When does almost everywhere convergence imply convergence in measure?

We can show the contrapositive.

Assume that $f_n$ doesn't converge to $f$ in measure. This means that there is $\delta\gt 0$, an $\varepsilon\gt 0$ and a subsequence $(f_{n_k})_k$ such that for each $k$, $$\mu\{|f_{n_k}-f|\gt \varepsilon\}\gt \delta.$$

If $(f_{m_k})_k$ is a subsequence of $(f_{n_k})_k$, we have to show that $f_{m_k}$ doesn't converge almost everywhere to $f$. Define $A_k:=\{|f_{m_k}-f|\gt \varepsilon\}$. Then we have, using finiteness of the measure space, $$\mu\left(\limsup_{k\to \infty}A_k\right)=\mu\left(\bigcap_{j=1}^\infty\bigcup_{k\geqslant j}A_k\right)\geqslant \delta.$$

This is an easy corollary from Egorov's theorem, which states:

Given some measure space $(X,\Sigma,\mu)$ Let $f_n: E\rightarrow \mathbb{R}$ be a sequence of measurable functions on some $E \in \Sigma, \mu(E)<\infty$. Where $f_n \rightarrow f^*$ pointwise for some $f^*: E \rightarrow \mathbb{R}$. Then for all $\epsilon > 0$ there is a set $F_\epsilon \in \Sigma, F_\epsilon \subset E$ such that $\mu(F_\epsilon) < \epsilon$ and $f_n \rightarrow f^*$ uniformly on $E\backslash F_\epsilon$.

Can you deduce the theorem from Egorov on your own?