Outer Lebesgue's measure

Question

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.

Answer 1

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.

Answer 2

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 $E\setminus A\subset B\subset E$.