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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.

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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, $\{\tau=\sigma\}=\{\tau\leq\sigma\}\setminus\{\tau<\sigma\}\in\mathcal{F}_{\sigma\land\tau}$.

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