Let $B\_t, t\geq 0$ be standard Brownian motion. 


Let $\big(\mathcal{G}\_t, t\geq 0\big)$ be the natural filtration, defined by $\mathcal{G}\_t=\sigma(B\_s, 0\leq s\leq t)$.


Define also a filtration $\big(\mathcal{F}\_t, t\geq 0\big)$ by $\mathcal{F}\_t=\bigcap\_{\epsilon>0} \mathcal{G}\_{t+\epsilon}$.



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> Let $\tau$ be a stopping time with respect to the filtration $(\mathcal{F}\_t)$. Does there always exist $\tau'$ which is a stopping time with respect to the filtration $(\mathcal{G}\_t)$ such that $\tau=\tau'$ with probability 1?
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Related: Blumenthal's 0-1 law says that, for fixed $t$, for any event $A\in\mathcal{F}\_t$, there is an event $\tilde{A}\in\mathcal{G}\_t$ such that the symmetric difference of $A$ and $\tilde{A}$ has probability 0. 


However, this on its own is not enough. For example, let $U$ be a uniform random variable on $[0,1]$, and define a process $C\_t, t\geq 0$ by $C\_t=0$ for $t\leq U$ and $C\_t=t-U$ for $t\geq U$. Then a similar 0-1 law holds, and $U$ itself is a stopping time for the filtration $\mathcal{F}\_t=\bigcap\_{\epsilon>0}\sigma(C\_s, 0\leq s\leq t+\epsilon)$, but there is no stopping time $V$ for the filtration $\mathcal{G}\_t=\sigma(C\_s, 0\leq s\leq t)$ such that $U=V$ with probability 1.



Stopping times for Brownian motion
There does, but it's not such an obvious fact.


Let $\tau$ be an $(\mathcal F\_t)$ stopping time. Then $\tau$ is also a stopping time of the augmented filtration $(\tilde {\mathcal F}\_t)$, obtained by adding all null sets in the completion of $\mathcal F\_\infty$ to each $\mathcal F\_t$. As noted already, it is a consequence of the Blumenthal zero-one law that $(\tilde{ \mathcal F}\_t)$ coincides with the analogous augmentation $(\tilde{\mathcal G}\_t)$ of $(\mathcal G\_t)$. Also, it is known (see section 5 of Chapter IV of the book of Revuz and Yor, for example) that each $(\tilde{ \mathcal F}\_t)$ stopping time is predictable. In particular, $\tau$ is $(\tilde{ \mathcal F}\_t)$-predictable. Thus the process $Z\_t:=1\_{\{\tau=t\}}$ is $(\tilde{ \mathcal G}\_t)$-predictable. By the "de-completion" result stated as a Lemma in section 6 of Appendix I of Volume B of *Probabilités et Potentiel* by Dellacherie and Meyer, $Z$ is indistinguishable from a $(\mathcal G\_t)$-predictable process $Z'$. The $(\mathcal G\_t)$ stopping time $\tau':=\inf\{t:Z'\_t=1\}$ is a.s. equal to $\tau$.


