## Measures on Real Line and open sets.

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

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

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

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

By a theorem, I also know that:

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

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

Proof so far.

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

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

Define $\displaystyle V = \bigcap G_n$.

Then we know the followings:

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

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

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

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

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

Any hints? Thank you.