# Inequality between stopping times in continuous time

Let $$(\mathcal{F}_t)_{t\geq0}$$ be a filtration and let $$\sigma$$ and $$\tau$$ be $$(\mathcal{F}_t)_{t\geq0}$$-stopping times. I would like to show that the events $$\{\tau\leq\sigma\}$$, $$\{\tau<\sigma\}$$ and $$\{\tau=\sigma\}$$ are elements of $$\mathcal{F}_{\sigma}\cap\mathcal{F}_\tau=\mathcal{F}_{\sigma\land\tau}$$. I can do this is discrete time, but I get stuck in the case of continuous time. So far I have reasoned that since $$\mathcal{F}_{\sigma\land\tau}=\{A\in\mathcal{F}\mid A\cap\{\sigma\land\tau\leq t\}\in\mathcal{F}_t\;\forall t\geq 0\},$$ I must e.g. show that $$\{\tau<\sigma\}\cap\{\sigma\land\tau\leq t\}\in\mathcal{F}_t$$ for all $$t\geq0$$. I have $$\{\tau<\sigma\}\cap\{\sigma\land\tau\leq t\}=\{\tau<\sigma\}\cap\{\tau\leq t\},$$ but I am not seeing how to proceed further. I would very much appreciate a hint from someone. Thank you in advance.

I believe that I have solved it! I will give my solution here for anyone with the same question in the future. Continuing from before, we have for an arbitrary $$t\geq0$$: $$\{\tau<\sigma\}\cap\{\tau\leq t\}=\bigcup_{q\in\mathbb{Q}\cap[0,1]} \{qt<\sigma\}\cap\{\tau\leq qt\}.$$ To show this, notice that the inclusion ''$$\supseteq$$'' is easy: if there exists some $$q\in\mathbb{Q}\cap[0,1]$$ satisfying $$\tau\leq qt<\sigma$$, then $$\tau<\sigma$$ and $$\tau\leq qt\leq t$$. On the other hand, if $$\tau<\sigma$$ and $$\tau\leq t$$, then we choose a $$q\in\mathbb{Q}\cap[0,1]$$ such that $$qt\in[\tau,\sigma)$$. This may be done by choosing a rational $$q$$ number close enough to $$(\tau+\sigma)/2t$$. This shows ''$$\subseteq$$''. Now, notice that since $$\tau$$ and $$\sigma$$ are stopping times, we have $$\{\tau\leq qt\}\in\mathcal{F}_t$$ and $$\{qt<\sigma\}=\{\sigma\leq qt\}^\mathsf{c}\in\mathcal{F}_t$$ for all $$q\in\mathbb{Q}\cap[0,1]$$, and thus, $$\bigcup_{q\in\mathbb{Q}\cap[0,1]} \{qt<\sigma\}\cap\{\tau\leq qt\}\in\mathcal{F}_t.$$ This shows the assertion $$\{\tau<\sigma\}\in\mathcal{F}_{\sigma\land\tau}$$. It follows that $$\{\tau\geq\sigma\}=\{\tau<\sigma\}^\mathsf{c}\in\mathcal{F}_{\sigma\land\tau}$$, and by renaming: $$\{\tau\leq\sigma\}\in\mathcal{F}_{\sigma\land\tau}$$. Finally, $$\{\tau=\sigma\}=\{\tau\leq\sigma\}\setminus\{\tau<\sigma\}\in\mathcal{F}_{\sigma\land\tau}$$.