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Let's say we are given a cadlag Levy process $X$, $X_0 = 0$ and a Borel set $\Gamma$ such that $0 \notin \bar{\Gamma}$, where $\bar{\Gamma}$ is the closure of $\Gamma$. Let's put $$ T = \inf\{t>0:\Delta X_t \in \Gamma\} $$ Question: Suppose the filtration is right continuous. By checking $(T \geq t) \in \mathscr{F}_{t+}$, show that $T$ is a stopping time.

Can anyone give a hint on the question? I attempted to consider either open or closed $\Gamma$ first, but just can't reduce $(T \geq t)$ to a simple form. Any hint will be greatly appreciated!

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    The proof of the Début theorem for arbitrary Borel set is quite involved. One might take advantage of the fact that $\Delta X_t$ are independent (and there is at most countable number of non-zero jumps). For me, this sounds doubtful. – 2017-02-03
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    @zhoraster Thanks very much for the reply! If I understand the debut theorem correctly, the conditions can be weakened, that is, $X$ doesn't need to be Levy, as $\Delta X_t$ is progressively measurable. Then, it also means that without supposing independence, there could possibly be some "elementary" proof that doesn't refer to Debut theorem.. And I am wondering if there is such a proof? – 2017-02-03
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    I've no idea of any elementary proof. The simplest idea which comes to my mind is first to prove this in the case where $\Gamma = \mathbb R\setminus [-A,A]$ (this is not hard) and then try to show that the family of sets such that debuts of such sets are stopping times is a $\sigma$-algebra. – 2017-02-04

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