# quadratic variation - convergence of martingale

Let $X_0,X_1,\dots\in L^2(\Omega,\mathcal{F}, \mathbb{P})$ be an $(\mathcal{F_n})$-Martingale. $(\langle X \rangle _n)$ quadratic variation of $X$. $\langle X \rangle _\infty :=lim_{n\to\infty}\langle X \rangle _n$ and $T_a:=inf \{n \ge 0:\langle X \rangle _{n+1}>a\}$ for $a>0$ stopping time.

a) Show that the Martingale $(X_{n\wedge T_a})$ converges a.s. and in $L^2$.

b) Show that $(X_n)$ converges a.s. on $\{\langle X \rangle _\infty < \infty \}$

For a) I think i have to show that $sup \, \mathbb{E} [ \vert X_{T_a \wedge n}\vert^2]<\infty$ Than i can use the $L^p$-martingale convergence theorem. But I dont know how to get there.

For b) i have no idea. A.Klenke says " $\mathbb{P}[T_a= \infty] \to 1$ for $a \to \infty$, so $X$ converges a.s" I don't get what is used/meant.

Any help or a hint is much appreciated. Thanks!

Hints for a): Since $(|X_{n \wedge T_a}|^2-\langle X \rangle_{n \wedge T_a})_{n \geq 1}$ is a martingale, we have $$\mathbb{E}(|X_{T_a \wedge n}|^2) = \mathbb{E}(\langle X \rangle_{T_a \wedge n}).$$ Show that $\langle X \rangle_{T_a \wedge n} \leq a$ and conclude that $(X_{n \wedge T_a})_{n \geq 1}$ is an $L^2$-bounded martingale.
$$\mathbb{P}(\{\omega; X_n(\omega) \, \, \text{does not converge as n \to \infty}\} \cap \{\langle X \rangle_{\infty} < \infty\}) \leq \mathbb{P}(\{\omega; \exists a \in \mathbb{N}: X_{n \wedge T_a}(\omega) \, \, \text{does not converge as n \to \infty}). \tag{1}$$ (Hint: If $\langle X \rangle_{\infty}(\omega)<\infty$, then we have $X_n(\omega) = X_{n \wedge T_a}(\omega)$ for $a=a(\omega)$ sufficiently large.) Conclude from part (a) that the probability on the right-hand side equals $0$.