General form of Hunt lemma Consider on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_n)_n,P),$ a sequence of random variable $(X_n)_n$ converging a.s to $X.$ $(Y_n)_n$ is a sequence of random variable in $L^1$, converging a.s to $Y \in L^1,$ such that $|X_n| \leq Y_n$ and $E[Y_n|\mathcal{F}_{\infty}]$ converges a.s to $E[Y|\mathcal{F}_{\infty}],$ where $\mathcal{F}_{\infty}=\sigma(\bigcup_{n \in \mathbb{N}}\mathcal{F}_n).$
Prove or disprove the following: $E[X_n|\mathcal{F}_n]$ converges a.s to $E[X|\mathcal{F}_{\infty}].$
$|E[X_n|\mathcal{F}_n]-E[X|\mathcal{F}_{\infty}]| \leq E[|X_n-X||\mathcal{F}_n]+|E[X|\mathcal{F}_n]-E[X|\mathcal{F}_{\infty}]|, E[X|\mathcal{F}_n]$ converges a.s to $E[X|\mathcal{F}_{\infty}],$ it remains to prove that $E[|X_n-X||\mathcal{F}_n]$ converges a.s to $0,$ it seems it's a version of the dominated convergence theorem.
Any suggestions ?
Remark: If $E[Y_n|\mathcal{F}_{n}]$ converges a.s to $E[Y|\mathcal{F}_{\infty}],$ where $\mathcal{F}_{\infty}=\sigma(\bigcup_{n \in \mathbb{N}}\mathcal{F}_n),$ then $E[X_n|\mathcal{F}_n]$ converges a.s to $E[X|\mathcal{F}_{\infty}].$
$|E[X_n|\mathcal{F}_n]-E[X|\mathcal{F}_{\infty}]| \leq E[|X_n-X||\mathcal{F}_n]+|E[X|\mathcal{F}_n]-E[X|\mathcal{F}_{\infty}]|.$
$E[X|\mathcal{F}_n]$ converges a.s to $E[X|\mathcal{F}_{\infty}].$
We can prove that $E[|X_p-X||\mathcal{F}_p]$ converges a.s to $0$ by writing for all $r,k \in \mathbb{N}^*,p \geq k,$
$$E[|X_p-X||\mathcal{F}_p] \leq \frac{1}{r}+E[(Y+Y_p)1_{\{|X_p-X|>\frac{1}{r}\}}|\mathcal{F}_p] \leq\frac{1}{r}+E[Y1_{\bigcup_{q \geq k}\{|X_q-X|>\frac{1}{r}\}}|\mathcal{F}_p]+E[Y_p1_{\bigcup_{q \geq k}\{|X_q-X|>\frac{1}{r}\}}|\mathcal{F}_p]$$
So $$E[|X_p-X||\mathcal{F}_p] \leq\frac{1}{r}+E[Y1_{\bigcup_{q \geq k}\{|X_q-X|>\frac{1}{r}\}}|\mathcal{F}_p]+E[Y_p1_{\bigcup_{q \geq k}\{|X_q-X|>\frac{1}{r}\}}|\mathcal{F}_p]-E[\min(r,Y_p)1_{\bigcup_{q \geq k}\{|X_q-X|>\frac{1}{r}\}}|\mathcal{F}_p]+E[\min(r,Y_p)1_{\bigcup_{q \geq k}\{|X_q-X|>\frac{1}{r}\}}|\mathcal{F}_p]$$
Hence,
$$E[|X_p-X||\mathcal{F}_p] \leq\frac{1}{r}+E[Y1_{\bigcup_{q \geq k}\{|X_q-X|>\frac{1}{r}\}}|\mathcal{F}_p]+E[Y_p|\mathcal{F}_p]-E[\min(r,Y_p)|\mathcal{F}_p]+E[\min(r,Y_p)1_{\bigcup_{q \geq k}\{|X_q-X|>\frac{1}{r}\}}|\mathcal{F}_p]$$
Take $p \to \infty:$ $$\limsup_p E[|X_p-X||\mathcal{F}_p] \leq \frac{1}{r}+E[Y1_{\bigcup_{q \geq k}\{|X_q-X|>\frac{1}{r}\}}|\mathcal{F}_\infty]+E[Y|\mathcal{F}_{\infty}]-E[\min(Y,r)|\mathcal{F}_{\infty}]+E[\min(r,Y)1_{\bigcup_{q \geq k}\{|X_q-X|>\frac{1}{r}\}}|\mathcal{F}_\infty],$$
Taking $k \to \infty $ and noticing that $P(\limsup_q\{|X_q-X|>\frac{1}{r}\})=0$ we obtain:
$$\limsup_p E[|X_p-X||\mathcal{F}_p] \leq \frac{1}{r}+E[Y1_{\limsup_{q}\{|X_q-X|>\frac{1}{r}\}}|\mathcal{F}_\infty]+E[Y|\mathcal{F}_{\infty}]-E[\min(Y,r)|\mathcal{F}_{\infty}]+E[\min(r,Y)1_{\limsup_{q}\{|X_q-X|>\frac{1}{r}\}}|\mathcal{F}_\infty] \leq \frac{1}{r}+E[Y|\mathcal{F}_{\infty}]-E[\min(Y,r)|\mathcal{F}_{\infty}]$$
To conclude by letting $r \to \infty.$
Additional facts:

*

*If $W_k$ is an increasing sequence of positive random variable, $W=\sup_k W_k$ then $$\lim_k E[W_k|\mathcal{F}_k]=E[W|\mathcal{F}_{\infty}]$$ since
$E[W_k|\mathcal{F}_k] \leq E[W|\mathcal{F}_k]$ so that $\limsup_k E[W_k|\mathcal{F}_k] \leq E[W|\mathcal{F}_{\infty}]$ $(E[W|\mathcal{F}_k]$ is a positive supermartingale converging to $E[W|\mathcal{F}_{\infty}]).$ Also $$\forall q,k, E[\min(W_k,q)|\mathcal{F}_k] \leq E[W_k|\mathcal{F}_k]$$ so for all $q$ $$E[\min(W,q)|\mathcal{F}_{\infty}] \leq \liminf_k E[W_k|\mathcal{F}_k]$$
then we let $q \to \infty$

*From the previous point we can prove the following version of Fatou lemma:
$$E[\liminf_kW_k|\mathcal{F}_{\infty}] \leq \liminf_k E[W_k|\mathcal{F}_k]$$ which holds for any sequence $(W_k)_k$ of positive random variable.

*The above exercise can be proved using the previous point $($take $W_k=|X|+Y_k-|X_k-X|)$

*More generally if $(U_k),(V_k),$ are sequence in $L^1$, converging a.s to $U \in L^1$ and $V \in L^1$ respectively, $(W_k)_k$ is such that $U_k \leq W_k \leq V_k,$ $W_k$ converges a.s to $W,$ $E[U_k|\mathcal{F}_k],E[V_k|\mathcal{F}_k]$ converges a.s to $E[U|\mathcal{F}_{\infty}]$ and $E[V|\mathcal{F}_{\infty}]$ respectively then $E[W_k|\mathcal{F}_k]$ converges a.s to $E[W|\mathcal{F}_{\infty}].$

*All the results remain true if $\mathcal{F}_k$ are decreasing

*All the results remain true if we have convergence in probability instead of a.s convergence (by taking subsequences)

 A: Suggestions:

*

*$\{Y_n\}$ is uniformly integrable. (Why?)


*Because of 1, $\{X_n\}$ is uniformly integrable. (Why?)


*Show that $M_n:=\sup_{m\ge n}|X_m-X|$ converges to $0$, as $n\to\infty$, in $L^1$ (not just a.s.).


*$E\left[|X_n-X|\mid\mathcal F_n\right]\le \sup_{m\ge n} E\left[|X_m-X|\mid\mathcal F_n\right]\le E[M_n\mid\mathcal F_n]$. By 3., the middle term in this inequality converges in $L^1$ to $0$; but that term is monotone decreasing in $n$, so it also converges a.s.!
A: I misread the statement of your question. In the usual statement of Hunt's Lemma, $Y_n=Y$ for all $n$; or, more generally, each $Y_n\in L^1$ and $Y_n\to Y$ a.s. and in $L^1$.
Here's an example to show that some additional condition on $(Y_n)$  is needed. Take $\Omega=[0,1)$ endowed with Lebesgue measure. $\mathcal F=\mathcal F_\infty$ is the Borel subsets of $[0,1)$, and $\mathcal F_n$ is the $\sigma$-algebra generated by the intervals $[(k-1)/n,k/n)$ $k=1,2,\ldots,n$. Now define $X_n$ to take the value $n$ on each interval of the form $[(k-1)/n, (k-1)/n+1/n^{2})$ for $k=1,2,\ldots, n$, and to equal $0$ elsewhere. Let $Y_n:=X_n$ for each $n$. Then $\lim_n X_n=0=:X$ a.s., but $E[X_n\mid\mathcal F_n]=1$ a.s. for each $n$. Notice that $E[Y_n\mid\mathcal F_\infty]=Y_n$ for each $n$. Your hypotheses are satisfied but the convergence of $E[X_n\mid \mathcal F_n]$ to $E[X\mid \mathcal F_\infty]$ fails.
A: The following example, it got from John Dawkins' answer on Apr. 16, may be helpful for consider your proof.
Let $ \{(U_n,V_n), n\ge 1\} $ be a sequence of i.i.d. 2d-rvs with independent uniform distribution on $ (0,1) $ and
\begin{gather*}
 \xi_n=n^21_{\{U_n\le 1/n^2\}}, \quad \eta_n=2V_n, \qquad
 X_n=Y_n=\xi_n\eta_n, \\
 \mathscr{F}_n=\sigma(\xi_{i-1},\eta_i, 1\le i\le n),
\end{gather*}
then
\begin{gather*}
 \mathsf{E}[\xi_n]=1,\quad \xi_n\stackrel{a.s.}{\longrightarrow}0,\qquad 
 |\eta_n|\le 2,\quad \mathsf{E}[\eta_n]=1,\\
 0\le X_n=Y_n\stackrel{a.s.}{\longrightarrow}0(=Y=X)
\end{gather*}
Meanwhile,
\begin{equation*}
 \mathsf{E}[Y_n|\mathscr{F}_\infty]=Y_n\stackrel{a.s.}{\longrightarrow}0
 (=\mathsf{E}[Y|\mathscr{F}_\infty]). 
\end{equation*}
But
\begin{equation*}
 \mathsf{E}[X_n|\mathscr{F}_n]=\mathsf{E}[Y_n|\mathscr{F}_n]=\eta_n\mathsf{E}[\xi_n|\mathscr{F}_n]
 =\eta_n\mathsf{E}[\xi_n]=\eta_n\not\to 0.
\end{equation*}
