Convergence in weak topology implies pointwise convergence for a subsequence

Question

Prove that if a sequence $\{x_n\}_{n \in \mathbb{N}} \subset L^p$, $1<p<\infty$ converges in the weak topology to an element $x \in L^p$, then there is a subsequence $\{x_{n_k}\}_{k \in \mathbb{N}}$ of it that converges almost everywhere to $x$. From Mazur's lemma, we can find convex combinations that converge strongly to $x$, but I couldnt finish the problem from there. Any hint is appreciated.

Answer 1

This is not true. The sequence $$ x_n := sign(\sin(n \pi t) $$ converges weakly in $L^p(0,1)$ to $x=0$. But $x_n\ne0$ almost everywhere.