A question about a sequence of sets with positive measure

Question

For each $n\in\mathbb{N}$, $E_n$ is a measurable subset of $[0,1]$. Let $m$ be the Lebesuge measure, suppose $$ m(E_n)\geq \delta>0, \quad \forall n\in \mathbb{N}. $$ My question is the following: is there a subsequence $\{n_k\}_{k\in\mathbb{N}}$ such that $n_k\to \infty(k\to\infty)$ and $$ m(\cap_{k=0}^\infty E_{n_k})>0? $$

Answer 1

No. I will give a proof using Probability Theory. There exist i.i.d random variables $\{X_n\}$ on $[0,1]$ with Lebesgue measure such that $P\{X_n=1\}=P\{X_n=-1\}=\frac 1 2$ for all $n$. Let $E_n=\{X_n=1\}$. Then $P(E_n)=\frac 1 2$ for all $n$ but $P\{ X_{n_k}=1,k=1,2...\}=0$ for any subsequence $(n_k)$.

Translated to a non-probabilistic setting this says the following: expand each $x \in (0,1)$ to base $2$ as $x= \sum \frac {a_n(x)} {2^{n}}$ with $a_n(x) \in \{0,1\}$. Then $E_n\{x:a_n(x)=1\}$ serves as counterexample to your statement.