# 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.

• 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 $E=G-(G-E)$ – Ofir Aug 8 '11 at 14:24
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 $$r\leq \lambda(U_n)\leq r+\frac{1}{n}$$. 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 $$\lambda^*(B)$$.