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

Is it the case that any stopping time $\tau$ with respect to the filtration $\mathcal{F}_t$ is also a stopping time with respect to the filtration $\mathcal{G}_t$?

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 is not a stopping time for the filtration $\mathcal{G}_t=\sigma(C_s, 0\leq s\leq t)$.

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.