Let $(E,\mathcal{A},\mu)$ be a finite measure space. Let $\{f\_n\}$ and $\{g\_n\}$ two uniformly integrable sequences such that:
$$
\qquad\forall n\geq 1~:~ f\_n(t)=g\_n(t)~~ \text{ a.e.}\tag1
$$
$$
\qquad f\_n\underset{n}{\to} f\_\infty\text{ weakly in } L^1 \tag2
$$
and for all subsequence $\{g\_{n\_i}\}$ of $\{g\_n\}$ we have
$$
\qquad\frac{1}{m}\sum\_{i=1}^{m}{g\_{n\_i}(t)}\underset{m}{\to} g\_\infty(t) \text{ a.e.} \tag3
$$
Can we say that $f\_\infty(t)=g\_\infty(t) \text{ a.e}$?



Can we say that $f_\infty(t)=g_\infty(t) \text{ a.e}$?
Yes: by uniform integrability and Vitali's convergence theorem, one concludes that the pointwise a.e. convergence (3) can be upgraded to strong $L^1$ convergence, hence also weak convergence.
Moreover, (2) implies that the Cesáro mean
$$
\frac{1}{m}\sum\limits\_{i=1}^mf\_{n\_i}\rightharpoonup f\_\infty
\qquad\mbox{ weakly in }L^1
$$
(Apply the usual fact that if some real sequence converges to a finite limit then its Cesáro mean also converges to the same limit, here the real sequence is $\int f\_{n\_i} \phi\to \int f\_\infty\phi$ for any fixed $\phi\in L^\infty$ as $i\to\infty$.)
Since $f\_n=g\_n$ as elements of $L^1$ the uniqueness of weak limits guarantees indeed $f\_\infty=g\_\infty$


