Proving weak convergence without Vitali's convergence theorem. 
Problem: Let $(f_{n})_{n=1}^{\infty}$ be sequence in $L^{2}([0,1])$ such that uniformly bounded, i.e., $\sup_{n \in \mathbb{N}}||f_{n}||_{L^{2}} =M < +\infty$ and $f_{n} \to f$ in measure. Then show that $f_{n} \to f$ weakly in $L^{2}$.

I think I solve this problem without using Vitali's convergence theorem, however, hint of this problem says use the Vitali's theorem. Could you check that my attempt is okay? Please let me know this argument is right or wrong.

My attempt: Since $L^{2}$ is reflexive (actually Hilbert space), it suffices to show that $\forall g \in L^{2}$, $\int f_{n}g \to \int fg$ as $n \to \infty$. Let $f_{n_{j}}$ be any subsequence of $f_{n}$. Then it also coverges in measure. So it has a subsequence $f_{n_{j_{k}}}$ converges to $f$ pointwise a.e. Also, by the Holder's inequality,
  $$ \int|f_{n_{j_{k}}}g| \leq ||f_{n_{j_{k}}}||_{L^{2}}||g||_{L^{2}} \leq M||g||_{L^{2}} < +\infty. $$
  Hence, by the Dominated Convergence theorem, $\lim_{k \to \infty}\int f_{n_{j_{k}}}g = \int fg.$ This shows that every subsequence of  $\int f_{n}g$ has convergent sub-subsequence to $\int fg$. Hence, $f_{n} \to f$ weakly in $L^{2}$.

 A: As pointed out by gerw, the problem is to find $z\in\mathbb L^1$ such that $|f_n g | \leqslant z$ a.e. for all $n$, and the control on the $\mathbb L^2$ of $f_n$ does not give this. 
Here is an approach without Vitali's convergence theorem. 


*

*Extract an a.e. convergent subsequence and use Fatou's lemma to get that $f$ is square integrable.

*Considering $f_n-f$ instead of $f_n$, it suffices to treat the case $f=0$.

*In order to prove weak convergence to $0$, let $g\in\mathbb L^2$. Then for all $R\gt 0$, 
$$\left\lvert\int f_n g\right\rvert\leqslant \left\lvert\int f_n g
\mathbf 1\left\{ \left\lvert g\right\rvert\leqslant R         \right\}        \right\rvert+\left\lvert\int f_n g
\mathbf 1\left\{ \left\lvert g\right\rvert\gt R         \right\}        \right\rvert \leqslant R\int\left\lvert f_n\right\rvert+M\left\lVert g
\mathbf 1\left\{ \left\lvert g\right\rvert\gt R         \right\} \right\rVert_2             $$
hence it suffices to prove that $\int\left\lvert f_n\right\rvert\to 0$. 

*To this aim, fix a positive $\varepsilon$, split the integral into two part ($\left\lvert f_n\right\rvert$ bigger or smaller than $\varepsilon). For the first part, use Cauchy-Schwarz inequality and convergence in measure.   
