• $d\in\mathbb N$
  • $(\mathcal G_t)_{t\ge 0}$ be a complete right-continuous filtration on $(\Omega,\mathcal A)$
  • $(Y_t)_{t\ge 0}$ be a $\mathbb R^d$-valued $\mathcal G$-adapted almost surely right-continuous stochastic process on $(\Omega,\mathcal A,\operatorname P)$
  • $H\subseteq\mathbb R^d$ be open

(1) How can we show that $$\tau:=\inf\left\{t\in[0,T]:Y_t\in H\right\}$$ is an $\mathcal F$-stopping time?

Let $t\in[0,T]$. Since $\mathcal G$ is complete and right-continuous, it's sufficient to show that $\left\{\tau<t\right\}\in\mathcal F_t$.

(2) Let $N\in\mathcal A$ with $\operatorname P[N]=0$ such that $Y(\omega)$ is right-continuous for all $\omega\in\Omega\setminus N$. If I'm not terribly wrong, it's even sufficent to show that $\left\{\tau<t\right\}\cap(\Omega\setminus N)\in\mathcal G_t$. Why? Well, $\left\{\tau<t\right\}\cap N$ is a $\operatorname P$-null set and hence $\mathcal G_0$-measurable by the completness assumption on $\mathcal G$. Am I right?

(3) By (2) we can assume that $Y$ is surely right-continuous and by the $\mathcal G$-adaptedness of $Y$ and the right-continuity of $\mathcal G$ all we need to show is $$\left\{\tau<t\right\}=\bigcup_{s\in\mathbb Q\cap[0,\:t)}\left\{Y_s\in H\right\}\;.$$ "$\supseteq$" is trivial. How can we show the other inclusion? I know that $$\inf_{x\in[a,b]}f(x)=\inf_{x\in[a,b]\cap\mathbb{Q}}f(x)$$ for a continuous function $f:[a,b]\to\mathbb R$ ($-\infty<a\le b<\infty$) and I suppose we need to use something similar here.


1 Answer 1


For fixed $\omega \in \{\tau<t\}$ there exists $\tau(\omega) \leq s < t$ such that $Y_s(\omega) \in H$. As $H$ is open, there exists some $\epsilon>0$ such that $B(Y_s(\omega),\epsilon) \subseteq H$. Since $r \mapsto Y_r(\omega)$ is right-continuous, there exists $\delta>0$ such that

$$\forall r \in [s,s+\delta] : |Y_r(\omega)-Y_{s}(\omega)| < \frac{\epsilon}{2},$$

which implies

$$Y_r(\omega) \in H \qquad \text{for all} \, \, r \in [s,s+\delta].$$

As $\tau(\omega)\leq s<t$ and $\mathbb{Q}$ is dense, we can find $q \in [0,t) \cap \mathbb{Q}$ such that $Y_q(\omega) \in H$. Hence,

$$\{\tau<t\} \subseteq \bigcup_{q \in \mathbb{Q} \cap [0,t)} \{Y_q \in H\}.$$

  • $\begingroup$ What do you say to my remarks about the almost sure right-continuity of $Y$ and completeness of $\mathcal G$? $\endgroup$
    – 0xbadf00d
    Oct 23, 2016 at 18:12
  • $\begingroup$ @0xbadf00d Nothing wrong about them. $\endgroup$
    – saz
    Oct 24, 2016 at 6:00
  • $\begingroup$ Since $H$ is open, $\tau$ is a hitting time does not imply $Y_\tau(\omega) \in H$, only $Y_\tau(\omega) \in \bar{H}.$ So I think that the statement "there exists some $\epsilon>0$ such that $B(Y_\tau(w),\epsilon) \subseteq H$" is not correct... $\endgroup$
    – MI70
    Nov 4, 2016 at 14:32
  • $\begingroup$ @MI70 You are right; I fixed it. $\endgroup$
    – saz
    Nov 4, 2016 at 22:03

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