$\mathcal{M}_\mu$ is the domain of the Lebesgue-Stieltjes measure $\mu = \mu_F$, where $F$ is an increasing, right continuous function on the Reals. I'm trying to prove the following:
If $E \subset \mathbb{R}$, then (a) implies (b) and (c), where:
(a) $E \in \mathcal{M}_\mu$
(b) $E = V \backslash N_1$ where $V$ is a $G_\delta$ set and $\mu(N_1) = 0$
(c) $E = H\cup N_2$ where $H$ is an $F_\sigma$ set and $\mu(N_2) = 0$.
(A countable intersection of open sets is called a $G_\delta$ set. A countable union of closed sets is called $F_\sigma$ set.)
Here is my proof (I get stuck at the end). Suppose that $E \in \mathcal{M}_\mu$. I make use of the theorem:
If $E \in \mathcal{M}_\mu$, then $$ \mu(E) = \inf \{ \mu(U): E \subset U \text{ and }U \text{ is open} \} $$ $$ = \sup \{ \mu(K): K \subset E \text{ and }K \text{ is compact} \} $$
Suppose $\mu(E) < \infty$. For $j \in \mathbb{N}$, we can choose open $U_j \supset E$ and compact $K_j \subset E$ such that $\mu(U_j) - 2^{-j} \le \mu(E) \le \mu(K_j) + 2^{-j}$. Let $V = \bigcap_1^\infty U_j$ and $H = \bigcup_1^\infty K_j$. Then, $H \subset E \subset V$. It is easy to prove that $\mu(H) = \mu(V)= \mu(E)$, thus as $\mu(E) < \infty$, $\mu(E\backslash H) = \mu(V\backslash E) = 0$. Noting $E = V \backslash (V \backslash E)$, $E = H \cup (E \backslash H)$ as $H \subset E$, we are finished.
However, I am not sure how to generalize this proof to the case $\mu(E)$ is not finite. Can anybody help me with this?