Why $\{\forall t\in [0,T], X_t\leq \alpha \}$ measurable if $(X_t)$ is càdlàg?

Question

Let $(\Omega ,\mathcal F,\mathbb P)$ a probability space and $(X_t)_{t\geq 0}$ a càdlàg stochastic process (this means continuous at right and limit at left). Why $$\mathbb P(\forall t\in [0,T]:X_t\leq \alpha )$$ makes sense. In other words, why $\{\forall t\in [0,T]:X_t\leq \alpha \}\in \mathcal F$ ? I know that $\mathcal F$ contain cylinders (and thus countable intersection of cylinder), but here we have a non-countable intersection of cylinders. So measurability doesn't look clear.

Answer 1

$X_t \leq \alpha$ for all $t \in [0,T]$ iff $X_t \leq \alpha$ for all rational numbers $t$ in $[0,T)$ and $X_T \leq \alpha$. Hence $\{X_t \leq \alpha \forall t \in [0,T]\}$ is measurable.

Answer 2

Hint

Using right-continuity one can prove that $$\big\{\forall t\in [0,T], X_t\leq \alpha \big\}=\left\{\sup_{t\in [0,T]}X_t\leq \alpha \right\}=\{X_T\leq \alpha \}\cap\left\{\sup_{t\in [0,T]\cap \mathbb Q}X_t\leq \alpha \right\}=\{X_T\leq \alpha \}\cap\bigcap_{\substack{0\leq t\leq T\\ t\in \mathbb Q}}\left\{X_t\leq \alpha \right\}\in \mathcal F.$$