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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 $\mathcal{F}_t=\bigcap_{\epsilon>0} \mathcal{G}_{t+\epsilon}$.

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?

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.

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    $\begingroup$ Is it true that there exists $A \in {\cal F}_0$ such that $A \not\in {\cal G}_0$ ? In this cas $\tau = {\boldsymbol 1}_A$ is a ${\cal F}$-stopping time but not a ${\cal G}$-stopping time. $\endgroup$ – Stéphane Laurent Jul 23 '16 at 13:17
  • $\begingroup$ Thankyou Stéphane. Good point! I did not ask the right question. I should instead ask whether there is a $\mathcal{G}$-stopping time which is equal to $\tau$ with probability 1. In the case you mention, every such $A$ must have probability 0 or 1. I will edit. $\endgroup$ – James Martin Jul 23 '16 at 13:24
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    $\begingroup$ I would try to combine the 0-1 law with the strong Markov property to show that the statement holds (and this would also show why the example is not a counterexample). $\endgroup$ – Martin Hairer Jul 24 '16 at 14:02
  • $\begingroup$ I guess the tough part is, how do we construct a candidate stopping time $\tau'$? @MartinHairer's approach seems useful in verifying that they are a.s. equal. $\endgroup$ – Nate Eldredge Jul 25 '16 at 2:23
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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 $\tau$.

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