Let $E$ be a topological space. Say that $x:[0,\infty)\to E$ is of pure-jump type if there is a nondecreasing $(\tau_n)_{n\in\mathbb N_0}\subseteq[0,\infty]$ with $\tau_0=0$ such that either
- there is a $n_0\in\mathbb N$ with $$\tau_0<\cdots<\tau_{n_0}=\infty;\tag1$$ or
- $(\tau_n)_{n\in\mathbb N_0}$ is increasing.
Moreover, $$\left|x\left([\tau_{n-1},\tau_n)\right)\right|\le1\;\;\;\text{for all }n\in\mathbb N\tag2$$ and $$x(\tau_{n-1})\ne x(\tau_n)\;\;\;\text{for all }n\in\mathbb N\text{ with }\tau_n<\infty\tag3.$$ In that case, $(\tau_n)_{n\in\mathbb N_9}$ is called jump sequence of $(X_t)_{t\ge0}$.
Now let $(\Omega,\mathcal A)$ be a measurable space and $(X_t)_{t\ge0}$ be an $(E,\mathcal E)$-valued pure-jump tpye process on $(\Omega,\mathcal A)$ with jumping sequence $(\tau_n)_{n\in\mathbb N_0}$. How can we show that each $\tau_n$ is a stopping time with respect to the filtration $\mathcal F_t:=\sigma(X_s,s\le t)$, $t\ge0$, generated by $(X_t)_{t\ge0}$?
This claim is made on p. 277 of Kallenberg's book Foundations of Modern Probability (Third Edition). Actually, he is claiming that each $\tau_n$ is a stopping time with respect to the right-continuous filtration $(\mathcal F^+_t)_{t\ge0}$, but I don't know why we need the right-continuity. He writes that it would follow by a "simple approximation based on Lemma 9.3":
But I don't see how this Lemma could be useful here.