Show that for a sequence \(\left(u_{n}ight)_{n \in \mathbb{N}}\) of measurable functions on a finite measure space

\[\lim _{k ightarrow \infty} \mu\left\{\sup _{n \geqslant k}\left|u_{n}ight|>\epsilonight\}=\mu\left(\limsup _{n ightarrow \infty}\left\{\left|u_{n}ight|>\epsilonight\}ight) \quad \forall \epsilon>0\]

and combine this with Problem 22.1 to give a new criterion for a.e. convergence.

Data from problem 22.1

 Let \((X, \mathscr{A}, \mu)\) be a finite measure space and \(\left(u_{n}ight)_{n \in \mathbb{N}} \subset \mathcal{M}(\mathscr{A})\). Prove that

\[
\lim _{k ightarrow \infty} \mu\left\{\sup _{n \geqslant k}\left|u_{n}ight|>\epsilonight\}=0 \quad \forall \epsilon>0 \Longrightarrow u_{n} \xrightarrow[n ightarrow \infty]{ } 0 \text { a.e. }
\]

[\(\left|u_{n}ight| ightarrow 0\) a.e. if, and only if, \(\mu\left(\bigcup_{n \geqslant k}\left\{\left|u_{n}ight|>\epsilonight\}ight)\) is small for all \(\epsilon>0\) and big \(k \geqslant k_{\epsilon}\).]