Weak and pointwise convergence in a $L^2$ space Let $I$ be a measured space (typically an interval of $\Bbb R$ with the Lebesgue measure), and let $(f_n)_n$ a sequence of function of $L^2(I)$.

Assume that the sequence $(f_n)$ converge pointwise and weakly. How to prove that the pointwise limit and the weak limit are the same ?
Here's a functional analytic approach:

A weakly convergent sequence in a Hilbert space $H$ is bounded, and by the Banach-Saks theorem has a subsequence whose Cesàro averages converge strongly in $H$ to the same limit. 
Almost sure convergence is preserved by taking subsequences and Cesàro averages.
So, without loss of generality you may assume that your weakly convergent sequence is actually strongly convergent. 

Both strong $L^2$ convergence and almost sure convergence imply convergence locally in measure, so you only need to show that such limits are unique, which  is easy.
Assume that $f_n \rightharpoonup g$ and $f_n \rightarrow f$ a.e. Then
\begin{equation}
 | \int_{I} (f-g)| \le \int_{I} |f_n - f| + \int_{I} |f_n - g|
\end{equation}
As $f_n \rightharpoonup g$, e $1 \chi \{I\} \in L^{2}(I)$ we have
\begin{equation}
\int_{I} |f_n - g| \rightarrow 0.
\end{equation}
Now, notice that
\begin{equation}
|f_n -f| \le |f_n| + |f| &lt; 2|f| + \varepsilon.
\end{equation}
for $n&gt;&gt;1$. Then by Dominated convergence Theorem
\begin{equation}
\int_{I} |f_n - f| \rightarrow 0.
\end{equation}
Hence $f=g$ a.e.