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Given an adapted cadlag Levy Process $X$ and a constant $C$ defined as the $\sup_{t} |\Delta X_t|$ where $\Delta X_t=X_t- X_{t^-}$ and the book defines:

$T_1=\inf\{t:|X_t| \geq C\} $

$\vdots$

$T_{n+1}=\inf \{t>T_n: |X_t-X_{T_n}| \geq C\}$

My attempt: I proceed by induction

Since X is cadlag adapted, $T_1$ is clearly a stopping time. Then I assume that $T_n$ is a stopping time and try to conclude that $T_{n+1}$ is a also a stopping time

$\{T_{n+1} \leq t\}= \bigcup_{s \in (T_n,t]\cap \mathbb{Q}} \big\{ \{T_n

If the above two sets are equal?(I am not absolutely sure if they are!!) then since by assumption that $T_n $ is a stopping time we have that $\{T_n

The countable union is still in $\mathcal{F}_t$ and we are done.

Is this proof valid? Can you point out where the problem is and how could I fix it?

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    I don't understand why $T_1$ is a stopping time. $C$ seems to be a (random) constant which depends on the entire path: it's the largest jump in your process.Asking yourself whether you are taller now than you will ever be able to jump does not seem a question you can answer with the information of today. (Maybe this comparison is clear only to me :D) – 2017-02-09
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    @Corn no it's not random . Its deterministic independent of the path , a universal constant for this stochastic process – 2017-02-09
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    @corn if we know the supremum of jumps is 5 , then asking yourself if the jump until today was less than equal to 5 is something that I know until .Does this make sense ? – 2017-02-10
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    You are definitely right, given that C is not random. But what I still don't understand is why $C (\omega)= sup_{t \ge 0} |\Delta X_t (\omega)|$ is actually not random. Or is it given in the exercise that the supremum is bounded by some deterministic constant? – 2017-02-10

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