Q: Why $\sup_{0\leq s\leq t}X_s$ is measurable if $(X_s)$ is continuous? I'm working for my exam, and in my lecture, they wrote that since $(X_s)$ is continuous, then $\sup_{0\leq s\leq t}X_s$ is measurable, and thus $\mathbb P\left(\sup_{0\leq s\leq t}X_s\leq \alpha \right)$ is well defined for alll $\alpha \in\mathbb R$.
I'm not so sure why the fact that $(X_s)$ is continuous implies that $\sup_{0\leq s\leq t}X_s$.
Attempts :
$$\left\{\sup_{0\leq s\leq t}X_s\leq \alpha \right\}=\{\exists s\in [0,t]X_s\leq \alpha \},$$
how can I continue ?

A: Hint
Using density of $\mathbb Q$ in $\mathbb R$ one can prove that
$$\sup_{0\leq s\leq t}X_s=\sup_{s\in [0,t]\cap \mathbb Q}X_s.$$
Now,
$$\left\{\sup_{s\in [0,t]\cap \mathbb Q}X_s\leq \alpha \right\}=\bigcap_{s\in [0,t]\cap \mathbb Q}\{X_s\leq \alpha \},$$
which is measurable since it's a countable intersection of measurable set.
