I am thinking of the next proposition:
Proposition.
Let $(\Omega, \mathcal{F}, P)$ a probability space, and $\{X_n\}_{n=1,\cdots}$ a uniformly integrable r.v. sequence s.t. $X_n \rightarrow X ~ \text{a.s.}$, where $X$ is an $L^1$ r.v. Then for any $\mathcal{G}$: a sub-$\sigma$-algebra of $\mathcal{F}$, we have
\begin{equation*}
\lim_{n\rightarrow \infty} \mathbb{E}[X_n | \mathcal{G}] = \mathbb{E}[X | \mathcal{G}] \quad \text{a.s.}
\end{equation*}
My quesiont is: Is this proposition true? I consider it indeed is true, for the following reason. For any $A\in \mathcal{G}$ and $R > 0$, $\{X_n 1_A\}$ is a u.i. sequence, which is shown by
\begin{equation*}
\sup_n \mathbb{E}[ |X_n 1_A|, |X_n 1_A| > R ] \leq \sup_n \mathbb{E}[ |X_n |, |X_n| > R ]
\end{equation*}
with the right hand side going to $0$ with $R \rightarrow \infty$ (we denote $\mathbb{E}[X,A] = \mathbb{E}[X1_A]$). Thus by the commutability of conditional expectation and limit in a u.i. sequence we have
\begin{equation*}
\mathbb{E}[X1_A] = \mathbb{E}[ \lim_{n \rightarrow \infty}X_n 1_A] = \lim_{n \rightarrow \infty} \mathbb{E}[X_n 1_A]. \qquad (1)
\end{equation*}
On the other hand,
\begin{align*}
& \mathbb{E}[ \lim_{n \rightarrow \infty} \mathbb{E}[X_n | \mathcal{G}], A] = \mathbb{E}[ ( \lim_{n \rightarrow \infty} \mathbb{E}[X_n | \mathcal{G}]) 1_A] = \mathbb{E}[ \lim_{n \rightarrow \infty} (\mathbb{E}[X_n | \mathcal{G}] 1_A)] \\
=~& \mathbb{E}[ \lim_{n \rightarrow \infty} (\mathbb{E}[X_n 1_A | \mathcal{G}])]\\
=~& \lim_{n \rightarrow \infty} \mathbb{E}[ \mathbb{E}[X_n 1_A | \mathcal{G}]]\\
=~& \lim_{n \rightarrow \infty} \mathbb{E}[X_n 1_A]. \qquad (2)
\end{align*}
Here to swap the expectation and the limit I used the fact that the sequence $\{ \mathbb{E}[Y_n | \mathcal{G}] \}_{n=1,\cdots}$ with a u.i. r.v. sequence $\{ Y_n \}_{n=1,\cdots}$ is u.i., which I presume is true.
Therefore by (1) and (2), the above proposition holds.
A previous post (https://mathoverflow.net/questions/124589/uniformly-integrable-sequence-such-that-a-s-limit-and-conditional-expectation-d) claims otherwise, which I guess is wrong: I think one can't compose a u.i. sequence $\{Z_n\} = \{X_n Y_n\}$ that satisfies the conditions written in the link above.
Best Answer
The proposition in the post does not hold. The problem arises from the implicit assumption that the a.s. limit of the conditional expectations exists.
The example described in the link [1] is correct, but could be made more concrete. Here is an explicit realization of that example. Let $\{U_j\}_{j \ge 0}$ be independent uniform variables in $[0,1]$. Let $X_n$ be the indicator $$X_n={\large\bf 1}_{\displaystyle \{\forall j\in \{1,\ldots,n-1\}, \quad U_n>U_j\}}$$ and let $$Y_n=n \cdot {\large \mathbf 1}_{\displaystyle \Bigl\{U_0 \in (0,\frac1n]\Bigr\}} \,.$$ Denote by $\mathcal G$ the $\sigma$-field generated by $\{U_j\}_{j \ge 1}$. Then $\mathbb{E}(X_n)=1/n \to 0$ but $X_n=1$ for infinitely many $n$ a.s.
The sequence $Z_n=X_nY_n$ tends to $0$ a.s. and in $L^1$, so it is uniformly integrable. However, $$\mathbb{E}(Z_n | \mathcal G)=X_n \mathbb{E}(Y_n)=X_n$$ does not tend to $0$ a.s. Indeed, this sequence almost surely diverges.
Remark (added to address comment by OP): Moving away from the example, note that in general, if $X_n \to X$ in $L^1$, then necessarily $\mathbb{E}[X_n|G] \to \mathbb{E}[X|G]$ in $L^1$, (whence a subsequence of $\mathbb{E}[X_n|G]$ converges a.s. to $\mathbb{E}[X|G]$.)
Indeed, $$\Bigl| \mathbb{E}[X_n|G] - \mathbb{E}[X|G] \Bigr| =\Bigl| \mathbb{E}\bigl[X_n-X\,\,|G\bigr] \Bigr| \le \mathbb{E}\Bigl[\, |X_n-X|\,\, \Big|G \Bigr] $$ implies that $$\mathbb{E}\Bigl| \mathbb{E}[X_n|G] - \mathbb{E}[X|G] \Bigr| \le \mathbb{E}\Bigl(\mathbb{E}\Bigl[\, |X_n-X|\,\, \Big|G \Bigr]\Bigr)= \mathbb{E} \Bigl[\, |X_n-X| \Bigr] \,.$$
In particular, if $X_n \to X$ in $L^1$, and the sequence $\mathbb{E}[X_n|G]$ converges a.s., then the limit must equal $\mathbb{E}[X|G]$ a.s.
[1] https://mathoverflow.net/questions/124589/uniformly-integrable-sequence-such-that-a-s-limit-and-conditional-expectation-d