Let $E \subset [0,1]$ be a Lebesgue measurable set. It is known that there exists a $G_{\delta}$ set $G \supset E$ such that $m(E)=m(G)$. But my problem is:
Does $E \subset [0,1]$ Lebesgue measurable set always contain a $G_{\delta}$ subset $G \subset E$ such that $m(E)=m(G)$?
Thanks for any help!
Let $E$ be a meager set of measure 1, which one can construct in various ways; e.g. There exist meager subsets of $\mathbb{R}$ whose complements have Lebesgue measure zero. If $G \subset E$ is $G_\delta$ then since $G$ is meager, by the Baire category theorem it is not dense. That means there is some nonempty open $U \subset [0,1]$ with $G \subset [0,1] \setminus U$. As such, $m(G) \le 1 - m(U) < 1$.
