Outer Lebesgue's measure I have  a problem to prove the following statement. 
Assume that $E$ is Lebesgue measurable and $A$ is arbitrary subset of $E$. Then there exists Lebesgue measurable set $B$ such that $E\setminus A\subset B\subset E$ and $\mu(B)=\mu^*(E\setminus A)$, where $\mu$ is Lebesgue's measure.
 A: By definition of the outer measure, for any $\epsilon>0$, there is an open set $O$ such that $E\setminus A\subset O$ and $\mu^*(E\setminus A)+\epsilon>\mu(O)$. 
Thus, we can take a sequence of open $O_n$ such that $E\setminus A\subset O_n$ and $\mu^*(E\setminus A)+\frac 1 n>\mu(O_n)$. If we let $E_n=O_n\cap E$, we still have $E_n$ is measurable as an intersection of measurable sets, and $E\setminus A\subset E_n\subset E$ and as $E_n\subset O_n$, we also have $\mu^*(E\setminus A)+\frac 1 n>\mu(E_n)$.
Set $B=\bigcap_{n=1}^\infty E_n$. Then $B$ is measurable, and $\mu(B)\le \mu^*(E\setminus A)+\frac 1 n$ for any $n\in\mathbb{N}$. Also, as $E\setminus A\subset B$, $\mu(B)\ge\mu^*(E\setminus A)$, so $\mu(B)=\mu^*(E\setminus A)$ and $E\setminus A\subset B\subset E$. 
A: We have
$$
\mu^* (E \setminus A)= \inf \{ \mu(X) \mid E \setminus A \subseteq X \wedge X \text{ is measurable} \}.
$$
Since $E \setminus A \subseteq E$ and $E$ is measurable it follows that
$$
\mu^* (E \setminus A)= \inf \{ \mu(X) \mid E \setminus A \subseteq X \subseteq E \wedge X \text{ is measurable} \}.
$$
(For any measurable $E \setminus A \subseteq X$ we have that $E \setminus A \subseteq X \cap E$, $X \cap E$ is measurable and $\mu (X) \le \mu(E \cap X)$.
Hence there is a sequence $(X_n \mid n \in \mathbb N)$ of measurable sets such that, for all $n \in \mathbb N$, 


*

*$E \setminus A \subseteq X_n \subseteq E$,

*$X_{n+1} \subseteq X_n$ and

*$\mu^*(E \setminus A) + \frac{1}{n} > \mu(X_n)$. 


Now show that $X := \bigcap_{n \in \mathbb N} X_n$ is as desired.
