Can we say that $f_\infty(t)=g_\infty(t) \text{ a.e}$?

Question

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 $f_\infty(t)=g_\infty(t) \text{ a.e}$?

Answer 1

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 $f_\infty=g_\infty$