My question is:
Let $E \subset [0,1]$ be a set on $[0,1]$ and let $S=\sup\{m(F)\mid F$ is closed and $F \subset E\}$.
Prove that $S=m^*(E)$ if and only if $E$ is measurable.
I have not found any help here or on other websites for my specific inquiries. We are not working with compact sets so any answer using that also doesn't help me understand. My main issue, I think, is how am I supposed to express this in a formal proof?
My understanding of the question is: $S$ is the largest that $m(F)$ can be, where $F$ is still closed and a subset of $E$. Since closed sets are measurable, $F$ is already always measurable (hence the $m(F)$ not $m^*(F$)).
$\color{red}{\Leftarrow}$ If $E$ is measurable, then by definition, we're taking the supremum of the inner approximation of $E$, hence it is equal to the $m^*(E)$. (Theorem: Let $E \in M$, $m(E) < \infty$. Then $\forall \epsilon>0, \exists F,$ a closed set, s.t. (1) $F\subset E$ and (2) $0 \leq m(E/F) < \epsilon$.) $M$ is the set of all measurable sets. Also, we know that $E$ cannot be infinite, since $E\subset [0,1]$.
The only issue I have with my proof here is that the theorem says that $F$ is existing based on the $\epsilon$, and here I think we're give a fixed set $F$. But then I also feel like it says $S$ contains any $F$ I want it to. So am I allowed to still used my (non-formally worded) proof?
$\color{red}{\Rightarrow}$ If $S=m^*(E)$, then it follows immediately that $E \in M$. This is because this means $m^*(E)=\sup\{m(F)\mid F\text{ is closed}, F\subset E\}$, in other words, the measure of $E$ is the measure of the largest subset of $E$ that's closed, and that closed subset is already measurable. So the measure of $E$ is equal to the measure of a measurable set.
Am I super off-base with my answers? Or are these valid, they just need a more refined proof?
Edit: For clarification, the $m^*$ refers to the Lesbegue outer measure.
