If $f: \mathbb{R} \rightarrow \mathbb{R}$ is Lebesgue measurable, prove that there exists a Borel measurable function $g$ s.t. $f = g$ a.e. 

Consider the four cases.

Let $f = \chi\_E$, where E is a measurable set.

Let $f = \sum\_{i=1}^{n}{a\_i\chi\_{E\_i}}$

Let $f$ be a non-negative measurable function. There exists a simple function $S\_n$ s.t. $\lim\_{n \rightarrow \infty}{S\_n} = f$

Let $f$ be any measurable function, $f = f^+ + f^-$

I have proven the first case.

There exists a set $F$ that is a countable union of closed sets contained in E, therefore F is Borel measurable and the $m(E\setminus F)=0$ If I set $g(x) = \chi\_F$. Then $$ m(\{x : f \neq g\}) \leq m(E \setminus F) = 0 $$ Therefore $f = g$ a.e.

My trouble is working on the next three cases.