# a nonmeasurable set $E$ of finite measure and a $G_{\delta}$ set $G$ that contains $E$

I understand that the measurability of a set is equivalent for the existence of a $G_{\delta}$ set $G$ that contains the set and has the same outer measure.

However, I do not know how to answer this question in my text: Let $E$ be a nonmeasurable set of finite outer measure. Show that there is a $G_{\delta}$ set $G$ that contain $E$ such that outer measure of $E$ is the same as the outer measure of $G$ while outer measure of $G\setminus E$ is greater than zero.

The Theorem of Vitali states that any set of real number with positive outer measure contains a subset that fails to be measurable but I do not know how to relate this theorem to the problem.

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are you talking about the Lebesgue measure (or in general, a complete measure)? because then $G-E$ must have outer measure greater than zero, or otherwise it would be measurable (with measure zero), and then $E$ would be measurable since $E=G-(G-E)$ – Prometheus Aug 8 '11 at 14:24
@ Prometheus: Yes. I am asking about Lebesgue measure. But how do you know that outer measure of the nonmeasurable set E is the same as outer measure of G? – Yeonjoo Yoo Aug 8 '11 at 14:27
Possible duplicate? math.stackexchange.com/questions/22282/… – user83827 Aug 8 '11 at 15:58

Let $E$ be any set with a finite outer measure $r=\lambda^*(E)$. From the definition of outer measure $r$ is the infimum of the measures of open sets containing E.
For each n you can find $U_n$ open such that $E\subseteq U_n$ and $r\leq \lambda(E)\leq r+\frac{1}{n}$. Taking $G=\bigcap U_n$ we get that G is a $G_\delta$ set, $E\subseteq G$ and therefore $r=\lambda^*(E)\leq \lambda^*(G)=\lambda(G)$, and for every n we also have $\lambda(G)\leq\lambda(U_n)\leq r+\frac{1}{n}$ so $\lambda(G)=r$.
If $\lambda^*(G-E)=0$ then $G-E$ would be measurable and then E would be measurable since $G-(G-E)=E$ and both G,G-E are measurable. This shows in particular that $\lambda^*(A)+\lambda^*(B-A)$ can be larger than $\lambda^*(B)$.