Suppose that $1<p<\infty,\ f,\ f_n\in L^p([0,1]),\ n\in\mathbb{N},\ ||f_n||_{L^p}\leq 1$ for all $n$. $f_n\to f$ a.e.
Show that $f_n\to f$ weakly and $||f||_{L^p}\leq 1$.
Attempt:
If $f_n\not\to f$ weakly, then there is some subsequence $\{f_{n_k}\}$, $\epsilon> 0$, $g\in L^q$ such that $|\int (f_{n_k}-f)g|\geq \epsilon$
By Alaoglu's theorem, the closed ball in $L^p$ is weakly compact, so there is some subsequence $\{f_{n_{k\ell}}\}$ that converges to $f$ weakly, which is a contradiction.
However, how would I justify that the sequence converges to $f$ and not something else? Is a.e. convergence sufficient?
For the second part, $1\geq \lim_n|\int f_n g| = |\int fg|$, and taking the $\sup$ over $g\in L^q,\ ||g||_q=1$ gives us $||f||_p\leq 1$
I think I have solved most of the problem, but there's are parts that I am unconvinced about.
$\{f_n\}$ is uniformly integrable and $f_n \to f$ a.e. so $f_n \to f$ in $L^{1}$. Hence $\int_A f_n \to \int_A f$. If $f_n \to g$ weakly where $g \in L^{q}$ then $\int_A f_n \to \int_A g$. This gives $\int_A f=\int_A g$ for all $A$, so $f=g$ a.e.. For the second part you can simply use Fatou's Lemma: $\int |f|^{p} \leq \liminf \int |f_n|^{p} \leq 1$.