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Let $X$ be a compact metric space, and $\mu$ a Borel probability measure. For $S\subset\mathbb{N}$ we denote the upper density with $\overline{D}(S).$

Let $f:X\rightarrow2^{\mathbb{N}}$ be a measurable function such that there exists positive $a$ such that $\overline{D}(f(x))\geq a$ for almost every $x.$

Does there exist $b$, $S_{1}\subset\mathbb{N}$ such that $\overline{D}(S_{1})>b$ and $\mu\left\{ x\in X:t\in f(x)\right\} \geq b$ for every $t\in S_{1}?$

Can $b$ be chosen so that it depends only on $a$ and not on the function $f?$

The way I think of this problem is this way, $X$ represent the members of a club, and the function represents the times when they visit the club. We have that almost every member visits the club with a rate of $a$, does this mean subset of times of positive upper density such that a good portion of the members will be in the club at that time.

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  • $\begingroup$ Is $f$ completely arbitrary, or should it at least be measurable? $\endgroup$ Dec 28, 2013 at 19:13
  • $\begingroup$ I didn't think it could help, but yes we can assume it is measurable with respect to the Borel sigma algebra of the product topology on the image. $\endgroup$
    – Rob
    Dec 28, 2013 at 20:18
  • $\begingroup$ I changed the conclusion to something a bit weaker. $\endgroup$
    – Rob
    Dec 28, 2013 at 21:07

1 Answer 1

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To answer the second part (Can $b$ be chosen so that it depends only on $a$ and not the function $f$):

No, it cannot.

First note that there exist infinitely many upper density one subsets $S_n\subseteq\mathbb N$, $n\in\mathbb N$ that are disjoint (to construct such sets, just let them take turns getting their density up, in a dove-tailing fashion).

Then, fixing a $b>0$, consider $f$ which equals $S_n$ with probability $p_n$, where each $p_n < b$ and of course $\sum p_n = 1$. (We can take $X=[0,1]$.)

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