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

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?

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 $$t), but $$\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)$$. This because $$\{\tauc\}\in \mathcal F_t$$.
• Thanks for your answer. Do you know if usual stochastic integrals (w.r.to cadlag processes) are well defined for stopping times of $(\mathcal{F}_{t+})$. I feel there can be some measurability issues. Jan 24, 2021 at 21:13