Finding a Lebesgue measurable set that contains a set and they have the same outer measure

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 $\mu^*(S\Delta B)<\epsilon$.

• Couldn't you just add a single point to the set? – Matt Samuel Nov 2 '14 at 22:29
• But is that make it Lebesgue measurable? – Haley13 Nov 2 '14 at 23:07
• It's the union of two Lebesgue measurable sets. $\sigma$-algebras are closed under finite union. So yes. – Matt Samuel Nov 2 '14 at 23:08
• We do not know whether $A$ is Lebesgue measurable or not. – Haley13 Nov 2 '14 at 23:10
• Oh, sorry I misunderstood. – Matt Samuel Nov 2 '14 at 23:11

Just use the definition of the outer measure. For each $n\ge 1$ let $A=\bigcup_{k\ge1} I_{n,k}$ be a union of open rectangles such that $\mu(\bigcup_{k\ge1}I_{n,k})\le\mu^*(A)+\frac1n$ and set $$A'=\bigcap_{n\ge1}\bigcup_{k\ge1} I_{n,k}$$ Surely $A'\supseteq A$ and $\mu(A')=\mu^*(A)$.
• I am sorry, but what do you mean by intervals here? Because $A$ is in a plane. – Haley13 Nov 2 '14 at 23:18