4.6.2. Levy's theorem
Consider a sequence of conditional expectations $E(X|\mathcal{F}_n)$ with fixed $X.$ By using the theorem from previous subsection we can determine convergence of this sequence as well.
Levy’s theorem
Let $X$ be an integrable random variable and $(\mathcal{F}_n)$ be a filtration. then $E(X|\mathcal{F}_n) \to E(X|\mathcal{F}_\infty)$ a.s. where $\mathcal{F}_\infty = \sigma\left(\cup_n \mathcal{F}_n \right).$
Let $X_n = E(X|\mathcal{F}_n)$ then $X_n$ is a closable, thus regular, martingale and there exists $X_\infty$ such that $X_n \to X_\infty$ a.s. It suffices to show that $X_\infty = E(X|\mathcal{F}_\infty)$ a.s. We show this with $\pi\text{-}\lambda$ theorem. Let $\mathcal{L} = \{A: \int_A X_\infty dP = \int_A X dP\}$ be a $\lambda$-system. Then $\cup_n \mathcal{F}n \subset \mathcal{L}$ and $\cup_n \mathcal{F}_n$ is a $\pi$-system. By $\pi\text{-}\lambda$ theorem $\mathcal{F}_\infty \subset \mathcal{L}$ thus $X_\infty = E(X|\mathcal{F}_\infty)$ a.s.
Similar result holds for a sequence $(X_n)$ uniformly dominated by an integrable random variable.
Suppose $X_n \to X$ a.s., $|X_n| \le Z, \forall n,$ $E|Z|<\infty$ then $E(X_n|\mathcal{F}_n) \to E(X|\mathcal{F}_\infty)$ a.s.
Let $W_n = \sup_{k,l \ge n} |X_k - X_l|$ then $W_n \downarrow 0$ a.s. and $|X_n-X| \le W_m$ for all $m \le n$ and $W_n \le 2Z.$ By the previous theorem we only need to show $E(\left|X_n-X\right||\mathcal{F}_n) \to 0$ a.s. Given $m,$ $$\limsup_n E(\left|X_n-X\right||\mathcal{F}_n) \le \lim_n E(W_m|\mathcal{F}_n) = E(W_m|\mathcal{F}_\infty).$$ By conditional DCT, $E(W_m|\mathcal{F}_\infty) \to 0$ a.s. as $m\to\infty.$ Thus $E(\left|X_n-X\right||\mathcal{F}_n) \to 0$ a.s. With Levy's theorem and triangle inequality the desired result follows.
Acknowledgement
This post series is based on the textbook Probability: Theory and Examples, 5th edition (Durrett, 2019) and the lecture at Seoul National University, Republic of Korea (instructor: Prof. Sangyeol Lee).