I have been attempting to solve the following question with no luck for a while now: Let $X_1,X_2,\dots \in L_1$ with $X_n \uparrow X$ a.s. (and $X \in L_1$) . Show that for any filtration $(F_n)$ we have $\mathbb{E}[X_n|F_n] \to \mathbb{E}[X|F_\infty]$ a.s. where $F_\infty = \sigma(\bigcup_{n\geq 1} F_n)$
My initial approach was to try to to prove $\mathbb{E}[X_n|F_n] \leq \mathbb{E}[X_{n+1}|F_{n+1}]$ and then apply the monotone convergence theorem but I have been unable to prove it and don't even know if it is true. I know for a fixed $m$ we have $\mathbb{E}[X_n|F_m] \leq \mathbb{E}[X_{n+1}|F_m]$ but I can't find a way to relate $\mathbb{E}[X_n|F_n]$ and $\mathbb{E}[X_{n+1}|F_{n+1}]$ since the $\sigma$-algebra we are conditioning on is changing . Perhaps the fact it is a filtration helps somehow? But I have been unable to find or prove any convergence results for conditional expectation when the $\sigma$-algebra changes with $n$.
I know it is a straightforward result of Doob Martingales that for a fixed $m$ we have $\mathbb{E}[X_m|F_n] \to \mathbb{E}[X_m|F_\infty]$. Further, for a fixed $m'$ the conditional monotone convergence theorem tells us that $\mathbb{E}[X_n|F_{m'}] \to \mathbb{E}[X|F_{m'}]$. However I don't know how to handle both things changing with $n$ at the same time.
Best Answer
One can notice that $(E[X_n|\mathscr{F}_n])_{n \in \mathbb{N}}$ is a submartingale: it is adapted, integrable and by the fact that $(\mathscr{F}_{n})_{n \in \mathbb{N}}$ is a filtration and monotonicity of conditional expectations we have $$E[E[X_{n+1}|\mathscr{F}_{n+1}]|\mathscr{F}_n]=E[X_{n+1}|\mathscr{F}_{n}]\geq E[X_n|\mathscr{F}_n]$$ Furthermore, one can see that $X_n^+\uparrow X^+$ a.s. so by Jensen's inequality and monotone convergence $$\sup_nE[(E[X_n|\mathscr{F}_n])^+]\le \sup_nE[X_n^+]=E[X^+]<\infty$$ because $X \in L^1$. Submartingale convergence yields that there exists $Y$ s.t. $Y=\lim_nE[X_n|\mathscr{F}_n]$ a.s. and $Y$ is $\mathscr{F}_\infty$-measurable. Note: $$\begin{aligned} E[|E[X|\mathscr{F}_\infty]-E[X_n|\mathscr{F}_n]|]&\leq E[|E[X|\mathscr{F}_\infty]-E[X|\mathscr{F}_n]|]\\ &+E[|E[X_n|\mathscr{F}_n]-E[X|\mathscr{F}_n]|] \end{aligned}$$ But then by Jensen's inequality and monotone convergence (and the fact that $X_n,X \in L^1$): $$E[|E[X_n|\mathscr{F}_n]-E[X|\mathscr{F}_n]|]\leq E[|X-X_n|]=E[X]-E[X_n]\to 0$$ And by the fact that the Doob martingale $(E[X|\mathscr{F}_n])_{n \in \mathbb{N}}$ is uniformly integrable: $$E[|E[X|\mathscr{F}_\infty]-E[X|\mathscr{F}_n]|]\to 0$$ Therefore, $E[X_n|\mathscr{F}_n]\stackrel{L^1}{\to }E[X|\mathscr{F}_\infty]$, so there exists a subsequence $(E[X_{n_k}|\mathscr{F}_{n_k}])_{k \in \mathbb{N}}$ s.t. $E[X_{n_k}|\mathscr{F}_{n_k}]\to E[X|\mathscr{F}_\infty]$ a.s. Therefore, we conclude: $$|Y-E[X|\mathscr{F}_\infty]|\leq |E[X|\mathscr{F}_\infty]-E[X_{n_k}|\mathscr{F}_{n_k}]|+|Y-E[X_{n_k}|\mathscr{F}_{n_k}]|\to 0$$ a.s.