# First hitting time of leftcontinuous process

Suppose that we are working with a Filtration which is right continuous. I know then, that the first hitting time of a right continuous process into an open set is a stopping time. Is the same true, if we replace right continuity with left continuity?

hulik

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So you have a left continuous process, with a left continuous filtration, and an open set $O$, and you want to see whether $T_O:=\inf_{t>0,X_t\in O}$ is a stopping time? – Davide Giraudo Aug 4 '12 at 10:35
@DavideGiraudo No, sorry if that wasn't clear. The filtration is again right continuous, just the process should be left continuous instead of right continuous. – user20869 Aug 4 '12 at 11:35

For a right continuous filtration $(\mathcal F_t)_{t\geq 0}$, a left continuous process and $O$ an open set, the map $$T_O\colon\omega\mapsto \inf\{t>0, X_t(\omega)\in O\}$$ is a stopping time.
Since the filtration is right continuous, it's enough to show that $\{T_O<t\}\in\mathcal F_t\}$ for all $t>0$. We have $$\{T_O<t\}=\bigcup_{q<t,q\in\Bbb Q}\{X_q\in O\}.$$ Indeed, let $\omega$ such that $T_O(\omega)<t$. We can find $t'<t$ such that $X_{t'}(\omega)\in O$. By left continuity and using openness of $O$, we can find a rational $q<t'$ such that $X_q(\omega)\in O$.
Conversely, if $X_q(\omega)\in O$ for some rational $q<t$, we have $T_O(\omega)\leq q<t$.
Since $\{X_q\in O\}\in\mathcal F_q\subset\mathcal F_t$ for each $q<t$, we are done.