Proof of Levy's zero-one law Let $(\Omega, \mathcal{F},\mathbb P)$ be a probability space and let $X$ be a random variable in $L^1$. Let $(\mathcal{F}_k)_k$ be any filtration, and define $\mathcal{F}_{\infty}$ to be the minimal $σ$-algebra generated by $(\mathcal{F}_k)_k$. Then
$$E[X|\mathcal{F}_k]\rightarrow E[X|\mathcal{F}_{\infty}],\ \ k\rightarrow\infty$$
both $\mathbb P$-almost surely and in $L^1$.
Could someone give me a proof of this proposition or a reference? Many thanks!
 A: Let $Y_n = E(X| \mathcal{F}_n)$. Then it $Y_n$ is a martingale, and
$$\sup_n E(|Y_n|) = \sup_n E(|E(X| \mathcal{F}_n)|) \leq \sup_n E(E(|X||\mathcal{F}_n)) = E(|X|) $$
where the bound in the middle is due the conditional Jensen inequality.
Now the heavy artillery, by Doob's Convergence theorem $Y_\infty := \lim_{n \to \infty} Y_n$ exists almost surely. And since the sequence is dominated by $X$ (again conditional Jensen) we conclude the $L^1$ convergence and thus the convergence in probability.
You can find the Doob's Convergence theorem in Williams' "Probability with Martingales" Thm. 11.5. Is a rather important result based on a "band argument" and it can be extended to continuous time martingales.
A: I decided this part was too long for a comment. We know that with $X_k = E[X \mid \mathcal{F}_k]$, we have
$$
X_k \rightarrow X_{\infty}
$$
in $L^1$ and almost surely for some $L^1$ random variable $X_{\infty}$.
Recall that $\mathcal{F}_{\infty} = \sigma \left(\cup_k \mathcal{F}_k \right)$. To show that $X_{\infty} = E[X \mid \mathcal{F}_{\infty}]$, we'll appeal to the defining property of conditional expectation: we must show that
$$
(*) ~~~~~E[X I_A ] = E[X_{\infty} I_A]
$$
for all $A \in \mathcal{F}_{\infty}$. Convince yourself that $(*)$ holds for all $A \in \mathcal{F}_k$ for any $k$ (Hint: use Dominated Convergence), and thus that $(*)$ holds for all $A \in\cup_k \mathcal{F}_k$, as this is an increasing union.
There's a technical problem here, that $\cup_k \mathcal{F}_k$ is merely an algebra, and not generally a sigma-algebra. Resolving this is a good exercise in applying the Dynkin $\pi-\lambda$ theorem. Try to work this out using the fact that $\cup_k \mathcal{F}_k$ is a $\pi$-system, and show that the set of $A \in \mathcal{F}_{\infty}$ for which $(*)$ holds is a $\lambda$-system.
A: Thanks so much for A Blumenthal and Bunder, I could accomplish the proof.
Firstly we show that $X_{\infty}$ is measurable with respect to $\mathcal{F}_{\infty}$. For every $X_n$ is measurable in $\sigma$-algebra $\mathcal{F}_n$, so in $\mathcal{F}_{\infty}$, then the limit a.s. $X_{\infty}$ is measurable in $\mathcal{F}_{\infty}$.
Then we have to show 
$$(1)\ \ E[1_{A}X]=E[1_{A}X_{\infty}],\ \ \forall A\in\cup_n\mathcal{F}_n$$
implies 
$$(2)\ \ E[1_{A}X]=E[1_{A}X_{\infty}],\ \ \forall A\in\mathcal{F}_{\infty}=\sigma(\cup_n\mathcal{F}_n)$$
Let $\mathcal{F}:=\{A\in\mathcal{F}_{\infty}: (1) \text{ holds}\}$, then we see that $\mathcal{F}$ is a $\lambda$-system:


*

*$\Omega\in\mathcal{F}$;

*if $A\subset B\in\mathcal{F}$, then $B\A\in\mathcal{F}$;

*if $A_n$ is a sequence of sets in $\mathcal{F}$ s.t $A_n\subset A_{n+1}$, then by dominated convergence theorem we have $\cup_nA_n\in\mathcal{F}$
and $\cup_n\mathcal{F}_n$ is a $\pi$-system:


*

*$\cup_n\mathcal{F}_n$ is not empty;

*if $A$, $B\in\mathcal{F}$, then $A\cap B\in\mathcal{F}$
By Dynkin System Theorem we have $\mathcal{F}=\mathcal{F}_{\infty}$, so by definition we prove that $X_{\infty}=E[X|\mathcal{F}_{\infty}]$.
