Let $A$ be a set in $E=[0,1]\times[0,1]$. Then, there exists a Lebesgue measurable set $A'$ such that $A\subseteq A' \subseteq E$ and $\mu^*(A)=\mu^*(A')$.
$\mu^*$ is Lebesgue outer measure.
I was searching for a proof for bigger theorem on Lebesgue measure, and I encountered this claim. But, I think it is too hard for me to prove it. I tried to find the proof in some books such as Real Analysis by Royden and Introductory Real Analysis by Kolmogorov, and some websites, but to no success.
I do hope that anyone could provide suitable reference for the proof of this claim. I just start to learn this Lebesgue measure, so maybe I do not have sufficient tools to prove it.
Additional info: The definition of set $S$ to be Lebesgue measurable is: for all $\epsilon > 0$, there exists an elementary set $B$ in $E$ such that $\mu^*(S\Delta B)<\epsilon$.