Does there exist a subsequence whose intersection has measure greater than $1/2$?

Question

I ran across the following problem on this review guide. It is problem 1.25, though I've changed the wording slightly. The measure is implicitly Lebesgue measure.

Let $E_n$ be a sequence of measurable sets in $[0,1]$ with $m(E_n)\rightarrow 1$ and $n\rightarrow 1$. Prove there exists a subsequence whose intersection has measure greater than $1/2$.

To show this, just take a subsequence with measures greater than $1-\epsilon/2$, $1-\epsilon/2^2$, $1-\epsilon/2^3$, $1-\epsilon/2^4$, $\dots$.

Does this statement remain true if we only assume $m(E_n)>\alpha$ for some fixed $\alpha \in (3/4, 1)$ and all $n$? I suspect not, but no counterexample comes to mind immediately.

Answer 1

Consider a sequence of sets $E_n$ of measure $t \in (0,1)$ whose indicator functions are all independent random variables. That is, given $E_1, \ldots, E_n$, consider each of the $2^n$ sets $G_j = F_1 \cap \ldots F_n$ where $F_i$ is either $E_i$ or its complement $E_i^c$, and let $E_{n+1}$ have fraction $t$ of each $G_j$. Then the intersection of any $k$ members of the sequence has measure $t^k$, and so the intersection of any infinite subsequence has measure $0$.

Answer 2

Heh. If anyone's wondering how to construct those independent sets, here's a cute way to organize the construction:

Say $0<\alpha<1$. For $S\subset[0,1]$ define $T(S)\subset[0,1]$ by $$T(S)=\alpha S\cup(\alpha+(1-\alpha)S).$$(Here $cS=\{ct:t\in S\}$ and $c+S=\{c+t:t\in S\}$.)

Define $$E_1=[0,\alpha]$$and $$E_{n+1}=T(E_n).$$