0
$\begingroup$

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 $$ \left|\bigcap_{n\in\mathbb{N}} A_{i_n}\right|>1-2\epsilon? $$

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $\epsilon n > 1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.