Let $f$ and $f_k$, $k = 1, 2, \ldots$, be measurable and finite a.e. in $E$. If $f_k \rightarrow f$ a.e. on $E$ and $|E| < +\infty$, then $f_k \rightarrow f$ in measure on $E$.
My proof is as follows: given $\epsilon > 0$. Since $f_k \rightarrow f$ a.e. on $E$, we have $$\{|f - f_k| \leq \epsilon \} \nearrow E$$ So $$|\{|f - f_k| \leq \epsilon \}| \rightarrow |E|$$ Note that $\{|f - f_k| > \epsilon \} = E - \{|f - f_k| \leq \epsilon \}$.
And since $|E| < \infty$, we have $$|\{|f - f_k| > \epsilon \}| = |E| - |\{|f - f_k| \leq \epsilon \}| \rightarrow 0$$ Hence $f_k \rightarrow f$ in measure on $E$.
Is my proof correct?