Show that if $E$ is not measurable, then there is an open set $O$ containing $E$ that has finite outer measure and for which $m^*(O-E)>m^*(O)-m^*(E)$ Let $E$ have finite (Lebesgue) outer measure. Now we need to show that if $E$ is not measurable, then there is an open set $O$ containing $E$ that has finite outer measure and for which $m^*(O-E)>m^*(O)-m^*(E)$. (Here $m^*(E)$ denotes the Lebesgue outer measure of $E$). The following is my attempt.
Suppose $E$ is not measurable. Assume that for all open sets $O$ containing $E$ that has finite outer measure we have $m^*(O-E)\leq m^*(O)-m^*(E)$. Let $\epsilon >0$. Since $m^*(E)=\inf\{m^*(U)|E\subseteq U\ \text{and}\ U\ \text{is open}\}$, $\exists U\ \text{open with}\ E\subseteq U\ \text{such that}\ m^*(U)-m^*(E)<\epsilon$. Then $U$ has finite outer measure. So by assumption it follows that $m^*(U-E)<\epsilon$. Therefore $E$ is measurable; contradiction. Hence the result.
Could someone please tell me if this proof is alright? Thanks.

How do you define $m^*(U)$ for $U$ open? In order to say that for any $\epsilon$, $\exists U \supseteq E$ open such that $m^*(U) - m^*(E) < \epsilon$, it seems like you need to make precise how you could find this $U$ with outer measure between $m^*(E)$ and $m^*(E) + \epsilon$.
Other than this point, the proof looks correct to me.
