simple zero. This means that there is a small ball B(ξ l , ρ) and a bi-Hölder map γ l from E ∩B(ξ l , ρ) to an open set V of a plane, such that h• γ -1 l is of class C 1 on V and the differential of h • γ -1 l at γ l (ξ l ) is of rank 2. But it was observed near (17.22) that if we take ρ small enough, the restriction of

At the same time, y l is a zero of q t , so by the proof of (17.49), By (17.50) and Example 5.4 in [Do], the degree at 0 of the restriction of q t to W l is ±1. We add up all these degrees and get that the degree of q t (on the sphere) is equal, modulo 2, to the number of zeroes of q t . Since q 0 and q 1 are homotopic, they have the same degree, and q 0 and q 1 have the same number of zeroes modulo 2.

We are now ready to prove (17.43). Since h k-1 and h k coincide out of B(x k , ρ 2 ) and have no zero in E ∩ A, we just need to consider the zeroes in (17.44) and (17.46), and q 0 (ψ -1 k (x)) = p 0 (x) = 0 by 0 by (17.48). So every zero of h k-1 gives a zero for q 0 . The converse is true, by (17.49). Similarly, the number of zeroes for h k in E ∩ B(x k , ρ 1 ) is the same as the number of zeroes for q 1 . This proves (17.43) and, as explained near (17.43), Proposition 16.24 follows.

M S-minimal sets in R 3

In this section we return to sets of dimension 2 in R 3 , and in particular prove Theorem 1.9, which says that every nonempty reduced M S-minimal set in R 3 is a plane, a cone of type Y, or a cone of type T.

We start with a variant of Proposition 16.24 for points of type T , which is valid for (standard) almost-minimal sets, but needs an extra separation assumption. This proposition yields an analogue of Theorem 1.9 for Al-minimal sets, again with an extra separation assumption when E looks like a minimal cone of type T near infinity (see Proposition 18.29).

We can then apply Proposition 18.29 to M S-minimal sets, because the separation assumption is easy to get in this context, and obtain Theorem 1.9.

The results of this section and the previous ones also apply to M S-almost-minimal sets, but to avoid new discussions about definitions, we simply mention this in Remark 18.44 at the end of the section.

We start with a variant of Proposition 16.24 that finds a T -point when the reduced almost-minimal set E is close enough to a set of type T in a ball.

Proposition 18.1. There is constant ε 7 > 0 with the following property. Let E be a reduced A-almost-minimal set of dimension 2 in an open set U ⊂ R 3 , with the gauge function h. Suppose that

by f z,t : B(z, 2t) → R 3 the corresponding biHölder mapping and by Z z,t the associated cone of type Y centered at z. Let ρ be as in (18.13); we can safely assume that ρ < ρ 2 , so f z,ρ and Z z,ρ are defined. Pick three points ζ j ∈ B(z, ρ/2), lying in the three different connected components of R 3 \ Z z,ρ , and set z j = f z,ρ (ζ j ). By the analogue of (18.7) the z j lie in B(z, ρ) \ E, and by (8.13) there is a path γ ⊂ B(z, ρ 2 ) \ E that connects two of the z j .

Since γ may get way out of B(z, ρ), we cannot find a contradiction immediately in B(z, ρ), and we need to zoom out again. But we want to do this carefully, because we want to keep the information that E should separate the three z j because Z z,ρ separates the ζ j .

Let us construct three new paths leaving from the z j . Denote by ζ 0,j the point of ∂B(z, ρ) that lies in the same connected component of R 3 \ Z z,ρ as ζ j , and as far as possible from Z z,ρ . Set l 0,j = [ζ j , ζ 0,j ]. Thus l 0,j does not meet Z z,ρ (the components of R 3 \ Z z,ρ are convex), so g 0,j = f z,ρ (l 0,j ) does not meet E (because it is contained in B(z, 3ρ/2) by the analogue of (18.7), and by the first part of the analogue of (18.10)). Now we iterate, and construct, for k ≥ 1 such that 2 k ρ ≤ 2ρ 2 , line segments l k,j and arcs g k,j , so that in particular (18.16)

Let k ≥ 1 be given, with 2 k ρ ≤ 2ρ 2 . Thus we can use f = f z,2 k ρ and Z = Z z,2 k ρ . Denote by z k-1,j the final endpoint of g k-1,j . If (18.16) holds for k -1 (which, by the way, is the case

) is defined and lies in B(z, 2 k ρ) \ Z, by the analogues for f of (18.7) and (18.10). Choose

and as far as possible from Z. As before (for l 0,j ), l k,j does not meet Z and is contained in B(z, 2 k ρ). Finally set g k,j = f (l k,j ); then (18.16) holds, with the same proof as for g 0,j . Thus we can construct the g k,j by induction.

Next we check that for each k, the three ζ k,j lie in different components of R 3 \ Z z,2 k ρ . This is true for k = 0, because the three ζ j lie in different components of R 3 \ Z z,ρ , and ζ 0,j lies in the same component as ζ j . Now suppose that k ≥ 1 and that this is true for k

ρ, by (18.7) for f and f z,2 k-1 ρ . In addition, Z is quite close to Z z,2 k-1 ρ , because they are both close to E in B(z, 2 k-1 ρ) (use the analogue of (18.10)). The three ζ k-1,j lie far from Z z,2 k-1 ρ (by construction), and in different components of R 3 \ Z z,2 k-1 ρ (by induction assumption). Then the ζ ′ k-1,j also lie far from Z, and in different components of R 3 \ Z. This is also true for the ζ k,j , because ζ k,j lies in the same component as ζ ′ k-1,j . This is what we wanted. Let m denote the largest exponent k such that 2 k ρ ≤ 2ρ 2 ; thus 2 m ρ > ρ 2 . We just proved that the three ζ m,j lie in different components of R 3 \ Z z,2 m ρ . Denote by g j the path from z j to z j,m obtained by concatenation of all the g j,k , 0 ≤ k ≤ m. Then g j ⊂ B(z, 3•2 m-1 ρ)\E, by (18.16). We also have a path γ ⊂ B(z, ρ 2 )\E that connects two of the z j , and when we add to it the two corresponding g j , we get a path γ ⊂ B(z, 3 • 2 m-1 ρ) \ E that connects two z j,m . Set Γ = f -1 z,2 m ρ ( γ). This path does not meet Z z,2 m ρ , by (18.10) and the usual argument. Yet it connects two different ζ m,j , a contradiction. This completes our proof of the first inclusion in (18.12). far from T and Y y , and on different sides of Y y (by (18.24)), and this stays true for the c i (by (18.25)); this proves (18.28).

Recall that w j ∈ ∂B(0, 1)∩γ j , and γ j = f j (g j ), where g j = ℓ j ∩B(y j , 10 -3 ) (see below (18.18)). Let ζ j ∈ g j be such that f j (ζ j ) = w j . Then |ζ j -1| ≤ τ 10 -3 , by (18.25). The distance from ζ j to the spine of T (and here this means the half line through z j ) is less than C(τ + ε 7 ), because ζ j ∈ g j and Y j is quite close to T near y j . Then |ζ jz j | ≤ C(τ + ε 7 ) (because |z j | = |-a j | = 1), so ζ j and w j both lie in B(y j , 10 -4 ) (recall that |y j -z j | ≤ 2ε 7 ).

For each i = j, c i ∈ B(y j , 10 -4 ) \ Y j and ζ j ∈ B(y j , 10 -4 ) ∩ ℓ j ; then I = [c i , ζ j ) ⊂ B(y j , 10 -4 )\Y j and f j (I) ⊂ B(0, 2)\E, by (18.27). Hence

Since f j (I) gets as close to w j = f (ζ j ) as we want, V (a i ) is one of the elements of A(w j ).

Conversely, let V be one of the elements of A(w j ), and pick x ∈ V ∩B(w j , 10 -4 ). Then f -1 j (x) is defined by (18.26), lies in B(y j , 10 -3 ) by ( 18.25) and because w j ∈ B(y j , 10 -4 ), and lies out of Y j by (18.27). Thus f -1 j (x) lies in some W i , by (18.28). Then J = [f -1 j (x), c i ] lies in W i by convexity of W i , and f (J) does not meet E by (18.27). Its extremities x and b

This completes our proof of Lemma 18.22.

We are ready to conclude. By assumption, at least three of the four V (a i ) are different. Let us assume that these are the V (a i ), 1 ≤ i ≤ 3. By (18.21), we can connect w 4 ∈ γ 4 to some other w k by a path in E Y ∩ B(0, 1); without loss of generality, we can assume that k = 3. Since A(x) is locally constant on E Y , A(w 4 ) = A(w 3 ) or, by Lemma 18.22,

Let us also apply (18.21) with j = 2; we get that A(w 2 ) = A(w l ) for some l = 2. Since V (a 2 ) ∈ A(w l ), it lies in A(w 2 ) as well, so it is equal to V (a 1 ) or V (a 3 ) (we already know that V (a 4 ) = V (a 3 )). This contradicts our assumption that three V (a j ) are different, and completes the proof of Proposition 18.1.

Next we prove a variant of Theorem 1.9 for Al-minimal sets, but with an additional assumption that will allow us to apply Proposition 18.1. Theorem 1.9 will then follow rather easily.

Proposition 18.29. Let E be a reduced Al-minimal set of dimension 2 in R 3 (as in Definition 1.6). If there is a minimal cone T of type T, centered at the origin, and a sequence {t k } such that (18.30) lim k→+∞ t k = +∞ and lim k→+∞ d 0,t k (E, T ) = 0, assume in addition that there is a subsequence {t k l } for which the following separation assumption holds. Denote by a i , 1 ≤ i ≤ 4, the four points of ∂B(0, t k l /2) that lie the furthest from T . Thus the -a i are the intersections of ∂B(0, t k l /2) with the spine of T . Denote by V (a i ) the connected component of a i in B(0, t k l ) \ E k l . Then (assume that) (18.31) at least three of the four components V (a i ) are distinct.

We claim (but we shall not check) that the following things are true. First, E * is a M S-almost-minimal set in U ⊂ R 3 , with the same gauge function as E. See "Coral pairs" on pages 58-60 of [D2], but the main point is our proof of (18.41).

Next, all the results from the previous sections apply to M S-almost-minimal sets. We cannot say directly that this is because all M S-almost-minimal sets are A ′ -almostminimal, even though we can localize the proof of Remark 1.8. Indeed the Al-competitors F = ϕ 1 (E), where the family {ϕ t } satisfies (2.2)-(2.6), are not necessarily M S-competitors as above, because even though W in (2.6) is compact and contained in U , it is not sure that it is contained in a compact ball B(x, r) ⊂ U . But we claim that this is not a problem, that all sets W used in and for this paper are contained in compact balls B(x, r) ⊂ U with a diameter that is not too large.

Finally, we can even remove the separation assumption in Proposition 18.29 when we deal with M S-almost-minimal sets, by the final argument in our proof of Theorem 1.9.

19.

A set E ⊂ R 3 that looks like a cone of type T but has no T -point

In this section we try to describe a closed set E ⊂ R 3 that coincides outside of the unit ball with a set of type T centered at the origin, but is locally C 1 -equivalent to a plane or a set of type Y near each of its points. This set is probably never minimal, so it is only a counterexample to what would be a wrong way to prove an analogue of Theorem 1.9 for Al-minimal sets. It is quite possible that this set is the same as the one depicted in Figure 1.1.1 in [LM] and Figure 3 of [Mo2], following a suggestion of R. Hardt.

Even though we shall try to give a reasonable description, it is probable that the best way for the reader to convince herself that nothing goes wrong with the topology is to construct her own paper model with scissors and glue.

So we start from the minimal set T centered at the origin, and call L its spine. Thus L is composed of four half lines L j , 1 ≤ j ≤ 4. Call x j the point of L j ∩ ∂B(0, 1). We first construct two faces of E and the set E Y of Y -points of E.

Call P 1 the plane that contains L 1 ∪ L 2 , and then pick a nice smooth curve Γ 1 ⊂ P 1 that has L 1 \B(0, 1/2) and L 2 \B(0, 1/2) as its ends. The simplest is to add to L 1 \B(0, 1/2) and L 2 \ B(0, 1/2) the arc of circle that goes from L 1 ∩ ∂B(0, 1/2) to L 2 ∩ ∂B(0, 1/2) and is tangent to each L i at the point of L i ∩ ∂B(0, 1/2). But the reader could take a smoother, equivalent curve. Call F 1 the convex component of P 1 \ Γ 1 ; this will be one of the faces of E; it is just a little smaller than the corresponding face of T .

Similarly, call P 4 the plane that contains L 3 ∪ L 4 , and pick a nice curve Γ 4 ⊂ P 4 that contains L 3 \ B(0, 1/2) and L 4 \ B(0, 1/2). Also call F 4 the convex component of P 4 \ Γ 4 .

At this point, we have two faces of E (F 1 and F 4 ), and we also decide that E Y = Γ 1 ∪Γ 4 . Observe that here Γ 1 and Γ 4 are neither knotted nor braided; this will make our discussion and the construction of paper models simplest, but the case when the Γ i are knotted and linked could be more interesting , because then it could be harder to prove that E is not minimal.

Before we continue with the construction, we should observe that when we do all the construction below, we shall be able to preserve the two symmetries, with respect to the planes P 1 and P 4 respectively.