Hitting time of an open set is not a stopping time for Brownian Motion Let $(B_t)$ be a standard Brownian motion and $\mathcal F_t$ the associated canonical filtration. It's a standard result that the hitting time for a closed set is a stopping time for $\mathcal F_t$ and the hitting time for an open set is a stopping time for $\mathcal F_{t+}$. 
Is there an elementary way to see that the hitting time for an open set is not in general a stopping time for $\mathcal F_t$? Say the hitting time for an open interval $(a,b)$?
I'm interested in this question because it would show the filtration generated by a right-continuous process need not be right-continuous. There are other counterexamples for this on M.SE but they're all somewhat artificial. 
My apologies if this is obvious. I just started learning about such things. 
 A: No, it's not at all obvious.
If we interpret $\mathcal{F}_t$ as "information up to time $t$", it's not surprising that the hitting time of an open interval $(a,b)$ is not an $\mathcal{F}_t$-stopping time. For instance if $B_t(\omega)=a$ for some $t>0$, then the information about the past, i.e. $(B_s(\omega))_{s \leq t}$, is not enough to decide whether $\tau(\omega)=t$; we need a small glimpse into the future. 
Making this intuition rigorous is, however, not easy. One possibility is to apply Galmarino's test which states that a mapping $\tau: \Omega \to [0,\infty]$ is a stopping time (with respect to $(\mathcal{F}_t)_{t \geq 0}$) if and only if $$\tau(\omega)=t, B_s(\omega) = B_s(\omega') \, \, \text{for all $s \leq t$} \implies \tau(\omega')=t, \tag{1}$$ see this question. If $\tau$ is the hitting time of an open interval $(1)$ is not satisfied; the easiest way to see this is to consider the canonical Brownian motion, i.e. consider $(B_t)_{t \geq 0}$ as a process on the space of continuous mappings.
Let me finally remark that there are other ways to prove that the filtration generated by a Brownian motion is not right-continuous, see, for instance, this answer.
A: Consider the open interval $(1,2)$ and two (extremely stylized) sample paths:
$$
\omega_1(t) =\cases{t,&$0\le t\le 1$,\cr 2-t,&$t>1$,\cr}
$$
and
$$
\omega_2(t)=t,\quad t\ge 0.
$$
These two paths agree up to time $t=1$. Letting $T$ denote the hitting time of $(1,2)$, you have $T(\omega_1) =1$ but $T(\omega_2)=+\infty$. If $T$ were a stopping time for $(\mathcal F_t)$ you would have $\{T\le 1\}\in\mathcal F_1$, which would in turn imply that if $T(\omega_1)\le 1$ and $\omega_1|_{[0,1]}=\omega_2|_{[0,1]}$ then also $T(\omega_2)\le 1$. As this fails to hold, $T$ cannot be a stopping time for $(\mathcal F_t)$.
