I'm trying to show that given any bounded measurable function $f:[0,1]\rightarrow\Bbb{R}$, there exists a sequence of continuous functions $f_n:[0,1]\rightarrow\Bbb{R}$ such that $f_n\rightarrow f$ a.e.
Since $f$ is measurable, by Lusin's theorem, for all $\varepsilon >0$ there exists $A\subseteq [0,1]$ such that $\mu(A)<\varepsilon$ and $f$ is continuous on $[0,1]\backslash A$. Since we are given that $f$ is bounded, we know that $f$ is integrable on $[0,1]\backslash A$... and I'm stuck.
I may be completely off but any help is appreciated.
Since $f$ is integrable on $[0,1]$ there exist continuous functions $f_n$ such that $\int_0^{1}|f_n(x)-f(x)| \, dx \to 0$. This implies $f_{n_k} \to f$ almost everywhere for some subsequence $f_{n_k}$.