# If $(X_n)_{n\in \mathbb{N}}$ is a martingale s.t. $\sup_n E[|X_n|]\leq M < \infty$, then $\sum_{n\geq 2}(X_n-X_{n-1})^2<\infty$ almost surely.

I need to prove the above statement. The hint provided was to consider the stopping time $$T_l=\inf\{n\in \mathbb{N}||X_n(w)|\geq l\}$$ where $$l\in \mathbb{N}$$ and then show that $$E[\sum_{n=2}^K(X_n-X_{n-1})^21_{T_l>K}]$$ is uniformly bounded in $$K$$.

I have 2 issues here:

1. $$E[\sum_{n=2}^K(X_n-X_{n-1})^21_{T_l>K}]$$ is uniformly bounded: I know that if $$T_l>K$$ this implies that $$|X_n-X_{n-1}|^21_{T_l>K}\leq (2l)^2$$.

But this just shows that $$E[\sum_{n=2}^K(X_n-X_{n-1})^21_{T_l>K}]\leq K(2l)^2$$ which is not uniform in K. I feel we have to use the fact that $$(X_n)$$ converges to a finite limit since it is $$L_1$$-bounded.

1. How does the hint then prove the original statement? I am stuck here. I think we have have to use Lebesgue Dominated Convergence Theorem but I don't see how.

Any ideas? I have been stuck for some time now. I would also appreciate another alternative method if anyone has one.

1. First suppose all $$X_n$$ are square-integrable. We have \begin{align} &E\left(\sum_{n=2}^{K}\left(X_{n}-X_{n-1}\right)^{2} 1_{T_l>K}\right)\notag\\ \leqslant& E\left(\sum_{n=2}^{K}\left(X_{n}-X_{n-1}\right)^{2} 1_{T_l>n}\right)\notag\\ =& E\left(\sum_{n=2}^{K}\left(X_{n}-X_{n-1}\right)^{2} 1_{T_l>n-1}\right)-E\left(\sum_{n=2}^{K}\left(X_{n}-X_{n-1}\right)^{2} 1_{T_l=n}\right) \end{align} (Note that the term after $$\leqslant$$ may not exist if we do not assume $$X_n$$ is square-integrable.) The first term is \begin{align*} & E\left(\sum_{n=2}^{K}\left(X_{n}-X_{n-1}\right)^{2} 1_{T_l>n-1}\right)\\ =&E\left(E\left(\sum_{n=2}^{K}\left(X_{n}-X_{n-1}\right)^{2}|\mathcal{F}_{n-1}\right) 1_{T_l>n-1}\right)\\ =&\sum_{n=2}^{K}E\left[\left(E\left(X_n^2|\mathcal{F}_{n-1}\right)+X_{n-1}^2-2E\left(X_nX_{n-1}|\mathcal{F}_{n-1}\right)\right)1_{T_l>n-1}\right]\\ =&\sum_{n=2}^{K}E\left((X_n^2-X_{n-1}^2) 1_{T_l>n-1}\right)\\ =&\sum_{n=2}^{K}E\left(X_n^21_{T_l=n}+X_n^21_{T_l>n}-X_{n-1}^21_{T_l>n-1}\right)\\ =&E\left(X_n^21_{T_l>n}\right)-E\left(X_1^21_{T_l>1}\right)+\sum_{n=2}^{K}E\left(X_n^21_{T_l=n}\right) \end{align*} The second term is \begin{align*} &E\left(\sum_{n=2}^{K}\left(X_{n}-X_{n-1}\right)^{2} 1_{T_l=n}\right)\\ =&\sum_{n=2}^{K}E\left(X_n^21_{T_l=n}\right)+\sum_{n=2}^{K}\left(E(X_{n-1}^21_{T_l=n})-2E(X_{n-1}X_n1_{T_l=n})\right) \end{align*} So (1) becomes\begin{align*} E\left(X_n^21_{T_l>n}\right)-E\left(X_1^21_{T_l>1}\right)-\sum_{n=2}^{K}E(X_{n-1}^21_{T_l=n})+2\sum_{n=2}^{K}E(X_{n-1}X_n1_{T_l=n}) \end{align*} By definition of $$T_l$$, the first two terms are both bounded by $$l^2$$. The second term is bounded by $$l^2\sum_{n=2}^{K}P(1_{T_l=n})\leqslant l^2$$ For the last term \begin{align*} &\left|2\sum_{n=2}^{K}E(X_{n-1}X_n1_{T_l=n})\right|\\ \leqslant&2\sum_{n=2}^{K}E(|X_{n-1}X_n|1_{T_l=n})\\ \leqslant&2l\sum_{n=2}^{K}E(|X_n|1_{T_l=n})\\ \leqslant&2lM\sum_{n=2}^{K}P(T_l=n)\\ \leqslant&2lM \end{align*} Therefore, (1) is bounded by $$3l^2+2lM$$, which is uniform in $$K$$. For general $$\{X_i\}$$ (not necessarily square integrable), take a number $$H>l$$. Then $$\{X_{n\wedge T_H}\}$$ is a square integrable martingale, so $$E\left(\sum_{n=2}^{K}\left(X_{n\wedge T_H}-X_{(n-1)\wedge T_H}\right)^{2} 1_{T_l>K}\right)$$ is bounded by $$3l^2+2lM$$. Notice that $$\sum_{n=2}^{K}\left(X_{n}-X_{n-1}\right)^{2} 1_{T_l>K}=\sum_{n=2}^{K}\left(X_{n\wedge T_H}-X_{(n-1)\wedge T_H}\right)^{2} 1_{T_l>K}$$ since $$H>l$$. This concludes the proof of the statement in the hint.
2. Denote $$A_l=\bigcap_{n=2}^{\infty}\{\omega:T_l(\omega)>n\}=\{\omega:|X_n(\omega)| Let $$K$$ goes to infinity, we have $$E\left(\sum_{n\geqslant 2}(X_n-X_{n-1})^21_{A_l}\right)<+\infty$$ due to the statement provided by the hint. So $$\sum_{n\geqslant 2}(X_n-X_{n-1})^21_{A_l}<+\infty \qquad a.e.$$ for every $$l>0$$. We then have $$\sum_{n\geqslant 2}(X_n-X_{n-1})^21_{A_{\infty}}<+\infty \qquad a.e.$$ where $$A_{\infty}=\bigcup_{l=1}^{\infty} A_l=\{\omega:\sup_n |X_n(\omega)|<\infty\}$$ Note that $$X_n$$ converges a.e. since it is $$L_1$$-bounded. So $$P(A_{\infty})=1$$.