Prove that if the outer measure of a set $E$ is the supremum of closed bdd sets, then $E$ is measurable. More formally, let $E \subset \mathbb{R}^d$ and let the outer measure $m_*(E) < +\infty$. Suppose
\begin{align} m_*(E) = \{m(F): F\subset E \,\, \text{is closed and bounded}\} \end{align}
Prove $E$ is measurable (Lebesgue).
Proof. First note that by properties of the supremum, that for any $\epsilon >0$, we can select $F\subset E$ such that
\begin{align} m_*(E) - m_*(F) < \epsilon \end{align}
Since $F\subset E$, we also have $E = (E-F) \cup F$. And, since $(E-F)$ is disjoint from $F$
\begin{align} m_*(E) =& m_*(E-F) + m_*(F) \\ \\ &\implies m_*(E-F) \leq m_*(E) - m_*(F) < \epsilon \end{align}
Which implies the outer measure of $(E-F)$ is zero, which implies $(E-F)$ is measurable. Therefore $E$ is the union of measurable sets, which makes it measurable.