Convergence in $L^p$ and convergence almost everywhere

Why $f_n$ converges to $f$ in $L^p$ space implies that exists subsequence of $f_n$ converging to $f$ almost everywhere?

The typical proof of this uses $$L^p \text{ convergence } \,\, \implies \text{ convergence in measure } \,\, \implies \text{ subseq. converges pointwise a.e.}$$

For the latter, if $f_n \to f$ in measure, then for each $k \in \mathbb N$ we can find $n_k \in \mathbb N$ so that $$\mu \left( \left\{ \lvert f_{n} - f \rvert > \frac{1}{k} \right\} \right) \le \frac{1}{2^k}$$ for all $n \ge n_k$. In particular, putting $$F_k = \left\{ \lvert f_{n_k} - f \rvert > \frac{1}{k} \right\} \text{ and } E_\ell = \bigcup^\infty_{k = \ell+1} F_k$$ we have $$\mu(F_k) < \frac{1}{2^k} \,\,\, \text{ and so } \,\,\, \mu(E_\ell) < \sum^\infty_{k=\ell+1} \frac{1}{2^k} = \frac{1}{2^\ell}. $$ Hence, put $$N = \bigcap_{\ell = 1}^\infty E_\ell.$$ Then continuity from above tells us that $\mu(N) = 0$ and for $x \not \in N$, we have $f_{n_k}(x) \to f(x)$.