Measure of sumsets and ejective sets I.

Miklós Laczkovich*

Department of Analysis, Eötvös Loránd University Múzeum krt. 6-8, Budapest, H-1088 Hungary.

Imre Z. Ruzsa*

Mathematical Institute of the Hungarian Academy of Sciences Budapest, Pf. 127, H-1364 Hungary.

1. Introduction

A classical theorem of Erdős [2] in number theory asserts that if a set H of nonnegative integers is a basis, that is, the set

kH = H + ... + H, k summands

contains every nonnegative integer for some k, then the density of a sumset A + H can be estimated as follows:

\underline{d}(A + H) ≥ \underline{d}(A) + \frac{\underline{d}(A)(1 − \underline{d}(A))}{2k}.  (1.1)

Raikov [3] proved analogous results for measures. Let G be the circle with the normed Lebesgue measure μ.

Definition 1.1. We say that H ⊂ G is a basis of order k for G if kH = G.

Raikov shows that for an arbitrary A ⊂ G and a basis H of order k we have

\underline{μ}(A + H) ≥ \underline{μ}(A) + \frac{\underline{μ}(A)(1 − \underline{μ}(A))}{k}.  (1.2)

If 0 ∈ H and A is measurable (the general case of (1.2) is easily reduced to the measurable case by considering subsets of A), then with an arbitrary h ∈ H we have A + H ⊃ A ∪ (A + h), hence

\underline{μ}(A + H) ≥ μ(A ∪ (A + h)) = μ(A) + μ((A + h)\A).  (1.3)

The Erdős-Raikov proof had the curious feature that in order to estimate the cardinality or measure of A + H, the cardinality or measure of (A + h)\A is shown to be large for a suitable h ∈ H. We interpret this as the amount by which the addition of h “ejects” A out of itself. This motivates the following definitions.

* Supported by Hungarian National Foundation for Scientific Research, Grant T7582.

Definition 1.2. By the measure of ejectivity of the set H we mean the function

ζ(x) = ζ_{H}(x) = inf_{μ(A)=x} sup_{h∈H} μ((A + h)\A).  (1.4)

(1.3) shows that

\underline{μ}(A + H) ≥ μ(A) + ζ(μ(A)).

Definition 1.3. If ζ(x) = 0 for all x, we say that H is nonejective; if ζ(x) > 0 for some x, then H is ejective; if ζ(x) > 0 for all x ∈ (0, 1), then H is essentially ejective.

So far we spoke about sets on the circle. Definitions 1.1-1.3 make sense in every structure with an addition and a measure. The most general situation we shall consider is when G is a unimodular locally compact group, and μ is the Haar measure. ζ is defined for those values of x that may occur as a value of μ; if G is not discrete, this means every 0 ≤ x ≤ μ(G). We shall be interested mainly in the classical cases, the line, the circle and finite dimensional spaces, but when a more general case can be settled with little extra effort, we shall do so.

The principal aim of this first part is to construct nontrivial examples of ejective sets, hence showing that our subject is nonvoid.

Raikov’s theorem admits the following generalization.

Statement 1.4.

If G is compact, μ is normed so that μ(G) = 1 and H is a basis of order k for G, then we have

ζ_{H}(x) ≥ \frac{x(1 − x)}{k}.  (1.5)

For sake of completeness, we give a proof of this result, which is essentially the same as the Erdős-Raikov argument, in Section 2.

Thus, by ‘nontrivial’ we mean ejective sets that are not bases. We shall construct examples that are too small to form a basis. This will be expressed in terms of the box dimension. Let G = ℝ/ℤ, that is, the unit interval with addition modulo one (this will be more comfortable to work with than the set of unimodular complex numbers). Let A ⊂ G, and for a positive integer n let b(n) denote the minimal number of intervals of length 1/n that together cover A. The number

β(A) = lim sup_{n→∞} \frac{log b(n)}{log n}

is called the box dimension of A. An easy argument shows that β(A) ≥ 1/k holds for any basis of order k (while the Hausdorff dimension of a basis may be 0). Our main result is the following.

Theorem 1.5. In ℝ/ℤ there exists an ejective set whose box dimension is 0, even

b(n) < c(log log n)^{3} log n  (1.6)

holds for large n.

2. The ejectivity of bases

Lemma 2.1. For every unimodular locally compact group G, arbitrary H ⊂ G, real x and

positive integer k we have

ζ_{kH}(x) ≤ kζ_{H}(x).  (2.1)

Proof. Consider the function

d(u) = μ((A + u)\A).

We have

d(u + v) = μ((A + u + v)\A) ≤ μ((A + u + v)\(A + u)) + μ((A + u)\A) = d(v) + d(u).

(We used the translation invariance of μ, but commutativity is not assumed here.) An induction yields

d(h_{1} + ... + h_{k}) ≤ d(h_{1}) + ... + d(h_{k}),

which immediately implies (2.1).

Lemma 2.2. Assume that G is compact and μ(G) = 1. We have

ζ_{G}(x) ≥ x(1 − x).  (2.2)

Proof. Take an A ⊂ G, μ(A) = x. An easy application of Fubini’s theorem yields

∫ μ((A + h) ∩ A) dμ(h) = x^{2},

consequently μ((A + h) ∩ A) ≤ x^{2} for a suitable h and then for this h we have

μ((A + h)\A) = μ(A) − μ((A + h) ∩ A) ≥ x(1 − x).

Proof of Statement 1.4. By the previous lemmas we have

ζ_{H}(x) ≥ \frac{1}{k} ζ_{kH}(x) = \frac{1}{k} ζ_{G}(x) ≥ \frac{x(1 − x)}{k}.

3. The measure of ejectivity of the whole group

In the previous section we estimated ζ_{G} from below. Here we prove that this estimate is often exact.

Theorem 3.1. Let G be an infinite compact commutative group and assume that μ(G) = 1. We

have

ζ_{G}(x) = x(1 − x).  (3.1)

Lemma 3.2. Let X_{1}, ..., X_{k} be independent real random variables, X_{i} − EX_{i} ≤ K and

∑ DX_{i} ≤ σ^{2}.

Write S = ∑ X_{i}. For λ ≤ 4σ/K we have

P(S − ES ≥ λσ) ≤ exp(−λ^{2}/4).  (3.2)

Note that only an one-sided bound is required for X_{i} − EX_{i}. Proof. This follows from Bennett’s Bernstein-type inequality [1]. His estimate for

the probability in (3.2) is exp(−λ^{2}h(λK/σ)), where

h(t) = \frac{(t + 1) log(t + 1) − t}{t^{2}}.

We obtain (3.2) by observing that h(t) ≥ 1/4 for t ≤ 4.

Lemma 3.3.

For every set H, its measure of ejectivity has the following properties:

a) 0 ≤ ζ(x) ≤ x for all x.

b) |ζ(x) − ζ(y)| ≤ |x − y| for all x, y.

c) If G is compact and μ(G) = 1, then ζ(x) = ζ(1 − x) for all x.

Proof. Property a) is obvious. To prove part b), assume that x < y. For a set A such that μ(A) = x and μ((A + h)\A) < ζ(x) + ε for every h ∈ H, take any B ⊃ A, μ(B) = y. We have

(B + h)\B ⊂ ((A + h)\A) ∪ ((B\A) + h)

which shows μ((B + h)\B) ≤ ζ(x) + ε + (y − x) and hence ζ(y) ≤ ζ(x) + y − x. The inequality ζ(x) ≤ ζ(y) + y − x can be proved similarly, by applying the inclusion

(A + h)\A ⊂ ((B + h)\B) ∪ (B\A).

To prove part c), take a set A such that μ(A) = x and consider the set B = G\A. It satisfies μ(B) = 1−x and (B +h)\B = A\(A+h), hence μ((B +h)\B) = μ((A+h)\A).

Lemma 3.4.

Let G be a finite group, |G| = m, μ(G) = 1. For every x = k/m, where 0 ≤ k ≤ m is an integer, we have

ζ_{G}(x) ≤ x(1 − x) (1 + c\sqrt{\frac{log m}{m}})  (3.3)

with an absolute constant c.

Proof. By property c) above, it is sufficient to consider x ≤ 1/2. Write x(1−x) = y. Let ξ_{g}, g ∈ G be independent 0-1 valued random variables that assume 1 with

probability x. Let

B = {g ∈ G : ξ_{g} = 1},

a random set. We have

|B| = ∑ ξ_{g},

which has a binomial distribution. Hence

P(|B| = k) = \binom{m}{k}x^{k}(1 − x)^{n−k} ≥ 1/(m + 1).

(The last inequality follows from the fact that this is the maximal term in the binomial distribution and the sum of all m + 1 terms is 1.)

Now we consider B ∩ (B + h) for h ∈ G, h ≠ 0. We have

η = |B ∩ (B + h)| = ∑_{g∈G} ξ_{g}ξ_{g−h}.

This is a sum of m random variables, each of which assumes 1 with probability x^{2} and 0 otherwise, hence Eη = mx^{2}. The terms are not all independent. First we show that we can decompose G as G = G_{1} ∪ G_{2} ∪ G_{3} so that in the corresponding decomposition of η,

η = η_{1} + η_{2} + η_{3}, η_{i} = ∑_{g∈G_{i}} ξ_{g}ξ_{g−h},

each η_{i} is a sum of independent variables. (This decomposition will depend on h.) This will happen if for all g, g′ ∈ G_{i}, g ≠ g′, the elements g, g − h, g′, g′ − h are all different, which is equivalent to saying that h ∉ G_{i} − G_{i}. Now let H be the subgroup generated by h, l = |H| the order of h, and consider the decomposition of G into left cosets of H:

G = ⋃_{k=1}^{m/l}(r_{k} + H).

We define for i = 1, 2

G_{i} = {r_{k} + jh : 1 ≤ k ≤ m/l, 1 ≤ j ≤ l − 1, j ≡ i (mod 2)}

and

G_{3} = {r_{k} : 1 ≤ k ≤ m/l}.

Now we apply Lemma 3.2 for the variables X_{g} = −ξ_{g}ξ_{g−h}. Recall that X_{g} assumes the values 0 and −1, 0 with probability 1 − x^{2} and 1 with probability x^{2}. Hence EX_{g} = −x^{2} and DX_{g} = x^{2}(1−x^{2}) ≤ x^{2}, and the assumptions are fulfilled with K = x^{2} and σ^{2} = mx^{2}, thus up to λ ≤ 4\sqrt{m}/x we have

P(η_{i} − Eη_{i} ≤ −λx\sqrt{m}) ≤ exp(−λ^{2}/4).

We put λ = 4\sqrt{log m}; this is in the permissible range since log m < m. We obtain that

P(η_{i} − Eη_{i} ≤ −4\sqrt{log m}x\sqrt{m}) ≤ m^{−4},

consequently

P(η − Eη ≤ −12\sqrt{log m}x\sqrt{m}) ≤ 3m^{−4}.

Recall that η − Eη = |B ∩ (B + h)| − x^{2}m. Since there are < m possible choices of h, the probability that

|B ∩ (B + h)| ≤ x^{2}m − 12x\sqrt{m log m}

for some h is less that 3m^{−3} < 1/(m + 1) if m ≥ 3. Consequently there is a choice of B that satisfies both |B| = k and

|B ∩ (B + h)| ≥ x^{2}m − 12x\sqrt{m log m}

for all h, whence

|(B + h)\B| = |B| − |B ∩ (B + h)| ≤ mx(1 − x) (1 + 24\sqrt{\frac{log m}{m}}) .

(We used the fact that x ≤ 1/2.)

Lemma 3.5.

Let G be the circle (additive group of reals modulo 1). We have ζ_{G}(x) = x(1 − x) for every 0 ≤ x ≤ 1.

Proof. Take two positive integers k ≤ m. Let G′ be the additive group of residue classes modulo m, which we identify with the integers 0, 1, ..., m−1. Take a set B′ ⊂ G′ with the properties |B′| = k and

|(B′ + h)\B′| ≤ k(1 − k/m) (1 + c\sqrt{\frac{log m}{m}})  (3.4)

for every h; the existence of such a set follows from the previous lemma.

Now define a set B ⊂ G by

B = ⋃_{j∈B′} [\frac{j}{m}, \frac{j + 1}{m}) .

For a number of the form h = r/m, 0 ≤ r ≤ m − 1 integer, we have

μ((B + h)\B) = \frac{1}{m}|(B′ + r)\B′|

and this measure is a linear function of h between r/m and (r + 1)/m. Hence (3.4) implies

μ((B + h)\B) ≤ \frac{k}{m} (1 − \frac{k}{m}) + c\sqrt{\frac{log m}{m}}

for all h. We conclude that

ζ_{G}(k/m) ≤ \frac{k}{m} (1 − \frac{k}{m}) + c\sqrt{\frac{log m}{m}}

for every k ≤ m. By property b) of Lemma 3.3 we conclude that ζ_{G}(x) ≤ x(1 − x) for every x ∈ [0, 1].

Proof of Theorem 3.1. Let G be an arbitrary compact group and G′ a factorgroup of G, with the natural homomorphism ϕ : G → G′. Take any B′ ⊂ G′ and let B = ϕ^{−1}(B′). For every h ∈ G we have

μ((B + h)\B) = μ′((B′ + ϕ(h))\B′),

hence we have ζ_{G}(x) ≤ ζ_{G′}(x) for every x assumed by μ′.

Now let G be an infinite commutative compact group. If G has a factor isomorphic to the circle, then we obtain ζ_{G}(x) ≤ x(1 − x) by the above observation and Lemma 3.5. If this is not the case, then G has arbitrarily large finite factors. Then the above observation and Lemma 3.4 yield

ζ_{G}(k/m) ≤ \frac{k}{m} (1 − \frac{k}{m}) + c\sqrt{\frac{log m}{m}}

for any m that can appear as a size of a finite factor, and for all k ≤ m. By property b) of Lemma 3.3 we conclude that ζ_{G}(x) ≤ x(1 − x) for every x ∈ [0, 1]. Together with Lemma 2.2 we get the desired conclusion.

4. A criterion for ejectivity

In this section let G be a commutative group. Unlike in the previous sections, we do not assume G to be compact, thus the Haar measure may be infinite.

For a measure ν on G, we use ν̂ to denote its Fourier transform, that is,

ν̂(γ) = ∫ γ dν

for any character γ of G.

We use G^{∗} to denote the group of characters, and γ_{0} to denote the principal character (identically 1). We use dx and dγ for integration on G and Γ with respect to their Haar measures μ and μ^{∗}. We assume that these are normed so that in Plancherel’s formula the constant is 1.

Theorem 4.1.

Assume that H ⊂ G and there exists a probability measure ν supported on H such that

μ^{∗}({γ : Re ν̂(γ) > 1 − η}) = v < ∞

with some constant η > 0. Then for every x ≤ 1/v we have

ζ_{H}(x) ≥ ηx(1 − vx);  (4.1)

in particular, H is ejective. If for every ε > 0 we can achive v < ε with a suitable ν and η (which may depend on ε), then H is essentially ejective.

Proof. Take a set A ⊂ G, μ(A) = x and let ι denote the indicator function of A. We have

ι(x − h) − ι(x) = \{ \begin{matrix}1 & if x ∈ (A + h)\A, \\ −1 & if x ∈ A\(A + h) , \\ 0 & otherwise. \end{matrix}

Consequently

∫ |ι(x − h) − ι(x)|^{2} dx = 2μ((A + h)\A).  (4.2)

The Fourier transform of ι is

F (γ) = ∫ ι(x)γ(x) dx = ∫_{A} γ dμ,

and that of the function ι(x − h) is

∫ ι(x − h)γ(x) dx = F (γ)γ(h).

Hence an application of Plancherel’s identity to (4.2) yields

2μ((A + h)\A) = ∫ |F (γ)(γ(h) − 1)|^{2} dγ.

Since |z − 1|^{2} = 2(1 − Re z) whenever |z| = 1, we obtain

μ((A + h)\A) = ∫ |F (γ))|^{2}(1 − Re γ(h)) dγ.  (4.3)

Now we integrate (4.3) with respct to ν; this yields

∫ μ((A + h)\A) dν(h) = ∫ |F (γ)|^{2} ∫ (1 − Re γ(h)) dν(h) dγ
= ∫ |F (γ)|^{2}(1 − Re ν̂(γ)) dγ
≥ η ∫_{Re ν̂(γ)≤1−η} |F (γ)|^{2} dγ.

(4.4)

Again by Plancherel’s formula we know that ∫ |F (γ)|^{2} dγ = μ(A) = x and |F (γ)| ≤ x for each γ, hence

∫_{Re ν̂(γ)≤1−η} |F (γ)|^{2} dγ ≥ x − vx^{2}.

By substituting this into (4.3) we obtain

∫ μ((A + h)\A) dν(h) ≥ ηx(1 − vx).

Since the supremum of μ((A+h)\A) is not less than its ν-mean, the theorem is proved.

We formulate separately the case of compact groups.

Theorem 4.2.

Let G be a compact commutative group and μ(G) = 1. Assume that H ⊂ G and there exists a probability measure ν supported on H such that Re ν̂(γ) ≤ 1−η for every character γ with at most k exceptions for some constant η > 0. Then for every x we have

ζ_{H}(x) ≥ ηx(1 − kx);  (4.5)

in particular, H is ejective. If k = 1, that is, the only exception is the principal character, then

ζ_{H}(x) ≥ ηx(1 − x)  (4.6)

and H is essentially ejective.

5. Proof of Theorem 1.5

We construct a set B satisfying (1.6).

We shall obtain B as the intersection of a decreasing sequence B_{k} of sets. We build B_{k} in the following way. We define a sequence m_{k} of integers such that m_{k}|m_{k+1} for all k, another sequence b_{k} such that 1 ≤ b_{k} ≤ m_{k}, a set J_{k} ⊂ [0, m_{k} − 1] of integers with |J_{k}| = b_{k} and put

B_{k} = ⋃_{j∈J_{k}} [\frac{j}{m_{k}} , \frac{j + 1}{m_{k}}) .

The numbers m_{k}, b_{k} will be given in the course of the proof.

To establish the ejectivity of B we are going to apply Theorem 4.2. The characters of the circle group G = ℝ/ℤ are the functions

γ_{r}(x) = e(rx) = e^{2πirx}.

Correspondingly, the Fourier transform of a measure ν is the function

ν̂(r) = ∫ e(rx) dν(x).

We shall construct measures ν_{k} on B_{k} satisfying

|ν̂_{k}(r)| ≤ η_{k} = \frac{1}{2} − \frac{c}{\sqrt{k}}  (5.1)

for k > k_{0} and r ≠ 0.

Our measure ν_{k} will be the normalized restriction of the Lebesgue measure to the sets B_{k}. The Fourier transform of this measure is

ν̂_{k}(r) = \frac{m_{k}}{b_{k}} ∑_{j∈J_{k}} ∫_{j/m_{k}}^{(j+1)/m_{k}} e(rx) dx.

This is connected to the sum

g_{k}(r) = \frac{1}{b_{k}} ∑_{j∈J_{k}} e(jr/m_{k})

via the formula

ν̂_{k}(r) = e (\frac{r}{2m_{k}} ) \frac{sin πr/m_{k}}{πr/m_{k}} g_{k}(r).

In particular, |ν̂_{k}(r)| ≤ |g_{k}(r)| for all r, and |ν̂_{k}(r)| = 0 if m_{k}|r and r ≠ 0. Observe that g_{k}(r) is periodic with period m_{k}. Hence to ensure (5.1) it is sufficient to achieve

|g_{k}(r)| ≤ η_{k} for 1 ≤ r ≤ m_{k} − 1.  (5.2)

Write n_{k} = m_{k+1}/m_{k}. The condition B_{k+1} ⊂ B_{k} holds if each element of J_{k+1} is of the form

j = n_{k}j′ + q, j′ ∈ J_{k}, 0 ≤ q ≤ n_{k} − 1.  (5.3)

We construct the sets J_{k} recursively. Suppose that J_{k} is given, and it satisfies (5.2). We find J_{k+1} so that (5.2) holds also for k + 1, assuming that certain growth conditions hold for m_{k} and b_{k}.

Let

ϱ = \frac{b_{k+1}}{b_{k}n_{k}} .

We select each number j of the form (5.3) into J_{k+1} independently, with probability ϱ. Then

E|J_{k+1}| = ϱn_{k}|J_{k}| = b_{k+1}

and

P(|J_{k+1}| = b_{k+1}) = \binom{b_{k}n_{k}}{b_{k+1}}ϱ^{b_{k+1}}(1 − ϱ)^{b_{k}n_{k}−b_{k+1}} ≥ (1 + b_{k}n_{k})^{−1} ≥ (1 + m_{k+1})^{−1}

by the same argument as we used in Section 3.

For any j of the form (5.3), let ξ_{j} be the indicator of J_{k+1}; these are independent random variables, each assuming 1 with probability ϱ and 0 with probablity 1 − ϱ. We can express the Fourier series of J_{k+1} as follows.

g_{k+1}(r) = \frac{1}{b_{k+1}} ∑_{j∈J_{k+1}} e (\frac{rj}{m_{k+1}} ) = \frac{1}{b_{k+1}} ∑ ξ_{j}e (\frac{rj}{m_{k+1}} ) .

For a number expressed in the form (5.3) we have

\frac{j}{m_{k+1}} = \frac{j′}{m_{k}} + \frac{q}{m_{k+1}} .

Hence the expectation of g_{k+1}(r) is

Eg_{k+1}(r) = \frac{ϱ}{b_{k+1}} ∑_{j′∈J_{k}} ∑_{q=0}^{n_{k}−1} e (\frac{rj′}{m_{k}} + \frac{rq}{m_{k+1}} ) = \frac{1}{n_{k}} ∑_{q=0}^{n_{k}−1} e (\frac{rq}{m_{k+1}} ) g_{k}(r).

Consequently |Eg_{k+1}(r)| ≤ |g_{k}(r)| for all r, and it is = 0 if m_{k}|r but m_{k+1} |̸ r. We can conclude by the inductive assumption that

|Eg_{k+1}(r)| ≤ η_{k} for 1 ≤ r ≤ m_{k+1} − 1.  (5.4)

To estimate the deviation from the mean we apply Lemma 3.2 separately for the four variables ±ξ_{j} cos(rj/m_{k+1}) and ±ξ_{j} sin(rj/m_{k+1}). The parameters will be chosen as K = 1, σ^{2} = ϱn_{k}b_{k} = b_{k+1} and λ = 4\sqrt{log m_{k+1}} so that exp(−λ^{2}/4) = m_{k+1}^{−4}. The condition λ ≤ 4σ/K becomes

log m_{k+1} ≤ b_{k+1}.  (5.5)

Since |z| ≤ | Re z| + | Im z|, these four inequalities imply

P(|g_{k+1}(r) − Eg_{k+1}(r)| ≥ δ) ≤ m_{k+1}^{−4},

where

δ = 2\frac{λσ}{b_{k+1}} = 8\sqrt{\frac{log m_{k+1}}{b_{k+1}}} .

Since there are m_{k+1} − 1 possible choices of r, the probability that

|g_{k+1}(r) − Eg_{k+1}(r)| ≥ δ

happens for at least one r is less than m_{k+1}^{−3}. This is less than the probability that |J_{k+1}| = b_{k+1}, thus there is a choice that satisfies both |J_{k+1}| = b_{k+1} and

|g_{k+1}(r) − Eg_{k+1}(r)| < δ

for all 1 ≤ r ≤ m_{k+1} − 1. Hence, using (5.4) we obtain

|g_{k+1}(r)| ≤ η_{k} + δ,

and the induction works if

η_{k} + δ ≤ η_{k+1}.  (5.6)

Now we put m_{k} = 2^{2^{k}} for all k and b_{k} = k^{3}2^{k} for k > k_{0}, b_{k} = m_{k} for 1 ≤ k ≤ k_{0}. Hence b_{k}/m_{k} will be decreasing and condition (5.5) is satisfied. We get

δ = 8\sqrt{log 2}(k + 1)^{−3/2},

while

η_{k+1} − η_{k} = \frac{c}{\sqrt{k}} − \frac{c}{\sqrt{k + 1}} ≥ \frac{c}{2}(k + 1)^{−3/2},

thus (5.6) holds if we take, say, c = 16.

We start the induction from k = k_{0}. Here J_{k} = {0, 1, ..., m_{k} − 1}, thus g_{k}(r) = 0 for all 1 ≤ r ≤ m_{k} − 1, thus we only need that η_{k_{0}} ≥ 0, that is, c/\sqrt{k_{0}} ≤ 1/2.

Having defined the sequences B_{k} and ν_{k}, we put B = ∩B_{k}. We can estimate b(n), the number of intervals of length 1/n that cover B, as follows. If m_{k−1} < n ≤ m_{k}, then the b_{k} intervals that constitute B_{k} cover B, thus

b(n) ≤ b_{k} = k^{3}2^{k} ≪ (log log n)^{3} log n.

To establish that B is ejective, take any convergent subsequence (in the weak topology) ν_{k_{j}} of the sequence ν_{k}. The limit ν of this sequence is a probability measure supported on B such that

|ν̂(r)| = lim |ν̂_{k_{j}}(r)| ≤ lim η_{k} = 1/2

for every r ≠ 0. Hence B is essentially ejective by Theorem 4.

[1] Bennett, G., Probability inequalities for the sum of independent random variables, J. American Statist. Assoc. 57(1962), 33–45.

[2] Erdős, P., On the arithmetical density of the sum of two sequences, one of which forms a basis for the integers, Acta Arithmetica 1(1936), 201–207.

[3] Raikov, On the addition of point sets in the sense of Schnirelmann (in Russian), Mat. Sbornik 5(47)(1939), 425–440. Printed 1999.2.28.