Let
- $(\Omega,\mathcal A)$ be a measurable space;
- $I\subseteq[-\infty,\infty]$;
- $(\mathcal F_t)_{t\in I}$ be a filtration on $(\Omega,\mathcal A)$ and$^1$ $$\mathcal F_{\sup I}:=\bigvee_{t\in I}\mathcal F_t;$$
- $\tau:\Omega\to I\cup\{\sup I\}$ be an $(\mathcal F_t)_{t\in I}$-stopping time and $$\mathcal F_\tau:=\left\{A\in\mathcal A:A\cap\{\tau\le t\}\in\mathcal F_t\text{ for all }t\in I\right\}.$$
Question 1: Are we able to show that $$\{\tau<t\}\in\mathcal F_t\tag0$$ for all $t\in I\cup\{\sup I\}$?
My argument would be $$\{\tau<t\}=\bigcup_{\substack{s\in I\cap\mathbb Q\\s<t}}\{\tau\le s\}$$ (which also works for $t=\sup I$), but is this argument really correct? What if $I=\mathbb R\setminus\mathbb Q$, for example? On the other hand, couldn't we similarly argue that since $[-\infty,\infty]$ is separable, $I$ is separable as well and hence contains a dense countable subset $D$?
If $\mathcal G$ is a $\sigma$-algebra on $\Omega$ and $A\subseteq\Omega$, then $$\left.\mathcal G\right|_A:=\{A\cap B:B\in\mathcal G\}.$$
Question 2: Are we able to show that $$\left.\mathcal F_\tau\right|_{\{\tau\:=\:t\}}=\left.\mathcal F_t\right|_{\{\tau\:=\:t\}}\tag1$$ for all $t\in I\cup\{\sup I\}$?
$^1$ Note that the definition of $\mathcal F_{\sup I}$ is consistent if already $\sup I\in I$.