Cumulative distribution of a martingale Let $(M_n)_{n\in\mathbb{N}}$ be a martingale with respect to a filtration $(\mathcal{F}_n)_{n\in\mathbb{N}}$. We can define the bracket of $M_n$ by
$$\langle M\rangle_n=\sum\limits_{k=0}^{n-1}\mathbb{E}\left[(M_{k+1}-M_k)^2|\mathcal{F}_k \right].$$
Morally, $\langle M \rangle_n$ is of the same order as $M_n^2$. This idea can be made precise thanks to the BDG inequality, which says that, for every $p>1$ (actually, any $p>0$ for continuous martingales.), there exists universal constants $c_p$ and $C_p$ such that for every $n\in\mathbb{N}$,
$$c_p\mathbb{E}\left[\langle M \rangle_n^{p/2}\right]\leq \mathbb{E}\left[(M_n^{*})^p \right]\leq C_p\mathbb{E}\left[\langle M \rangle_n^{p/2}\right]$$
where $M_n^*=\underset{1\leq k\leq n}\max|M_k|$.
Therefore, my question is the following one:
Is this possible to compare $\mathbb{P}\left(M_n^*> \lambda\right)$ and $\mathbb{P}(\langle M\rangle_n>\lambda^2)$?
Or in the case where $M$ and its bracket converge almost surely, $\mathbb{P}\left(|M_{\infty}-M_n|> \lambda\right)$ and $\mathbb{P}\left(\langle M\rangle_{\infty}-\langle M\rangle_n>\lambda^2 \right)$?

Regarding your question, the following Lenglart's inequality is useful(cf.  J. Jacod, and  A. N. Shiryayev, Limit Theory for Stochastic Processes, 2ed. Springer, 2003, Lemma I.3.30, P.35.).
From Lenglart inequality, Since $M^2$ is a càdlàg adapted process which is L-dominated by an predictable increasing process $\langle M\rangle$(i.e. $\mathsf{E}[M^2_T] = \mathsf{E}\langle M \rangle_T$ for every bounded stopping times), then for all stopping times $T$ and all $\epsilon, \eta>0$,
\begin{gather*}
 \mathsf{P}(M^*_T>\epsilon)\le \frac{\eta}{\epsilon^2} + \mathsf{P}(\langle M\rangle_T \ge \eta),\\
 \mathsf{P}\Big(\sup_{m\ge n}|M_m-M_n|>\epsilon\Big)\le \frac{\eta}{\epsilon^2} + \mathsf{P}(\langle M\rangle_\infty-\langle M\rangle_n \ge \eta).
\end{gather*}
