Is the supremum of a right continuous process right continuous? Defining a stopping time as the first hitting time of a supremum process

probability theoryreal-analysisstochastic-calculusstochastic-processesstopping-times

This is a question that came to my mind while reading Lemma 13 of the notes here : https://almostsuremath.com/2009/12/23/localization/

In this lemma, the writer assumes $X$ is a right-continuous adapted process such that $X_t^*$ is almost surely finite for all $t$ and the filtration is only a completed filtered space with no assumptions on right-continuity. $X_t^*$ is defined as $\sup_{s\in [0,t]} |X_s|$. The proof itself is simple but states without justification that $$\tau_n = \inf \{t\ge 0: X_t^* \ge n\}$$ is a sequence of stopping times.

I believe this follows from the Debut theorem given here in Theorem 2 :https://almostsuremath.com/2009/11/15/stopping-times-and-the-debut-theorem/

But then we would need that $X_t^*$ is a right continuous adapted process. However, I cannot prove this fact. Is this true or is there another justification for this being a stopping time here? I would greatly appreciate any help.

Best Answer

Since $X_t^*$ is nondecreasing, $\{\tau_n\le t\}=\{X_t^*\ge n\}\in \mathcal{F}_t$ because $\{X_t^*\}$ is adapted.