Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space and $M=(M_t, t\ge 0)$ a continuous martingale with respect to a filtration $(\mathcal F_t, t\ge 0)$. Let $T>0, \lambda >0$, and $p \ge 1$.
I'm reading a proof of below theorem from this link.
Theorem If $\mathbb E [ |M_T|^p] < \infty$ then $$ \mathbb P \bigg [ \sup_{t \in [0, T]} |M_t| \ge \lambda \bigg ] \le \frac{\mathbb E [ |M_T|^p ]}{ \lambda^p}. $$
Proof For $p \ge 1$, the map $x \mapsto |x|^p$ is convex. It follows from $\mathbb E [ |M_T|^p] < \infty$ and Jensen's inequality that $(|M_t|^p, t \in [0, T])$ is a sub-martingale w.r.t. $(\mathcal F_t, t\ge 0)$. Let $$ \tau := T \wedge \inf \{t \ge 0 : |M_t| \ge \lambda\}. $$
Then $\tau$ is a stopping time w.r.t. $(\mathcal F_t, t\ge 0)$ such that $\tau \le T$. By Doob’s optional stopping theorem for a sub-martingale, $$ \mathbb E [|M_{\tau}|^p] \le \mathbb E [|M_{T}|^p]. $$
By definition of $\tau$, $$ \begin{align} |M_{\tau}|^p &= |M_{\tau}|^p 1_{\{\sup_{t \in [0, T]} |M_t| \ge \lambda\}} + |M_{\tau}|^p 1_{\{\sup_{t \in [0, T]} |M_t| < \lambda\}} \\ &\ge \lambda^p 1_{\{\sup_{t \in [0, T]} |M_t| \ge \lambda\}} + |M_T|^p 1_{\{\sup_{t \in [0, T]} |M_t| < \lambda\}}. \end{align} $$
Thus $$ \mathbb P \bigg [ \sup_{t \in [0, T]} |M_t| \ge \lambda \bigg ] \le \frac{\mathbb E [|M_T|^p 1_{\{\sup_{t \in [0, T]} |M_t| \ge \lambda\}}]}{\lambda^p} \le \frac{\mathbb E [|M_T|^p]}{\lambda^p}. $$
My understanding Let $(X_t, t \in [0, T])$ be a stochastic process. Then $\sup_{t \in [0, T]} |X_t|$ is not necessarily Borel measurable because $[0, T]$ is uncountable. Let $(t_n, n \in \mathbb N)$ be a dense subset of $[0, T]$. Because $M$ has continuous paths, $$ \sup_{t \in [0, T]} |M_t| = \sup_n |M_{t_n}|. $$
The supremum over a countable collection of Borel measurable functions is again Borel measurable. So $\sup_{t \in [0, T]} |M_t|$ is Borel measurable.
Could you confirm if my above understanding is correct?
