is first hitting time of a general non-cadlag process a stopping time?

Question

Suppose I have an increasing random process $A_t$ with non-right-continuous filtration ($A_t$ can have double jumps). $A_t$ is predictable and its right limit $A_{t+}$ is optional. Consider $$\tau=\inf \{t>0 : A_{t+}>c\} ,$$ $c$ is some positive constant. Is the random time $\tau$ a stopping time? I guess, we can write $\{\tau>t\}= \cap_{s\le t}\{A_{s+}\le c\} \in \mathcal{F}_t$, so it is a stopping time, am I correct?

Answer 1

Your description of the event $\{\tau>t\}$ is not quite right. Consider, for example, deterministic $A_t(\omega) = t$ for all $t$, $\omega$. Then $\tau(\omega) = c$ so $\{\tau>t\}$ is either empty (if $t\ge c$) or the entire sample space (if $t<c$), but $\cap_{s\le t}\{A_{s+}\le c\}$ is either empty (if $t>c$) or the entire sample space (if $t\le c$).

The example hints at the truth: In general, $\tau$ is a stopping time of $(\mathcal F_{t+})$, but not necessarily of $(\mathcal F_t)$. This because $\{\tau<t\}=\cup_{s<t, s\in\Bbb Q}\{A_{s+}>c\}\in \mathcal F_t$.