If $X_n$ is an $L_1$-bounded martingale, show that $\sum_n(X_n-X_{n-1})^2< \infty$ a.s.

Question

I had asked this question previously in the following post but there were no replies.

Recently, I found a two page article(free download) with a possible alternate line of proof to the one suggested in the first link

Note: $X_n$ is an $L_1$-bounded martingale means that $\sup_nE[|X_n|]<\infty$

However in the link, the author shows that $\sum_n(X_n-X_{n-1})^2< \infty$ a.s. on a set $E$ which is defined as follows:$$E_n:=\bigcap_{k\leq n}\{\omega||X_n(\omega)|\leq M\}\text{ and }E=\bigcap_n E_n.$$ He then states that for large $M$ the measure of $E$ is near 1, since by the martingale convergence theorem, $\sup_n|X_n|<\infty$ which completes the proof.

However, I have two issues.

1. Why can the measure of $E$ be made close to 1?
2. Shouldn't we require a set $E$ with measure exactly 1 to show a.s convergence. If so how do we find a set E of measure 1?

Can someone please explain how to rectify these issues?

Answer 1

1. In Austin's note, the following was proved: \begin{align*} &\{\omega: \sup_n|X_n(\omega)|<M\}\\ &\subset \Big\{\omega: \sum_n(X_n(\omega)-X_{n-1}(\omega))^2<\infty\Big\}, \quad \forall M>0 \tag{1} \end{align*} (or exactly \begin{equation*} \mathsf{P}\Big(\{\sup_n|X_n|<M\}\setminus \Big\{ \sum_n(X_n-X_{n-1})^2<\infty\Big\}\Big)=0,\quad \forall M>0.) \end{equation*} Using (1), \begin{align*} \{\omega: \sup_n|X_n(\omega)|<\infty\}&= \bigcup_{M=1}^\infty\{\omega: \sup_n|X_n(\omega)|<M\}\\ &\subset \Big\{\omega: \sum_n(X_n(\omega)-X_{n-1}(\omega))^2<\infty\Big\}, \end{align*} This is the reason of last sentence in Austin's note(the measure of $ E $ is near 1(if the $ M $ is large enough)).

2. In following two texts, the conclusion you are interesting in is proved also:

Y. S. Chow & H. Teicher, Probability Theory, 3Ed, Springer Verlag, 1997, Theorem 11.1.2, p.408.

P. Malliavin, Integration and Probability, Springer Verlag, 1995. Prop. IV.5.6.2, p.214.

Answer 2

$(E_n)_n$ is a nonincreasing sequence (for the inclusion order) of events, hence $\mathbb P(E_n)\underset{n\to+\infty}{\to}\mathbb P(E)$. Let $\varepsilon>0$. Then there exists $N$ such that $\mathbb P(E)\ge\mathbb P(E_N)-\varepsilon$. Since $\mathbb P(\sup_n\vert X_n\vert<+\infty)=1$, it is clear that $\mathbb P(E_n)\underset{M\to+\infty}{\to}1$. You can see the latter as a direct consequence of the dominated convergence theorem, for example. Indeed, $$ \mathbb P(E_n)=\mathbb E[\prod_{k\le n}1_{\vert X_n\vert\le M}]\underset{M\to+\infty}{\longrightarrow}1. $$

So for $M$ large enough, $\mathbb P(E_N)\ge1-\varepsilon$, hence $\mathbb P(E)\ge1-2\varepsilon$. It is not written, but the latter event $E$ depends on $M$. Let us then write $E^M$. We just showed that $$ \forall\varepsilon>0,\quad\exists M>0,\quad\mathbb P(E^M)\ge1-2\varepsilon. $$

Take any sequence $(\varepsilon_n)_n$ converging to $0$. Then there exists $(M_n)_n$ such that for all $n$, $\mathbb P(E^{M_n})\ge1-2\varepsilon_n$. Let $E=\bigcup_{n}E^{M_n}$. Then $\mathbb P(E)=1$ (since $\mathbb P(E)\ge\mathbb P(E^{M_n})\ge1-2\varepsilon_n$ for all $n$) and $\sum_n(X_n-X_{n-1})^2<+\infty$ on $E$ (since it is finite on all $E^{M_n}$ for all $n$).