If $E$ is a measurable set, how to prove that there are Borel sets $A$ and $B$ such that $A\subset E$, $E\subset B$ and $m(A)=m(E)=m(B)$? If $E$ is a measurable set, then  how to prove that there are Borel sets $A$ and $B$ such that $A$ is a subset of $E$, $E$ is a subset of $B$ and $m(A)=m(E)=m(B)$?

Let $E$ be a Lebesgue-measurable set. By definition of the Lebesgue-measure, for each $m\in \Bbb N$ there is a sequence of intervals $(I_{m,n})_n$ such that $E\subseteq \bigcup_{i=1}^{\infty}I_{m,n}=:B_m$ and $m(B_m)\le m(E)+\frac 1m$. Let $B:=\bigcap_{m=1}^{\infty}B_m$, then $B$ is borel, $E\subseteq B$ and by consruction $m(B)=m(E)$.
Now do the same construction with $\Bbb R\setminus E$ instead of $E$ to find a borel set $A^*$ with $\Bbb R\setminus E\subseteq A^*$ and $m(A^*)=m(\Bbb R\setminus E)$ and let $A:=\Bbb R\setminus A^*$. It follows from the measurability of $E$ that $m(A)=m(E)$.
