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 $\infty$