Let $(\Omega,\mathcal A)$ be a measurable space, $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A)$, $$\mathcal F_\infty:=\sigma\left(\bigcup_{t\ge0}\mathcal F_t\right),$$ $\tau:\Omega\to[0,\infty]$, $$\mathcal G:=\{A\cap\{t<\tau\}:A\in\mathcal F_t\text{ and }t\ge0\}$$ and $$\mathcal F_{\tau-}:=\sigma(\mathcal G).$$
Are we able to show that $$A\cap\{\tau=\infty\}\in\mathcal F_{\tau-}\tag1$$ for all $A\in\mathcal F_\infty$?
If $s\ge0$ and $A\in\mathcal F_s$, then $$A\cap\{\tau=\infty\}=\bigcap_{\substack{t\in\mathbb N\\t\ge s}}\left(A\cap\{t<\tau\}\right)\in\mathcal F_{\tau-}\tag2.$$ Does it immediately follow that $(1)$ must hold? How do we need to argue?
If $n \geq s$ then $A \cap \{n <\tau\} \in \mathcal G \subseteq \mathcal F_{\tau-}$ (since $A \in \mathcal F_n$). Taking intersection over all $ n \geq s$ we get $ A\cap \{\tau=\infty\} \in \mathcal F_{\tau-}$.