I am trying to solve Exercise 17 from Section 2.4 of Royden's Real Analysis. The goal is to show that a set $E$ is (Lebesgue) measurable iff for all $\epsilon>0$, there exists an open set $O$ and a closed set $F$ such that $F \subseteq E \subseteq O$ and $m^*(O-F)< \epsilon$.
I intend on using this result from Royden which was proved earlier. I will paraphrase it below.
The following are equivalent:
$E$ is measurable
For all $\epsilon>0$, there is an open set $O \supseteq E$ such that $m^*(O-E)< \epsilon.$
For all $\epsilon>0$, there is a closed set $F \subseteq E$ such that $m^*(E-F)< \epsilon.$
From the highlighted result, we can prove the forward direction very easily. Assume $E$ is measurable and let $\epsilon >0$. Then we know (from the result) that there is $O$ open and $F$ closed such that $F \subseteq E \subseteq O$ and $m^*(E-F)< \epsilon/2$ and $m^*(O-E)< \epsilon/2$. Using subadditivity, one sees $m^*(O-F)< \epsilon$ as needed.
I am stuck, however, proving the reverse direction. I have tried a proof by contradiction, assuming $m^*(O-F)< \epsilon$ but that $m^*(O-E) \geq \epsilon/2$ and $m^*(E-F) \geq \epsilon/2$ for sufficiently small epsilon. The problem is, subadditivity does seem to get us very far.
I am aware that this problem Lebesgue measurability criterion has been solved before, but the OP uses a different definition of measurability and does not take advantage of the highlighted equivalence theorem. I want to see if there is a short, beautiful way to show prove Exercise 17 as a corollary.
To prove the reverse direction, use that $m^*(E-F)\leq m^*(O-F)< \epsilon$.