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 $f_\infty(t)=g_\infty(t) \text{ a.e}$?
My effort:
Since $\{g_n\}$ is U.I then there exists a subsequence $\{g_{n_i}\}$ of $\{g_n\}$ such that weakly convergence in $L^1$ . It follows from $(1)$ and $(2)$ that $$ \qquad g_{n_i}\underset{i}{\to} f_\infty\text{ weakly in } L^1 $$ then for all $A\in \mathcal{A}$ $$ \lim_i \int_A g_{n_i} \,d\mu=\int_A f_\infty \,d\mu $$ then, by Cesàro lemma, we have
$$ \lim_m\int_A \frac{1}{m} \sum_{i=1}^m g_{n_i} \,d\mu=\int_A f_\infty \, d\mu $$
The sequence $\{s_m\}$, defined by $s_m:=\frac{1}{m} \sum_{i=1}^m g_{n_i}$, is uniformly integrable and converge weakly to $g_\infty$ then $$ \lim_m\int_A s_m \, d\mu=\int_A g_\infty \, d\mu\qquad\forall A\in\mathcal{A} $$ then $$ \int_A f_\infty \, d\mu=\int_A g_\infty \, d\mu\qquad\forall A \in \mathcal{A} $$ then $f_\infty(t)=g_\infty(t) \text{ a.e.}$
Is what I wrote correct? If not, please help.