Is it true that for any $A \subset \mathbb{R}^n$ we can find a set $B$ such that $A \subset B$, $B$ is a Lebesgue measurable and $\mu^*(A)=\mu(B)$? I am trying to figure out the following question: Is it true that for any $A \subset \mathbb{R}^n$ we can find a set $B$ such that 
$A \subset B$, $B$ is a Lebesgue measurable and $\mu^*(A)=\mu(B)$? 
(Where $\mu$ is Lebesgue measure, $\mu$ is Lebesgue outer measure). 
I found it trivial for $A$ Lebesgue measurable, but what happens for example for $A$ Vitali set? And in general?
 A: Recall, for any $A$, the Lebesgue outer measure is defined via
$$
\mu^{*}(A) = \inf_{\{I_j\}_{j=1}^{\infty} : E \subseteq\bigcup_j I_j}\sum_{j=1}^{\infty}\mu(I_j)
$$
where $I_j$'s are open intervals and $\mu(I_j)$ is simply the length of the interval. Now, note that for every $\epsilon>0$, there exists a collection of open intervals $\{I_{j,\epsilon}\}_{j=1}^{\infty}$ such that
$$
\sum_{j=1}^{\infty}\mu(I_{j,\epsilon})<\mu^{*}(A) + \epsilon.
$$
Now, denote $O_{\epsilon} = \bigcup_j I_{j,\epsilon}$. $O_{\epsilon}$ is an open set, and hence is Lebesgue measurable. Furthermore, from subadditivity of measure,
$$
\mu(O_{\epsilon})\leq \sum_{j=1}^{\infty}\mu(I_{j,\epsilon})<\mu^{*}(A)+\epsilon.
$$
Take $\epsilon_n = 2^{-n}$ (or whatever the sequence you'd prefer, as long as $\epsilon_n \searrow 0$), and recover open sets $O_n$ corresponding to each $\epsilon_n$. It suffices to take $B$ to be
$$
B = \bigcap_{n=1}^{\infty}O_n.
$$
Clearly, $B \supset A$, hence $\mu(B) = \mu^{*}(B) \geq \mu^{*}(A)$. Furthermore, since $B \subset O_n$, for each $n$, we have
$$
\mu(B) \leq \mu(O_n) \leq \mu^{*}(A) + 2^{-n} \implies \mu(B) \leq \mu^{*}(A).
$$
Hence, 
$\mu(B) = \mu^{*}(A)$, as desired.
Techinical Remark 
Since $B$ is a $G_{\delta}$ set, namely, it is obtained by taking countable intersection of open sets, and since open sets are measurable and measurability is preserved under countable intersections, $B$ itself is measurable. This allowed us to write
$$
\mu(B) = \mu^{*}(B).
$$
