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.