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.
Best Answer
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 $\lambda^*(B)$.