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.
Best Answer
Hint for the second part: as you showed in the first part, for each $i$ there is a Borel set $F_i \subset E_i$ with $m(E_i \setminus F_i) =0$. Take $g = \sum_{i=1}^n a_i \chi_{F_i}$. Now show that $f=g$ a.e.