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 $E$ and $E$ is a subset of $B$ and $m(A)=m(E)=m(B)$.

  • 1
    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 $m(A)=m(E)$.

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.