# If $X$ is a Lévy process, why is $t\mapsto\sum_{\substack{s\in[0,\:t]\\\Delta X_s(\omega)}}1_B(\Delta X_s(\omega))$ càdlàg?

Let $$E$$ be a normed $$\mathbb R$$-vector space, $$(X_t)_{t\ge0}$$ be an $$E$$-valued càdlàg Lévy process on a filtered probability space $$(\Omega,\mathcal A,(\mathcal F_t)_{t\ge0},\operatorname P)$$, $$B\in\mathcal B(E)$$ with $$0\not\in\overline B$$ and $$N_t(\omega):=\left|\left\{s\in(0,t]:\Delta X_s(\omega)\in B\right\}\right|=\sum_{\substack{s\in[0,\:t]\\\Delta X_s(\omega)}}1_B(\Delta X_s(\omega))$$ for $$\omega\in\Omega$$ and $$t\ge0$$.

How do we see that $$t\mapsto N_t(\omega)$$ is càdlàg?

In order for $$N$$ to be cadlag, it should be assumed that $$X$$ has finite jump activity in $$B$$, i.e. $$\nu(B)<\infty$$, 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 $$N_t = \sum_{n\ge 1} \mathbf{1}_{[\tau_n,\infty)} (t)$$, where $$\tau_n$$ is the time of $$n$$th jump in $$B$$; this is clearly cadlag.