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I wonder if a set of integers $A$ given by an increasing sequence of integers $A=(a_k)_k$ with positive upper density $\limsup_n\frac{\sharp A\cap [0,n]}{n}>0$ satisfies the following property.

For any $\epsilon>0$ there exists an integer $L>0$ such that we can find arbitrarily large intervals $J=[a_p,a_q]$ for which the union $J_L$ of intervals $[a_k,a_{k+1}]$ with $a_{k+1}-a_k>L$ lying inside $J$ represents a proportion less than $\epsilon$ in $J$ :

$$\frac{|J_L|}{|J|}<\epsilon.$$

The example of not piecewise syndetic set with positive upper density that I have in mind satisfies this property. This is certainly false but I have no idea about a counterexample.

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  • $\begingroup$ Take $(a_n)$ to be the increasing enumeration of the union of the squares and the numbers starting with a 7. Then there are long dense stretches and long sparse stretches (where your conclusion is violated). $\endgroup$ Commented Dec 12, 2019 at 23:15
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    $\begingroup$ Dear Anthony, your set is piecewise syndetic and it satisfies therefore my property... $\endgroup$ Commented Dec 13, 2019 at 12:38
  • $\begingroup$ What does piecewise syndetic mean? $\endgroup$ Commented Dec 13, 2019 at 19:04
  • $\begingroup$ en.wikipedia.org/wiki/Piecewise_syndetic_set $\endgroup$ Commented Dec 13, 2019 at 21:08
  • $\begingroup$ Do we agree that (1) $d^*(A)>0$ implies $\lim_{L\to\infty}d^*(A+\{1,\ldots,L\})=1$; and (2) that this resolves your question? $\endgroup$ Commented Dec 17, 2019 at 5:13

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OK. So now I understand the question (I think). And the answer is yes.

I think it is helpful to introduce the upper Banach density, $d^*(S)=\lim_{n\to\infty}\sup_{|I|\ge n}|S\cap I|/|I|$, where $I$ runs over intervals in $\mathbb N$. Clearly $d^*(S)$ is at least as large as the upper density.

Let $S$ be a set with positive upper density. Let $a=d^*(S)$. Let $\epsilon>0$ and let $4\eta/(a+\eta)<\epsilon$. By definition of $d^*(S)$, there exists an $N$ such that $|S\cap I|\le (a+\eta)|I|$ for any $I$ with $|I|\ge N$. Now fix any $M>0$. Again by definition of $d^*(S)$, there exists an interval $J$ of length at least $M$ such that $|S\cap J|\ge (a-\eta)|J|$. We may suppose without loss of generality that $|J|$ is a multiple of $N$. Let $L=2N$. Divide $J$ into sub-intervals of length $N$ (call these grid sub-intervals) and consider those which do not intersect $S$. Suppose there are $k$ of these. On the other $|J|/N-k$ intervals, we have $|S\cap I|\le (a+\eta)N$, so that $|S\cap J|\le (|J|/N-k)(a+\eta)N$. Combining the two inequalities gives $$ (a-\eta)|J|\le (|J|/N-k)(a+\eta)N. $$ Rearranging gives $Nk\le \epsilon|J|/2$. However each interval $[a_i,a_{i+1}]$ forming $J_L$ must contain grid sub-intervals whose total length is at least $(a_{i+1}-a_i)/2$. In particular, $|J_L|$ is at most twice the total length of the empty grid sub-intervals, that is $Nk$. Hence $|J_L|\le \epsilon|J|$.

By the way, I should say a little about how I arrived at this answer. First, I think what you are trying to prove is the equivalent to $d^*(A+\{1,\ldots,L\})\to 1$ as $L\to\infty$. In fact, the slightly more general result is $d^*(A)>0$ implies $d^*(A+\{1,\ldots,L\})\to 1$. My first proof of this went through the Furstenberg correspondence, where the result translates to a well known fact in ergodic theory: if $T$ is an ergodic probability measure-preserving transformation and $B$ is any set of positive measure, then $\mu(\bigcup_{n=1}^N T^{-i}B)\to 1$ as $N\to\infty$. I then tried to give an elementary proof of the same thing.

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  • $\begingroup$ Great. Many Thanks. In fact this is not true that $d^*(A+\{1,\cdots, L\})\rightarrow 1$ when $L$ goes to infinity for $d^*(A)>0$ (take the natural numbers starting with a $7$...) so that I first believed my property was false. $\endgroup$ Commented Dec 14, 2019 at 10:48
  • $\begingroup$ For the natural numbers starting with 7, $d^*(A)=1$. This is upper Banach density, the density in long sub-intervals; not the density in initial sub-intervals. $\endgroup$ Commented Dec 14, 2019 at 17:10
  • $\begingroup$ Yes, I thought about the density. $\endgroup$ Commented Dec 14, 2019 at 19:33

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