I've read that the first jump time of the Poisson process is totally inaccessible (definition at the bottom for anyone interested). This made me wonder if the first jump time is a stopping time. I think the answer is yes. (If the answer was no, the first jump time would automatically be totally inaccessible and people probably wouldn't bother mentioning totally inaccessibility.) More generally, I think the answer is yes for any right-continuous process.
EDIT: Lost1 has pointed out that that the answer is easily yes for a Poisson process. That still leaves general right-continuous processes.
Let $(\mathcal{F}_t)_{t \geq 0}$ be a filtration of a probability space. A random variable $T$ is called an stopping time if $\left\{T \leq t \right\} \in \mathcal{F}_t$ for all $t \geq 0$. For us, $\mathcal{F}_t = \sigma(X_{s} | s \leq t)$.
Here's my attempt to show the first jump time for a right-continuous process is a stopping time. The first jump time is defined to be $T = \inf\{t>0 | X_t \neq X_0 \}$. We need $\{ T \leq t \} \in \mathcal{F}_{t}$. I don't know how to prove this, but I do know that
$$
\{ T < t \} = \{ X_s \neq X_0 \text{ for some $s < t$} \} = \bigcup_{r < t, \, r \in \mathbb{Q}} \{ X_r \neq X_0 \} \in \mathcal{F}_{t},
$$
where the last equality is by right-continuity. It's not clear to me how to write $\{ T \leq t \}$ in a way that shows it belongs to $\mathcal{F}_{t}$.
By the way, here is the definition of a totally inaccessible stopping time.
If $T_n$ are stopping times and $T_n \uparrow T$, then $T$ is an stopping time and we call it predictable. If $P(S = T < \infty) = 0$ for all predictable stopping times $S$, we say $T$ is totally inaccessible.
Best Answer
First, I cannot understand what you are trying to prove? Are you trying to prove $T$ is a stopping time? Why is the notion of totally inaccessible here?
Poisson first jump times are not predictable. If $H_t$ is predictable, then it $H_t\in\mathcal{F}_{t-}$. This is clearly not the case for jump processes.
The intuition is that, if a stopping time is predictable, you will know that you are getting close to the stopping time before it occurs (for example for a Brownian motion), whereas Poisson jumps do not happen this way. If it hasn't happened for 5 minutes, it is not any more likely to happen than 5 minutes ago
Also why are you trying to show $\{T<t\}\in\mathcal{F_t}$ when you have given $\{T\leq t\}\in\mathcal{F_t}$ as your definiton as a stopping time. The former is not equivalent to $T$ being a stopping time, without extra knowledge about the filtration etc.
Am I insane or isn't $\{T\leq t \} = \{ X_t>X_0 \}$ which looks like a totally measurable with respect to $\mathcal{F}_t$?