Let $\{E_{n}\}_{n=1}^{\infty}$ be a sequence of non-empty Lebesgue measurable subsets of the interval $[0,1]$ which satisfies the following.
$\lim_{n \rightarrow \infty}m(E_{n})=1$.
Prove that for each $\epsilon \in [0,1]$ there is a subsequence -- call it $\{E_{n_{k}}\}_{k=1}^{\infty}$ -- of $\{E_{n}\}_{n=1}^{\infty}$ such that $m(\cap_{k=1}^{\infty}E_{n_{k}}) \geq \epsilon$.
Sorry I don't have a lot of effort to show because I'm not sure how to go about proving it. My textbook gives a few properties about similar families of measurable functions, but they aren't seeming that useful to me. They are more for proving functions and the sequences' limits are measurable, which is less complex than this.
I do know that if some $\{f_{n}\}_{n=1}^{\infty}$ converges to $f$ in $L^{1}$ then there exists a subsequence $\{f_{n_{k}}\}_{k=1}^{\infty}$ such that $f_{n_{k}}\rightarrow f(x)$ a.e. $x$, and that a family of integrable functions is dense in $L^{1}$ if for any $f \in L^{1}$ and $\epsilon > 0$ there exists $g$ in the family so that $||f-g||_{L^{1}} < \epsilon$. It seems helpful but I am not sure how to use. Thanks.
Lemma: if $A$ and $B$ are measurable sets contained in $[0,1]$. Then $m(A\cap B) \geq m(A)+m(B)-1$.
Proof: Note that $m(A\cup B) = m(A)+m(B)-m(A\cap B)\implies m(A\cap B)=m(A)+m(B)-m(A\cup B)\geq m(A)+m(B)-1$.
Given $\epsilon>0$ let $\chi=1-\epsilon$
We take a subsequence $B_1,B_2,\dots$ such that $m(B_n)\geq 1-\frac{\chi}{2^n}$.
We now prove by induction that $m(\bigcap \limits_{i=1}^n B_n)\geq 1- (\chi/2+ \chi/4+\dots \chi/2^n)$.
All we have to do is apply the lemma to $B_{n+1}$ and $(\bigcap \limits_{i=1}^\infty B_n)$. Notice that the sum of their measures is at least $2-(\chi/2+\dots+\chi/2^{n+1})$, so $m(\bigcap \limits_{i=1}^{n+1} B_n)\geq 1- (\chi/2+ \chi/4+\dots \chi/2^{n+1})$.
Now use the theorem of convergence of measure, we have:
$m(\bigcap \limits_{i=1}^\infty B_n)=\lim\limits_{n\to \infty}m(\bigcap \limits_{i=1}^n B_n)\geq \lim\limits_{n\to \infty}1- (\chi/2+ \chi/4+\dots \chi/2^n)=1-\chi=\epsilon$