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

(i) Let \(\left(x_{n}ight)_{n \in \mathbb{N}} \subset \mathbb{R}\). Show that \(\lim _{n ightarrow \infty} x_{n}=0\) if, and only if, every subsequence \(\left(x_{n_{k}}ight)_{k \in \mathbb{N}}\) satisfies \(\lim _{k ightarrow \infty} x_{n_{k}}=0\).

(ii) Show that \(u_{n} \xrightarrow{\mu} u\) if, and only if, every subsequence \(\left(u_{n_{k}}ight)_{k \in \mathbb{N}}\) has a sub-subsequence \(\left(\tilde{u}_{n_{k}}ight)_{k \in \mathbb{N}}\) which converges a.e. to \(u\) on every set \(A \in \mathscr{A}\) of finite \(\mu\)-measure.

[use Lemma 22.4 for necessity. For sufficiency show that \(\tilde{u}_{n_{k}} ightarrow u\) in measure, hence the sequence of reals \(\mu\left(A \cap\left\{\left|u_{n_{k}}-uight|>\epsilonight\}ight)\) has a subsequence converging to 0 ; use (i) to conclude that \(\mu\left(A \cap\left\{\left|u_{n}-uight|>\epsilonight\}ight) ightarrow 0\).]

(iii) Use part (ii) to show that \(u_{n} \xrightarrow{\mu} u\) entails that \(\Phi \circ u_{n} \xrightarrow{\mu} \Phi \circ u\) for every continuous function \(\Phi: \mathbb{R} ightarrow \mathbb{R}\).