Proposition 14.6.4. Let $X$ be a LCH space, $\mu: \cb_{X} \to [0, \infty]$ be a $\sigma$-finite Radon measure, and $E \in \cb_{X}$, then
For every $\eps > 0$, there exists $U \in \cn^{o}(E)$ and $F \subset E$ closed such that $\mu(U \setminus F) < \eps$.
There exists a $F_{\sigma}$ set $A$ and a $G_{\delta}$ set $B$ such that $A \subset E \subset B$ and $\mu(B \setminus A) = 0$.
Proof. (1): Let $\seq{E_n}\subset \cb_{X}$ such that $\mu(E_{n}) < \infty$. By outer regularity, there exists $\seq{U_n}$ open such that $U_{n} \in \cn^{o}(E_{n})$ and $\mu(U_{n}) < \mu(E_{n}) + \eps/2^{n}$ for all $n \in \natp$. In which case,
so $U = \bigcup_{n \in \natp}U_{n}$ is the desired open set.
Applying the above result to $E^{c}$, there exists $V \in \cn^{o}(E^{c})$ such that $\mu(V \setminus E^{c}) < \eps$. Let $F = V^{c}$, then $F \subset E$ is closed and