Let's assume that we have a discrete martingale $(M_n)_{n\in\mathbb{N}}$ with respect to a filtration $(\mathcal{F}_n)_{n\in\mathbb{N}}$. It takes only real values. We define the discrete bracket $\langle M\rangle_n=\sum\limits_{i=1}^{n-1} \mathbb{E}(M_{k+1}^2-M_k^2|\mathcal{F}_k)$.
Of course, if $\langle M\rangle_n$ is bounded in $L^2$ then $M_n$ converges in $L^2$. However if we only assume that $\langle M\rangle_n$ converges almost surely toward a finite random variable, is it true that $M_n$ converges almost surely?
If it is true in the discrete case, is it true in the continuous case?
Yes, this is true in both the discrete and continuous setting, assuming $M$ is continuous. Let $\tau_m := \inf \{t : \langle M \rangle_t > m\}$ and $\lim_{t \rightarrow \infty} \langle M \rangle_t = \langle M \rangle_\infty$. Then $\langle M \rangle^{\tau_m}$ is bounded and in particular in $L^2$ so $\lim_{t \rightarrow \infty}M_t^{\tau_m}$ exists almost surely. But since for almost every $\omega$ we have $\langle M \rangle_\infty(\omega) < \infty$, we know $\tau_m(\omega) =\infty$ for all $m$ sufficiently large, so $\lim_{t \rightarrow \infty} M_t$ exists as well.