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.
- Why can the measure of $E$ be made close to 1?
- 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?
Best Answer
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.