On the density or measure of sets and their sumsets in the integers or the circle

TL;DR

This paper investigates the possible densities and measure properties of sets of integers and their sumsets, providing solutions to existence questions and exploring diophantine constraints within random sets and the circle.

Contribution

It characterizes the existence of integer sequences with prescribed densities and sumset densities, and analyzes the structure of these sets in both integers and the circle, introducing diophantine constraints.

Findings

Existence of integer sequences with specified density and sumset density.

Characterization of k-tuples of densities for sets in integers and the circle.

Introduction of diophantine constraints within pseudo s-th power sets.

Abstract

Let $\mathrm{d}(A)$ be the asymptotic density (if it exists) of a sequence of integers $A$. For any real numbers $0\leq\alpha\leq\beta\leq 1$, we solve the question of the existence of a sequence $A$ of positive integers such that $\mathrm{d}(A)=\alpha$ and $\mathrm{d}(A+A)=\beta$. More generally we study the set of $k$-tuples $(\mathrm{d}(iA))_{1\leq i\leq k}$ for $A\subset \mathbb{N}$. This leads us to introduce subsets defined by diophantine constraints inside a random set of integers known as the set of ``pseudo $s$th powers''. We consider similar problems for subsets of the circle $\mathbb{R}/\mathbb{Z}$, that is, we partially determine the set of $k$-tuples $(\mu(iA))_{1\leq i\leq k}$ for $A\subset \mathbb{R}/\mathbb{Z}$.

Equations

$$\mathrm{d}\big{(}A\big{)}=\lim_{t\to\infty}\frac{A(t)}{t}.$$

$${\mathrm{d}}_B(A)=\underset{t\to \infty }{\lim }\frac{A(t)}{B(t)}.$$

$$(\mathrm{d}\big{(}A\big{)},\mathrm{d}\big{(}2A\big{)},\ldots,\mathrm{d}\big{(}kA\big{)})$$

$$\underset{x\to +\infty }{\lim }\frac{1}{x}\sum _{n\le x}f(\{\lambda n\})=\int f.$$

$$B_{\lambda ,{\sum }_i{A}_i^{ϵ}}\subset \sum _i{B}_{\lambda ,A_i}\subset B_{\lambda ,{\sum }_i{A}_i}.$$

$$\mu\Big{(}\sum_{i}sA_{i}^{\epsilon}\Big{)}\leq\underline{\mathrm{d}}\Big{(}\sum_{i}B_{\lambda,A_{i}}\Big{)}\leq\overline{\mathrm{d}}\Big{(}\sum_{i}B_{\lambda,A_{i}}\Big{)}\leq\mu\Big{(}\sum_{i}A_{i}\Big{)}.$$

$$\frac{\beta }{2}<\alpha \le \frac{\beta }{2}+\frac{1}{2g_0}.$$

$$T_{k,\theta}=\Big{\{}n\geq 1\,\mid\,0<\{\theta n\}<\frac{1}{k+1}\Big{\}}.$$

$$A^{\prime }=\{\lfloor \theta a\rfloor ,a\in A\}.$$

$$\forall a_1,\dots ,a_{k+1}\in A,\lfloor \theta a_1\rfloor +\lfloor \theta a_2\rfloor +\cdots +\lfloor \theta a_{k+1}\rfloor =\lfloor \theta (a_1+a_2+\cdots +a_{k+1})\rfloor .$$

$$\mathrm{d}\big{(}(k-1)A_{q}\big{)}=0<\mathrm{d}\big{(}kA_{q}\big{)}=\frac{\beta}{q}<\mathrm{d}\big{(}(k+1)A_{q}\big{)}=\frac{1}{q}.$$

$$\left\{\begin{aligned} &\mathrm{d}\big{(}(k-1)A^{\prime}\big{)}=|(k-1)U|\times\mathrm{d}\big{(}(k-1)A_{q}\big{)}=0,\\ &\mathrm{d}\big{(}kA^{\prime}\big{)}=|kU|\times\mathrm{d}\big{(}kA_{q}\big{)}=\frac{kr\beta}{q}=\frac{k}{k+1}\beta\gamma^{\prime},\\ &\mathrm{d}\big{(}(k+1)A^{\prime}\big{)}=|(k+1)U|\times\mathrm{d}\big{(}(k+1)A_{q}\big{)}=\frac{(k+1)r}{q}=\gamma^{\prime}.\end{aligned}\right.$$

$$\left|\frac{1}{n}\left|\{1\le j\le n:s_j\in A\}\right|-\mu (A)\right|\ll \frac{1}{m}+\frac{1}{n}\sum _{k=1}^m\frac{1}{k}\left|\sum _{j=1}^n{e}^{2i\pi s_{j}k}\right|,$$

$$\left|\sum _{j=1}^m{e}^{2i\pi j\theta }\right|\le \frac{1}{2\Vert \theta \Vert },$$

$$\left|\frac{1}{n}{B}_{\theta ,A}(n)-\mu (A)\right|\ll \frac{1}{m}+\frac{1}{n}\sum _{k=1}^m\frac{1}{k\Vert \theta k\Vert }.$$

$$\frac{1}{n}\sum _{k=1}^{m(n)}\frac{1}{k\Vert \theta k\Vert }\asymp \frac{1}{m(n)}\to 0$$

$$\left|\frac{1}{n}{B}_{\theta ,A}(n)-\mu (A)\right|\le \eta (n).$$

$$\left|\frac{1}{n}(B_{\theta ,A}(n+X)-B_{\theta ,A}(X))-\mu (A)\right|\le \eta (n)$$

$$X_{\theta}=\Big{\{}n\in\mathbb{N}\,:\,2\eta(n/2)<\{\theta n\}<1-2\eta(n/2)\Big{\}}.$$

$$\mathrm{d}\big{(}X_{\theta}\big{)}=1.$$

$$P({\xi }_k=1)={\beta }_k,k\ge 1$$

$$K_n=\{0<k<n/2:k\in T_{1,\theta }\cap (n-T_{1,\theta })\},$$

$$R(n)=\sum _{k\in K_n}{\xi }_k{\xi }_{n-k}.$$

$$P(R(n)=0)=\prod _{k\in K_n}P({\xi }_k{\xi }_{n-k}=0)\le (1-{\beta }_n^2{)}^{|K_n|}\le \exp (-\left|K_n\right|{\beta }_n^2).$$

$$K_n\supset \left\{0<k<n/2:\left\{\theta k\right\}<\left\{\theta n\right\}\right\},$$

$$\sum _{n\in X_{\theta }}P(R(n)=0)<\infty .$$

$$\mathrm{d}\big{(}2A\big{)}\geq 2\mathrm{d}\big{(}A\big{)}-\frac{1}{g}.$$

$$A_g$$

$$A_g^{\prime }$$

$$A_g^{\prime \prime }$$

$$\tilde{A}=\bigcup _{\overline{x}\in A_g^{\prime }}(x+g\mathbb{N})\cup \{x:\overline{x}\in A_g^{\prime \prime }\}.$$

Full text

This work was performed within the framework of the LABEX MILYON (ANR-10-LABX-0070) of Université de Lyon, within the program “Investissements d’Avenir” (ANR-11-IDEX-0007) operated by the French National Research Agency (ANR)

1. Introduction

For $A\subset\mathbb{N}$ and $t>1$, we let $A(t)=|A\cap[1,t]|$. We define if it exists the so-called asymptotic density of $A$ by

[TABLE]

Otherwise we define the lower and the upper asymptotic densities $\underline{\mathrm{d}}(A)$ and $\overline{\mathrm{d}}(A)$ using $\liminf$ and $\limsup$ instead of limits. More generally, if $A\subset B\subset\mathbb{N}$, we define if it exists the density of $A$ inside $B$ as

[TABLE]

The density of $A$ inside $\mathbb{N}$ is therefore simply the density, and if $B$ has a density, we have \mathrm{d}_{B}(A)=\mathrm{d}\big{(}A\big{)}/\mathrm{d}\big{(}B\big{)}.

For a subset $A$ of a semigroup $G$, let $A+A=\{a+b:a,b\in A\}$. For $k\geq 1$, we denote by $kA$ its $k$-fold sumset. From Kneser’s Theorem [10], we know that for subsets $A\subset\mathbb{N}$, the inequality \underline{\mathrm{d}}\big{(}A+A\big{)}<2\underline{\mathrm{d}}\big{(}A\big{)} may only hold when \underline{\mathrm{d}}\big{(}A+A\big{)} is a rational number. Similarly, for any subset $A$ of the circle $\mathbb{T}=\mathbb{R}/\mathbb{Z}$ equipped with its Haar probability measure $\mu$, a theorem of Raikov [15] implies that $\underline{\mu}(2A)\geq\min(1,2\underline{\mu}(A))$ where $\underline{\mu}(A)=\sup_{\begin{subarray}{c}F\subset A\\ F\text{ closed}\end{subarray}}\mu(F)$.

In this paper, we determine the possible values $(\alpha,\beta)$ of pairs (\underline{\mathrm{d}}\big{(}A\big{)},\underline{\mathrm{d}}\big{(}2A\big{)}) and $(\underline{\mu}(A),\underline{\mu}(2A))$. We first completely settle the case $\beta\geq\min(1,2\alpha)$.

Theorem 1.1**.**

Let $(\alpha,\beta)\in[0,1]^{2}$. Suppose $\beta\geq\min(2\alpha,1)$. Then the following statements both hold.

1. a)

There exists $A\subset\mathbb{N}$ such that \mathrm{d}\big{(}A\big{)} and \mathrm{d}\big{(}2A\big{)} exist and equal $\alpha$ and $\beta$ respectively. 2. b)

There exists a measurable subset $A\subset\mathbb{T}$ such that $2A$ is measurable and $\mu(A)=\alpha$ and $\mu(2A)=\beta$. Further, for $\alpha>0$, we may take $A$ to be open (in fact a finite union of open intervals).

The case $\beta=2\alpha$ is obvious for the second item (with an interval $A$), and is a special case of a theorem by Faisant et al [6] for the first item, whereas allowing different summands, Volkmann [18] proved that, given positive real numbers $\alpha_{1}$, $\alpha_{2}$ and $\gamma$ such that $\alpha_{1}+\alpha_{2}\leq\gamma<1$, there exist111In Volkmann’s construction, the sets of integers are sets of relative integers and not necessarily positive integers though. $A_{1},A_{2}$ such that \mathrm{d}\big{(}A_{i}\big{)}=\alpha_{i}, $i=1,2$, and \mathrm{d}\big{(}A_{1}+A_{2}\big{)}=\gamma; he actually proved the corresponding result for subsets of the circle too. A similar result was obtained by Nathanson [14], including a version for Schnirelmann’s density.

More generally, we investigate the set $\mathcal{D}_{k}$ of possible values of the tuple

[TABLE]

when $A$ ranges over the set of sequences for which all of these densities exist. In parallel, we consider the similar problem in the circle $\mathbb{T}=\mathbb{R}/\mathbb{Z}$ equipped with its Haar measure $\mu$. Thus let $\mathcal{E}_{k}$ be the set of all the possible values of $(\mu(A),\ldots,\mu(kA))$ for $A\subset\mathbb{T}$ for which these measures exist. We may sometimes need to work with the subset $\mathcal{E}_{k}^{o}\subset\mathcal{E}_{k}$ of all the possible values of $(\mu(A),\ldots,\mu(kA))$ for $A\subset\mathbb{T}$ open and Riemann-measurable and similarly $\mathcal{E}_{k}^{c}$, where we consider closed sets $A$.

There is a close connection between $\mathcal{E}_{k}$ and $\mathcal{D}_{k}$ because of Weyl’s criterion for equidistribution, of which we now state a direct consequence. For $A\subset\mathbb{T}$ and $\lambda\in\mathbb{R}\setminus\mathbb{Q}$, let $B_{\lambda,A}=\{n\in\mathbb{N}:\{\lambda n\}\in A\}$, where $\{x\}=x-\lfloor x\rfloor$ denotes the fractional part of the real number $x$.

Theorem 1.2**.**

For any irrational number $\lambda$ and any Riemann-measurable function $f:[0,1]\rightarrow\mathbb{R}$, we have

[TABLE]

In particular, for any Riemann-measurable subset $A\subset\mathbb{T}$, we have \mathrm{d}\big{(}B_{\lambda,A}\big{)}=\mu(A). The latter equality may be extended to open sets $A$.

The extension to open sets is [18, Lemma 4]. Further, Theorem 1.2 and a simple compactness argument shows that for any $\epsilon>0$, there exists a constant $C=C(\theta,\epsilon)$ such that for any interval $I$ of length at least $\epsilon$ we have $B_{\theta,I}(C)\geq 1$. Finally, the operation $A\mapsto B_{\lambda,A}$ behaves well with respect to set addition.

Lemma 1.3**.**

Let $k\geq 2$ and $A_{i}\subset\mathbb{T}$ be open for $i=1,\ldots,k$ and $\lambda$ irrational. Then \mathrm{d}\big{(}B_{\lambda,\sum_{i=1}^{k}A_{i}}\big{)}=\mu(\sum_{i=1}^{k}A_{i}).

Proof.

For $A\subset\mathbb{T}$ open, let $A^{\epsilon}=\{x\in A:\textrm{dist}(x,\partial A)>\epsilon\}$. Thus $A=\bigcup_{\epsilon>0}A^{\epsilon}$ and $\mu(A)=\lim_{\epsilon\rightarrow 0}\mu(A^{\epsilon})$. Further, $\sum_{i}A_{i}=\bigcup_{\epsilon>0}\sum_{i}A_{i}^{\epsilon}$. We observe that

[TABLE]

The rightmost inclusion is easy; for the leftmost one, let $x\in B_{\lambda,\sum_{i}A_{i}^{\epsilon}}$, thus $x=\sum_{i}a_{i}$ where $a_{i}\in A_{i}^{\epsilon}$. Consequently, $(a_{i}-\epsilon/k,a_{i}+\epsilon/k)\subset A_{i}$ for $i\in\{1,\ldots,k-1\}$. If $n$ is large enough (larger than some constant $C(\epsilon,k)$), there exists $n_{1},\ldots,n_{k-1}\leq n/k$ such that $\{n_{i}\lambda\}\in(a_{i}-\epsilon/k,a_{i}+\epsilon/k)$. Let $n_{k}=n-n_{1}-\cdots-n_{k-1}>0$. Then $\{n_{k}\lambda\}=\{n\lambda\}-\{n_{1}\lambda\}-\cdots-\{n_{k-1}\lambda\}\text{ mod }1$, which implies $\{n_{k}\lambda\}\text{ mod }1\in(a_{k}-\epsilon,a_{k}+\epsilon)\subset A_{k}$, in other words $n_{k}\in B_{\lambda,A_{k}}$. Thus $n\in\sum_{i}B_{\lambda,A_{i}}$.

Taking densities and applying Theorem 1.2 in equation (1), we find that

[TABLE]

Letting $\epsilon\rightarrow 0$, we conclude. ∎

Consequently, $\mathcal{E}_{k}^{o}\subset\mathcal{D}_{k}$; in particular, the second item of Theorem 1.1 implies the first one when $\alpha>0$, but we will provide another proof for it. Further, Raikov’s theorem together with Theorem 1.1 means that $\mathcal{E}_{2}=\mathcal{E}_{2}^{o}=\mathcal{E}_{2}^{c}=\{(\alpha,\beta)\in[0,1]^{2}:\beta\geq\min(1,2\alpha)\}$.

To complete our description of $\mathcal{D}_{2}$, we need to understand which pairs $(\alpha,\beta)$ with $\beta<2\alpha$ belong to it, which we do in the next theorem. For an integer $n$, let $v_{2}(n)$ be its dyadic valuation; we extend it to rational numbers by letting $v_{2}(p/q)=v_{2}(p)-v_{2}(q)$.

Theorem 1.4**.**

Let $\beta\in\mathbb{Q}\cap(0,1)$ such that $v_{2}(\beta)\leq 0$, let $\alpha\in(0,1)$ satisfy $\beta<2\alpha$ and $g_{0}$ denote $\min\{g\geq 1\,:\,g\beta\text{ is odd}\}$. Then there exists a sequence $A\subset\mathbb{N}$ such that \mathrm{d}\big{(}A\big{)}=\alpha, \mathrm{d}\big{(}2A\big{)}=\beta<2\alpha if and only if

[TABLE]

Example 1.5**.**

The pair $\alpha=4/9$, $\beta=5/9$ enforces $g_{0}\leq 3$ and $1\leq r\leq 2$, whence $\beta=1$, $1/2$ or $1/3$, a contradiction.

Example 1.6**.**

The pair $\alpha=1/5$, $\beta=3/10$ yields $g_{0}\leq 10$ and $1\leq r\leq 5$. Choosing $r=2$ gives the required condition.

We briefly discuss iterated sumsets. It is not clear what constraints a tuple $(\alpha_{i})_{i\in[k]}$ must satisfy for a set $A\subset\mathbb{R}/\mathbb{Z}$ satisfying $\mu(iA)=\alpha_{i}$ to exist; we certainly need $\alpha_{i}\geq\min(1,\alpha_{j}+\alpha_{i-j})$ for any $j<i$ due to Raikov’s theorem but it may not be sufficient. In particular, we will deduce the following constraint from a theorem of Gyarmati, Konyagin and Ruzsa [8].

Theorem 1.7**.**

*There exists a constant $c>0$ such that the following holds. Let $A\subset\mathbb{T}$ be closed, and suppose that $\mu(2A)<c$. Then $\mu(3A)\geq\frac{3}{2}\mu(2A)$. *

In view of Lev’s analogous result [12] on finite sets of integers, one may more generally imagine that $\mu((k+1)A)\geq\frac{k+1}{k}\mu(kA)$ under certain restrictions on $\mu(kA)$. Note that another result from [8] implies that the constant $c$ may not be taken to be 1. Gyarmati et al. conjecture that its optimal value is 1/2. Note that for any finite set $A$ of integers, we have $2\left\lvert 3A\right\rvert\geq 3\left\lvert 2A\right\rvert-1$. On the other hand, due to the Plünnecke-Ruzsa inequalities, we know that if \mathrm{d}\big{(}2A\big{)}\leq K\mathrm{d}\big{(}A\big{)}, we must have \mathrm{d}\big{(}3A\big{)}\ll K^{3}\mathrm{d}\big{(}A\big{)}. Similarly, in the circle, if $\mu(2A)<3\mu(A)$ and $\mu(A)$ is small enough, Moskvin et al. [13] showed that $A$ must satisfy strict structural conditions that imply that $\mu(3A)\leq 3(\beta-\alpha)$.

We solve partially the problem with $k=3$.

Theorem 1.8**.**

Let $(\alpha,\beta,\gamma)\in(0,1]^{3}$, and suppose that $\beta<\min(3\alpha,1)$ and $\gamma\in[\min(1,3\beta/2),\min(1,2\beta-\alpha)]$ or that $\beta=3\alpha$ and $\gamma\in[3\beta/2,2\beta]$. Then $(\alpha,\beta,\gamma)\in\mathcal{E}_{3}$.

For general $k$, our understanding of $\mathcal{E}_{k}$ and $\mathcal{D}_{k}$ is yet poorer. Note that in general, our sets $A\subset\mathbb{N}$ satisfy \mathrm{d}\big{(}(k+1)A\big{)}\geq\frac{k+1}{k}\mathrm{d}\big{(}kA\big{)}, which, in view of the aforementioned result of Lev, may be inevitable.

Theorem 1.9**.**

Let $\alpha=(\alpha_{1},\ldots,\alpha_{k+1})\in[0,1]^{k+1}$, where $k\geq 1$.

1. a)

If $\alpha_{1}=\cdots=\alpha_{k-1}=0$ and $\alpha_{k+1}\geq\frac{k+1}{k}\alpha_{k}$, or $\alpha_{k+1}\geq\alpha_{k}$ and $\alpha_{k+1}$ is the inverse of an integer, then $\alpha\in\mathcal{D}_{k+1}$. 2. b)

If $\alpha_{1}=\cdots=\alpha_{k}=0$, then $\alpha\in\mathcal{E}_{k+1}$. 3. c)

If $\alpha_{i}=i\alpha$ for each $i$ and some $\alpha\leq 1/(k+1)$, then $\alpha\in\mathcal{E}_{k+1}^{0}\subset\mathcal{D}_{k+1}$.

The last item is obvious by taking an interval of length $\alpha$, and was also proven somewhat differently for $\mathcal{D}_{k}$ in [6].

In the next section, we prove the complete description of $\mathcal{D}_{2}$ and $\mathcal{E}_{2}$ given in Theorems 1.1 and 1.4.

2. Sumsets in the integers

2.1. A preliminary reduction

We show that Theorem 1.9 a) follows from the special case below, where $\alpha_{k+1}=1$ in the notation of that theorem.

For a real number $\theta>1$, let

[TABLE]

Note that \mathrm{d}\big{(}T_{k,\theta}\big{)}=1/(k+1) if $\theta$ is irrational, while $T_{k,\theta}=\mathbb{N}$ if $\theta$ is an integer. In any case, $(k+1)T_{k,\theta}=\mathbb{N}$.

Proposition 2.1**.**

Let $\beta\in[0,1)$ and integer $k\geq 1$. There exists a set $A\subset T_{k,\theta}$ such that $iA$ has density 0 for any $i<k$, whereas $kA$ has density $\beta$ inside $kT_{k,\theta}$ and $(k+1)A$ has density 1 in $\mathbb{N}$.

In particular, we have \mathrm{d}\big{(}kA\big{)}=\beta k/(k+1) if $\theta$ is irrational while \mathrm{d}\big{(}kA\big{)}=\beta if $\theta$ is an integer.

We now deduce Theorem 1.9 a) from Proposition 2.1. Let $\alpha\in[0,1]^{k+1}$ be as in the hypothesis of the former theorem, and let $\beta^{\prime}=\alpha_{k}$ and $\gamma^{\prime}=\alpha_{k+1}$. We distinguish several cases.

1. a)

We first assume that $\gamma^{\prime}$ is an irrational number. Let $A$ be the set given in Proposition 2.1 with parameters $\theta=\frac{1}{\gamma^{\prime}}$ and $\beta=\frac{\beta^{\prime}}{\gamma^{\prime}}$ and $A^{\prime}$ be defined by

[TABLE]

Since $A\subset T_{k,\theta}$ we have

[TABLE]

Since $\theta>1$, we get \mathrm{d}\big{(}jA^{\prime}\big{)}=\theta^{-1}\mathrm{d}\big{(}jA\big{)}, $j=1,2,\dots,k+1$.

This yields Theorem 1.9 a) when $\gamma^{\prime}$ is an irrational number. 2. b)

If $\gamma^{\prime}$ is the inverse of a positive integer $q$, we use again Proposition 2.1 with parameters $\theta=\frac{1}{\gamma^{\prime}}$ and $\beta=\frac{\beta^{\prime}}{\gamma^{\prime}}$ to generate a set $A$ and define a set $A_{q}=\{qa,\ a\in A\}$ satisfying

[TABLE] 3. c)

We finally assume that $\gamma^{\prime}=\frac{s}{q}$ is a rational number with $2\leq s<q$. Upon multiplying numerator and denominator by appropriate numbers, we may assume that $s=(k+1)r$ for some integer $r$ satisfying $3\leq r<\frac{q}{k+1}$. Let $U=\{0,1,\dots,r-2,r\}.$ Then $|jU|=jr$ for any $j$. Letting $A^{\prime}=U+A_{q}$, we thus obtain

[TABLE]

This concludes the proof of Theorem 1.9 a), assuming Proposition 2.1. We will now prove the latter, focussing first on the case $k=1$ (so concerning twofold sumsets, that is Theorem 1.1), since it is much more simple than, while retaining some important features of, the general case, which we handle later.

2.2. Twofold sumsets

Before embarking on the proof of Proposition 2.1 in the case $k=1$, we need a quantitative version of Weyl’s criterion (Theorem 1.2), due to Erdős and Turán [5, Theorem III].

Theorem 2.2**.**

For any sequence $s_{j}$ of elements of the torus and any interval $A$, we have for any integers $n$ and $m$ the bound

[TABLE]

where the implied constant is absolute.

Applying this with $s_{j}=\{\theta j\}$ for some irrational number $\theta$ and using the standard exponential sum bound

[TABLE]

where $\left\lVert\theta\right\rVert=\min_{k\in\mathbb{Z}}|\theta-k|$, we obtain

[TABLE]

The series $\sum_{k=1}^{m}\frac{1}{k\left\lVert\theta k\right\rVert}$ diverges as $m$ tends to infinity, but selecting $m=m(n)$ as a sufficiently slowly increasing function of $n$, one may achieve

[TABLE]

as $n$ tends to infinity, and thus there exists a function $\eta:\mathbb{N}\rightarrow\mathbb{R}_{+}$ (depending on $\theta$ only) that tends to zero such that

[TABLE]

Note that the bound (3) is uniform in $A$; in particular, it is still valuable if $A$ is replaced by a sequence $A_{n}$ of intervals of sufficiently slowly decaying measure (e.g. $\mu(A_{n})\geq 2\eta(n)$). Also we note that using the sequence $s_{j}=\left\{\theta(j+X)\right\}$, we may obtain the more general bound

[TABLE]

for any integers $X$ and $n$.

We now start the proof of Proposition 2.1 in the case $k=1$. We will adopt a probabilistic construction. Let $\theta$ be an irrational number and $\eta$ be a function for which equation (3) holds and

[TABLE]

Equation (3) and the ensuing remarks imply that

[TABLE]

We now define our desired random sequence $A$. Let $(\xi_{k})_{k\geq 1}$ be a sequence of mutually independent Bernoulli random variables such that

[TABLE]

where $\beta_{k}$ is the constant sequence equal to $\beta$ if $\beta>0$ and the decaying sequence $k^{-1/5}$ if $\beta=0$. Let $A$ be the random sequence consisting of the integers $k\in T_{1,\theta}$ such that $\xi_{k}=1$. It is easy to see that the density of $A$ inside $T_{1,\theta}$ satisfies $\mathrm{d}_{T_{1,\theta}}(A)=\beta$ almost surely as required.

Now we prove that $A+A\supset X_{\theta}\setminus F$, where $F$ is almost surely a finite set. This would imply that \mathrm{d}\big{(}A+A\big{)}=1, as desired. Let $n\in X(\theta)$. We define

[TABLE]

and

[TABLE]

Then by the independence of the $\xi_{k}$’s

[TABLE]

We now estimate $\left\lvert K_{n}\right\rvert$ from below. By definition $k<n/2$ belongs to $K_{n}$ if and only if $\{\theta k\}<1/2$ and $\{\theta(n-k)\}<1/2$.

Let $I=(2\eta(n/2),1/2)$. Since $n\in X_{\theta}$, we have $\{\theta n\}\in I\cup(1-I)$. Suppose for instance $\{\theta n\}\in I$, the case $\{\theta n\}\in 1-I$ being similar. Then for any $k$ such that $\{\theta k\}<\left\{\theta n\right\}<1/2$, we have $\left\{\theta(n-k)\right\}=\left\{\theta n\right\}-\left\{\theta k\right\}<1/2$. Thus $k\in K_{n}$. This means that

[TABLE]

whence $\left\lvert K_{n}\right\rvert\geq n/2(\left\{\theta n\right\}-\eta(n/2))\geq\frac{n}{2}\eta(n/2)$ by equation (3). If $\left\{\theta n\right\}\in 1-I=(1/2,1-2\eta(n))$ instead, it suffices to replace the condition $\left\{\theta k\right\}<\left\{\theta n\right\}$ by $\frac{1}{2}-\left\{\theta k\right\}<1-\left\{\theta n\right\}$ to obtain the same result.

One can choose $\eta(n)$ to be arbitrarily slowly decaying, say $\eta(n)\geq n^{-1/2}$. This way $\left\lvert K_{n}\right\rvert\gg\sqrt{n}$, so that $\beta_{n}^{2}\left\lvert K_{n}\right\rvert\gg n^{1/10}$ and from (6) we get

[TABLE]

We conclude by the Borel-Cantelli lemma (cf. [17, Lemma 1.2]) that almost surely, all but finitely many integers of $X_{\theta}$ are sums of $2$ terms from the random sequence $A$. The result follows from (5). This finishes the proof of Proposition 2.1 in the case where $k=1$, and thus of Theorem 1.1.

We now determine which pairs $(\alpha,\beta)\in\mathbb{R}^{2}$ with $0<\alpha\leq\beta<2\alpha$ belong to $\mathcal{D}_{2}$, that is, we prove Theorem 1.4.

Let $A\subseteq\mathbb{N}$ such that \beta=\mathrm{d}\big{(}2A\big{)}<2\mathrm{d}\big{(}A\big{)}=2\alpha. By Kneser’s theorem for infinite sequences, there exists a (minimal) positive integer $g$ such that $\ (2A+g\mathbb{N})\setminus 2A$ is finite and

[TABLE]

Let

[TABLE]

We have $A_{g}=A^{\prime}_{g}\cup A^{\prime\prime}_{g}$. Let

[TABLE]

Then

[TABLE]

Since $g$ is minimal we have $|A_{g}+A^{\prime}_{g}|\geq|A_{g}|+|A^{\prime}_{g}|-1$ since otherwise $A_{g}+A^{\prime}_{g}$ has a nontrivial period. Hence from \mathrm{d}\big{(}2A\big{)}<2\mathrm{d}\big{(}A\big{)} we get $|A_{g}|+|A^{\prime}_{g}|-1<2|A^{\prime}_{g}|$ giving $A^{\prime}_{g}=A_{g}$ and finally $|2A_{g}|=2|A_{g}|-1$.

Let $r=|A_{g}|$. Then $\beta=\frac{2r-1}{g}$ with $1\leq r\leq\frac{g+1}{2}$. We get

[TABLE]

We proved the following.

Proposition 2.3**.**

Let $A$ such that \mathrm{d}\big{(}2A\big{)}<2\mathrm{d}\big{(}A\big{)}. Then there exist two positive integers $g$ and $r\leq\frac{g+1}{2}$ such that

[TABLE]

Conversely, let $\beta\in[0,1]\cap\mathbb{Q}$ have nonpositive dyadic valuation, and $g$ be the smallest positive integer for which $g\beta$ is odd, thus $\beta=\frac{2r-1}{g}$ and let $\alpha$ satisfy

[TABLE]

Then let $R=\{0,\ldots,r-1\}\in\mathbb{Z}/g\mathbb{Z}$, so that $\left\lvert 2R\right\rvert=2r-1$. Let $\gamma=\alpha\frac{g}{r}$, thus $\gamma\in(0,1]$. Take $A_{g}\subset g\mathbb{N}$ constructed in the proof of Proposition 2.1 (with $k=1$), so that \mathrm{d}\big{(}A_{g}\big{)}=\gamma/g and \mathrm{d}\big{(}2A_{g}\big{)}=1/g and let $A=\cup_{x\in R}x+A_{g}$, which has density $\alpha$. Consequently $2A=\cup_{x\in 2R}x+g\mathbb{Z}$, which yields \mathrm{d}\big{(}2A\big{)}=\beta as desired.

This completes the proof of Theorem 1.4.

3. Measures of sumsets in the circle

3.1. Twofold sumsets

To start with, we show that in order to achieve a large ratio $\mu(2A)/\mu(A)$, a large number of connected components will be necessary.

Proposition 3.1**.**

Let $A$ be a disjoint union of $k$ intervals. Then $\mu(2A)\leq(k+1)\mu(A)$. If the intervals are open, the equality case happens when all the $\binom{k+1}{2}$ intervals of the sum are pairwise disjoint.

Proof.

Let $A=\bigcup_{j=1}^{k}I_{j}$. So $A+A=\bigcup_{i\leq j}(I_{i}+I_{j})$. Let $\mu(I_{i})=m_{i}$, so $\mu(I_{i}+I_{j})=m_{i}+m_{j}$ and $\mu(A+A)\leq\sum_{i\leq j}(m_{i}+m_{j})=(k+1)\sum_{i}m_{i}$. The equality case is clear. ∎

We now attempt to prove the first item of Theorem 1.1 in the case $\alpha>0$. Let $(\alpha,\beta)\in(0,1]^{2}$ satisfy $\beta\geq\min(2\alpha,1)$. If $\beta=\min(2\alpha,1)$, the set $A=(0,\alpha)$ satisfies $\mu(A)=\alpha,\mu(2A)=\beta$. So we now suppose $0<\alpha<1/2$ and $\beta>2\alpha$.

First, note that for any $k$, if $A=[0,\ell]\cup\{2\ell\}\cup\cdots\cup\{(k-1)\ell\}$, then $A+A=[0,k\ell]$ so we can achieve a duplication ratio $\mu(2A)/\mu(A)=k$. The idea is then to somewhat “thicken” the singletons, in order to reduce the duplication ratio of the set.

Let $k=\lfloor\beta/\alpha\rfloor$, thus $k\leq\beta/\alpha<k+1$ and $k\geq 2$.

Then let $A=(0,x)\cup(\{2x,\ldots,kx\}+(-\epsilon,0))$, for some $x\leq\alpha$ and $\epsilon\leq x/2$ to be determined later. Note that

[TABLE]

Thus $\mu(A)=x+(k-1)\epsilon$ and $\mu(2A)=(k+1)x+2(k-1)\epsilon$. The doubling ratio is therefore

[TABLE]

We have $f(0)=k+1$ and while $f(1/2)=4k/(k+1)\leq k$. Therefore by continuity of $f$, there is a value of the ratio $y=\epsilon/x$ for which the doubling ratio is the desired $\beta/\alpha$.

Then there remains to pick $x$ such that $\alpha=x+(k-1)\epsilon=x(1+(k-1)y)$, namely $x=\frac{\alpha}{1+(k-1)y}$, and then the corresponding $\epsilon$.

In the case $\alpha=0$, a radically different construction will be necessary. Let $C\subset[0,1]$ be the classical ternary Cantor set. It is well known that $C+C=[0,2]$. For the sake of completeness, we reproduce a short proof. It suffices to prove $C+C\supset[0,2]$. Let $u\in[0,2]$ and let $(\epsilon_{i})_{i\geq 1}\in\{0,1,2\}^{\mathbb{N}\setminus\{0\}}$ be the digits of $u/2$ in its ternary expression, thus

[TABLE]

We construct sequences $\alpha$ and $\beta$ in $\{0,2\}^{\mathbb{N}\setminus\{0\}}$ such a way that for each $i\geq 1$, we have $\alpha_{i}+\beta_{i}=2\epsilon_{i}$. Thus if $\epsilon_{i}=0$ we take $\alpha_{i}=\beta_{i}=0$; if $\epsilon_{i}=1$ we define $\alpha_{i}=0$ and $\beta_{i}=2$; otherwise $\alpha_{i}=\beta_{i}=2$. Letting $x=\sum_{i\geq 1}\alpha_{i}3^{-i}$ and $y=\sum_{i\geq 1}\beta_{i}3^{-i}$, we see that $x$ and $y$ are in $C$ and $x+y=2\cdot u/2=u$, which concludes.

Scaling $C$ it by a factor $\beta/2$ and projecting it to the circle, we obtain a set $A=(\beta/2)C$ of measure 0 such that $\mu(2A)=\beta$.

3.2. Threefold sumsets

First we prove Theorem 1.7. We will derive it from the following theorem of Gyarmati, Konyagin and Ruzsa [8].

Proposition 3.2**.**

There exists an absolute constant $c>0$ such that the following holds. Let $p\geq 29$ be a prime. Let $A\subset\mathbb{Z}/p\mathbb{Z}$ and let $(n,s)=(\left\lvert 2A\right\rvert,\left\lvert 3A\right\rvert)$. If $n<cp$, then $s\geq\frac{3n-1}{2}$.

We derive the analogous result for measures in the circle by a standard method. We first prove Theorem 1.7 for simple sets, that is, the union of finitely many closed intervals. Let $A\subset\mathbb{T}$ be a simple set. Let $c$ be the constant given by Proposition 3.2 and suppose that $\mu(2A)<c$. Let $p\geq 29$ be a prime, that we will let tend to infinity ultimately. Let

[TABLE]

This notation should not conflict with the notation $A(t)$ defined in the introduction. One may check that $\left\lvert A(p)\right\rvert=p\mu(A)+O(1)$ as $p$ tends to infinity. Further note that $(kA)(p)=kA(p)$ for any $k\in\mathbb{N}$. Since $2A$ and $3A$ are simple, one has $\left\lvert(kA)(p)\right\rvert=p\mu(kA)+O(1)$ for $k=2,3$; thus we have $\left\lvert(2A)(p)\right\rvert<cp$ for $p$ sufficiently large, so we can apply Proposition 3.2 and conclude in the case of simple sets.

Now if $A$ is closed (that is, compact), writing $I_{\delta}=(-\delta,\delta)$, we have $A=\bigcap_{\delta>0}(A+I_{\delta})$, in fact $kA=\bigcap_{\delta>0}(kA+I_{k\delta})$ for any integer $k\geq 1$. So for any fixed $\epsilon>0$, we can chose $\delta$ such that $\mu(kA+I_{k\delta})\leq\mu(kA)+\epsilon$. Further, by compacity, there exists a simple set $A^{\prime}$ (the union of finitely many translates of $I_{\delta}$) such that $A\subset A^{\prime}\subset A+I_{\delta}$. We have

[TABLE]

Letting $\epsilon$ tend to zero, we conclude the proof of Theorem 1.7.

We prove Theorem 1.8. If $\alpha\geq 1/3$, the triplets $(\alpha,\beta,\gamma)$ that belong to $\mathcal{E}_{k}$ are the ones for which $\beta\geq\min(1,2\alpha)$ and $\gamma=1$.

We now consider triplets where $\alpha<1/3$; we prove the following proposition, which implies Theorem 1.8.

Proposition 3.3**.**

The set of triplets $(\mu(A),\mu(2A),\mu(3A))$ for sets $A\subset[0,1/3]\subset\mathbb{T}$ having at most two connected components is

[TABLE]

Proof.

We may take $A$ of the form $(0,x)\cup(y,z)$ for some $0\leq x\leq y\leq z\leq 1/3$. So $A+A=(0,2x)\cup(y,x+z)\cup(2y,2z)$ and $3A=(0,3x)\cup(y,2x+z)\cup(2y,2z+x)\cup(3y,3z)$.

We are seeking for which triplets $(\alpha,\beta,\gamma)$ the system

[TABLE]

admits solutions. We now discuss the existence of solutions according to the number of connected components of $2A$ and $3A$, that is, for each max above, whether it is positive or not. In the following discussion, the necessary conditions we provide may always easily be seen to be sufficient, although we do not always explicitly state it.

If $2A$ is an interval, then so is $3A$ so $\gamma=3\alpha=3\beta/2$. 2. 2)

If $2A$ has two connected components, so exactly one overlap between the intervals of $2A$, we distinguish.

1. a)

If $2x>y$ and $x+z<2y$, so $2A=(0,x+z)\cup(2y,2z)$, we have $\beta=x-2y+3z$. We have necessarily $3x>y$ and $2x+z>2y$, so $3A=(0,2z+x)\cup(3y,3z)$ where the last two intervals may overlap or not.

1. i)

If they do, so $2z+x>3y$, we have $\gamma=3z$. So $\beta=x-2y+\gamma$ and $\alpha=x-y+\gamma/3$. Get $\alpha-\beta=y-2\gamma/3$ so $y=\alpha-\beta+2\gamma/3$ while $x=2\alpha-\beta+\gamma/3$. We check that the inequalities are satisfied: $2x-y=3\alpha-\beta>0$ so $\beta<3\alpha$, $2y-x-z=-\beta+2\gamma/3>0$ implies $\gamma>3\beta/2$. Further, we need $2z+x-3y=-\alpha+2\beta-\gamma>0$ which amounts to $3\beta/2<\gamma<2\beta-\alpha<5\alpha$. Conversely, whenever these conditions are satisfied, the system has solutions. 2. ii)

If they don’t, so $2z+x<3y$, we have $\gamma=3(z-y)+2z+x$. Thus a solution exists if and only if $\gamma=2\beta-\alpha$. 2. b)

Now if $2x<y$ and $x+z>2y$, so $2A=(0,2x)\cup(y,2z)$, we have $\beta=2x-y+2z$. We have necessarily $2x+z>2y$ and $2z+x>3y$, so $3A=(0,3x)\cup(y,3z)$, where the two intervals may or not overlap.

1. i)

If they do, so $2x<y<3x$, we have $\gamma=3z$. Further we find $y=\beta-2\alpha$, and $x=\beta-\alpha-\gamma/3$. So $y-2x=-\beta+2\gamma/3>0$ implies yet again $\gamma>3\beta/2$. Further $y-3x=-2\beta+\alpha+\gamma<0$ implies $\gamma<2\beta-\alpha$. Also $x+z-2y=3\alpha-\beta>0$ amounts to $\beta<3\alpha$. 2. ii)

Otherwise, so $y>3x$, we find $\gamma=3z-y+3x=3\alpha+2y$ and again $\gamma=2\beta-\alpha$. 3. 3)

If $2A$ has three connected components (no overlap), then $\beta=3\alpha$. We have $2x<y$ and $x+z<2y$. We distinguish according to the presence of overlaps or not in $3A$.

1. a)

If there is no overlap, we have $\gamma=6\alpha$. It is realisable, just take $x$, then $y>3x$, then $y<z<\min((3y-x)/2,1/3)$, then all constraints are realised. We can achieve that for any value of $\alpha\leq 1/6$. 2. b)

If $3A$ is connected, $\gamma=3z$. Now the conditions $2x<y$ and $x+z<2y$ imply $z<3(y-x)$, which is equivalent to $2z>3(x+z-y)$, and finally $\gamma>3\beta/2$. 3. c)

If there is exactly one overlap, that is, if $3A$ has three connected components, we distinguish.

1. i)

Suppose $3x>y$. And $2x+z<2y$ and $2z+x<3y$. So $\gamma=6\alpha-3x+y$. This imposes $\gamma\in(5\alpha,6\alpha)=(5\beta/3,2\beta)$. 2. ii)

Now suppose $2x+z>2y$. And $y>3x$ and $2z+x<3y$. Then $\gamma=6\alpha-2x-z+2y=5\alpha-x+y>5\alpha$. 3. iii)

If only the last gap is overcome, $\gamma=6\alpha-2z-x+3y=5\alpha-z+2y>5\alpha$. 4. d)

If $3A$ has two connected components, we distinguish.

1. i)

If all but the last gap are overcome, $\gamma=6\alpha-3x+y-2x-z+2y=5\alpha-4x+2y>5\alpha.$ 2. ii)

If all but the middle gaps are overcome, $\gamma=6\alpha-3x+y-2z-x+3y=5\alpha-3x-z+3y>5\alpha$. 3. iii)

If all but the first gap are overcome, $\gamma=6\alpha-2x-z+2y-2z-x+3y>5\alpha$.∎

Regarding sets with $k$ connected components when $k\geq 3$, the determination of the possible triplets $(\alpha,\beta,\gamma)$ becomes untractable by this method. Nevertheless, we can easily see that the structure of the set of the possible triplets remains similar, that is, a connected union of finitely many (in fact $O_{k}(1)$ many) polytopes, where a polytope is the intersection of finitely many half-spaces.

3.3. Further iterated sumsets

We now prove Theorem 1.9 b). Let $\beta\in(0,1]$ and $k\geq 2$ an integer. Let $C_{k+1}$ be the Cantor set of initial segment $[0,1]\subset\mathbb{R}$ and ratio of dissection $1/(k+1)$. It is known [2, Corollary 2.3] that $\mu((k-1)C)$ has measure 0 whereas $kC=[0,k]$. A suitable scaling of $C_{1/(k+1)}$ provides the desired construction.

Note that this does not imply the first point of Theorem 1.9: the openness condition of Lemma 1.3 may not be removed. Indeed, if $A\subset\mathbb{R}/\mathbb{Z}$ has measure zero, one may see that $B_{\lambda,A}$ is empty for almost all $\lambda\in\mathbb{R}/\mathbb{Q}$, since the map $\lambda\mapsto n\lambda$ on the circle is measure-preserving for any integer $n$. So we need to provide a specific proof, which we do in the next section.

4. Iterated sumsets in the integers

We now prove Proposition 2.1 for $k\geq 2$. The (probabilistic) argument we will use subsumes, but is significantly more complicated than, the one used in Section 2, which is why we preferred to present it separately. First of all we collect a number of useful but technical results.

4.1. Preliminary lemmas

First we need to somewhat generalise the bound (4) obtained via the Erdős-Turán theorem.

Proposition 4.1**.**

Let $k,D,M,X$ be integers. Let $f=\sum_{i=1}^{k}P_{i}\mathbf{1}_{I_{i}}$ where $(I_{i})_{i\leq k}$ is a family of pairwise disjoint intervals in $[0,1)$ and $P_{i}$ a polynomial of degree less than $D$ whose coefficients are all at most $M$. Then

[TABLE]

A function $f$ satisfying the above hypothesis will naturally be referred to as piecewise polynomial.

Proof.

It suffices to prove it for monomials and for $k=1$, the general case following by linear combinations (incurring an extra factor $Mk$). Thus let $a<b$ be in $[0,1)$, and let $d\leq D$ and $f$ be defined by $f(x)=x^{d}\mathbf{1}_{(a,b)}$. Using the bound (4), we note that

[TABLE]

Further, observe that

[TABLE]

Hence

[TABLE]

Given that $b^{d}-a^{d}\leq d(b-a)$, we find that

[TABLE]

Then splitting the interval $[a,b]$ into $O(\sqrt{\eta(N)}^{-1})$ consecutive intervals of size $\lfloor\sqrt{\eta(N)}\rfloor$, we obtain, for each of these intervals, an error term of size $O(d\eta(N))$, and so in total, an error term of size $O(D\sqrt{\eta(N)})$. ∎

A certain type of sums will appear in the sequel, for which we now give an asymptotic.

Lemma 4.2**.**

Let $0<\alpha,\beta<1$ and

[TABLE]

Then

[TABLE]

where $B(\cdot,\cdot)$ denotes the Euler beta function defined by

[TABLE]

and $\zeta(\cdot)$ is the Riemann zeta function.

This can be proven by considering Riemann sums; we omit the standard details. The beta function satisfies the following functional equation involving Euler’s gamma function:

[TABLE]

By induction, we may achieve the following simple lemma.

Lemma 4.3**.**

Let $(\alpha_{1},\ldots,\alpha_{s})\in(0,1)^{s}$. Then

[TABLE]

Further, let $\epsilon:\mathbb{N}\rightarrow\mathbb{R}_{+}$ tend to 0. Then there exists a sequence $\epsilon^{\prime}$ depending only on $\epsilon$ that tends to zero such that

[TABLE]

Proof.

We prove the second part for $s=2$, the rest following by a simple induction. Let $K_{\delta}$ be such that for all $k\geq K_{\delta}$, we have $\epsilon(k)\leq\delta$. Further let $M$ be an upper bound for $\epsilon$. Then

[TABLE]

The right-hand side is $O(K_{\delta}^{1-\alpha_{1}-\alpha_{2}}+\delta n^{1-\alpha_{1}-\alpha_{2}})$ by Lemma 4.2. We have $K_{\delta}\rightarrow\infty$ (unless $\epsilon(k)=0$ eventually) as $\delta\rightarrow 0$, but choosing $\delta$ as a sufficiently slowly decaying function of $n$, we can make the error term as small as $o(n^{1-\alpha_{1}-\alpha_{2}})$ as desired. ∎

For any real number $0\leq x\leq 1$ and any integer $1\leq j\leq k-1$, let

[TABLE]

and $I_{j}(x)$ be the open interval

[TABLE]

Let $f_{1}=\mathbf{1}_{[0,1[}$ and

[TABLE]

Then for any $1\leq j\leq k-1$

1. i)

$a_{j}$ and $b_{j}$ are piecewise affine. Further $a_{j}(x)+b_{j}(x)=x$. 2. ii)

$\mu(I_{j}(x))=b_{1}(x)-a_{j}(x)=\max\left(0,\min\left(x,\frac{1}{k+1},\frac{j+1}{k+1}-x\right)\right)$. As a result, $f_{j}$ is supported on $(0,\frac{j}{k+1})$. 3. iii)

$f_{j}$ is a non negative, nonzero piecewise polynomial function. In fact $f_{j}$ has only finitely many zeros on $(0,j/(k+1))$.

We will need the following estimate.

Lemma 4.4**.**

Let $(\alpha,\beta)\in(0,1)^{2}$. Let $\theta>1$ be irrational and $x\in(0,1)$. Then for any $j$, we have

[TABLE]

where $\eta^{\prime}$ is a function $\mathbb{N}\rightarrow\mathbb{R}_{+}$ that tends to zero and that depends only on $\theta$.

Proof.

We decompose the interval of summation $[1,N)$ into subintervals of some length $m=f(N)$ tending to infinity rather slowly, $m=o(N)$ at any rate, even $m\ll N^{o(1)}$ but not too slowly either; we fix $m=\lfloor\eta(N)^{-1/2}\rfloor$ for definiteness. We write

[TABLE]

where the last interval has at most $m$ elements.

Let $K=\lfloor\frac{N}{m}\rfloor$. Let $a=-\alpha$ and $b=-\beta$. We note that

[TABLE]

Denoting by $S$ the sum to estimate, this implies that

[TABLE]

Also we note that when $n\in(km,(k+1)m]$, the expression $n^{a}(N-n)^{b}$ may be regarded as approximately constant, more precisely

[TABLE]

We may restrict the sum over $k$ to reasonably large $k$, like between $\sqrt{K}$ and $K-\sqrt{K}$; indeed, we have

[TABLE]

which is negligible to $N^{a+b+1}$. We may argue analogously to discard the sum over $k\geq K-\sqrt{K}$. This way $(1+O(1/k))(1+O(1/(K-k))=1+O(1/\sqrt{K})$ for any $k$ considered. Thus $S$, up to an error $O(N^{a+b+1}/\sqrt{K}))$, equals

[TABLE]

We now apply Proposition 4.1 to the inner sum, and by definition of $f_{j+1}$, we obtain

[TABLE]

Injecting that in (8), we find that

[TABLE]

for some $c<a+b+1$. Now we have from (7)

[TABLE]

by the same arguments as above. Finally, upon gathering all error terms together (whereby the term in $O(\eta(\sqrt{m}))$ provides the largest one), we obtain the desired conclusion. ∎

We are now ready to state this subsection’s main result.

Lemma 4.5**.**

For any integer $n$, we have

[TABLE]

where $\lambda_{k}=\dfrac{\Gamma(\frac{1}{k})^{k}}{k!}$ and $\eta^{\prime\prime}$ is a function decaying to zero (depending on $\theta$ and $k$).

Proof.

Let

[TABLE]

and

[TABLE]

so that $S^{\prime}_{k}(n)=O(E_{k}(n))+k!S_{k}(n)$. We observe that $E_{k}(n)=O(n^{-1/k})$. Further, reformulating the diophantine constraints using the intervals $I_{j}$, we have the decomposition

[TABLE]

To simplify the notation, let us denote $n_{j}=n-u_{1}-\cdots-u_{k-j}$, thus $n_{k}=n$ and $n_{j}=n_{j+1}-u_{k-j}$. We shall prove by induction on $j\leq k$ that

[TABLE]

where C_{j}=\prod_{i=1}^{j-1}B\Big{(}\frac{1}{k},\frac{i}{k}\Big{)} and $\epsilon_{j}$ tends to 0. When $j=k$, there is no more summation at all and (11) boils down to $C_{k}f_{k}(\{\theta n\})+\epsilon_{k}(n)$, which is the desired result since

[TABLE]

Equation (10) is the $j=1$ case. We now suppose that (11) holds for some $j\leq k-1$. Let $A_{j}(n)$ be the main-term of the right-hand side of (11). Using Lemma 4.4 on the innermost sum, and reparametrising by writing $n_{j+1}=v_{1}$ and $u_{i}=v_{i+1}$ in the error term, we find

[TABLE]

Now the error term is certainly $o(1)$ using the fact that $\eta^{\prime}$ tends to 0 and Lemma 4.3. This concludes the induction step and therefore the proof of the lemma. ∎

4.2. The construction

We argue by the probabilistic method (see [17, Chapter 1] for a brief introduction or [1] for a detailed one). Let $c>0$ and $\xi_{m}$, $m\geq 1$, be a sequence of independent Boolean random variables such that

[TABLE]

Let $S$ be the random increasing sequence of the $m$’s such that $\xi_{m}=1$. This is essentially a sequence of pseudo $k$-th powers. These objects have been well studied since their introduction by Erdős and Renyi [4]. In particular Goguel [7] computed the (almost sure) density of $kS$ and Deshouillers and Iosifescu [3] found that the density of $(k+1)S$ is almost surely 1. Now we let $A=S\cap T_{k,\theta}$, where $T_{k,\theta}$ was defined by equation (2). From now on we will suppose $\theta$ is irrational; if $\theta$ is an integer, $T_{k,\theta}=\mathbb{N}$ so $A=S$ and the previous references apply. The treatment of this simpler case may still be read out from our proofs by discarding all the (then vacuous) diophantine conditions. The next proposition implies Proposition 2.1.

Proposition 4.6**.**

*Almost surely we have *

1. a)

\mathrm{d}\big{(}jA\big{)}=0, for any $j=1,\dots,k-1$, 2. b)

\mathrm{d}\big{(}(k+1)A\big{)}=1, 3. c)

\mathrm{d}\big{(}kA\big{)}=\frac{k}{k+1}-F_{k}(c)* where $F_{k}(c)$ is a continuous function and increasing from [math] to $k/(k+1)$ when $c$ is decreasing from $\infty$ to [math].*

Proof.

a) By an appropriate version of the strong law of large numbers (cf. [9, chapter III, Theorem 11]) we know that with probability 1, $A(x)\sim x^{1/k}$ when $x\to\infty$, thus for any $1\leq j\leq k-1$

[TABLE]

It follows that \mathrm{d}\big{(}jA\big{)}=0 almost surely.

b) Let $n$ be a positive integer and observe that $0<\{\theta n\}<1$. We denote $I(t,k)$ the open interval

[TABLE]

and

[TABLE]

Then $R_{k+1}(n)>0$ implies that $n\in(k+1)A$. Moreover

[TABLE]

We denote by $\mathcal{U}(n)$ the set of the ordered $(k+1)$-uples $\underline{u}$ such that $n=\sum_{i=1}^{k+1}u_{i}$ and $\{\theta u_{i}\}\in I(\{\theta n\},k)$, $i=1,\dots,k$.

The events $\mathcal{A}(\underline{u})=\{\xi_{u_{1}}\dots\xi_{u_{k}}\xi_{u_{k+1}}=1\}$, $\underline{u}\in\mathcal{U}(n)$, are not necessarily pairwise independent: for distinct $(k+1)$-tuples $\underline{u},\underline{v}$, the events $\mathcal{A}(\underline{u})$ and $\mathcal{A}(\underline{v})$ are not independent if and only if $\underline{u}\sim\underline{v}$, where the notation $\sim$ means $u_{i}=v_{j}$ for some $i,j$. Let

[TABLE]

By Janson’s inequality [17, Theorem 1.28]

[TABLE]

We firstly have

[TABLE]

The summand in the inner-sum is at least \big{(}\frac{n}{k+1}\big{)}^{-k+1/k}, hence

[TABLE]

where $I=I(\{\theta n\},k)$. By equation (3),

[TABLE]

Hence if $2k\eta(n)<\{\theta n\}<1-2k\eta(n)$, we have

[TABLE]

Now we examine $\Delta_{n}$. By a discussion according to the number $s\leq k-1$ of positions where two distinct $(k+1)$-tuples in $\mathcal{U}(n)$ agree, and ignoring the diophantine conditions, we get

[TABLE]

where

[TABLE]

Applying Lemma 4.3, we see that the inner sum is $\ll(n-u_{1}-\dots-u_{s})^{-1+r/k}$. For every fixed tuple $(u_{1},\ldots,u_{s-1})$ in the sum above, we now apply Lemma 4.2 on the sum

[TABLE]

If $1-r/k\geq 1/2$, we obtain

[TABLE]

where we used Lemma 4.3 for the second inequality. If $1-r/k<1/2$ then by Lemmas 4.2 and 4.3 again

[TABLE]

Notice that if $s+r=k+1$ with $s>0$, then $s-2(k-r)>0$ implies $t=2-s=1$ and $s=1$. We can now inject our upper bounds for $\Delta_{n}(s,r)$ in equation (14), in which the main contribution is given by $s=1$, from the above discussion. We get

[TABLE]

By (12) and (13) with the Borel-Cantelli lemma, we infer that almost surely, all but finitely many integers $n$ such that $2k\eta(n)<\{\theta n\}<1-2k\eta(n)$ are sums of $k+1$ members of $A$ and that \mathrm{d}\big{(}(k+1)A\big{)}=1 since their complementary set in $\mathbb{N}$, namely

[TABLE]

has density [math].

c) Let $n$ such that $0<\{\theta n\}<k/(k+1)$. We consider

[TABLE]

that is the random variable counting the number of representations of $n$ as a sum of $k$ distinct members of $A$. The key result is Lemma 4.5.

As in the study of $R_{k+1}(n)$ in the previous paragraph we need to show that the dependency of the events $\{\xi_{u_{1}}\dots\xi_{u_{k}}=1\}$ is not too high. We shall use Landreau’s work on sums of $k$ pseudo $k$-th powers (cf. [11, Lemme 1 (i) and Lemme 5 (iii)]):

[TABLE]

Since $\eta^{\prime\prime}(t)\to 0$ when $t\to\infty$, we deduce from Lemma 4.5 that

[TABLE]

When $k/(k+1)\leq\{\theta n\}<1$ we clearly have $R_{k}(n)=0$, hence $\mathbb{P}(R_{k}(n)=0)=1$.

Let $\zeta_{n}$, $n\geq 1$, be the sequence of Boolean random variables defined by

[TABLE]

and

[TABLE]

By (16) we have

[TABLE]

Hence,

[TABLE]

We get by Theorem 1.2 and the fact that $f_{k}$ is supported on $(0,k/(k+1))$ the asymptotic

[TABLE]

We follow the arguments used in the proof of [9, chapter III, Theorem 4*′* (iii)] or alternatively [11, Section 4] to estimate the variance $\mathbb{V}(X_{N})$. We may ignore the diophantine conditions in (15), the only resulting effect being to increase the related variance. We finally get $\mathbb{V}(X_{N})=O(N^{-1/k})$ and consequently by [9, chapter III, lemma 34] that

[TABLE]

Hence almost surely \mathrm{d}\big{(}kA\big{)}=\frac{k}{k+1}-F_{k}(c). Observing that $f_{k}$ is a non negative piecewise polynomial function that has finitely many zeros on $(0,k/(k+1))$, we see that $F_{k}(c)$ is a decreasing continuous function satisfying $\lim_{c\rightarrow 0}F_{k}(c)=k/(k+1)$ and $\lim_{c\rightarrow+\infty}F_{k}(c)=0$; this ends the proof of Proposition 4.6. ∎

Acknowledgments

The authors are thankful to Georges Grekos for useful conversations.