Conisder Lusin's Theorem (Folland Chap 2 Ex 44):
If $f: [a,b] \rightarrow \mathbb{C}$ is Lebesgue measurable and $\epsilon > 0$, there is a compact subset $E \in [a,b]$ s.t. $\mu(E^C) < \epsilon$ and $f$ restricted to $E$ is continuous.
To prove this, I basically:
- Defined $E_k = \{x | x \in [a,b], |f(x)|<k\}$ and $f_k' = f \chi_{E_k}$. Then $f_k' \rightarrow f$ p.w.
- Approximated each $f_k'$ is a continuous function $f_k$ (one can show each $f_k'$ is $L^1$)
- $f_k \rightarrow f$ p.w. so in particular, $f_k \rightarrow f$ a.e
- By Egoroff's Theorem $\exists F \subset [a,b]$ s.t. $\mu(F) < \epsilon$ and $f_k \rightarrow f$ unifomly on $F^c$. Ergo $f$ is continuous on $F^C$
To complete, I think I can get a compact set by finding one inside $F^C$ because $F^C$ is Borel. But I really didn't use compactness in an essential way to obtain the conclusion. Am I missing something?