measurable set columns $\left\{E_{n}\right\}$ in $[0,1]$ , and $ \varlimsup_{n \rightarrow \infty} m\left(E_{n}\right)=1 $
Prove that for $0<a<1$, $\left\{E_{n_{k}}\right\}$ exists, s.t. $$ m\left(\bigcap_{k=1}^{\infty} E_{n_{k}}\right)>a .$$
I am not sure how to use the condition $ \varlimsup_{n \rightarrow \infty} m\left(E_{n}\right)=1 $ to get the results.
Anybody could help?
Many Thanks.
Choose $(n_{k})_{k \in \mathbb{N}}$ such that $\lim_{k \to \infty} m(E_{n_{k}}) = 1$. Passing to a further subsequence, we can assume that $1 - m(E_{n_{k}}) < 2^{-k}$ for each $k \in \mathbb{N}$. Hence \begin{equation*} m \left( \bigcup_{k = K}^{\infty} [0,1] \setminus E_{n_{k}}\right) < \sum_{k = K}^{\infty} 2^{-k} = 2^{-K}. \end{equation*} This implies \begin{equation*} m \left( \bigcap_{k = K}^{\infty} E_{n_{k}} \right) \geq 1 - 2^{-K}. \end{equation*}
Edit: Fix $a \in (0,1)$. There is a $K \in \mathbb{N}$ such that $1 - 2^{-K} > a$. Define $(i_{\ell})_{\ell \in \mathbb{N}}$ by \begin{equation*} i_{\ell} = n_{K + \ell - 1}. \end{equation*} The above arguments imply that \begin{equation*} m \left( \bigcap_{\ell = 1}^{\infty} E_{i_{\ell}} \right) \geq 1 - 2^{-K} > a. \end{equation*}