If $f_n \xrightarrow{\mu} f$ and $f_n$ is increasing $f_n \xrightarrow{a.e}f$

Question

I have this problem
Let $f$, $f_n:X \to \mathbb{R}$ be measurable and $f_n \xrightarrow{\mu} f$. Show that if $f_n$ is monotone increasing then $f_n \xrightarrow{a.e}f$
My idea is by contradiction, to suppose that $f_n \not\xrightarrow{a.e}f$ so, $\{x:\lim f_n(x)\neq f\}$ it is not of measure zero, and how $f_n \xrightarrow{\mu} f$ there is a subsequence such that $f_{n_k} \xrightarrow{a.e}f$. But I don't know how finish. Could someone help me? Thank you in advance.

Answer 1

As you already pointed out, the hypothesis of convergence in measure implies there is a subsequence $(n_k)$ such that $f_{n_k} \to f$ almost everywhere. Now fix any $x$ such that $ f_{n_k}(x) \to f(x)$. The sequence $(f_n(x))$ is a monotonically increasing sequence, so it must converge to something (a priori, possibly $\infty$). But we already know that it has a subsequence converging to $f(x)$ so therefore the whole sequence $(f_n(x))$ must converge to $f(x)$.