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.