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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.

Thank you in advance.

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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
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Possible duplicate? math.stackexchange.com/questions/22282/… –  user83827 Aug 8 '11 at 15:58
    
Yeonjoo: An accept rate of 0% (at time of commenting) is less than acceptable in this community. In previous questions please mark the most helpful answer as the accepted answer by clicking the check mark by the voting arrows. If no answer is helpful, please edit the question to indicate what is missing from the given answers. –  Asaf Karagila Sep 7 '11 at 23:56

1 Answer 1

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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 $\lambda^*(B)$.

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