Let $\Omega\subset\mathbb{R}^n$ be a bounded domain. Suppose that $p\in (1,\infty)$. Assume that the sequence $u_n\in L^p(\Omega)$ satisfies:
There is $u,w\in L^p(\Omega)$ such that $u_n\to u$ a.e. in $\Omega$ and $u_n$ weakly converge to w in $L^p(\Omega)$. Can we conclude that $u=w$?
I was trying to prove this result by using Mazur lemma, which says that the sequence$$v_n=\sum_{k=n}^{f(n)}\alpha_{k,n}u_k\to w\ \mbox{in}\ L^p(\Omega)$$
where $f(n)\geq n$ and $\sum_{k=n}^{f(n)}\alpha_{k,n}=1$. We can assume without loss of generality that $v_n(x)\to w(x)$ a.e. in $\Omega$. Now, I want to use the fact that $u_n$ to $u$ a.e. everywhere to conclude that $v_n\to u$ a.e. Can I do this?
Update: I think I could finish the proof, please verify it to me. Fix $x\in \Omega$ such that $u_n(x)\to u(x)$. Note that $$|v_n(x)-u(x)|=\left|\sum_{k=n}^{f(n)}\alpha_{k,n}(u_k(x)-u(x))\right|\leq \sum_{k=n}^{f(n)}|\alpha_{k,n}(u_k(x)-u(x))|\tag{1}$$
For any $\delta>0$ choose $N$ um such a way that if $n\geq N$ then $|u_n(x)-u(x)|\leq\delta$ then, from $(1)$ we conclude that $$|v_n(x)-u(x)|\leq \delta$$
