Let $1<p<\infty$. Let $f_n$ be a sequence of (a.e.) positive functions, and $f$ a positive function, both in $\mathcal{L}^p$.
Is it true that: If $f_n$ converges weakly to $f$ then there exists a sub-sequence $f_{n_k}$ such that $f_{n_k}$ converges a.e. to $f$.
I know that this is false without the positive constraint. Thanks for the help :).
Consider $L^p(-\pi,\pi)$ for $1< p<\infty.$ The sequence $e^{inx}$ is weakly convergent to $0,$ when $|n|\to \infty$, due to the Riemann-Lebesgue lemma, as $L^q(-\pi,\pi)\subset L^1(-\pi,\pi),$ for $q=p/(p-1).$Therefore the sequence $f_n(x)=2+2\cos nx=2+e^{inx}+e^{-inx}\ge 0$ is weakly convergent to $2.$ However the sequence $f_n$ does not admit a subsequence convergent to $2$ almost everywhere. Indeed if $f_{n_k}(x)\to 2$ a.e., then $\cos^2(n_kx)\to 0$ a.e. But $${1\over 2\pi}\int\limits_{-\pi}^\pi \cos^2(n_kx)\,dx ={1\over 2}$$ We get a contradiction by the Lebesgue dominated convergence theorem.