This seems to be a simple problem, but I can not come up with a counterexample. I would be happier if it were true!

Set up: Consider $\{A_i\}_{i\in\mathbb{N}}$ a sequence of measurable subsets of $[0,1]$. Fix $\epsilon\in (0,1/2)$ and assume that the Lebesgue measure of each set is bounded below by $|A_i|>1-\epsilon$ for every $i\in\mathbb{N}$.

Question: Can we find an infinite set of indexes $\{i_n\}_{n\in\mathbb{N}}$, such that the Lebesgue measure of the intersection of all these sets is bounded below by $$ \left|\bigcap_{n\in\mathbb{N}} A_{i_n}\right|>1-2\epsilon? $$

  • $\begingroup$ I'm unclear on your quantifiers. Are you asking if we can find such a sequence for each $A_i$, or if there is an $A_i$ such that we can find such a sequence? $\endgroup$ – Mees de Vries Jul 18 '18 at 9:30
  • $\begingroup$ For any sequence $\{A_i\}_{i\in \mathbb{N}}$ can we find a subsequence $\{A_{i_n}\}_{n\in \mathbb{N}}$. Does this clear it up? $\endgroup$ – Sloth-Meister Jul 18 '18 at 9:34

No, you cannot do it in general.

Take $1/3 < \epsilon < 1/2$. Let $A_i$ be the subset of elements of $[0, 1]$ whose ternary expansion does not have a $2$ at the $i$th position (in case of two possible decimal expansions, always use the one which terminates). Trivially $\mu(A_i) = 2/3 > 1 - \epsilon$, and each of the $A_i$ are independent, in the sense that for any finite set $I \subseteq \mathbb N$ we have $$ \mu\left(\bigcap_{i \in I} A_i\right) = \prod_{i \in I}\mu(A_i). $$ Therefore, for any infinite $I \subseteq \mathbb N$ we have that $$ \mu\left(\bigcap_{i \in I} A_i\right) = 0 < 1 - 2\epsilon. $$ Edit: And obviously this approach can be generalized to any $\epsilon > 0$, you just have to take $n$-ary expansions for sufficiently large $n$ that $\epsilon n > 1$.


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