I'm not quite sure I understand it completely. Help is appreciated.
From what I can tell, these are the definitions: $(\Omega,\mathcal{F},\mathbb{P})$ is the probability space and let $\{ \mathcal{F}_t : t\in\mathbb{R}_+ \}$ be a filtration of $\mathcal{F}$, i.e. $\mathcal{F}_s\subset\mathcal{F}_t \subset \mathcal{F}$ for $s\leq t$ and every $\mathcal{F}_t$ is a sub-$\sigma$-Algebra. A stopping time $\tau$ is a non-negativ r.v. with $\{\tau\leq t\}\in\mathcal{F}_t$ for every $t$, it's $\sigma$-Algebra $\mathcal{F}_\tau$ the set of all $A\in\mathcal{F}$ s.t. $A\cap\{\tau\leq t\}\in \mathcal{F}_t$ for all $t$.
My question: does the Definition of $\mathcal{F}_\tau$ depend on $\mathcal{F}$? If one enlarges $\mathcal{F}$ to $\bar{\mathcal{F}} \supsetneq \mathcal{F}$, the collection $\{ \mathcal{F}_t \}$ is still a filtration, but $\mathcal{F}_\tau$ is now defined with respect to all $A\in\bar{\mathcal{F}}$.
Is $\mathcal{F}_\tau \subset \mathcal{F}_\infty = \sigma\bigcup_t \mathcal{F}_t$ true or false?
If $\tau$ is a stopping time taking only finite values, i.e. $\tau: \Omega \to [0,\infty)$, then
$$A = \bigcup_{n \in \mathbb{N}} \underbrace{(A \cap \{\tau \leq n\})}_{\in \mathcal{F}_n \subseteq \mathcal{F}_{\infty}} \in \mathcal{F}_{\infty}$$
for any $A \in \mathcal{F}_{\tau}$; hence $\mathcal{F}_{\tau} \subseteq \mathcal{F}_{\infty}$.
If $\tau$ takes the value $\infty$, then the inclusion $\mathcal{F}_{\tau} \subseteq \mathcal{F}_{\infty}$ does, in general, not hold true. To see this just consider the stopping time $\tau := \infty$: Then $\mathcal{F}_{\tau} = \mathcal{F}$, and therefore $\mathcal{F}_{\tau} \subseteq \mathcal{F}_{\infty}$ is, in general, not satisfied.