Limit and conditional expectation commute in a uniformly integrable sequence 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.
 A: 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
