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

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 $$\mathcal{F}_t$$.

The usual trick is to use $$\mathbb Q$$'s density in $$\mathbb R$$: $$\{S>T\}\cap\{T\le t\}=\bigcup_{q\in(\mathbb Q\cap[0,t])\cup\{t\}}\left(\{T\le q\}\cap\{q
For any $$q\in(\mathbb Q\cap[0,t])\cup\{t\}$$, $$\{T\le q\}\in\mathcal F_q\subset\mathcal F_t$$, and $$\{q. Hence $$\{S>T\}\cap\{T\le t\}\in\mathcal F_t$$ and $$\{S>T\}\in\mathcal F_T$$. By stability under complement, $$\{S\le T\}\in\mathcal F_T$$.
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 $$\{S\le T\}\in\mathcal F_T$$.