Measures on Real Line and open sets.

Let $\mu$ be a Borel measure on $\mathbb {R}$.

Denote $\mathbb {M} _ \mu$ as the domain of $\mu$

Then for any $E \in \mathbb {M} _ \mu$, we have:

$\mu (E) = inf \{ \sum _{1}^ \infty [F(b_n)-F(a_n)]:E \subset \bigcup _1^ \infty (a_n,b_n] \}$

By a theorem, I also know that:

$\mu (E)=inf \{ \mu (U) : E \subset U,U \ open \}$

Question: If $E \subset \mathbb {R}$, and if $E \in \mathbb {M} _{ \mu }$, prove that $E=V \backslash N$ where $V$ is a countable union of open subset and $\mu (N) = 0$

Proof so far.

Case 1. If $\mu (E) < \infty$:

Given $\epsilon > 0$, there exists open sets $G_n$ with $E \subset G_n$, $\mu (G_n) < \mu (E) + \frac { \epsilon }{2^n}$ since $\mu (E)$ is the infimum of these sets.

Define $V = \bigcap G_n$.

Then we know the followings:

i. $E \subset V$, implies that $\mu (V) \geq \mu (E)$

ii. $\mu (V) \leq \mu (G_n) < \mu (E) + \frac { \epsilon }{2^n}$
implies that $\mu (V) \leq \mu (E)$

Using both i and ii, I then conclude that $\mu (V \backslash E ) = 0$, in which implies $V \backlash E = N$, implies that $V \backslash N = E$ with $\mu (N) = 0$

Case 2. $\mu (E) = \infty$:

In here, I want to do the same trick. So can I assume that $\exists F \subset E \ s.t. \ 0 < \mu (F) < \infty$? I know I can do that for any semi-finite measures.

Any hints? Thank you.