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
- $\tau_1\in(0,\infty]$;
- 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 $\{\tau_1<t\}\in\mathcal F_t$ for all $t>0$. Can we show this?
Best Answer
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.