Can the following be proved using only the outer measure and the definition of a measurable set? I would like to prove this using $m_*(E)$ defined as the infimum over all open sets $O \supset E$ and the definition that $E$ is measurable if $\exists O \supset E$ open such that $m(O \setminus E) < \epsilon$.
Prove that if $E$ is measurable then $\forall \epsilon > 0$ $\exists F \subset E$ closed such that $m(E \setminus F) < \epsilon$.
All proofs I've found so far are much more involved and I'm curious if this can be done from a more basic setting (i.e. without using the fact that the complement of a measurable set is measurable or that closed sets are measurable).
Best Answer
Let $\varepsilon>0$. If $m^*(E)$ is finite then $m^*(E^C)$ is infinite. Thus there exists a collection of disjoint sets of finite outer measure $\{E_k\}$ such that $E^C=\bigcup_{k=1}^\infty E_k$. Since each $E_k$ has finite outer measure, there exists an open set $O_k\supseteq E^C_k$ such that $$ m^*(O_k)\le m^*(E_k)+\frac{\varepsilon}{2^k}\,\,\Leftrightarrow\,\, m^*(O_k\sim E_k)<\frac{\varepsilon}{2^k}. $$
Let $O=\bigcup_{k=1}^\infty O_k\supseteq\bigcup_{k=1}^\infty E_k\supseteq E^C$ and define $F=O^C$. Then
$$ \begin{align*} m^*(E\sim F)&=m^*(E\sim O^C)\\ &=m^*\left(\bigcup_{k=1}^\infty E_k\sim\bigcup_{k=1}^\infty O_k\right)\\ &=m^*\left(\bigcup_{k=1}^\infty E_k\sim O_k\right)\\ &\le \sum_{k=1}^\infty m^*(E_k\sim O_k)\\ &< \sum_{k=1}^\infty \frac{\varepsilon}{2^k}\\ &=\varepsilon. \end{align*} $$
The case when $m^*(E^C)$ is finite should mirror the proof of $m^*(O\sim E)$ is finite similar to how my proof mirrors the infinite case.