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?
I think that we are talking about the Lebesgue-meausre $m$. This argumentation of yours does not work then (because of many reasons), but most importantly, because $\mathbb{R}$ has infinite measure. Check the requirements of Lusin's theorem.
Try something else: It is well known that there exists a sequence of continuous functions $(f_n)_{n \in \mathbb{N}}$ such that $$ \int_\mathbb{R} \lvert f_n - f \rvert~\mathrm{d}x \overset{n \rightarrow \infty}{\longrightarrow} 0. $$ You can prove this using approximation of $f$ by a series of characteristic functions that themselves you approximate by continuous functions.
Now choose a subsequence $(f_{k_n})_{n \in \mathbb{N}}$ such that $$ \int_\mathbb{R} \lvert f_{k_n} - f \rvert~\mathrm{d}x \leq 2^n $$ for all $n \in \mathbb{N}$. Therefore $$ \lim_{m \rightarrow \infty} \sum_{n = 1}^m \int_\mathbb{R} \lvert f_{k_n} - f \rvert~\mathrm{d}x < \infty. $$ By the monotone convergence theorem: $$ \lim_{m \rightarrow \infty} \sum_{n = 1}^m \int_\mathbb{R} \lvert f_{k_n} - f \rvert~\mathrm{d}x = \int_{\mathbb{R}} \sum_{k = 1}^\infty \lvert f_{k_n} - f \rvert~\mathrm{d}x < \infty $$ Clearly, we then must have $$ \sum_{k = 1}^\infty \lvert f_{k_n}(x) - f(x) \rvert < \infty $$ a.e., so $f_{k_n} \overset{n \rightarrow \infty}{\longrightarrow} f$ pointwise a.e.. Thus $(f_{k_n})_{n \in \mathbb{N}}$ is your desired sequence.