In "Credit Risk: Modeling, Valuation and Hedging" they say, that any stopping time $\tau$ w.r.t the Brownian filtration is predictable.
Since predictable means that there exists a sequence of $(\tau_n)_n$ announcing $\tau$ i.e.
- $\tau_n<\tau$ on $\{\tau>0\}$
- $\lim_{n\rightarrow\infty}\tau_n=\tau$
I didn't manage to find such an announcing stopping time only based on the fact that $\{\tau<t\}\in\sigma(B_u:u\leq t)$ for all $t$.
Furthermore they say, that e.g. when $\tau$ is a stopping time w.r.t the Brownian Filtration $\mathbb{F}$, that it's not predictable w.r.t a strictly larger filtration containing $\mathbb{F}$. I don't get how one couldn't just use the announcing stopping time which has been found for $\mathbb{F}$.
The predictability of Brownian stopping times follows from the fact that all (local) martingales in the filtration of a Brownian motion have continuous sample paths — a consequence of the Martingale Representation Theorem. (A nice discussion of such things can be found here: https://almostsuremath.com/2011/05/26/predictable-stopping-times-2/.)
A different way to see this is through Markov process theory: Brownian motion is a path continuous strong Markov process; as such all of its stopping times are predictable.
Your suspicion of the assertion made by "CR:M,VaH", reported in your last paragraph, is well-founded: any $\Bbb F$ predictable stopping time is $\Bbb G$ predictable, for the reason you state.