If $S$ and $T$ are stopping times, is it true that $(S\leq T)\in \mathcal{F}_T$?

Question

Suppose $S$ and $T$ are stopping times with respect to the continuous filtration $(\mathcal{F}_t)_{t\geq 0}$. Is it true that $(S\leq T)\in \mathcal{F}_T$?

If the filtration is assumed to be discrete instead of continuous, it is easy to show that the answer is indeed yes, by considering, for all $k\in\mathbb{N}$, $(S\leq T)\cap(T=k)=(S\leq k)\cap (T=k)\in \mathcal{F}_k$. In the continuous case however this reasoning doesn't seem to work since it would lead to thinking $(S\leq T)\cap(T\leq t)$ as the uncountable union $\cup_{s\leq t} \bigg( (S\leq s)\cap(T=s)\bigg)$, which is not necessarily in $\mathcal{F}_t$.

Answer 1

The usual trick is to use $\mathbb Q$'s density in $\mathbb R$: $$ \{S>T\}\cap\{T\le t\}=\bigcup_{q\in(\mathbb Q\cap[0,t])\cup\{t\}}\left(\{T\le q\}\cap\{q<S\}\right).$$

For any $q\in(\mathbb Q\cap[0,t])\cup\{t\}$, $\{T\le q\}\in\mathcal F_q\subset\mathcal F_t$, and $\{q<S\}\in\mathcal F_t$. Hence $\{S>T\}\cap\{T\le t\}\in\mathcal F_t$ and $\{S>T\}\in\mathcal F_T$. By stability under complement, $\{S\le T\}\in\mathcal F_T$.

Answer 2

In discrete time, if an event $A$ is an element of $\mathcal F_T$ then there is an adapted sequence $Z=(Z_n)_{0\le n\le\infty}$ such that $1_A = Z_T$. (Conversely, given an adapted sequence $Z$, $Z_T$ is an $\mathcal F_T$ measurable random variable.)

In continuous time, something similar is true: If $(Z_t)_{0\le t\le\infty}$ is progressively measurable and $T$ is a stopping time, then $Z_T$ is $\mathcal F_T$ measurable.

If $S$ is a second stopping time, the above observation can be applied by noticing that $Z_t:=1_{\{S\le t\}}$ is adapted and right continuous, hence progressively measurable. And $Z_T=1_{\{S\le T\}}$ in this case, implying that $\{S\le T\}\in\mathcal F_T$.