# Proving a stopping time

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.

• Proper notation is $\langle a,b,c \rangle,$ not $<a,b,c>.$ See my edits to the question. Commented Sep 12, 2017 at 17:11
• Yes, that is what I meant. Thank you for editing it Commented Sep 12, 2017 at 17:28
• Also note that when you don't know something like this, you can usually find the answer by googling "latex symbols". Commented Sep 12, 2017 at 17:30
• 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? Commented Sep 12, 2017 at 18:17

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.