# How can we show that $\mathcal F_\tau\cap\left\{\tau=t\right\}=\mathcal F_t$? for any $\mathcal F$-stopping time $\tau$?

Let

• $(\Omega,\mathcal A)$ be a measurable space
• $I\subseteq\mathbb R$
• $(\mathcal F_t)_{t\in I}$ be a filtration of $\mathcal A$

If $\tau$ is an $\mathcal F$-stopping time, then $$\mathcal F_\tau:=\left\{A\in\mathcal A:A\cap\left\{\tau\le t\right\}\in\mathcal F_t\text{ for all }t\in I\right\}\;.$$ If $A\subseteq\Omega$ and $\mathcal G$ is a $\sigma$-algebra on $\Omega$, then $$\left.\mathcal G\right|_A:=\left\{A\cap B:B\in\mathcal G\right\}\;.$$

It's easy to show that $$\left.\mathcal F_\sigma\right|_{\left\{\:\sigma\:\le\:\tau\:\right\}}\subseteq\mathcal F_{\sigma\:\wedge\:\tau}=\mathcal F_\sigma\cap\mathcal F_\tau\tag1$$ for any $\mathcal F$-stopping times $\sigma,\tau$. Now, let $\tau$ be an $\mathcal F$-stopping time and $t\in I$. I want to show that $$\left.\mathcal F_\tau\right|_{\left\{\:\tau\:=\:t\:\right\}}=\mathcal F_t\tag2\;.$$

If $\tau\equiv t$, then it's easy to observe that $$\mathcal F_\tau=\mathcal F_t\;.\tag3$$

How can we conclude $(2)$ from $(1)$ and $(3)$?

This is not true because $\{\tau\ne t\}\notin\mathcal{F}_{\tau}\mid_{\{\tau=t\}}$ but $\{\tau\ne t\}\in \mathcal{F}_t$. You may show instead that $$\mathcal{F}_{\tau}\mid_{\{\tau=t\}}=\mathcal{F}_t\mid_{\{\tau=t\}} .^1$$
$^1$ For any $(\mathcal{F}_t)$-stopping times $\tau$ and $\sigma$, $\mathcal{F}_{\sigma}\mid_{\{\sigma=\tau\}}=\mathcal{F}_{\tau}\mid_{\{\tau=\sigma\}}$ (if $A\in\mathcal{F}_{\sigma}\cap \{\sigma=\tau\}$, then for each $t\in I$, $A\cap\{\sigma\le t\}=A\cap\{\tau\le t\}\in \mathcal{F}_t$. Thus $A\in \mathcal{F}_{\tau}$). Now take $\sigma\equiv t$.