# Is a stopping time still a stopping time when the time set of the filtration is extended?

Let $$I\subseteq\overline{\mathbb R}$$ and $$(\mathcal F_t)_{t\in I}$$ be a filtration on a measurable space $$(\Omega,\mathcal A)$$.

Remember that $$\tau:\Omega\to I\cup\{\sup I\}$$ is called $$(\mathcal F_t)_{t\in I}$$-stopping time if $$\forall t\in I:\{\tau\le t\}\in\mathcal F_t\tag1.$$

Now it's natural (think about $$I=\mathbb N_0$$ and $$J=[0,\infty)$$, for example) ask whether we can show the following: Let $$J\subseteq\overline{\mathbb R}$$ with $$I\cup\{\sup I\}\subseteq J\cup\{\sup J\}$$ and $$(\mathcal G_t)_{t\in I}$$ be a filtration on $$(\Omega,\mathcal A)$$ with $$\forall t\in I:\mathcal F_t=\mathcal G_t\tag2.$$ Assuming $$\tau$$ is an $$(\mathcal F_t)_{t\in I}$$-stopping time, is it an $$(\mathcal G_t)_{t\in I}$$-stopping time as well?

Let $$t\in J$$. We weed to show $$\{\tau\le t\}\in\mathcal G_t$$.

1. If $$t\in I$$, then $$\{\tau\le t\}\in\mathcal F_t=\mathcal G_t$$.
2. If $$t\ge\sup I$$, then $$\{\tau\le t\}=\Omega\in\mathcal G_t$$.
3. If $$t<\inf I$$, then $$\{\tau\le t\}=\emptyset\in\mathcal G_t$$.

Now, I struggle to show $$\{\tau\le t\}\in\mathcal G_t$$ in the remaining cases. Clearly, it will boil down to argue that there must be a $$s\in I$$ with $$s and $$\{\tau\le t\}=\{\tau\le s\}\in\mathcal F_s=\mathcal G_s\subseteq\mathcal G_t\tag3.$$ But can we really find such an $$s$$? (If we can but only under some additional assumption, please feel free to add this assumption.)

(I'm going to assume that $$\sup I\in I$$ and $$\sup J\in J$$; this entails no real loss of generality.)
Let $$I_0$$ be a countable dense subset of $$I$$. If $$t\in J\setminus I$$ then $$\{\tau\le t\}=\cup_{\{s\in I_0: s