let $\{E_{n}\}_{n\in\Bbb N}$ be a sequence of measurable subsets such that $E_n \subseteq (0,1)$and let $\limsup_{n\to\infty} m(E_{n})=1$, prove that exists a subsequence {$E_{n_{k}}$} such that $m(\bigcap_{k=1}^{\infty} E_{n_{k}}) > 0$, where m is the measure of Lebesgue.
We are in a metric space so by the definition of limsup i know (thanks to Bolzano-Weierstrass theorem) that exists a subsequence $E_{n_{k}}$ such that $\lim_{k\to\infty}m(E_{n_{k}})=1$, that means that $\forall\epsilon>0, \exists h\in \Bbb N : \forall k>h, 1-\epsilon<m(E_{n_{k}})<1+\epsilon$. I thought it was a good starting point but i don't know how to go on. Any ideas?
Let $\epsilon>0$ and choose $E_{n_k}$ such that $m E_{n_k} \ge 1-\frac{1}{2^k} \epsilon$.
Then we see that $m E_{n_k}^c \le \frac{1}{2^k} \epsilon$, and so $m (\cup_k E_{n_k}^c) \le \epsilon$. This means that $m (\cup_k E_{n_k}^c)^c \ge 1-\epsilon$, and since $(\cup_k E_{n_k}^c)^c = \cap_k E_{n_k}$, we can take any $\epsilon<1$ to get the desired result.