I quote Jacod-Protter
THEOREM Let $M=(M_n)_{n\geq0}$ be a martingale or a positive submartingale and $M_n^*=\sup_{j\leq n}|M_j|$. Then \begin{equation} \mathbb{P}(M_n^* \geq \alpha) \leq \frac{\mathbb{E}(|M_n|)}{\alpha} \end{equation}
During the proof of the above theorem, one defines $T=\min\{j: |M_j| \geq \alpha\}$ (with the convention according to which the minimum of an empty subset of natural numbers $\mathbb{N}$ is $+ \infty$).
It is stated that "the set $\{T\leq n, |M_T|\geq \alpha\}$ is equal to the set $\{M_n^*\geq \alpha\}$".
However, I have some counterexamples in mind. To put it in a very simple fashion, if for example one has $n=2$ and $|M_0|\leq|M_1|\leq |M_2|$ such that $|M_0|\geq \alpha$ (and clearly $|M_1|\geq \alpha$ and $|M_2|\geq \alpha$ will hold true as well), one will have that, by definition of $T$ and of $M_n^*$: \begin{equation} \{T\leq n, |M_T|\geq \alpha\}=\{|M_0|\geq \alpha\} \end{equation} and \begin{equation} \{M_n^*\geq \alpha\}=\{|M_2|\geq \alpha\} \end{equation}
I know for sure that there is some mistake in my above counterexample, but I cannot get it. What is wrong with it? Why can I state that "the set $\{T\leq n, |M_T|\geq \alpha\}$ is equal to the set $\{M_n^*\geq \alpha\}$"?
In your example, if $|M_0| \ge \alpha$, then no matter what happens after that, we'll take $T=0$, we'll have $|M_T| = |M_0| \ge \alpha$, and we'll have $$M_n^* = \max\{|M_0|, |M_1|, |M_2|\} \ge |M_0| \ge \alpha.$$ Therefore $\{T\leq n, |M_T|\geq \alpha\}=\{|M_0|\geq \alpha\}$ because both events are guaranteed to happen.
In general, if $M_n^* \ge \alpha$, then for $\sup_{j \le n} |M_j| \ge \alpha$ we must have $|M_j| \ge \alpha$ for some $j \le n$. Then $T$, the first $j$ such that $|M_j| \ge \alpha$, will also satisfy $T \le n$. Whenever we have $T < \infty$, we always have $|M_T| \ge \alpha$.
Going the other way, if $T \le n$ and $|M_T| \ge \alpha$, then $$ M_n^* = \sup_{j \le n} |M_j| \ge |M_T| \ge \alpha $$ as well.