I found this exercise online:
I am stuggling with the last part of the second exercise, that is I am not able to show that $\tau = \sup_i \tau_i$. Obviously we have that $\tau \ge \sup_i \tau_i$, but I struggle to show the other inequality. Any tips?
Also, I have heard that the second case also holds under the restriction from the first case, that is right continuity, or maybe we need to assume cadlag-trajectories. Do you see how we can show the second case assuming this?
My idea for proving this was this: I was able to show it easily if $\tau=0,\infty$. So I can exclude those cases, and look at any $\epsilon>0$. Then I must have that in the interval $[0,\tau-\epsilon]$ atleast one $G_i$ is not hit.
Can you please help me?
Since $G_i \supseteq B$, we have $\tau_i \leq t$ and therefore, as you already noted,
$$\sup_{i \in \mathbb{N}} \tau_i \leq \tau.$$
It remains to show that
$$\tau \leq \sup_{i \in \mathbb{N}} \tau_i.$$
Fix $\omega \in \Omega$. Without loss of generality, we may assume that the right-hand side is finite (otherwise there is nothing to prove), i.e.
$$T(\omega):= \sup_{i \in \mathbb{N}} \tau_i(\omega) <\infty.$$
Since $\tau_1 \leq \tau_2 \leq \ldots$, it follows from the very definition of "sup" that $\lim_{i \to \infty} \tau_i(\omega) = T(\omega)$. Hence, as $(X_t)_{t \geq 0}$ has continuous sample paths,
$$X_{T(\omega)}(\omega) = \lim_{i \to \infty} X_{\tau_i(\omega)}(\omega).$$
Since
$$|X_{T(\omega)}-b| \leq |X_{T(\omega)}-X_{\tau_i(\omega)}(\omega)| + |X_{\tau_i(\omega)}(\omega)-b|$$
for any $b \in B$, we find
$$d(X_{T(\omega)}(\omega),B) \leq |X_{T(\omega)}-X_{\tau_i(\omega)}(\omega)| + \underbrace{d(X_{\tau_i(\omega)}(\omega),B)}_{\leq i^{-1}} \xrightarrow[]{i \to \infty} 0,$$
i.e.
$$d(X_{T(\omega)}(\omega),B)=0.$$
As $B$ is closed, this entails $X_{T(\omega)}(\omega) \in B$. By the very definition of $\tau$, this means that $\tau(\omega) \leq T(\omega)$.
As the proof shows we just need left-continuity of the sample paths and not necessarily continuity.