Jumping times of a Lévy Process If one has a Levy-process, are the times when the process has a jump of size exceeding a positive $\varepsilon$ actually stopping times w.r.t. the canonical filtration?
In more detail: Let $X=(X_t)_{t\geq0}$ be a Levy process (it especially has a.s. cadlag paths) and $\varepsilon>0$. Then define the times $T_{\varepsilon}^{(0)}(X):=0$ and $T_{\varepsilon}^{(i+1)}(X):=\inf\{t>T_{\varepsilon}^{(i)}(X):|\Delta X_t|>\varepsilon\}$, where $\Delta X_t:=X_{t}-X_{t-}$ denotes a possible jump and $\inf\emptyset:=\infty$. Take the canonical (wlog. augmented) filtration $\mathcal{F}_t=\sigma(X_s,s\leq t)$. Question: Are the times $T_{\varepsilon}^{(i)}(X)$ stopping times?
My idea so far: Given $T_{\varepsilon}^{(1)}$ is a stopping time, one can write $T_{\varepsilon}^{(2)}(X)=T_{\varepsilon}^{(1)}(X)+T_{\varepsilon}^{(1)}(X\circ\theta_{T_{\varepsilon}^{(i)}(X)})$, where $\theta_T$ denotes the shift-operator (tbh I'm not sure if this is true at all, but I guess one could use the fact that $X$ has independent increments to prove this?) If that was true, it would suffice to show that $T_{\varepsilon}^{(1)}(X)$ is a stopping time. At this point I'm pretty clueless, since the condition $|\Delta X_t|>\varepsilon$ is rather hard to handle I guess?
Maybe some Markov-arguments may apply? Though it bothers me that the paths are not continuous. Some ideas? References? Thank you!
 A: It follows indeed from the Markov property (...which is a consequence of the independence of the increments) that it suffices to show the claim for $i=1$.
Let $t>0$, $\omega \in \Omega$ such that $|\Delta X_t(\omega)| > \varepsilon$. Since $X$ has (a.s.) càdlàg sample paths, we can find for any $k \in \mathbb{N}$ sufficiently large some $r_k, s_k \in \mathbb{Q}$, $t - 1/k \leq r_k \leq t \leq s_k \leq t+ 1/k$, such that
$$|X_{s_k}(\omega)-X_{r_k}(\omega)| > \varepsilon.$$
Consequently,
$$\{|\Delta X_t|>\varepsilon\} \subseteq \bigcup_{\ell \geq 1} \bigcap_{k \geq \ell} \bigcup_{\substack{r_k,s_k \in \mathbb{Q} \\ t-1/k \leq r_k \leq t \leq s_k \leq t+1/k}} \{|X_{s_k}-X_{r_k}|>\varepsilon\}. \tag{1}$$
Hence,
$$\{T_{\varepsilon}^{(1)}(X) < t\} = \bigcup_{\ell \geq 1} \bigcap_{k \geq \ell} \bigcup_{\substack{r_k,s_k \in \mathbb{Q} \\ |r_k-s_k| \leq 2/k, r_k \leq s_k <t}} \{|X_{s_k}-X_{r_k}|>\varepsilon\}=:A. \tag{2}$$
On the other hand, it is not difficult to see that $\omega \in A$ implies $\omega \in \{T_{\varepsilon}^{(1)} < t\}$. Since $A \in \mathcal{F}_t$ (it is a countable union of $\mathcal{F}_t$-measurable sets), we conclude that
$$\{T_{\varepsilon}^{(1)}(X) \leq t\} = \{T_{\varepsilon}^{(1)}(X) < t\} \cup \{|\Delta X_t|>\varepsilon\} \in \mathcal{F}_t.$$
(Note that $(1)$ shows that $\{|\Delta X_t|>\varepsilon\} \in \mathcal{F}_{t+}$; but if do not require $s_k \in \mathbb{Q}$, we can set $s_k = t$ and find that $\{|\Delta X_t|>\varepsilon\} \in \mathcal{F}_t$.)
Remark: The proof shows that the claim does not only hold for Lévy processes, but for càdlàg (strong) Markov processes.
