# Let $\xi_{n}=X_{n}-X_{n-1}$ where $EX_{0}^{2}, \sum\limits_{m}E\xi_{m}^{2} < \infty$ then $X_{n} \to X_{\infty}-$a.s. and $L^{2}$

Let $$X_{n}, n \in \mathbb N$$, be a martingale and $$\xi_{n}=X_{n}-X_{n-1}$$ where $$EX_{0}^{2}, \sum\limits_{m}E\xi_{m}^{2} < \infty$$.

Show that $$X_{n} \to X_{\infty}-$$a.s. and in $$L^{2}$$

It is clear we need to use the Orthogonality of Martingale Increments to fulfill the $$L^{p}$$ Convergence Theorem.

Let's define $$\tau_{K}:=\inf\{n: X_{n}\geq K\operatorname{or} X_{n}\leq -K\}$$ where $$K >0$$, we know that the stopped process $$(X_{n \land \tau_{K}})_{n}$$ is a martingale.

This means that $$\vert X_{n \land \tau_{K}}\vert\leq K+\sup\limits_{m}\vert \xi_{m}\vert$$

It is clear from $$\sum\limits_{m}E\xi_{m}^{2}<\infty$$ that $$\sup\limits_{m} E\xi_{m}^{2}<\infty$$. Can I imply from $$\sum\limits_{m}E\xi_{m}^{2}<\infty\Rightarrow \sup\limits_{m}\vert \xi_{m}\vert <\infty$$? This would be needed to show that $$\sup\limits_{n} E[\vert X_{n\land \tau_{K}}\vert ]<\infty$$

From there I can say that for any $$K>0$$: $$\sup\limits _{n}E[\vert X_{n \land \tau_{K}}\vert]\leq K+E[\sup\limits_{m}\vert \xi_{m}\vert]< \infty$$

We thus let $$K\to \infty$$ but I am not sure if this helps since then our upper bound is $$\infty$$

Since $$X_n-X_m = \sum_{j=m+1}^n (X_j-X_{j-1})$$ it follows from the orthogonality of the increments that $$\mathbb{E}((X_n-X_m)^2) = \sum_{j=m+1}^n \mathbb{E}((X_j-X_{j-1})^2).$$ By assumption, the series on the right-hand side is a Cauchy sequence, and so is the left-hand side. By the completeness of $$L^2$$, this implies that the limit $$X_{\infty} = \lim_{n \to \infty} X_n$$ exists in $$L^2$$. Conclude that $$(X_n)_{n \in \mathbb{N}}$$ is bounded in $$L^1$$ and apply the martingale convergence theorem to obtain that $$X_n \to X_{\infty}$$ almost surely.