If for each $\epsilon > 0$, there is an open set $O$ containing $E$ for which $m^*(O-E) < \epsilon$, then $E$ is measurable set.

Question

Let $E$ be a set of real numbers. Prove that if for each $\epsilon > 0$, there is an open set $O$ containing $E$ for which $m^*(O-E) < \epsilon$, then $E$ is measurable set.

$E$ is measurable set if for all set $A$, $m^{*}(A)=m^{*}(A\cap E)+m^{*}(A\cap E^{c})$ holds. I know how to prove converse. But I don't understand how to prove this statement. How do I start? Any hint will be sufficient. Thanks.

Answer 1

Let $m^{*} (O_n\setminus E)<\frac 1 n$ with $O_n$ open and $E \subset O_n$. Then $m^{*} (\cap_n O_n\setminus E)<\frac 1 k$ for each $k$ so $m^{*} (\cap_n O_n\setminus E)=0$ Thus $E=\cap_n O_n \setminus F$ where $F=\cap_n O_n\setminus E$ is a set of outer measure $0$. Now verify that $m^{*}(A\cap E)+m^{*}(A\cap E^{c})\leq m^{*}(A\cap V)+m^{*}(A\cap V^{c})$ where $V=\cap_n O_n$. Use the fact that $V$ is measurable to get $m^{*}(A\cap E)+m^{*}(A\cap E^{c})\leq m^{*}(A)$. The reverse inequality always holds.