Question about the expectation given a filtration of sigma algebras

Question

Suppose $\mathcal{F_n}$ is a filtration of $\sigma-$algebras, and let $A$ be an event in the $\sigma$-algebra generated by all $\mathcal{F_n}'s$

Prove: $E(\mathbb{1}_A|\mathcal{F_n})\to\mathbb{1}_A$

My intuition is to use the tower property combined with another claim I proved that given $X_n=E(\mathbb{1}_A|\mathcal{F_n}), $then $ E(X_n|\mathcal{F}_{n-1})=X_{n-1}$ (meaning that $X_n$ is a martingale). Can anyone help please?

Answer 1

Hints:

1. Since $X_n := \mathbb{E}(1_A \mid \mathcal{F}_n)$ is a non-negative martingale which is nicely bounded, it follows from standard martingale convergence theorems that there exists a non-negative random variable $Y \in L^1$ such that $X_n \to Y$ almost surely and in $L^1$. It remains to identify $Y$.
2. Using that $X_n \to Y$ in $L^1$ prove that $\mathbb{E}(Y \mid \mathcal{F}_j)= X_j$.
3. Using the tower property, conclude that $$\mathbb{E}(Y \mid \mathcal{F}_j) = \mathbb{E}(1_A \mid \mathcal{F}_j). $$ Deduce that $$\mathbb{E}(Y \mid \mathcal{F}_{\infty}) = \mathbb{E}(1_A \mid \mathcal{F}_{\infty}). \tag{1}$$
4. Since both $Y$ and $1_A$ are $\mathcal{F}_{\infty}$-measurable, $(1)$ implies $Y= 1_A$ almost surely.