Let $\mathbb F = \left( \mathcal F_n\right)_{n \in \mathbb N}$ be a filtration, and let $\mathcal F_\infty = \sigma\left(\mathcal F_n : n \geq 1\right)$. I want to try proving the following result:
If $X \in \mathcal L^1\left(\mathcal F_\infty\right)$, then $\mathbb E\left[X|\mathcal F_n\right] \to X$ as $n \to \infty$ almost surely.
It’s easy to show that $\left(\mathbb E\left[X|\mathcal F_n\right]\right)_{n \in \mathbb N}$ is a martingale with respect to $\mathbb F$, and since $X$ is integrable, one can show that
$$
\sup_{n \geq 1} \mathbb E\left[\mathbb E\left[X|\mathcal F_n\right]^+\right] \leq \mathbb E[|X|] < \infty
$$
So by the martingale convergence theorem, $\left(\mathbb E\left[X|\mathcal F_n\right]\right)_{n \geq 1}$ converges almost surely, say to the random variable $\tilde X$. But why does $X = \tilde X$ a.s.? How can I show that $\left| X – \mathbb E\left[X|\mathcal F_n\right]\right| \to 0$? I’m wondering if there’s a functional analytic reason, considering the projections $\mathbb E\left[\cdot|\mathcal F_n\right] : \mathcal L^1\left(\mathcal F_\infty\right) \to \mathcal L^1\left(\mathcal F_n\right)$, but I haven’t gotten very far with this approach. Or is there an obvious reason for this that I’m just overlooking?
Best Answer
Let $X_n=\mathbb{E}\left[X\mid\mathcal{F}_n\right]$. This is a closed martingale so it converges in $L^1$ and almost surely to $\tilde{X}$. For all $n\geq 1$ and $A\in\mathcal{F}_n$, the convergence in $L^1$ together with the continuity of the conditional expectation give $$\mathbb{E}\left[\mathbf{1}_AX\right]=\mathbb{E}\left[\mathbf{1}_AX_n\right]=\mathbb{E}\left[\mathbf{1}_A\tilde{X}\right]$$ This equality holds for any $A\in\bigcup_{n=1}^\infty\mathcal{F}_n$. If we can show that it holds for $A\in\mathcal{F}_\infty$ we are done because both $X$ and $\tilde{X}$ are $\mathcal{F}_\infty$-measurable. Now, the set $\bigcup_{n=1}^\infty\mathcal{F}$ is stable by finite intersection and the class $$\mathcal{G}=\left\{A\in\mathcal{F}_\infty:\mathbb{E}\left[\mathbf{1}_AX\right]=\mathbb{E}\left[\mathbf{1}_A\tilde{X}\right]\right\}$$ is a monotone class. Since $\bigcup_{n=1}^\infty\mathcal{F}_n\subset\mathcal{G}$ the monotone class lemma implies $\mathcal{F}_\infty\subset\mathcal{G}$.