# How do we show that the jumping times $\inf\{t:\Delta X_t\in B\}$ of a Lévy process $X$ are stopping times?

We can show the following: If $$E$$ is a normed $$\mathbb R$$-vector space, $$x:[0,\infty)\to E$$ is càdlàg, $$B\subseteq E\setminus\{0\}$$, $$\tau_0:=0$$ and $$\tau_n:=\inf\underbrace{\{t>\tau_{n-1}:\Delta x(t)\in B\}}_{=:\:I_n}$$ for $$n\in\mathbb N$$, then

1. $$\tau_1\in(0,\infty]$$;
2. If $$n\in\mathbb N$$, then either $$I_n=\emptyset$$ and hence $$\tau_n=\infty$$ or $$\tau_n\in I_n$$.

Now if $$X$$ is any $$E$$-valued càdlàg Lévy process on a filtered probability space $$(\Omega,\mathcal A,(\mathcal F_t)_{t\ge0},\operatorname P)$$, we can similarly define $$I_n$$ and $$\tau_n$$ with $$x$$ replaced by $$X$$.

$$(\Delta X_t)_{t\ge0}$$ is clearly $$\mathcal F$$-adapted and $$(X_{t_1+s}-X_{t_1})_{s\ge0}$$ is a Lévy process with respect to the filtration $$(\mathcal F_{t_1+s})_{s\ge0}$$ with the same distribution as $$X$$ for all $$t_1>0$$.

Question: Are we able to show that $$\tau_1$$ is $$\mathcal A$$-measurable? Or are we even able to show that $$\tau_1$$ is measurable with respect to the right-continuous filtration $$\mathcal F_{t+}:=\bigcap_{\varepsilon>0}\mathcal F_{t+\varepsilon}$$? The latter is equivalent to showing that $$\{\tau_1 for all $$t>0$$. Can we show this?

• What is $\Delta X(t)$? Commented Oct 24, 2021 at 11:34
• @KaviRamaMurthy $\Delta x(t):=x(t)-x(t-)$ and $x(t-):=\lim_{s\to t-}x(s)$. Commented Oct 24, 2021 at 13:56

The process of jumps $$\Delta X$$ is progressively measurable since it is the difference of two progressively measurable processes. Then, for a measurable subset $$B\subseteq E$$ the hitting time $$\tau_1=\inf\{t>0:\Delta X(t)\in B\}$$ is a stopping time according to the Début theorem. Note: The theorem assumes that the probability space is complete.
• Oh, and might there be a more elementary proof available, if we assume $0\not\in\overline B$? Commented Oct 31, 2021 at 13:40