Show that $f_{n}\longrightarrow f$ both weakly in $L^{p}$ and almost everywhere implies that $\|f_{n}\|_{L^{p}}\longrightarrow \|f\|_{L^{p}}$. I am in the middle of a proof of a harmonic analysis exercise, and I'd like to claim that:

For $1<p<\infty$, If $f_{n}\longrightarrow f$ both weakly in $L^{p}$ and almost everywhere, then $\|f_{n}\|_{L^{p}}\longrightarrow \|f\|_{L^{p}}$ as $n\longrightarrow\infty$. 

Especially, I need this statement to be true in the case of $p\neq 2$. (I don't really even know if this claim is true.)
If it is true, how could I prove this? 
I tried several ways to use the almost everywhere property, for instance, using dominated convergence theorem, but given the current conditions, there is nothing for me to dominate $|f_{k}|$. I even tried Vitali, but we do not know if $f_{n}$ is uniformly integrable. 
Also, I don't really know how to use the conditional the $L^{p}$ weak convergence.
Is there any other way for me to approach this statement? 
The statement is perhaps wrong, but is there any other alternative, similarly close statement?
Thank you so much!
 A: Not in the general case. Consider the set $\mathbb{N}$ with the “cardinality” measure with $p=2$. 
Define $f_n(k)=1$ if $k=n$ and $0$ otherwise. Then, $f_n$ goes ae to $0$. It is, furthermore, clear that $f_n$ goes weakly to zero. 
That example should generalize to the case of “sufficiently nice” infinite measures (any measure with nonnegative non-integrable density with respect to any Lebesgue measure should work) for $1 < p < \infty$.

Edit: I found a finite-measure very simple counterexample for $1 < p < \infty$, which does not seem to indicate an easy fix. 
Take the space to be $[0,1]$ with the Lebesgue measure. Take $f_k(x)=k^{s+1}(1/k-x)$ where $s=1/p$ if $0 \leq x \leq 1/k$ and $f_k(x)=0$ otherwise. 
Clearly $f_k \rightarrow 0$ almost everywhere and an easy computation yields $\|f_k\|_p^p=(p+1)^{-1}$. 
Weak convergence is trickier to obtain (but holds): let $g \in L^q$ and define $P_n(g)=\left(\int_0^{1/n}{|g|^q}\right)^{1/q}$. Then, by Hölder, if $k \geq 1$, $$\left|\int_0^1{f_kg}\right| \leq k^{1+1/p}\int_0^{1/k}{(1/k-x)|g|(x)\,dx} \leq k^{1+1/p}P_k(g)\left(\int_0^{1/k}{x^p\,dx}\right)^{1/p} = CP_k(g) \rightarrow 0.$$
