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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?

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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.
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    So lets see if I got this right.. From hint #1 we get $E[\mathbb{1}_A|\mathcal{F_{\infty}}]=Y$. Now to show #2 and #3: $E[Y|\mathcal{F_j}]=E[E[\mathbb{1}_A|\mathcal{F_{\infty}}]|\mathcal{F_j}]=E[\mathbb{1}_A|\mathcal{F_j}]=X_j$ (in the second equality we used tower property). Now if we take $j\to\infty$ we get $E[Y|\mathcal{F_{\infty}}]=E[\mathbb{1}_A|\mathcal{F_{\infty}}]$. I'm not sure how from that we get $\mathbb{1}_A=Y$ though, if you (or anyone) can elaborate how the measurability proves that, that would be great, thank you.. – 2017-01-16
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    @lfc #3: You cannot simply let $j \to \infty$ (at least I don't see why). To prove $(1)$ note that the measures $$\nu(F) := \int_F Y \, d\mathbb{P}$$ and $$\mu(F) := \int_F 1_A \, d\mathbb{P}$$ coincide on $\bigcup_j \mathcal{F}_j$, and therefore, by a classical uniqueness of measure result, also on $\sigma(\bigcup_j \mathcal{F}_j) = \mathcal{F}_{\infty}$. #4: $(1)$ implies $$\int_F Y \, d\mathbb{P} = \int_F 1_A \, d\mathbb{P}$$ for any $F \in \mathcal{F}_{\infty}$. Consider the events $\{Y>1_A\}$ and $\{Y<1_A\}$ (which are both in $\mathcal{F}_{\infty}$). – 2017-01-17
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    I see, I only know understand the meaning of $\mathcal{F_{\infty}}$, so in this case it's also wrong to say that $Y=E[\mathbb{1}_A|\mathcal{F_{\infty}}]$ right? I'm a bit confused then about what $X_n\to Y $ a.s means..Can you please explain? Thanks for your help so far – 2017-01-17
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    @lfc "a.s." is short for "almost surely"... so $X_n \to Y$ a.s. means that $\mathbb{P}(\lim_n X_n = Y)=1$. – 2017-01-17
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    So the way I proved 2 is also not ok? Is it ok for 2 to just say that $E[Y|\mathcal{F_j}]=Lim_n E[E[\mathbb{1}_A|\mathcal{F_n}]|\mathcal{F_j}]=E[\mathbb{1}_A|\mathcal{F_j}]=X_j$ (using the tower property in the second equality)? – 2017-01-17
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    @lfc Yes, that's correct. – 2017-01-17
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    Great, thanks a lot.. Now I only am not sure about the last part "Consider the events $\{Y>1_A\}$ and $\{Y<1_A\}$". Do you mean that because$ \forall F \in \mathcal{F}_{\infty} \int_F Y \, d\mathbb{P} = \int_F 1_A \, d\mathbb{P} $ then these events must only happen a finite amount of times? – 2017-01-17
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    @lfc By #3, we have $$\int_{\{Y>1_A\}} (Y-1_A) \, d\mathbb{P}=0$$ and also $$\int_{\{1_A>Y\}} (1_A-Y) \, d\mathbb{P}=0$$ and this implies $$\mathbb{P}(Y>1_A) = \mathbb{P}(1_A>Y)=0.$$ – 2017-01-17
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    I'm sorry I'm a bit slow with this, this is kind of new to me, I don't understand how do we get this from #3? why doesn't $\int_{\{1_A=Y\}} (1_A-Y) \, d\mathbb{P}=0$? – 2017-01-17
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    @lfc Well, that's also true, but doesn't help us. What I want you to use is the following statement: Let $Z$ be a random variable which is measurable with respect to a $\sigma$-algebra $\mathcal{F}$ and $$\forall F \in \mathcal{F}: \int_F Z \, d\mathbb{P}=0.$$ Then $Z=0$ almost surely. – 2017-01-18
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    Oh so in that case after we get $\forall F \int_F Y \, d\mathbb{P} = \int_F 1_A \, d\mathbb{P} \Rightarrow \forall F \int_F Y-1_A \, d\mathbb{P} = 0\ \Rightarrow 1_A=Y \ a.s$ and we're done, aren't we? I don't see the need to consider the events $\{Y>1_A\}, \{Y<1_A\}$ – 2017-01-18
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    @lfc Well, if you know this statement, then you are done, yes; if you don't know the statement, then you can prove it by considering the sets $\{Z>0\}$ and $\{Z<0\}$ (... that's why I mentioned $\{Y>1_A\}$ and $\{Y<1_A\}$). – 2017-01-18