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Let $X$ be a Levy Process. $\Lambda=\{(\omega,u) \in \Omega \times \mathbb{R}:e^{\iota u X_t(\omega)},t \in \mathbb{Q}_{+}, \text{is not a restriction of a cadlag function} \}$. Please find below the source attached .enter image description here

Also why is $\int \mathbb{1}_{\Lambda}(\omega,u)P(d \omega)=0$ for each $u \in \mathbb{R}$?

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    I'm reading: "For any real $u$, $(M_t^u)_{t\ge 0}$ is a complex-valued martingale and thus for a.a. $\omega$ the functions $t\mapsto M_t^u(\omega)$ and $t\mapsto e^{itX_t(\omega)}$, with $t\in\mathbb{Q}_+$, are the restrictions to $\mathbb{Q}_+$ of cadlag functions". Are you reading this too? Does this answer your last question? Concerning the measurability question (which you must add to the question body), this is a standard exercise, so please write your own thoughts on that. – 2017-01-26
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    @zhoraster I think I understand very little in this theorem. in Particular I do not understand why the maps $t \mapsto M_t^{u}(\omega) \text{ with } t \in \mathbb{Q}_+$ are restrictions of cadlag functions. I mean if the complex valued martingale were cadlag it would have been clear but since X is just a levy process and the martingale composed using the Levy process might not be cadlag. I apologize for the late reply, I didnt have access to a computer but just my phone so couldn't reply – 2017-01-31
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    "I don't understand" is not a question. Perhaps you should read the book, starting from definitions, more carefully. – 2017-01-31
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    @zhoraster I am trying my best . So can I say that since every martingale has a right continuous modification and since in the book the usual conditions are assumed , I can consider that the martingale M to be cadlag and therefore the levy process X ( or rather its modification corresponding the cadlag modification of M)would be cadlag too ( the way M is defined ) . I understand why the integral has a zero measure after rereading your first comment . – 2017-01-31
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    @zhoraster Ah this I see it. Is it because all martingales have a right continous modification(in particular cadlag) and we consider the right-continuous modification of $M^u$ and this gives us that $X$ is also cadlag (or there exisits a modification of $X$ which is cadlag. If what i said is true I see why the last integral is equal to 0 – 2017-01-31

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