# 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 $$m(A)=m(E)=m(B)$$?

• Measurable by what measure? Have you tried using the definitions of $m$? – Asaf Karagila Sep 7 '13 at 17:19

I assume by $m$ you mean Lebesgue measure on $\mathbb R^n$. Use that this measure is regular. This gives us that, if $m(E)<\infty$, then for any $n$ there are a compact set $K_n$ and and open set $U_n$ with $K_n\subset E\subset U_n$, and $m(E)-1/n<m(K_n)$ and $m(U_n)<m(E)+1/n$.
This implies that $A=\bigcup_n K_n$ and $B=\bigcap U_n$ have the same measure as $E$, and they are clearly a Borel subset and a Borel superset of $E$, respectively.
If $E$ has infinite measure, it is even easier: Take as $B$ the set $\mathbb R^n$. As before, regularity gives us for each $n$ a compact set $K_n$ with $K_n\subset E$ and $m(K_n)\ge n$. Then we can again take as $A$ the set $\bigcup_n K_n$.
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 $m(A)=m(E)$.