Let $$\mathcal F_s = \{ A : A\cap \{S \leq t\} \in \mathcal F_t, \forall t \geq 0\}$$

where $\left(\mathcal F_t\right)_{t\geqslant 0}$ is a right continuous filtration.

Let $S$ be a stopping time, let $A \in\mathcal F_s$ and let $R=S$ on $A$ and $R = \infty $ on $A^c$. Show that $R$ is a stopping time.

I'm really lost on this question, it's question 7.3.4 in Rick Durrett's Probability Theory book.

  • $\begingroup$ Proper notation is $\langle a,b,c \rangle,$ not $<a,b,c>.$ See my edits to the question. $\endgroup$ Commented Sep 12, 2017 at 17:11
  • $\begingroup$ Yes, that is what I meant. Thank you for editing it $\endgroup$ Commented Sep 12, 2017 at 17:28
  • $\begingroup$ Also note that when you don't know something like this, you can usually find the answer by googling "latex symbols". $\endgroup$ Commented Sep 12, 2017 at 17:30
  • $\begingroup$ Sure, I think I was just being careless but I'll make a note of it for next time. In any case, do you think the approach or answer to my problem is incorrect? Should I be solving it a different way? $\endgroup$ Commented Sep 12, 2017 at 18:17

1 Answer 1


Fix $t\ge0$. Then \begin{align*} \{R \le t\} &= (\{R \le t\} \cap A) \cup (\{R \le t\} \cap A^c)\\ &= (\{S \le t\} \cap A) \cup \emptyset\\ &= \{S \le t\} \cap A \in \mathcal{F}_t, \end{align*} since $A\in\mathcal{F}_S$. Since $t$ was arbitrary, this shows $R$ is a stopping time.


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