$(A_{n})_{n}$ are $\lambda$_lebesgue measurable sets in $[0,1]$ such that $\limsup\lambda(A_{n})=1$. I have to prove that for every $a\in (0,1)$ there exists a subsequence $(A_{n_{k}})_{k}$ of $(A_{n})_{n}$ such that $\lambda(\cap_{k}A_{n_{k}})>a$.
Any idea on how to prove this ?? I found that we have to construct $(A_{n_{k}})_{k}$ such that $\lambda([0,1]\setminus A_{n_{k}})<(1-a)/2^{k}$
But how we construct such subsequence and where does $\lambda([0,1]\setminus A_{n_{k}})<(1-a)/2^{k}$ helps us in order to prove $\lambda(\cap_{k}A_{n_{k}})>a$?
Since $\limsup\limits_{n \to \infty}\lambda(A_n)=1$ and $\lambda(A_n)=1-\lambda(A_n^c)$ we get $$\liminf_{n \to \infty}\lambda(A_n^c)=\lim_{n\to \infty}\inf_{k\geq n}\lambda(A_k^c)=0$$
Now am using the well known fact which states for a bounded sequence $\alpha_n$ that $$\liminf_{n \to \infty}\alpha_n \leq x \text{ if and only if the set } \{n \in \mathbb{N} : \alpha_n < x+\epsilon\} \text{ is infinite for every } \epsilon>0$$
Here $\alpha_n = \lambda(A_n^c)$ and $x= 0 $. So by using the above
preposition for $\epsilon_k = \dfrac{1-\alpha}{2^k}>0$ we get that the sets $$C_k = \{n \in \mathbb{N} : \lambda(A_n^c) < \epsilon_k\}$$ are infinite subsets of $\mathbb{N}$.So we can pick $$n_1 < n_2< ...< n_k<n_{k+1}<...$$ so that $n_k \in C_k$ for every $k \in \mathbb{N}$, which means that $$\lambda(A_{n_k}^c)< \epsilon_k = \dfrac{1-\alpha}{2^k}$$ Now observe that, $$\lambda\biggl(\bigcup_{k=1}^{\infty}A_{n_k}^c\biggr)\leq\sum_{k=1}^{\infty}\lambda(A_{n_k}^c)\leq 1-\alpha$$ and since $$\lambda\biggl(\bigcap_{k=1}^{\infty}A_{n_k}\biggr)+\lambda\biggl(\bigcup_{k=1}^{\infty}A_{n_k}^c\biggr)=1$$ we get that $$\lambda\biggl(\bigcap_{k=1}^{\infty}A_{n_k}\biggr)\geq \alpha$$