Weak convergence and subsequential limits Let $f_n:[0,1]\rightarrow \mathbb{R}$ be functions for $n\in\mathbb{N}$ such that
(1) for some $M>0$, $\vert f_n\vert_{L_\infty([0,1])}\leq M$ for all $n\in\mathbb{N}$ and (2) $(f_n)$ weakly converges to some $f$ in $L_2([0,1])$. Is it true that for almost all $x\in[0,1]$, $f(x)\in[\liminf \, f_n(x),\limsup \,f_n(x)]$?
 A: If a sequence $\left(g_n\right)_{n\geqslant 1}$ of non-negative functions converges  weakly on $\mathbb L^2\left(\left[0,1\right]\right)$ to $g$, then $g$ is non-negative. Indeed, for a fixed integer $N$ let $A_N$ be the set of $x\in [0,1]$ such that $g\left(x\right)\leqslant -1/N$. Then 
$$
0\leqslant \int_{\left[0,1\right]}g_n\left(x\right)\mathbf 1_{A_N}\left(x\right)\mathrm d\lambda\left(x\right)
$$
and letting $n$ going to infinity, we derive that 
$$
0\leqslant   \int_{\left[0,1\right]}g\left(x\right)\mathbf 1_{A_N}\left(x\right)\mathrm d\lambda\left(x\right)\leqslant -N^{-1}\lambda\left(A_N\right)
$$
hence $\lambda\left(A_N\right)=0$.
Now use the result with $g_n\colon x\mapsto \sup_{k\geqslant 1}f_k\left(x\right)-f_n\left(x\right)$ to get that for almost every $x$,
$$
f\left(x\right)\leqslant \sup_{k\geqslant 1}f_k\left(x\right).
$$
Apply this what we have done to the sequence $\left(f_{n+n_0}\right)_{n\geqslant 1}$ instead of $\left(f_{n}\right)_{n\geqslant 1}$ to get that for each integer $n_0$ and almost every $x$, 
$$
f\left(x\right)\leqslant \sup_{k\geqslant n_0}f_k\left(x\right).
$$
This gives the wanted result for $\limsup$. Apply this to $-f$ to obtain the other inequality.
