Emmanuel Lesigne

Anthony Quas

Joseph M Rosenblatt

Maté Wierdl

Generation of measures on the torus with good sequences of integers

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Generation of measures on the torus with good sequences of integers E. Lesigne, A. Quas, J. Rosenblatt, M. Wierdl November 12, 2023 Let S := (s 1 < s 2 < . . . ) be a strictly increasing sequence of positive integers and denote e(β) := e 2πiβ . We say S is good if for every real α the limit lim N 1 N ∑ n≤N e(s n α) exists. By the Riesz representation theorem, a sequence S is good iff for every real α the sequence (s n α) possesses an asymp- totic distribution modulo 1. Another characterization of a good sequence follows from the spectral theorem: the sequence S is good iff in any probability measure preserving system (X, m, T) the limit lim N 1 N ∑ n≤N f (T s n x) exists in L 2 -norm for f ∈ L 2 (X).

Of these three characterization of a good set, the one about limit measures is the most suitable for us, and we are interested in finding out what the limit measure µ S,α := lim N 1 N ∑ n≤N δ s n α on the torus can be. In this first paper on the subject, we investigate the case of a single irrational α. We show that if S is a good set then for every irrational α the limit measure µ S,α must be a continuous Borel probability measure. Using random methods, we show that the limit measure µ S,α can be any measure which is absolutely continuous with respect to the Haar-Lebesgue probability measure on the torus. On the other hand, if ν is the uniform probability measure supported on the Cantor set, there are some irrational α so that for no good sequence S can we have the limit measure µ S,α equal ν. We leave open the question whether for any continuous Borel probability measure ν on the torus there is an irrational α and a good sequence S so that µ S,α = ν. Representing References 43

Introduction, main results

Throughout the paper we will use the arithmetic average operator A: for a finite index set S, a vector space V and a S → V function f we define A S f (s)

A S f (s) = A s∈S f (s) := 1 #S ∑ s∈S f (s) (2.1)
where #S denotes the number of elements in S.

We use the convention that if an interval appears as an index set in a summation then we consider only the integers in the interval. For example, ∑ n∈[0,N) a n = ∑ n∈{0,1,...,N-1} a n .

We also use Weyl's notation e(β) := e 2πiβ . Note that e p (β) = e(pβ) for every integer p.

We denote by T the torus R/Z and we represent it as the unit closed interval [0, 1] with 0 = 1.

Good sequences, main question

Definition ▶ Good sequence

We say that a sequence S = (s n ) n∈N of integers is good if the limit lim N A n∈[1,N] e(s n α) exists for every real number α.

Good sequences have been extensively studied in many parts of mathematics, such as in number theory and ergodic theory.

In this paper we restrict our attention to strictly increasing sequences S of positive integers in which case we can and will consider S as a subset of N, and we'll use the concept of good sequence and good set interchangeably.

Among the wellknown good sequences are the full set N of positive integers 1 , the sequence (n 2 ) n∈N of squares 2 and the sequence (p n ) n∈N of 1 Weyl 1916. 2 Weyl 1916. primes 3 where p n denotes the nth prime number. For these sequences the 3 Vinogradow 1937. limits lim N A n∈[1,N] e(s n α) are as follows

lim N A n∈[0,N) e(nα) = 1 if α = 1 0 if α ̸ = 0 lim N A n∈[0,N) e(n 2 α) =      A b∈[1,q] e b 2 a q if α = a q , gcd(a, q) = 1 0 if α is irrational lim N A n∈[1,N] e(p n α) =      A b∈[1,q]
gcd(b,q)=1 e(b/q) if α = a q , gcd(a, q) = 1

0 if α is irrational (3.1)
In case of a good sequence S = (s n ) and a fixed α, the existence of lim N A n∈[1,N] e(s n pα) for every p ∈ Z implies, by uniform approximation of a continuous T → C function by trigonometric polynomials, that for every continuous T → C function ϕ the limit lim N A n∈[1,N] ϕ(s n α) exists. By the Riesz representation theorem, this implies that the weak limit lim N A n∈[1,N] δ s n α of discrete measures A n∈[1,N] δ s n α on T exists.

By this argument, the existence of lim N A n∈[1,N] e(s n α) for every α implies the existence of the limit measure lim N A n∈[1,N] δ s n α for every α. Denote the Haar-Lebesgue probability measure on the torus T by λ and recall that the Fourier coefficients λ(e p ) of λ satisfy λ(e p ) =

1 for p = 0 0 for p ∈ Z, p ̸ = 0 (3.2)
where for a given measure ν and ν-integrable function ϕ, we use 4 the 4 and will use troughout the paper functional notation ν(ϕ) for the integral of ϕ with respect to ν, ν(ϕ) = ϕ dν (3.3)

For our three good sets the limit measures are as follows.

lim

N A n∈[1,N] δ nα = A b∈[1,q] δ b/q if α = a/q, gcd(a, q) = 1 λ if α is irrational lim N A n∈[1,N] δ n 2 α =    A b∈[1,q] δ b 2 a q if α = a q , gcd(a, q) = 1 λ if α is irrational lim N A n∈[1,N] δ p n α =     
A b∈ [1,q] gcd(b,q)=1 δ b/q if α = a q , gcd(a, q) = 1

λ if α is irrational (4.1)
What we see in these three examples is that in case of irrational α the limit measure is the Haar-Lebesgue measure λ and in case of rational α = a/q, gcd(a, q) = 1, the limit measure is supported on a subset of the qth roots of unity and appears to be quite uniform on its support. In case of irrational α, the simplest question is if it's possible that the limit measure is not λ. In case of rational α, we can ask if the limit measure always has to show some kind of uniformity.

Let us consider a good sequence S = (s n ). The existence of the limit lim N A n∈[1,N] e(s n α) for every α implies that the weak limit lim

N A n∈[1,N] δ s n α of discrete measures A n∈[1,N] δ s n α
on T exists for every α. Let us denote this weak limit measure by µ S,α ,

µ S,α := lim N A n∈[1,N] δ s n α (4.2)
The main question we want to investigate in this paper is

Question ▶ Main question

What can the limit measure µ S,α be? Can it be any Borel probability measure on T?

Main results

As we stated earlier, we try to answer question 4.1 for strictly increasing sequences, and unless we say otherwise, we assume from now on that S = (s n ) is a strictly increasing sequence of positive integers which we often consider as a subset of N.

Our first observation is that the answer to question 4.1 will depend on α. If α is a rational number, say, α = a q with gcd(a, q) = 1, then the limit measure is clearly supported on the set

T q := { b/q : b ∈ [1, q] } (4.3)
of qth roots of unity. So the question is if the limit measure µ S,a/q can be any probability measure supported on T q ? The answer is yes. First a terminology.

Definition ▶ Representable measure at α

Let S be a good set, and let ν be a nonzero, finite Borel measure on T.

We say that S represents ν at α ∈ T if µ S,α = 1 ν(T) ν. We say ν is representable at α if there is a good set which represents ν at α.

5.2

Theorem ▶ Every probability measure on T q can be represented Let q and a be positive integers with gcd(a, q) = 1, and let ν be a probability measure supported on the set T q of qth roots of unity. Then ν can be represented at a q , that is, there is a good set S so that µ S, a q = ν.

Before discussing the limit measure µ S,α for irrational α, let us note the following fact which will help us appreciate the concept of a good set.

Suppose we are given an irrational number α ∈ T and a Borel probability measure ν on T. We claim that there exists a sequence (x n ) in T with asymptotic distribution ν, i. e. such that lim

N A n∈[1,N] δ x n = ν.
Considering such a sequence and using the density of the sequence (nα) n in T, we can select a strictly increasing sequence (s n ) of integers so that lim n (s n αx n ) = 0 mod 1, and we have lim

N A n∈[1,N] δ s n α = ν. Taking S = { s n : n ∈ N }, we could say that µ S,α = ν, but nothing insures us that the set S is good.
There are different ways to prove the preceding claim. For example we can pick the numbers x n randomly and independently with law ν, and the strong law of large numbers asserts that the sequence (x n ) has, almost surely, the right asymptotic distribution.

It is particularly simple to get a point-mass as a limit measure. For example, to get the Dirac measure at 1/2, so ν = δ 1/2 , take a strictly increasing sequence (s n ) of natural numbers so that s n α converges to 1/2 mod 1, and let S := { s n : n ∈ N }. In contrast to this example, for good sets we have a dramatic departure from the case of rational α.

Theorem ▶ µ S,α is continuous for irrational α

Only continuous measures can be represented at an irrational number.

To spell this out, let S = (s n ) be a good sequence and α be an irrational number.

Then the limit Borel probability measure µ

S,α = lim N A n∈[1,N] δ s n α is a continuous measure.
The obvious question in turn is if any given continuous Borel probability measure can be represented at any irrational number. The answer is no, as the next result shows.

Theorem ▶ Some continuous measures cannot be represented at every irrational point

Let ν be a Borel probability measure on T so that its Fourier coefficients do not converge to 0, so

lim sup p→∞ |µ(e p )| > 0 (6.1)
Then there is a set A ⊂ T of full Lebesgue measure so that ν cannot be represented at any α ∈ A.

Since a measure ν is called a Rajchman measure 5 if its Fourier coeffi- 5 Lyons 1995.

cients vanish at infinity, that is, lim p ν(e p ) = 0, we can rephrase theorem 6.1 by saying that if ν is representable at every irrational α then it must be a Rajchman measure. A well known non-Rajchman continuous measure is the uniform measure on the triadic Cantor set. While theorem 6.1 doesn't exclude the possibility that A = T, that is, a non-Rajchman measure cannot be represented anywhere, Christophe Cuny and François Parreau 6 constructed a non-Rajchman measure which 6 Parreau and Cuny 2022. is representable at uncountably many α's. Nevertheless, the following question remains open.

Question ▶ Is every continuous measure representable somewhere?

Let ν be a continuous Borel probability measure on T.

Is there an irrational α so that ν is representable at α?

The next result says that if ν is absolutely continuous with respect to the Lebesgue probability measure λ on the torus T, then it can be represented at every irrational α.

Theorem ▶ Absolutely continuous measures are representable at every irrational point

Let ν be a Borel probability measure on T which is absolutely continuous with respect to the Lebesgue probability measure on T. Let α be an irrational number.

Then ν is representable at α.

Our proof of theorem 6.3 is flexible and enables us to show a more general result, namely it turns out that a given absolutely continuous measure can be represented by a good subset of any given good set, provided it doesn't increase too fast, it is sublacunary. For a given set R ⊂

N let R(N) denote the Nth initial segment of R, R(N) := R ∩ [1, N] (6.2)
We say R is sublacunary 7 if it satisfies the growth condition We will see that the proof of theorem 7.1 reveals a close connection between the Radon-Nikodym derivative ρ of ν with respect to µ R,α and the relative mean 8 of the set S representing ν. For a given R ⊂ N and 8 The usual terminology is relative density instead of relative mean, but we will use the more general concept of the mean of a R → C function in section 1.3 and we prefer to use a single terminology and notation for economical reasons.

S ⊂ R, the relative mean M R (S) of S in R is defined by M R (S) := lim N #S(N) #R(N) (7.3)
provided the limit on the right exists. The relative upper mean M R (S) of S in R is defined by

M R (S) := lim sup N #S(N) #R(N) (7.4)
In case R = N, we suppress the base set in our notation, and we write M(S) for M N (S) and M(S) for M N (S).

∈ L 1 (µ R,α ) with µ R,α (ρ) = 1 be bounded so ∥ρ∥ L ∞ (µ R,α ) < ∞.
Then there is a good set S ⊂ R representing the measure ρ • µ R,α at

α and satisfying M R (S) = 1 ∥ρ∥ L ∞ ( µ R,α ) .
b) Let S be a good subset of R with positive upper density in R, so M R (S) > 0.

Then for every irrational β the limit measure µ S,β is absolutely continuous with respect to µ R,β . Furthermore, the Radon-Nikodym S) .

derivative ρ β := dµ S,β dµ R,β is a bounded function satisfying ∥ρ β ∥ L ∞ (µR,β) ≤ 1 M R (
We see that theorem 7.1 gives a full characterization of the limit measure for sets with positive upper mean 9 , giving an exact relationship between 9 so now R = N the upper mean of the set and the bound of the RN derivative: On the one hand if M(S) > 0, the limit measure µ S,β for every β must be absolutely continuous with respect to λ with bounded RN derivative ρ β satisfying ∥ρ β ∥ L ∞ (λ) ≤ 1 M(S) . On the other hand, any Borel probability measure ν which is absolutely continuous with respect to λ with bounded, nonzero RN derivative ρ is representable at any irrational α with a set of positive mean satisfying M(S) =

1 ∥ρ∥ L ∞ (λ)
. Theorem 7.2 (b) has the following consequence.

Corollary

▶ If the RN derivative ρ is unbounded, then M R (S) = 0
Let R be a good set and α an irrational number. Suppose the unsigned function ρ ∈ L 1 (µ R,α ) with µ R,α (ρ) = 1 is unbounded, and that the good set S ⊂ R represents the measure ρ • µ R,α at α. Then S must have 0 mean in R, so M R (S) = 0.

Weighted averages

Our results in theorems 7.1 and 6.3 will be consequences, via a random procedure, of results on weighted averages. We need to fix some terminology and notation. We define the Besicovitch type seminorm ∥∥ 1 for all complex valued sequences f ∈

C N by ∥ f ∥ 1 := lim sup N A [1,N] | f |, f ∈ C N (8.1)
The number 1 in the subscript of ∥∥ 1 expresses the similarity of this norm to the L 1 norm. For a set S ⊂ N, we may use the notation ∥S∥ 1 instead of ∥1 S ∥ 1 , though in this case we do not get a new concept, since ∥S∥ 1 = M(S).

For an infinite set R ⊂ N we define the relative

1-norm ∥ f ∥ 1,R of a complex valued R → C function by ∥ f ∥ 1,R := lim sup N A R(N) | f |, f ∈ C R (8.2)
If the set R is given as a strictly increasing sequence (r n ) and for an f ∈

C R we define F by F(n) := f (r n ), then ∥ f ∥ 1,R = ∥F∥ 1 .
Let R ⊂ N be an infinite set. The R → R function w is called a Rweight if w is unsigned, so w ≥ 0, and ∑ r∈R w(r) = ∞. We may refer to an R-weight as "a weight supported on R".

An R-weight w can be considered a measure on the set R and in that case for S ⊂ R we may briefly write w(S) for the sum ∑ s∈S w(s).

For a finite set S ⊂ N let σ be a real valued, unsigned function defined on S. We can consider σ a measure on S, and as such, we assume σ(S) > 0. For a vector space V and S → V function f , define the σ-weighted average A σ S f of f on S by

A σ S f = A σ s∈S f (s) := 1 σ(S) ∑ s∈S σ(s) f (s) (9.1)
9.1 Definition ▶ Good weights and represented measures by them Let R ⊂ N be infinite. Let w be an R-weight.

We say w is a good R-weight if the weak limit lim N A w r∈R(N) δ rβ exists for every β ∈ T. We denote this limit by µ w,β ,

µ w,β := lim N A w r∈R(N) δ rβ (9.2)
Let ν be a Borel probability measure on T and let α ∈ T.

We say the R-weight w represents ν at α if w is good and µ w,α = ν.

Note the following form of the definition of the limit measure µ w,α when we consider R as the strictly increasing sequence (r n ):

µ w,α = lim N A w n∈[1,N] δ rn β , so now we have A w n∈[1,N] δ rn β = 1 ∑ n∈[1,N] w(rn ) ∑ n∈[1,N] w(r n )δ rn α .
Note the following characterization of good weights: The R-weight w is good iff the limit lim N A w r∈R(N) e(rα) exists for every α. In the special case of a good set S ⊂ N, we have µ S,α = µ 1 S ,α since the weighted averages with weight w := 1 S correspond to the averages along S.

In contrast to good sets, the representation of absolutely continuous measures by weights can always be accomplished by weights with positive, finite mean. In fact, the representing weight has an additional property.

Definition ▶ Integrable weight

Let R ⊂ N be infinite.

We call the R-weight w integrable if it can be approximated arbitrary closely in the seminorm ∥∥ 1,R by bounded, good weights: for every ϵ > 0 there is a good R-weight v with ∥v∥ ∞ < ∞ so that ∥v -w∥ 1,R < ϵ.

Theorem ▶ Representation by weights

Let R be a good set.

a) For an irrational α let the unsigned function

ρ ∈ L 1 (µ R,α ) satisfy µ R,α (ρ) = 1. Then there is an integrable R-weight w with M R (w) = 1 which represents the measure ρ • µ R,α at α. If ρ ∈ L ∞ (µ R,α ) then the R-weight w representing the measure ρ • µ R,α can also satisfy ∥ρ∥ L ∞ (µ R,α ) = ∥w∥ ∞ .
b) Let w be a good, integrable R-weight which satisfies ∥w∥ 1,R > 0.

Then for every β the limit measure µ w,β is absolutely continuous with respect to µ R,β .

Applications in ergodic theory

Besides the intrinsic interest of our main question, question 4.1, there may be several applications of studying limit measures. One major application is in ergodic theory.

Recall that a measure preserving dynamical system is a probability space (X, m), where m(X) = 1, equipped with a measurable, measure preserving transformation T of X. By the spectral theorem, a good set has the following characterization: the sequence S = (s n ) of positive integers is good iff the limit lim

N A n∈[1,N] f (T s n x) exists in L 2 (X)-norm in any measure preserving dynamical system (X, m, T) for any f ∈ L 2 (X).
This means that our work in describing the possible limit measures in case of a good set yields an identification of the limit in mean ergodic theorems. Identification of the limit is often the crucial step in some applications, and here we just mention two of these, recurrence and almost sure convergence. In case of studying recurrence, the identification of the limit readily tells us whether a given set is a set of recurrence. In case of trying to see if some ergodic averages converge almost everywhere, after the identification of the L 2 -limit, we usually want to see if there is some kind of rate with which the averages converge to the L 2 -limit. For example, this is the case when one proves that the ergodic averages along the squares converge almost surely. The application of the circle method here is exactly a quantitative expression of how the averages converge in L 2 -norm.

Future work

The techniques developed in this paper allow one to address the simultaneous representability of probability measures at several different points of the torus, and we plan to explore this in a future work. But which family { ν α : α ∈ T } of measures can be represented by a single good set remains open even if we restrict the family to absolutely continuous measures with respect to the Lebesgue probability measure λ. What we can say at this point is that for a given good set S, the set of α ∈ T where the limit measure µ S,α is not the Lebesgue measure is small: it is both of first Baire category and of 0 measure under every Rajchman measure 10 on 10 Lyons 1985, Theorem 3; see also Lyons 1995. T, so ν{ α : µ S,α ̸ = λ } = 0 for every Rajchman measure ν.

GENERATION OF MEASURES ON THE TORUS WITH GOOD SEQUENCES OF INTEGERS 11

Summary of notation

We realize that we use quite extensive notation, many of which are new, so we give a summary of our notations in table 1.

Symbol

Definition Parameters Name

N {1, 2, 3, . . . } Natural numbers T torus λ Haar-Lebesgue measure on T e(θ) exp(2πiθ) θ ∈ T e p (θ) e(pθ) p ∈ Z S(N) S ∩ [1, N] S ⊂ N initial segment of S #S(N) ∑ s∈S(N) 1 S ⊂ N counting function of S A S f 1 #S ∑ s∈S f (s) set S is finite average of f on S A w S f 1 w(S) ∑ s∈S w(s) f (s) w is a weight on S w-average of f on set S µ S,α lim N A s∈S(N) δ sα S ⊂ N, α ∈ T limit measure of S at α µ w,α lim N A w s∈S(N) δ sα weight w on S, α ∈ T limit measure of w at α ν(ϕ) T ϕ dν M( f ) lim N A [1,N] f f ∈ C N mean of f M R ( f ) lim N A R(N) f R ⊂ N, f ∈ C R relative mean of f M { f : f ∈ C N , M( f ) exists and is finite } sequences with mean M( f ) lim sup N A [1,N] f f ∈ C N upper mean M R ( f ) lim sup N A R(N) f R ⊂ N, f ∈ C R relative upper mean ∥ f ∥ 1 lim sup N A [1,N] | f | f ∈ C N 1-seminorm ∥ f ∥ 1,R lim sup N A R(N) | f | R ⊂ N, f ∈ C R relative 1-seminorm C + { ϕ : ϕ : T → [0, 1], continuous } ∥ν 1 -ν 2 ∥ V sup ϕ∈C + |ν 1 (ϕ) -ν 2 (ϕ)| ν i finite Borel measures on T variation distance

Basic example for representation

In this section we want to work out a rather simple but instructive example, which will then motivate and form the basis of many of our constructions later on. When we are done with presenting this example, we in fact proved theorem 7.1 in case the Radon-Nikodym derivative is the indicator of a Jordan measurable set.

Let α be irrational and let I ⊂ T be an interval. We want to show that if a probability measure ν is absolutely continuous with respect to λ with the Radon-Nikodym derivative equal 1 I , the indicator of I, then there is a set S which represents ν at α. Probably the simplest way 11 to define such a 11 We could also define such a set by taking { n : n ∈ N, n 2 α ∈ I (mod 1) } or { p : p ∈ P, pα ∈ I (mod 1) } where P is the set of primes. set S is by taking

S = { n : n ∈ N, nα ∈ I } (11.1)
There are two things to verify. First, that S is indeed a good set, and to do that, we need to show that the weak limit µ S,β = lim N A s∈S(N) δ sβ exists for every β. Second, we then have to verify that µ S,α = 1 λ(I) 1 I • λ. The second one, in fact, is almost instantaneous to do since it follows from the uniform distribution of (nα) n∈N (mod 1). To see how it follows, it's enough to show that for every interval J ⊂ T we have µ S,α (J) =

λ 1 J • 1 λ(I) 1 I , that is lim N A s∈S(N) 1 J (sα) = 1 λ(I) λ(J ∩ I) (12.1)
The left hand side can be written as

lim N A s∈S(N) 1 J (sα) = lim N N #S(N) A n∈[1,N] 1 I (nα)1 J (nα) since lim N #S(N) N = λ(I) by the uniform distribution of (nα) n∈N (mod 1), = 1 λ(I) lim N A n∈[1,N] 1 I∩J (nα)
again by the unifom distribution of (nα) n∈N (mod 1)

= 1 λ(I) λ(I ∩ J)
To show that the weak limit µ S,β = lim N A s∈S(N) δ sβ exists for every β, it's enough to show that lim N A s∈S(N) e(sβ) exists for every β. Since

A s∈S(N) e(sβ) = N #S(N) A n∈[1,N] 1 I (nα) e(nβ) (12.2)
and since lim N #S(N) N

= λ(I), it's enough to show that the limit lim N A n∈[1,N] 1 I (nα) e(nβ) exists for every β ∈ T. To see this, first note that if we replace 1 I by the character e k the limit of

A n∈[1,N] e k (nα) e(nβ) = A n∈[1,N] e(n(kα + β))
as N → ∞ exists and is as follows

lim N A n∈[1,N] e k (nα) e(nβ) = 1 if β = -kα (mod 1) 0 otherwise (12.3)
From this we get that if we replace 1 I by a trigonometric polynomial ϕ, the limit of A n∈[1,N] ϕ(nα) e(nβ) exists and can be given explicitly as 12

12 Notice that in eq. ( 12.4) λ ϕ e k is the kth Fourier coefficient of ϕ.

lim N A n∈[1,N] ϕ(nα) e(nβ) =    λ ϕ e k if β = -kα (mod 1) 0 otherwise (12.4)
Using Weierstrass' theorem on being able to uniformly approximate a continuous function by trigonometric polynomials, we can verify that in eq. ( 12.4) we can take ϕ to be any continuous function. Now, to go from continuous functions to the indicator 1 I of any interval I, it is enough to know that the indicator 1 I can be sandwiched between two unsigned continuous functions whose integrals (with respect to λ) are arbitrarily close. We thus have

lim N A n∈[1,N] 1 I (nα) e(nβ) = λ(1 I e k ) if β = -kα (mod 1) 0 otherwise. (13.1)
We finally get, since µ S,β (e) = 1

λ(I) lim N A n∈[1,N] 1 I (nα) e(nβ), µ S,β (e) =    1 λ(I) λ(1 I e k ) if β = -kα (mod 1) 0 otherwise (13.2)
The above shows that µ S,β (e) can be nonzero only if β is an integer multiple of α, and we recognize λ 1 I e k as the kth Fourier coefficient of the function 1 I , that is, 1 λ(I) λ 1 I e k is the kth Fourier coefficient of the measure 1 λ(I) 1 I λ. One can rather easily extend this example in two ways. First, the proof can be repeated almost verbatim for the case when we take any Jordan measurable set B in place of the interval I. Indeed, all we need to remark is that a set B is Jordan measurable iff, for every given ϵ > 0, its indicator function 1 B can be sandwiched between two unsigned, continuous functions ϕ a and ϕ b so that ϕ b ≤ 1 B ≤ ϕ a and λ(ϕ aϕ b ) < ϵ. Another way of expressing that the indicator of a set can be sandwiched between two continuous functions is that the boundary of the set has zero Lebesgue measure.

Definition ▶ ν-Riemann integrability

Let ν be a finite Borel measure on T and let ϕ be a Borel measurable T → C function.

We call the function ϕ ν-Riemann integrable if it's continuous at ν-almost every point. We call the Borel measurable set B ν-Jordan measurable if its indicator function

1 B is ν-Riemann integrable.
As it is well known, the equivalence of approximability by continuous functions and the boundary having zero measure carries over to the setting of any finite Borel measure on the torus. We can thus extend the example to the setting when the Lebesgue measure is replaced by an arbitrary finite Borel measure.

We record our findings in the following result.

Proposition ▶ The Radon-Nikodym derivative can be the indicator of a Jordan measurable set

Let R be a good set, α be an irrational number and let B ⊂ T be µ R,α -Jordan measurable with µ R,α (B) > 0.

Then the measure 1 B µ R,α , which is absolutely continuous with respect to µ R,α , can be represented at α by the good set S defined by

S := { r : r ∈ R, rα ∈ B } (14.1) so we have µ S,α = 1 µ R,α (B) 1 B µ R,α . We also have µ R,α (B) = M R (S).
Let us go back to trying to represent measures which are absolutely continuous with respect to the Lebesgue measure λ. New ideas are needed to cover the case when we want to represent the measure 1 B λ when B is a Borel set which is not Jordan measurable. What is the new difficulty? We'd like to think that we could just again take the "visit set" S = { n : n ∈ N, nα ∈ B }, but this is not the case anymore. Indeed, take B to be an open set with λ(B) < 1 and containing all integer multiples of our irrational α. This open set is not Jordan measurable anymore. The set S cannot represent the measure 1 B λ anymore since S = N. In fact, we show in section 10.3 that for any given irrational α, one can construct an open set B so that the visit set of B doesn't even have mean. So we definitely need new ideas.

We also need new ideas even for the case when we try to represent a measure which is absolutely continuous with respect to the Lebesgue measure with a Radon-Nikodym derivative which is not an indicator function. We need these new ideas even if this Radon-Nikodym derivative is a continuous function.

Proof of theorem 7.1 for indicators

Strictly speaking, we have already begun the proof of theorem 7.1 in the previous section, when we proved that at an irrational number every measure with Jordan measurable Radon-Nikodym derivative can be represented. Our fixed set up in this section is that we are given a good "base" set R ⊂ N and an irrational number α. Since the set R is fixed throughout the section, we suppress the set R from our notation for the limit measure,

µ β := µ R,β , for every β (14.2)
Since our focus is to widen the class of the Radon-Nikodym derivatives with respect to the base limit measure µ α , the following definition will simplify our language.

Definition ▶ Representing a function, a Borel set

Let ρ ∈ L 1 (T, µ α ) be unsigned and µ α (ρ) > 0. We say that the good set S ⊂ R represents ρ at α if it represents the measure ρ • µ α , that is, µ S,α = 1 µ α (ρ) ρ • µ α . If ρ is the indicator of a Borel measurable set B ⊂ T, we then say S represents B at α.

The sets S ⊂ R we consider in this section have positive mean in R. For such a set, the non-normalized averages A n∈[1,N] 1 S (r n )δ r n β are easier to handle than the normalized ones A s∈S(N) δ sβ . The convergence or divergence properties of the two averages are identical since they are connected by

lim N A n∈[1,N] 1 S (r n )δ r n β = M R (S) lim N A s∈S(N) δ sβ (15.1)
as can be seen from writing

A s∈S(N) δ sβ = #R(N) #S(N) A r∈R(N) 1 S (r)δ rβ and noting that lim N #S(N) #R(N) = M R (S) and lim N A r∈R(N) 1 S (r)δ rβ = lim N A n∈[1,N] 1 S (r n )δ r n β .
In section 2 we proved that if B is µ α -Jordan measurable, then it can be represented by the set S B defined by

S B = { r n : r n α ∈ B } (15.2)
and we have the relation

M R (S B ) = µ α (B) (15.3)
We also indicated that this definition of S B may not give a good set if B is not Jordan measurable. The idea of extending the representation to any Borel measurable set is via a limit procedure.

To explain what we mean by "a limit procedure", consider the case when B is an open set, and write it as a disjoint union of open intervals, B = ∪ j I j . Defining B k := ∪ j∈[1,k] I j for every k ∈ N, each B k is Jordan measurable and the sequence (B k ) increases monotonically to B. We have

lim k µ α (B k ) = µ α (B). Denoting S k := S B k , the sequence (S k ) also increases to a set S ⊂ R, but M R (S) not only may not be equal lim k M R (S k ) but M R (S)
may not even exist 13 . 13 See also section 10.3.

The limit procedure which is suitable for our purposes is determined by the seminorm ∥ f ∥ 1 which is defined by

∥ f ∥ 1 := lim sup N A [1,N] | f (n)|, f ∈ C N (15.4)
Our main tools will be two lemmas. The first one is modeled after a result of Marcinkiewicz 14 on the completeness of Besicovitch spaces. 14 Marcinkiewicz 1939. 16.1 Lemma ▶ Cauchy sequence is convergent in the seminorm ∥∥ 1

For each k ∈ N, let f k ∈ C N . Suppose that ( f k ) is a Cauchy sequence in the seminorm ∥∥ 1 , so we have

lim k sup l≥k ∥ f l -f k ∥ 1 = 0 (16.1) Then there is f ∈ C N satisfying lim k ∥ f k -f ∥ 1 = 0 (16.2)
The f in eq. ( 16.2) is pasted together from the f k 's in the following way: there are indices

N 1 < N 2 < . . . so that f = f k on the interval (N k , N k+1 ], f = ∑ k f k • 1 (N k ,N k+1 ] (16.3) 16.2 Remark ▶ f inherits properties of ( f k )
Since f is pasted together from the f k 's the way we can see it in eq. ( 16.3), f inherits some common properties the f k may have. For example

a) If f k ≥ 0 for every k then f ≥ 0. b) If | f k | ≤ c for a constant c for every k then | f | ≤ c. c) If each f k is 0 -1 valued then so is f . d)
If each f k is a weight, then the construction can be adjusted so that f also becomes a weight.

Only remark 16.2 (d) requires some explanation since we need to have ∑ n∈N f (n) = ∞. For this, we observe a flexibility in the choice of the sequence N 1 < N 2 < . . . in the upcoming proof of lemma 16.1. Namely the sequence (N k ) is defined recursively, and once

N 1 < N 2 < • • • < N k-1 are chosen, the index N k , N k > N k-1 , is chosen "large enough"
to satisfy some criteria. So it can always be chosen to be "even larger" to satisfy additional criteria. For our case the single additional criterion is to ensure

∑ n∈(N k-1 ,N k ] f k-1 (n) > 1, which is possible since f k-1 is assumed to be a weight, so ∑ n∈(N k-1 ,∞) f k-1 (n) = ∞.
Proof of lemma 16.1. For the recursive definition of the (N k ), define first the sequence (ϵ k ) by

ϵ k := 2 sup l≥k ∥ f l -f k ∥ 1 (16.4)
We can assume, without loss of generality, that ϵ k > 0 for every k, since

ϵ k = 0 for some k would imply ∥ f l -f k ∥ 1 = 0 for l ≥ k hence we could take f = f k .
In the first step of the recursion, let N 1 = 1.

In the second step, let N 2 > N 1 to be large enough to satisfy

N 1 N 2 < ϵ 1 (17.1) A [1,N] | f 1 -f 2 | < ϵ 1 for every N ≥ N 2 (17.2)
and

A [1,N] | f 1 -f 3 | < ϵ 1 for every N ≥ N 2 (17.3)
Complete the second step of the recursion by defining f to be equal f 1 on the interval (N 1 , N 2 ]. Let k > 2 and assume that we have defined

N 1 < N 2 < • • • < N k-1 and f to be equal f j on the interval (N j , N j+1 ] for j ∈ [1, k -2]. For step k of the recursion let N k > N k-1 be large enough to satisfy 1 N k ∑ [1,N k-1 ] f j -f < ϵ j , for every j ∈ [1, k -2] (17.4) A [1,N] f j -f k-1 < ϵ j for every N ≥ N k , j ∈ [1, k -2] (17.5)
and

A [1,N] f j -f k < ϵ j for every N ≥ N k , j ∈ [1, k -2] (17.6)
Complete the kth step of the recursion by defining f to be equal f k-1 on the interval (N k-1 , N k ].

Let us fix j and let N be large enough so that for some k ≥ j + 2 we have

N k ≤ N < N k+1 (17.7)
We want to show that

A [1,N] f j -f < 3ϵ j (17.8) Let us estimate A [1,N] f j -f as, A [1,N] f j -f = 1 N ∑ [1,N k-1 ] f j -f (17.9) + 1 N ∑ (N k-1 ,N k ] f j -f (17.10) + 1 N ∑ (N k ,N] f j -f (17.11)
We can estimate the term in eq. (17.9), using eq. ( 17.4) and that N ≥ N k , as 1

N ∑ [1,N k-1 ] f j -f < ϵ j (17.12)
For the term in eq. ( 17.10) we have

1 N ∑ (N k-1 ,N k ] f j -f < ϵ j (18.1)
This follows from eq. (17.5) since f = f k-1 on the interval (N k-1 , N k ].

For the term in eq. ( 17.11) we have

1 N ∑ (N k ,N] f j -f < ϵ j (18.2)
This follows from eq. ( 17.6) since f = f k on the interval (N k , N].

Putting the estimates in eqs. (18.1), (18.2) and (17.12) together we obtain eq. (17.8).

The second lemma shows that the family M of sequences f for which M( f ) = lim N A [1,N] f exists is closed with respect to the upper mean M() defined by

M( f ) := lim sup N A n∈[1,N] f (n) , f ∈ C N (18.3) 18.1 Lemma ▶ M is closed with respect to M() Let ( f j ) be a sequence from M. Suppose that ( f j ) converges to f ∈ C N in the seminorm M(), so lim j M( f j -f ) = 0 (18.4) Then f ∈ M and M( f ) = lim j M( f j ) (18.5)
Proof. First note that, as a consequence of eq. ( 18.4), the sequence ( f j ) is a Cauchy sequence, meaning that for a given ϵ > 0 there is J so that

M( f j -f J ) < ϵ for every j ≥ J (18.6) Since M( f j ) -M( f J ) = M( f j -f J ) = M( f j -f J ) we see, M( f j ) -M( f J ) < ϵ for every j ≥ J (18.7)
so the sequence M( f j ) of means is a Cauchy sequence of numbers. Denote L := lim j M( f j ). We want to show that M( f ) = L. For a given ϵ > 0, choose a j so that M( f j ) -L < ϵ and M( ff j ) < ϵ. We then have, for an arbitrary N,

A [1,N] f -L ≤ A [1,N] ( f -f j ) + A [1,N] f j -L (18.8)
Taking lim sup N of both sides, we get

lim sup N A [1,N] f -L ≤ M( f -f j ) + M( f j ) -L (18.9) Since M( f -f j ) < ϵ and M( f j ) -L < ϵ, we get lim sup N A [1,N] f -L < 2ϵ. Since ϵ > 0 was arbitrary, we have lim N A [1,N] f -L = 0 which means M( f ) = L = lim j M( f j ).
How do we now show that every open set can be represented? Let B ⊂ T be open with positive µ α measure, let B = ∪ j I j be its decomposition into pairwise disjoint open intervals I j and set B k := ∪ j∈[1,k] I j . Since µ α (B) > 0, we have µ α (B k ) > 0 for large enough k. For simplicity, we assume that µ α (B k ) > 0 for every k. The sets B k increase to B monotonically, hence, in particular, we have lim k µ α (B k △B) = 0. According to proposition 14.1, the set B k can be represented by the set S k ⊂ R defined by

S k := { r n : r n α ∈ B k } (19.1)
and we have M R (S k ) = µ α (B k ). Since for every k, l the set B k △B l is Jordan measurable, we also have

M R (S k △S l ) = µ α (B k △B l ) (19.2)
For each k let us define the sequence f k by

f k (n) := 1 S k (r n ), n ∈ N (19.3) We have M( f k ) = M R (S k ) for every k ∈ N (19.4)
and we can rewrite eq. (19.2) as

M| f k -f l | = µ α (B k △B l ) (19.5)
Since (B k ) is a Cauchy sequence, so lim k sup l≥k µ α (B k △B l ) = 0, eq. (19.5) implies that ( f k ) is also a Cauchy sequence in ∥∥ 1 , so we have

lim k sup l≥k M| f k -f l | = 0. Since M| f k -f l | = ∥ f k -f l ∥ 1 , accord- ing to lemma 16.1, there is f to which the ( f k ) converges, that is, so lim k ∥ f k -f ∥ 1 = 0, and by lemma 18.1, M( f ) = lim k M( f k ). Since M( f k ) = µ α (B k ) and lim k µ α (B k ) = µ α (B), we have M( f ) = µ α (B) > 0.
According to remark 16.2 (c) f is 0 -1 valued hence we can define a set S ⊂ R by its indicator as

1 S (r n ) := f (n), n ∈ N (19.6) We have M R (S) = M( f ) (19.7)
We want to show that S is good and it represents B at α. To this end, let β ∈ T be arbitrary and define the sequences f

β k and f β by f β k (n) := f k (n) e(r n β) for n ∈ N (19.8) f β (n) := f (n) e(r n β) for n ∈ N (19.9) Since M( f ) = lim k M( f k ) > 0 and M( f k ) = M R (S k ), we have M(S) > 0.
It follows that, by eq. ( 15.1), to show that S is good, it's enough to show that M( f β ) exists for every β and to show that S represents B at α it's enough to show that M( f pα ) = µ α (e p 1 B ) for every p ∈ Z.

Let us first show that M( f β ) exists for every β. Since each set S k is good with M(S k ) > 0, we have, as a consequence of eq. ( 15.1), that f β k ∈ M for every k, β. The fact that for every β, the sequence ( f β k ) converges to f β in the norm M() follows from the uniform estimate

M( f β k -f β ) ≤ ∥ f k -f ∥ 1 for every β (20.1) By lemma 18.1, f β ∈ M and M( f β ) = lim k M( f β k ) (20.2) Let us now show that S represents B at α, that is, M( f pα ) = µ α (e p 1 B ) for every p ∈ Z. Since the sequence (B k ) converges to B in L 1 (µ α )-norm we have lim k µ α e p 1 B k = µ α (e p 1 B ) for every p ∈ Z (20.3) Since M( f pα k ) = µ α e p 1 B k and, by eq. (20.2), lim k M( f pα k ) = M( f pα ), eq. (20.3) implies that M( f pα ) = µ α (e p 1 B ).
We record the general idea we used as proposition 20.1 (b) below.

Proposition ▶ Limit of good sets with positive mean is good

Let (S k ) be a sequence of good subsets of R with mean which converge to S ⊂ R in ∥∥ 1,R -seminorm, that is,

lim k ∥S k △S∥ 1,R = 0. Assume that lim sup k M R (S k ) > 0.
Then we have the following.

a) lim k M R (S k ) exists and M R (S) = lim k M R (S k ) > 0.
b) S is a good set.

c) The sequence µ S k ,β k of limit measures converge to µ S,β in variation distance and uniformly in β,

lim k sup β µ S k ,β -µ S,β V = 0 (20.4) d)
Let ν be a Borel measure on T.

If for some α, µ S k ,α is absolutely continuous with respect to ν with Radon-Nikodym derivative ρ k for every k, then µ S,α is also absolutely continuous with respect to ν with Radon-Nikodym derivative ρ which satisfies

lim k ∥ρ k -ρ∥ L 1 (ν) = 0 (20.5)
Proof. The proof of proposition 20.1 (a) follows from the triangle inequality for the ∥∥ 1 -seminorm, since we then have

|M R (S k ) -M R (S)| = ∥S k ∥ 1,R -∥S∥ 1,R ≤ ∥S k △S∥ 1,R
and just use the assumption that lim k ∥S k △S∥ 1,R = 0.

The argument we gave just before the enunciation of our proposition proves that S is a good set.

For the proof of proposition 20.1 (c) note that in the argument preceding our proposition we proved that the sequence µ S k ,β k of measures converges weakly to µ S,β for every β but an estimate similar to eq. ( 20.1) enables us to draw the stronger conclusion of eq. ( 20.4).

The following lemma gives us the estimates we need.

Lemma

▶ ∥∥ 1,R dominates ∥∥ V and ∥∥ L 1 Let v 1 , v 2 be good R-weights. Assume that max ∥v 1 ∥ 1,R , ∥v 2 ∥ 1,R > 0 (21.1)
Then we have the following. a)

sup β µ v 1 ,β -µ v 2 ,β V ≤ 2 max ∥v 1 ∥ 1,R , ∥v 2 ∥ 1,R ∥v 1 -v 2 ∥ 1,R (21.2)
b) If, for some α, the limit measures µ v 1 ,α and µ v 2 ,α are absolutely continuous with respect to a Borel measure ν on T with Radon-Nikodym derivatives ρ 1 and ρ 2 , respectively, then

∥ρ 1 -ρ 2 ∥ L 1 (ν) ≤ 4 max ∥v 1 ∥ 1,R , ∥v 2 ∥ 1,R ∥v 1 -v 2 ∥ 1,R (21.3)
Proof. To prove lemma 21.1 (a), that is, the inequality in eq. ( 21.2), fix β and ϕ ∈ C + , so ϕ is a continuous T → C function with 0 ≤ ϕ ≤ 1.

We can assume without loss of generality that max ∥v

1 ∥ 1,R , ∥v 2 ∥ 1,R = ∥v 1 ∥ 1,R .
Let (N l ) l be a strictly increasing sequence of indices so that

lim l A [1,N l ] v 1 = ∥v 1 ∥ 1,R (21.4)
Let us estimate as

A v 1 n∈[1,N l ] ϕ(r n β) -A v 2 n∈[1,N l ] ϕ(r n β) = 1 A [1,N l ] v 1 A n∈[1,N l ] v 1 (r n )ϕ(r n β) - 1 A [1,N l ] v 2 A n∈[1,N l ] v 2 (r n )ϕ(r n β) adding 0 = -1 A [1,N l ] v 1 A n∈[1,N l ] v 2 (r n )ϕ(r n β) + 1 A [1,N l ] v 1 A n∈[1,N l ] v 2 (r n )ϕ(r n β)
inside the absolute value and using the triangle inequality,

≤ 1 A [1,N l ] v 1 A n∈[1,N l ] v 1 (r n )ϕ(r n β) -A n∈[1,N l ] v 2 (r n )ϕ(r n β) + 1 A [1,N l ] v 1 - 1 A [1,N l ] v 2 A n∈[1,N l ] v 2 (r n )ϕ(r n β) ≤ 1 A [1,N l ] v 1 A [1,N l ] |v 1 -v 2 | + A [1,N l ] |v 1 -v 2 | A [1,N l ] v 1 A [1,N l ] v 2 A [1,N l ] v 2 = 2 A [1,N l ] v 1 A [1,N l ] |v 1 -v 2 | so we have A v 1 n∈[1,N l ] ϕ(r n β) -A v 2 n∈[1,N l ] ϕ(r n β) ≤ 2 A [1,N l ] v 1 A [1,N l ] |v 1 -v 2 | (22.1) Since lim l A v i n∈[1,N l ] ϕ(r n β) = µ v i ,β (ϕ), lim l A [1,N l ] v 1 = ∥v 1 ∥ 1,R and lim sup l 2 A [1,N l ] v 1 A [1,N l ] |v 1 -v 2 | ≤ 2 ∥v 1 ∥ 1,R ∥v 1 -v 2 ∥ 1,R , we get µ v 1 ,β (ϕ) -µ v 2 ,β (ϕ) ≤ 2 ∥v 1 ∥ 1,R ∥v 1 -v 2 ∥ 1,R (22.2)
which is independent of β and ϕ ∈ C + , proving eq. ( 21.2).

To prove lemma 21.1 (b), observe first that, since µ v i ,α = ρ i ν and ρ i are probability densities with respect to ν, we have ∥ρ

1 ν -ρ 2 ν∥ V = 1 2 ∥ρ 1 -ρ 2 ∥ L 1 (ν) . It follows that ∥µ v 1 ,α -µ v 2 ,α ∥ V = 1 2 ∥ρ 1 -ρ 2 ∥ L 1 (ν) (22.3)
and now just use eq. ( 21.2). Now, let us come back to the proof of proposition 20.1. Using eq. ( 21.2)

with v 1 = 1 S k and v 2 = 1 S , we get sup β µ S k ,β -µ S,β V ≤ 2 max ∥S k ∥ 1,R , ∥S∥ 1,R ∥S k △S∥ 1,R (22.4)
Using the assumption that lim k ∥S k △S∥ 1,R = 0 and that, by proposition 20.1 (a), we have

lim k ∥S k ∥ 1,R = lim k M R (S k ) = M R (S) = ∥S∥ 1,R > 0,
we get eq. ( 20.4).

For the proof of proposition 20.1 (d), by proposition 20.1 (a), we can assume, without loss of generality that M R (S k ) > 0 for every k. Using eq. ( 21.3) with v 1 = 1 S k and v 2 = 1 S l we get

∥ρ l -ρ k ∥ L 1 (ν) ≤ 4 max ∥S k ∥ 1,R , ∥S l ∥ 1,R ∥S k △S l ∥ 1,R (23.1)
This implies, since the sequence (S k ) is convergent in ∥∥ 1,R -seminorm and hence is Cauchy, that the sequence

(ρ k ) is Cauchy in L 1 (ν)-norm. Since L 1 (ν) is complete and ν(ρ k ) = 1 for every k, there is a ρ ∈ L 1 (ν) with ν(ρ) = 1 so that lim k ∥ρ k -ρ∥ L 1 (ν) = 0 (23.2) Since ∥ρ k -ρ∥ L 1 (ν) = 2∥ρ k ν -ρν∥ V and ρ k ν = µ S k ,α , we get lim k µ S k ,α -ρν V = 0 (23.3)
But by proposition 20.1 (c) we also have lim k µ S k ,αµ S,α V = 0 hence we must have µ S,α = ρν.

We can use proposition 20.1 in an argument similar to the one we used to show that any open set can be represented at α to prove that if a G δ set B has positive µ α -measure then it can be represented at α. Only the initial setup of the proof is different. This time let (B k ) be a decreasing sequence of open sets which converges to B. Let S k ⊂ R represent B k at α. We again have the isometry eq. (19.2) from which everything follows: the existence of a good set S which represents B at α and M(S) = µ α (B).

Since every Borel measurable set differs from a G δ set on a set of µ αmeasure zero, we in fact showed that every Borel set of positive µ αmeasure can be represented. So we proved the following more precise version of theorem 7.1 for the case when the Radon-Nikodym derivative of a measure with respect to µ α is an indicator.

Proposition ▶ Theorem 7.1 for indicators

Let R ⊂ N be a good set, α be an irrational number, and let B be a Borel set with µ α (B) > 0.

Then B can be represented at α by a set S ⊂ R which satisfies

M R (S) = µ α (B) > 0 (23.4)
4 Measures that cannot be represented at every irrational α

For this section, we suspend the proof of theorem 7.1 just to see how proposition 23.1 can be used to prove theorem 5.3. We will also prove theorem 6.1.

Proof of theorem 5.3

In this section we want to prove that if the Borel probability measure ν has a point-mass at a point γ ∈ T and α is irrational then ν cannot be represented at α. The proof is by contradiction: let us assume that for some γ ∈ T, ν({γ}) > 0 and that ν can be represented by the set R at α, so µ R,α = ν.

Then the Dirac mass δ γ is absolutely continuous with respect to µ R,α with Radon-Nikodym derivative equal 1 ν(γ) 1 {γ} . By proposition 23.1 there is a good set S ⊂ R which represents δ γ at α, so µ S,α = δ γ . Let us define the function ϕ : T → C as

ϕ(β) := µ S,β (e) (24.1)
Then, by the definition of µ S,β (e), ϕ is the limit of the sequence (ϕ N ) of continuous functions defined by ϕ N (β) := A n∈[1,N] e(s n β) where (s n ) is the elements of S arranged in increasing order. Since for every p ∈ Z we have µ S,pα (e) = µ S,α (e p ) and µ S,α (e p ) = e p (γ), we have |ϕ| = 1 on the dense set { pα : p ∈ Z } (24.2) By Weyl's theorem,15 ϕ = 0 on a set of full Lebesgue measure, so, as a consequence, ϕ = 0 on a dense set.

(24.3) By Baire's theorem, 16 eqs. (24.2) and ( 24.3) together are impossible to 16 Baire 1905, Page 83.

hold simultaneously for the limit of continuous functions.

Proof of theorem 6.1

So in this section we want to prove that if ν is a Borel probability measure on T with lim sup p→∞ |ν(e p )| > 0 then there is an irrational α where ν cannot be represented. In fact the set of such α's is of full Lebesgue measure.

From the assumption that lim sup p→∞ |ν(e p )| > 0 it follows that there is an ϵ > 0 and a infinite sequence p 1 < p 2 < . . . of indices so that A := α : { p k α : k ∈ N } has nonempty interior (mod 1) (24.5) has full λ measure. We want to show that A is a subset of those α's at which the measure ν cannot be represented. Let α ∈ A, and suppose the measure ν can be represented at α, say, by the set S = (s n ), that is, µ S,α = ν. Let us define the function ϕ : T → C as

ϕ(β) := µ S,β (e) (24.6)
Then, by the definition of µ S,β (e), ϕ is the limit of the sequence (ϕ N ) of continuous functions defined by ϕ N (β) := A n∈[1,N] e(s n β). Since for every p ∈ Z we have µ S,pα (e) = µ S,α (e p ) and µ S,α (e p ) = ν(e p ), by eq. ( 24.4) we have

µ S,p k α (e) > ϵ for every k ∈ N (25.1)
By the definition of ϕ, we can write the above as

|ϕ| > ϵ on the set { p k α : k ∈ N } (25.2) Since α ∈ A, the set { p k α : k ∈ N } is dense in a nondegenerate interval I ⊂ T.
By Weyl's theorem, 18 ϕ = 0 on a set U of full Lebesgue measure 18 Weyl 1916, Satz 21; Kuipers and Niederreiter 1974, Theorem 4.1.

ϕ = 0 on U (25.3)
Since both { p k α : k ∈ N } and U are dense in the interval I, by Baire's theorem,19 eqs. (25.2) and ( 25.3) cannot be true together for the limit ϕ of we continue in the tradition of section 3 suppressing the set R in our notation for the limit measure, so µ α = µ R,α .

In trying to extend the class of representable functions ρ from indicators, we first consider an easier problem. Instead of representing by sets, we represent by R-weights.

Definition ▶ Function represented by a weight

Let ρ be an unsigned L 1 (T, µ α ) function with µ α (ρ) > 0. We say the R-weight w represents ρ at α if w is good and it represents the measure ρ

• µ α , that is, µ w,α = 1 µ α (ρ) ρ • µ α .
The R-weights w we consider in this section have positive mean in R, so M R (w) > 0. For such a weight, the non-normalized averages A n∈[1,N] w(r n )δ r n β are easier to handle than the normalized ones A w n∈[1,N] δ r n β . The convergence or divergence properties of the two averages are identical since they differ only by the nonzero factor M R (w),

lim N A n∈[1,N] w(r n )δ r n β = M R (w) lim N A w n∈[1,N] δ r n β (25.4)
as can be seen from writing

A w n∈[1,N] δ r n β = N ∑ n∈[1,N] w(r n ) A n∈[1,N] w(r n )δ rβ .
In section 2 we have already seen that if ρ is an unsigned continuous function with µ α (ρ) > 0 then the R-weight w defined by

w(r n ) := ρ(r n α) (25.5)
is good, unsigned and it represents ρ at α. Since every unsigned µ αintegrable function can be approximated arbitrary closely by unsigned continuous functions in L 1 (T, µ α )-norm, the proof of theorem 9.3 (a) requires only an approximation argument similar to what we had in section 3. We restate theorem 9.3 (a) in the following form for the readers convenience.

Proposition ▶ Any integrable function is representable with weights

Let ρ be an unsigned function from L 1 (T, µ α ) with µ α (ρ) > 0.

Then there is an R-weight w which represents ρ at α. In particular, we have

M R (w) = µ α (ρ) (26.1)
Furthermore, if ρ is a bounded function then the representing R-weight w can be chosen to be bounded.

The proof of proposition 20.1 can be easily adjusted to obtain the following analog for weights.

Proposition ▶ Limit of good weights with positive mean is good

Let (w k ) be a sequence of good R-weights with mean which converge to the R-weight w in ∥∥ 1,R -seminorm, so lim

N ∥w k -w∥ 1,R = 0. Assume that lim sup k M R (w k ) > 0.
Then we have the following.

a) lim k M R (w k ) exists and lim k M R (w k ) = M R (w) > 0. b) w is a good R-weight.
c) The sequence µ w k ,β k of limit measures converge to µ w,β in variation distance and uniformly in β,

lim k sup β µ w k ,β -µ w,β V = 0 (26.2) d)
Let ν be a Borel measure on T.

If for some α, µ w k ,α is absolutely continuous with respect to ν with Radon-Nikodym derivative ρ k for every k then µ w,α is also absolutely continuous with respect to ν with Radon-Nikodym derivative ρ which satisfies

lim k ∥ρ k -ρ∥ L 1 (ν) = 0 (26.3)
With this proposition, we can complete the proof of proposition 26.1 exactly as we proved proposition 23.1, using a sequence (ρ k ) of unsigned continuous functions that converge to ρ in L 1 (µ α )-norm. We need to remark only that if ρ is a bounded function, then the sequence (ρ k ) of continuous functions can be chosen to be uniformly bounded.

Proof of theorem 7.1 for bounded ρ

In this section, we still are working with a fixed good set R of positive integers, an irrational number α, but now we also fix a bounded Borel measurable, unsigned function ρ with µ α (ρ) > 0. We proved in section 5 that ρ can be represented at α by a good, bounded R-weight w. In this section we will show that there is a good set S ⊂ R which also represents ρ at α, hence proving theorem 7.1 for bounded ρ. It follows from the definition of representation that if the good R-weight w represents ρ then so does the R-weight cw for every positive constant c. In particular, we can assume that the R-weight w representing ρ is bounded by 1. We will show that then there is a set S ⊂ R so that

lim N sup β A n∈[1,N] 1 S (r n ) e(r n β) -A n∈[1,N] w(r n ) e(r n β) = 0 (27.1)
The "construction" of S satisfying eq. ( 27.1) is done randomly. Our random method requires that we limit the growth of the set R; we need to assume that R is sublacunary. 21 We need the concept of a sublacunary 21 Jones, Lacey, and Wierdl 1999, Theorem B.

weight.

Definition ▶ Sublacunary weight

The R-weight w is called sublacunary if it satisfies

lim N w(R(N)) log N = ∞ (27.2)
We often consider the sequence (r n ) instead of the set R in which case we can use the following more convenient version of eq. ( 27.2).

lim N ∑ n∈[1,N] w(r n ) log r N+1 = ∞ (27.3)
Our main tool in this section is the following. Proof. Since we can always assume that the bound of the R-weight w is 1, proposition 27.2 follows from the following lemma.

Proposition

Lemma ▶ Random selection of a good set

Let σ be an R-weight bounded by 1. We assume that for a constant b > 0 we have

lim inf N σ R(N) log N > b (28.1)
Let (Ω, P) be a probability space and and let (X r ) r∈R be a sequence of totally independent Ω → {0, 1} random variables indexed by R and with distribution P(X r = 1) = σ(r) (so P(X r = 0) = 1σ(r)).

Then we have

P      ω : sup N max β∈T ∑ r∈R(N) X r (ω) -σ(r) e(rβ) (log N)σ R(N) < ∞      = 1 (28.2)
To see that proposition 27.2 indeed follows from lemma 28.1, let

σ = w ∥w∥ ∞
, so σ is bounded by 1. Here we make a bit more complicated argument than needed to show that there is a rate of convergence in eq. ( 27.4).

The sublacunarity assumption on w implies that σ is sublacunary. We then have, as a consequence of eq. ( 28.2), that there is a measurable subset Ω 1 of Ω with P(Ω 1 ) = 1 so that for every ω ∈ Ω 1 there is a finite positive constant C ω with

max β∈T 1 σ(R(N)) ∑ r∈R(N) X r (ω) e(rβ) - 1 σ(R(N)) ∑ r∈R(N) σ(r) e(rβ) ≤ C ω log N σ R(N) (28.3) For β = 0, we then have 1 σ(R(N)) ∑ r∈R(N) X r (ω) -1 ≤ C ω log N σ R(N) (28.4) This implies that if we replace σ(R(N)) by ∑ r∈R(N) X r (ω) in 1 σ(R(N)) ∑ r∈R(N) X r (ω) e(rβ)
we make a

O log N σ R(N)
error, hence eq. ( 28.3) implies

max β∈T 1 ∑ r∈R(N) X r (ω) ∑ r∈R(N) X r (ω) e(rβ) - 1 σ(R(N)) ∑ r∈R(N) σ(r) e(rβ) ≤ C ω log N σ R(N) (28.5) Defining S ω ⊂ R by S ω := { r : r ∈ R, X r (ω) = 1 } (28.6)
we can write eq. (28.5) as

max β∈T A s∈S ω (N) e(sβ) -A σ r∈R(N) e(rβ) ≤ C ω log N σ R(N)
for every ω ∈ Ω 1 (29.1) Since σ is a constant multiple of w, we can replace σ by w in eq. ( 29.1),

max β∈T A s∈S ω (N) e(sβ) -A w r∈R(N) e(rβ) ≤ C ω ∥w∥ ∞ log N w R(N) for every ω ∈ Ω 1 (29.2) Since lim N ∥w∥ ∞ log N w R(N)
= 0, due to the sublacunarity assumption on the R-weight w, we get eq. ( 27.4) if we take S = S ω for any ω ∈ Ω 1 .

Proof of lemma 28.1. To see clearly what we need to do, denote

Z N (β) := ∑ r∈R(N) X r (ω) -σ(r) e(rβ)
and

t N := c • (log N)σ R(N)
where we'll choose the constant c appropriately later. By the Borel-Cantelli lemma, it's enough to prove

∑ N P max β∈T |Z N (β)| ≥ t N < ∞ (29.
3)

The first idea in proving eq. ( 29.3) is that we do not have to take the maximum over all β ∈ T, but over a finite subset B of T which contains N 3 elements 22 . Since the degree of the trigonometric polynomial Z N (β) is at 22 In fact, we can take a set B with as few elements as 10N, but in our applications, 10N won't improve anything over N 3 .

most

P   ∑ k∈[1,K] Y k ≥ t   ≤ 4 max exp - t 2 /8 ∑ k∈[1,K] E|Y k | 2 ,
exp(-t/3) for every t > 0 (30.2) Take K = #R(N) and Y r (β) := (X rσ(r))e(rβ) for r ∈ R(N). Then |Y r (β)| ≤ 1 so the Y r satisfy the assumption in Bernstein's inequality, hence, with t = t N /2, we get the estimate

P |Z N (β)| ≥ t N /2 ≤ 4 max exp - t 2 N /32 ∑ r∈R(N) E|Y r | 2 , exp(-t N /6) (30.3) Since E|Y r (β)| 2 = σ(r)(1 -σ(r)) we have ∑ r∈R(N) E|Y r (β)| 2 ≤ σ R(N) (30.4) Using that t N = c • (log N)σ R(N) , we get t 2 N /32 ∑ r∈R(N) E|Y r (β)| 2 = (c 2 /32)(log N)σ R(N) ∑ r∈R(N) E|Y r (β)| 2
using the estimate in eq. (30.4)

≥ (c 2 /32)(log N)σ R(N) σ R(N) = (c 2 /32)(log N) hence exp - t 2 N /32 ∑ r∈R(N) E|Y r (β)| 2 ≤ e -(c 2 /32)(log N) (30.5)
In order to get e -(c 2 /32)(log N) ≤ N -5 = e -5 log N , we need to have c 2 /32 ≥ 5, so it enough to have, since √ 160 < 13, c ≥ 13 (30.6)

We also have

t N /6 = (c/6) • (log N)σ R(N) (31.1)
We again need to have e -(c/6) √ b log N ≤ N -5 = e -5 log N which poses the requirement (c/6)

√ b ≥ 5, that is, c ≥ 30 √ b (31.2)
Thus choosing the constant c large enough to satisfy both eqs. (31.2) and (30.6), the estimate in eq. ( 30.3) implies the one in eq. ( 30.1).

Notes to lemma 28.1

The type of method we used in lemma 28.1 to estimate trigonometric polynomials goes back to Salem-Zygmund. 24 Recent developments have 24 Salem and Zygmund 1954, Chapter IV.

been given for example by Weber 25 and by Cohen-Cuny. 26 25 Weber 2000. 26 Cohen and Cuny 2006.

Absolute continuity and positive mean

The general theme of this section is that if a good set or weight has positive mean then it can represent only an absolutely continuous measure.

To be specific, we want to prove theorems 7.2 (b) and 9.3 (b).

Our standing assumption is that R is a sublacunary good set, and hence we suppress it in our notation for the limit measure, so we write µ α instead of µ R,α .

Proof of theorem 7.2 (b)

Theorem 7.2 (a) says that if ρ is an unsigned L ∞ (µ α ) function with µ α (ρ) > 0 and α is an irrational number then ρ can be represented at α with a good set

S ⊂ R satisfying M R (S) = µ α (ρ) ∥ρ∥ L ∞ (µα )
. We have proved this in section 6.

Theorem 7.2 (b) says that the converse is also true: if the good set S ⊂ R satisfies ∥S∥ 1,R > 0 then the limit measure µ S,β is absolutely continuous with respect to µ β with a bounded Radon-Nikodym derivative ρ β which must satisfy

ρ β L ∞ (µβ) ≤ 1 ∥S∥ 1,R for every β (31.3)
This is what we intend to prove now. Since β ∈ T is fixed, we suppress it in our notation, so for example we write µ for µ β and µ S for µ S,β . Let S ⊂ R be such that ∥S∥ 1,R > 0. Let us first show that for every β, the limit measure µ S is absolutely continuous with respect to µ. This will follow if we show that for every Borel set B we have

µ S (B) ≤ 1 ∥S∥ 1,R µ(B) (32.1)
To see this, it's enough to show that for every unsigned, continuous function ϕ on T we have

µ S (ϕ) ≤ 1 ∥S∥ 1,R µ(ϕ) (32.2)
Let ϕ be such a function and let N 1 < N 2 < . . . be a sequence of indices for which lim k A r∈R(N k ) 1 S (r) = ∥S∥ 1,R . We can then estimate as

µ S (ϕ) = lim N A s∈S(N) ϕ(sβ) = lim k A s∈S(N k ) ϕ(sβ) = lim k 1 A r∈R(N k ) 1 S (r) A r∈R(N k ) 1 S (r)ϕ(rβ) ≤ lim sup k 1 A r∈R(N k ) 1 S (r) A r∈R(N k ) ϕ(rβ) since lim k 1 A n∈[1,N k ] 1 S (n) = 1 ∥S∥ 1,R and lim N A r∈R(N k ) ϕ(rβ) exists, = 1 ∥S∥ 1,R lim N A r∈R(N k ) ϕ(rβ) since lim N A r∈R(N k ) ϕ(rβ) = µ(ϕ), = 1 ∥S∥ 1,R • µ(ϕ) proving eq. (32.2). Now, inequality µ ρ β 1 B ≤ 1 ∥S∥ 1,R µ(B) applied to the Borel set B = ρ β > 1 ∥S∥ 1,R
readily gives eq. (31.3).

Proof of theorem 9.3 (b)

Since the good set R is fixed, we suppress it in our notation for the limit measures, so we write µ α instead of µ R,α .

In this section, we need to prove that if the good R-weight w has positive relative 1-norm and it is integrable, that is, it can be approximated arbitrary closely by bounded, good R-weights in ∥∥ 1,R -seminorm, then for every irrational β the limit measure µ w,β is absolutely continuous with respect to µ β .

Let (w k ) be a sequence of good, bounded R-weights which converges to w in ∥∥ 1,R -seminorm, lim k ∥w k -w∥ 1,R = 0. Since ∥w k ∥ 1,R -∥w∥ 1,R ≤ ∥w k -w∥ 1,R , we have lim k ∥w k ∥ 1,R = ∥w∥ 1,R > 0, and hence we can assume without loss of generality that ∥w k ∥ 1,R > 0 for every k. That for every k the measure µ w k ,β is absolutely continuous with respect to µ β for every β follows from

µ w k ,β (B) ≤ ∥w k ∥ ∞ ∥w k ∥ 1,R µ β (B) for every Borel set B (33.1)
The proof of this inequality is almost identical to the proof of the inequality in eq. (32.1), hence we omit it. Now the rest of the proof of theorem 9.3 follows from lemma 21.1.

8 Proof of theorem 7.1 for unbounded ρ

In this section we again work with a fixed, sublacunary good set R ⊂ N which we view as a sequence (r n ) arranged in increasing order. We omit R from our notation for the limit measures, so we write µ β instead of µ R,β . We also fix an irrational number α. Let ρ ∈ L 1 (µ α ). We want to find a good set 27 S ⊂ R which represents ρ at α. According to proposi-27 Which can be shown to be sublacunary as a consequence of the sublacunarity of the weight v below.

tion 26.1 there is a good R-weight w which represents ρ at α. Since this weight w has positive relative mean with respect to R, it's a sublacunary weight. The problem is that, as per construction, w is not a bounded weight if ρ is unbounded, hence we cannot use our proposition 27.2 to construct the desired set S.

Our main job in this section hence will be to construct a good Rweight v satisfying the following properties • v is bounded by 1;

• v is sublacunary;

• v represents the same measure at every β as w, so µ v,β = µ w,β for every β.

Once we have such a good weight v, we can use proposition 27.2 to "construct" the desired good set S.

The weight v will be of the form σ • w where the weight σ is a decreasing weight, that is, σ(r n ) ≥ σ(r n+1 ) for every n ∈ N. That a weight v of this form represents the same measures everywhere is a consequence of a general but probably familiar result-our main new tool in this section. Not to get bugged down with unnecessary notation, we will state the result for weights with the reindexing w(n) = w(r n ) with which R weights become N-weights.

First recall the definition of a dissipative sequence of measures on N.

a) v(R) = ∞.
b) The R-weight σ is decreasing σ(r 1 ) ≥ σ(r 2 ) ≥ . . . . c) The R-weight w is good.

Then v is a good R-weight and it represents the same measures everywhere as w, µ v,β = µ w,β for every β (37.1)

Proof. We need to show that for a given β we have According to the proof of lemma 16.1, we obtained w as the limit of a sequence (w k ) of bounded good weights by pasting the w k together piece by piece in a sense that after choosing indices N 1 < N 2 < . . . , we define w to be equal w k on the interval (N k , N k+1 ]

w(r n ) := ∑ k w k (r n )1 (N k ,N k+1 ] (n) (37.7)
Now, in order to obtain a good weight v which is bounded by 1 and would represent the same measures as w, we could do the following. Define σ by

σ(r n ) := 1 max j∈[1,k] w j ∞ • 1 (N k ,N k+1 ] (n) (37.8)
Then σ is decreasing and v := σw is bounded by 1. The remaining issue is to ensure that v is sublacunary, and to do that it's enough to ensure

lim N ∑ n∈[1,N] v(r n ) log r N+1 = ∞ (38.1)
as we noted in eq. ( 27.3). This would also ensure that both σ and v are weights. It turns out that in the recursive process of choosing the indices (N k ) if we choose N k large enough compared to N k-1 we can ensure that v is sublacunary. We want to show that we can choose the indices N k so that we will have eq. (38.1). Let us note that in the proof of lemma 16.1 the choice of N k is flexible, since it just has to be large enough to staisfy some criteria. So we now add one additional criterion, namely we want to choose N k large enough to also satisfy

N max j∈[1,k] w j ∞ > k log r N+1 for every N ≥ N k (38.2)
This is possible because of the sublacunarity condition in eq. (37.6), and eq. ( 38.2) ensures the sublacunarity of v, that is, eq. (38.1). That v represents the same measures as w at every β follows from corollary 36.1. As in the last step of our proof of theorem 7.1, we use proposition 27.2 to show the existence of a good set S ⊂ R which represents the same measures as v at every β, hence at β = α we have µ S,α = ρµ α .

The limit measure at rational points

In this section we want to prove theorem 5.2. The base set is N which we suppress in our notation, so we write µ β instead of µ N,β .

Given the probability measure ν on T q and the rational number a q , gcd(a, q) = 1, let us see what properties a good set S would need to have so that µ S,a/q = ν.

Introducing the sets S j by S j := { s : s ∈ S, sa ≡ j (mod q) }, for every j ∈ [1, q] (38.3) let us write, using that the S j are pairwise disjoint,

A s∈S(N) = 1 #S(N) ∑ s∈S(N) δ sa/q = 1 #S(N) ∑ j∈[1,q]
∑ s∈S j (N)

δ j/q = ∑ j∈[1,q]
#S j (N) #S(N) δ j/q P(X n = 0) = 1/2. Let us also consider another sequence of random variables (Y n ) defined by

Y n =    X n if n ∈ [2 k , 2 k+1 ) for even k 1 -X n if n ∈ [2 k , 2 k+1 ) for odd k (40.1)
The (Y n ) is also an iid sequence with the same distribution as the (X n ).

Define the sets R ω , S ω by R ω := { n : X n (ω) = 1 } and S ω := { n : Y n (ω) = 1 }. By lemma 28.1 both R ω and S ω are good sets almost surely with M(R ω ) = M(S ω ) = 1/2. We claim that M(R ω ∩ S ω ) almost surely doesn't exists.

To see this, denote T ω := R ω ∩ S ω and observe that if M(T ω ) existed then lim k

T ω ∩[2 k ,2 k+1 ) 2 k
would exist. But, denoting by O the odd numbers and by E the even numbers, we almost surely have

lim k∈O T ω ∩ [2 k , 2 k+1 ) 2 k = 0 lim k∈E T ω ∩ [2 k , 2 k+1 ) 2 k = 1 2 10.2 R 1 ∪ R 2 and R 1 ∩ R 2 have

means but are not good

Here is an example of two good sets R 1 and R 2 each with mean 2/3,

M(R 1 ∩ R 2 ) = 1/2 but R 1 ∩ R 2 is not good and M(R 1 ∪ R 2 ) = 5/6 but R 1 ∪ R 2 is not good.
Both sets will be defined in blocks of intervals. Partition N into a sequence of disjoint intervals I n so that their lengths go to infinity but slower than the left endpoints go to infinity. For example, I n = [n 2 , (n + 1) 2 ) will do.

The first good set R 1 will contain all iNtegers from I 1 , then only Odd numbers from I 2 then Even numbers from I 3 then repeat this pattern for I 4 , I 5 , I 6 etc:

NOENOE . . . (40.2)
The set R 2 is defined similarly, except it will have one pattern in intervals J k := [3 k , 3 k+1 ) for even k and another for odd k.

7. 2

 2 Theorem ▶ Connection between dν dµ R,α , M R (S) and M R (S)Let R be a sublacunary good set. a) For an irrational α let the unsigned function ρ

  |ν e p k | > ϵ for k ∈ N (24.4) By Weyl's result, 17 the set A ⊂ T defined by 17 Weyl 1916, Satz 21; Kuipers and Niederreiter 1974, Theorem 4.1.

▶

  There is a set representing the same measures as a bounded weight Let w be a bounded, sublacunary R-weight. Then there is a set S ⊂ R so thatlim N max β∈T A s∈S(N) e(sβ) -A w r∈R(N) e(rβ) = 0 (27.4)As a consequence, if the R-weight w is good then so is the set S and we have µ S,β = µ w,β for every β (27.5)

  N] e(r n β) = µ w,β (e)(37.2) to do this, use proposition 34.2 with σ N defined byσ N (n) := σ(r n )1 [1,N] (n) (37.3)and (x n ) defined byx n := e(r n β) (37.4)Let us now go back to our good R-weight w which represents ρ at α. Since we now consider R as the sequence (r n ), its sublacunarity assumption is expressed more conveniently as lim in eq. (7.2). Since the weight w satisfies M R (w) > 0, eq. (37.5) implies that w is also sublacunary. Writing N+1

EONEON

  sets are good and they represent the same (uniform) measure at every β. The intersection R 1 ∩ R 2 has the patterns EOEEOE . . . for even k (40.5) OOEOOE . . . for odd k (40.6)

Theorem ▶ Absolutely continuous measures can be represented by subsets of a good set Let

  R be a sublacunary good set. Let α be an irrational number, and let the Borel probability measure ν be absolutely continuous with respect to µ R,α . Then there is a good set S ⊂ R which represents ν at α.

	GENERATION OF MEASURES ON THE TORUS WITH GOOD SEQUENCES OF INTEGERS 7
	lim N	#R(N) log N	= ∞	(7.1)
	In case we consider the sequence (r n ) instead of the set R, it's more useful
	to write eq. (7.1) in the form
	lim N	N log r N	= ∞	(7.2)
	7.1 As a consequence of theorem 7.1, every
				measure which is absolutely continuous
				with respect to the Lebesgue measure
				can be represented at any given irrational
				α by a subset of the primes, squares, or
				n 2 log n : n ∈ N .
				it satisfies lim inf n	lacunary if rn > 1, and such a r n+1
				sequence satisfies #R(N) = O(log N).
				Traditionally, a sublacunary sequence is one that satisfies lim n r n+1 rn = 1 and such a
				sequence satisfies lim N	#R(N) log N = ∞. Our
				definion of a sublacunary sequence in
				eq. (7.1) describes sequences which satisfy lim inf n r n+1 rn = 1 but may not satisfy lim n r n+1 rn = 1.

7

Traditionally, (r n ) is called

Table 1 :

 1 Notations

  N, we can readily see that sup β∈T |Z ′ N (β)| ≤ N 2 sup β∈T |Z N (β)|. It follows that if we take B N ⊂ T to be an arithmetic progression with |B N | = N 3 then max To prove eq. (30.1), we use the Bernstein-Chernoff exponential estimate. 23 This estimate says that if Y k , k ∈ [1, K], are totally independent,

	This follows if we prove
		P |Z N (β)| ≥ t N /2 <	2 N 5 for every β ∈ T	(30.1)
					23 Tao and Vu 2006, Exercise 1.3.4 with
	mean zero, complex valued random variables with |Y k | ≤ 1, then	t = λσ.
				β∈T	|Z N (β)| ≤ 2 max β∈B N	|Z N (β)|	(29.4)
	Hence we have
	P max β∈T	|Z N (β)| ≥ t N ≤ P max β∈B N	|Z N (β)| ≥ t N /2	(29.5)
	Using the union estimate, we get
	P max β∈B N	|Z N (β)| ≥ t N /2 ≤ N 3 max β∈B N	P(|Z N (β)| ≥ t N /2)	(29.6)
	Thus eq. (29.3) follows from
			∑ N	N 3 max β∈B N	P(|Z N (β)| ≥ t N /2) < ∞	(29.7)

Weyl 1916, Satz 21; Kuipers and Niederreiter 1974, Theorem 4.1.

Baire 1905, Page 83. continuous functions.5 Representing by weightsIn this section, we fix the good set

R and the irrational number α, and 20 Note that we make no further assumption on R, such as sublacunarity

Definition ▶ Dissipative sequence of measures

Let (v N ) N∈N be a sequence of finite measures on N.

We say, the sequence (v N ) N∈N is dissipative if lim N v N (j) v N (N) = 0, for every j ∈ N (34.1)

Proposition ▶ Decreasing weights preserve limits

Let w be a weight, (σ N ) N∈N be a sequence of finite measures on N and let x = (x n ) be a sequence from a normed space (X, ∥∥). Denoting v N := σ N • w, we assume the following a) Each σ N has finite support.

b) The sequence (v N ) is dissipative.

c) For each N the measure σ N is decreasing, σ N (1) ≥ σ N (2) ≥ . . . .

d)

The sequence A w n∈ [1,N] x n N converges to some y ∈ X,

Then, the sequence A v N j∈N x j N of averages converge to the same limit as the w-weighted averages,

At the heart of this result is the following quantitative estimate: For a

for every N ≥ K.

Note that the estimate in eq. (34.4) indeed implies the conclusion of the proposition in eq. ( 34.3). To see this, let N → ∞ in eq. (34.4). Then, since

Proof of proposition 34.2. The main idea of the proof is to write A ,j] x n for all N (35.1)

These measures q N will also satisfy

The measure q N appears during performing summation by parts: setting σ N (0) := 0, w(0) := 0 and x 0 := 0, we have

Thus, defining the measure q N by q N (j

we get the identity in eq. ( 35.1) once we show that q N really is a measure satisfying eq. (35.2). That q N (j) is unsigned follows from the assumption that the sequence (σ N (j)) j∈N is decreasing for fixed N. That q N (N) = v N (N) follows by setting x j = 1 for every j in the summation by parts argument above since then we get exactly q N (N) = v N (N):

Using the now obvious identity y = A q N j∈N y together with eq. ( 35.1), we can now write

GENERATION OF MEASURES ON THE TORUS WITH GOOD SEQUENCES OF INTEGERS 36

Splitting the summation on j in A q N j∈N A w n∈ [1,j] x ny into two parts at K and using the triangle inequality, we get the estimate

We can estimate the first term as

2) Using the definition of q N (j) as given in eq. ( 35.3), we can estimate

Using this estimate and that q N (N) = v N (N) in eq. ( 36.2) we get

3) The second term in eq. (36.1) can be estimated, using eq. (35.5), as

Putting the estimates in eqs. (36.3) and (36.4) into eq. ( 36.1) and using the identity in eq. ( 35.4) we get eq. (34.4).

Corollary ▶ Decreasing weights preserve limit measures of weights

Let w and σ be R-weights. Denoting v := σ • w, we assume the following If we make the assumption 28 that lim N #S j (N)

#S(N) exists for every j then, 28 In fact, the existence of lim N #S j (N) #S(N)

follows from S being a good set.

letting N → ∞, we get

Since µ S,a/q is supposed to be equal ν, we get

This gives us the idea how to construct S: we start out from the set R j defined by

Note that R j is a full residue class mod q, namely, if j ′ denotes the unique solution to the congruence j ′ a ≡ j (mod q), then R j is the arithmetic progression { kq + j ′ : k ∈ N }. Note that R j is a good set, as are all arithmetic progressions. We clearly have

Now what remains is to find a set S j ⊂ R j with relative mean ν j q and make sure that S j is a good set. Let γ be an irrational number and consider S j := r : r ∈ R j , rγ ∈ 0, ν j q for every j ∈ [1, q] (39.5)

Using proposition 14.1 with α = γ and R = R j , we deduce that S j is a good set with M R j (S j ) = ν j q , as desired. We finally define S as

The set S is good since it's the finite union of pairwise disjoint good sets with mean. Indeed, we have M(S j ) = 1 q • ν j q and hence M(S) = 1 q .

10 Examples 10.1 Two good sets, but their intersection has no mean.

Here we construct randomly two good sets, R, S with M(R) = M(S) = 1/2 but M(R ∩ S) doesn't exist. Let (X n ) be a iid sequence of random variables on the probability space (Ω, P), modeling fair coin flipping, so with distribution P(X n = 1) = Clearly M(R 1 ∩ R 2 ) = 1/2 but the average of e(n/2) is different on J k for even k from those on odd k: for even k the average will go to 1/3 while for odd k it goes to -1/3.

As for the union R 1 ∪ R 2 , it has the patterns NONNON . . . for even k (41.1)

Clearly M(R 1 ∪ R 2 ) = 5/6 but the average of e(n/2) is different on J k for even k from those on odd k: for even k the average will go to -1/3 while for odd k it goes to 1/3.

Open set

Let α be an irrational number in the torus T. We show that there exists an open subset U of the torus such that the sequence

does not converge when N goes to infinity.

The construction does not use at all the group structure or the dimensional properties of the torus. This can be extended in a general context of a sequence in a compact metric space with a non purely atomic asymptotic distribution.

We want to construct an open subset U of the torus and an increasing sequence of positive integers (N k ) k≥0 such that the averages A n∈[1,N 2k ] 1 U (nα), k = 0, 1, 2, . . . , with even indices are large whereas the averages A n∈[1,N 2k+1 ] 1 U (nα), k = 0, 1, 2, . . . with odd indices are small.

The sequence (N k ) will be constructed by induction and each N k will be associated to ϵ k := 1/(2 k+4 N k ). In this induction process, we construct also a sequence of open subsets (U k ) k≥0 .

We start with N 0 > 1 fixed and we define

We have of course

This is the initial step of our construction. In order to be understandable, let us describe the two next steps.

By the uniform distribution of the sequence (nα) n in the torus, there exists a number N 1 > N 0 such that

GENERATION OF MEASURES ON THE TORUS WITH GOOD SEQUENCES OF INTEGERS 42

We have

Note also that by construction U 0 ∩ U 1 = ∅.

By the uniform distribution of the sequence (nα) n in the torus, there exists a number N 2 > N 1 such that

We fix such a N 2 . To any n ∈ [1, N 2 ] with nα / ∈ U 1 we associate a real δ n satisfying

Note that the values of the δ n 's are reinitialized.

We define

Let us state now our induction hypothesis. Suppose that, for a fixed integer k > 0 we have already constructed two sequences

• If ℓ is even and ℓ ′ is odd, then U ℓ and U ℓ ′ are disjoint,

• Each U ℓ is a finite union of open intervals,

GENERATION OF MEASURES ON THE TORUS WITH GOOD SEQUENCES OF INTEGERS 43

Here begins the induction process. By the uniform distribution of the sequence (nα) n in the torus, there exists a number N k+1 > N k such that

We fix such a N k+1 . To any n ∈ [1, N k+1 ] with nα / ∈ U k we associate a real δ n that 0 < δ n ≤ ϵ k+1 and (nαδ n , nα + δ n ) ∩ U k = ∅

We define

Note that the values of δ n 's are reinitialized at each induction step.

The items of the induction hypothesis are now satisfied by the sequences (U ℓ ) 0≤ℓ≤k+1 and (N ℓ ) 0≤ℓ≤k+1 .

We can consider these sequences as infinite, and we define U := k≥0 U 2k .

Recalling our choice N k ϵ k = 2 -k-4 , we obtain