Let \(\left(u_{n}ight)_{n \in \mathbb{N}}\) be a sequence of measurable functions on a \(\sigma\)-finite measure space \((X, \mathscr{A}, \mu)\) and assume that \(u_{n} \stackrel{\mu}{ightarrow} u\).

(i) Show that \(\lim _{n ightarrow \infty} \int\left|u_{n}-uight| \wedge \mathbb{1}_{A} d \mu=0\) for every \(A \in \mathscr{A}\) with finite measure \(\mu(A)

(ii) Show that every subsequence \(\left(u_{n}^{\prime}ight)_{n \in \mathbb{N}} \subset\left(u_{n}ight)_{n \in \mathbb{N}}\) contains a further subsequence \(\left(u_{n}^{\prime \prime}ight)_{n \in \mathbb{N}} \subset\left(u_{n}^{\prime}ight)_{n \in \mathbb{N}}\) such that \(\lim _{n ightarrow \infty} u_{n}^{\prime \prime}=u\) almost everywhere.

[ apply (i) with \(A=A_{i}\) from a sequence \(A_{i} \uparrow X, \mu\left(A_{i}ight)

(iii) Use (ii) to give an alternative proof of Lemma 22.5 .

Data from corollary 13.8

Data from lemma 22.5

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Corollary 13.8 Let (Un)nEN CLP (u), pe [1, ) with LP-limn Un=u. Then there exists a subsequence (un(k))keN such that limk un(k)(x) = u(x) holds for almost every x EX. Proof Since (un)neN converges in LP (), it is also an LP-Cauchy sequence and the claim follows from (the remarks following) formula (13.10). As we have already remarked, pointwise convergence alone does not guar- antee convergence in LP, not even of a subsequence, see Problem 13.8. Let us repeat the following sufficient criterion, which we have already proved on page 119.