If $f:\mathbb{R}\rightarrow \mathbb{R}$ is Lebesgue measurable, then there exists a sequence of continuous functions $\{f_n \}$ converges pointwise to $f$ almost everywhere on $\mathbb{R}$.
My work:
I take $n\in\mathbb{N}$ be fixed, then by Lusin's theorem there exists a closed subset $F_n $ such that $m(\mathbb{R} - { F_n}) < \frac{1}{n}$ and a continuous function $f_{n}=f$ on $F_n$. Then, we can define $F=\cup_{n=1}^{\infty}F_n$ so that for any $x\in \mathbb{R} - F$, there exists $N \in \mathbb{N}$ such that$|f-f_n| =0 < \epsilon$ for any $\epsilon >0$ on $F_n$ for all $n> \mathbb{N}$.I think this shows the pointwise convergence on $\mathbb{R}$ a.e as $m(\mathbb{R} - { F_n}) < \frac{1}{n}$.
I think I need to select $F_n$ to be increasing sequence of closed sets, right?
