Product of stopping times in discrete time

Question

Let $T$ and $S$ be stopping times wrt a filtration $(\mathcal{F}_n)$ and assume that $S \leq T<\infty$ a.s. The question is whether $TS$ is also a stopping time. In continuous time, I believe it isn't, but, in my course, we have only been considering discrete time, in which case I think I have a proof that it is a stopping time, but I'm not sure. The $\{ST=0\}$ case is straightforward. For any $t\in\mathbb{N}$, $$\{ST=t\}=\bigcup_{d|t \\ d<t}\{S=d\}\cap\{T=\frac{t}{d}\}.$$ Clearly $\{S=d\}\in\mathcal{F}_d\subset\mathcal{F}_t$ and $\{T=\frac{t}{d}\}\in\mathcal{F}_{\frac{t}{d}}\subset\mathcal{F}_t$, so $\{ST=t\}\in\mathcal{F}_t$ hence $ST$ is a stopping time. This seems to contradict the fact that $ST$ is not always a stopping time. Have I made a mistake somewhere or is it just that $ST$ is a stopping time in the discrete time case but not the continuous time case?

Answer 1

Your argument seems fine. What distinguishes the discrete-time setting is that you have $ST\ge T\vee S$ unless $ST=0$. By contrast, in continuous time you can have $ST\le t$ when $S$ is very small (but $>0$) while $T$ is much larger than $t$.

I guess it's true in continuous time that if $S$ and $T$ are stopping times with both $S\ge 1$ and $T\ge 1$, then $ST$ is a stopping time.This because for $t>0$, $$ \{ST\le t\}=\{ST\le t, S\le t, T\le t\}, $$ and $S1_{\{S\le t\}}$ and $T1_{\{T\le t\}}$ are $\mathcal F_t$ measurable.