# Let E be a normed <mrow class="MJX-TeXAtom-ORD"> <mi mathvariant="double-struck">R

Let $E$ be a normed $\mathbb{R}$-vector space, $\left({X}_{t}{\right)}_{t\ge 0}$ be an $E$-valued càdlàg Lévy process on a filtered probability space $\left(\mathrm{\Omega },\mathcal{A},\left({\mathcal{F}}_{t}{\right)}_{t\ge 0},\mathrm{P}\right)$, $B\in \mathcal{B}\left(E\right)$ with $0\notin \overline{B}$ and
${N}_{t}\left(\omega \right):=|\left\{s\in \left(0,t\right]:\mathrm{\Delta }{X}_{s}\left(\omega \right)\in B\right\}|=\sum _{\begin{array}{c}s\in \left[0,\phantom{\rule{mediummathspace}{0ex}}t\right]\\ \mathrm{\Delta }{X}_{s}\left(\omega \right)\end{array}}{1}_{B}\left(\mathrm{\Delta }{X}_{s}\left(\omega \right)\right)$
for $\omega \in \mathrm{\Omega }$ and $t\ge 0$.

How do we see that $t↦{N}_{t}\left(\omega \right)$ is càdlàg?
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Alexzander Bowman
In order for $N$ to be cadlag, it should be assumed that $X$ has finite jump activity in $B$, i.e. $\nu \left(B\right)<\mathrm{\infty }$, where $\nu$ is the Lévy measure of $X$ (otherwise, it blows up to infinity immediately). And in such case, this is rather straightforward: with probability 1, the number of jumps in $B$ is locally finite, and ${N}_{t}=\sum _{n\ge 1}{\mathbf{1}}_{\left[{\tau }_{n},\mathrm{\infty }\right)}\left(t\right)$, where ${\tau }_{n}$ is the time of $n$th jump in $B$; this is clearly cadlag.