How can we show that the jump measure of a càdlàg process is a *counting* measure?

Question

Let $E$ be a normed $\mathbb R$-vector space, $(X_t)_{t\ge0}$ be an $E$-valued càdlàg process on a probability space $(\Omega,\mathcal A,\operatorname P)$ and $$\pi_\omega(B):=\sum_{\substack{s\:\ge\:0\\\Delta X_s(\omega)\:\ne\:0}}1_B\left(s,\Delta X_s(\omega)\right)\;\;\;\text{for }B\in\mathcal B([0,\infty)\times E)$$ for $\omega\in\Omega$.

Using that a regular function on a bounded interval has countably many jumps, we know that $\{s\ge0:\Delta X_s(\omega)\ne0\}$ is countable and hence $\pi_\omega(B)\in\mathbb N_0\cup\{\infty\}$.

But how do we see that $\pi_\omega$ is a counting measure$^1$?

I don't think that this is necessary, but I've only seen this claim under the assumption that $X$ is a Lévy process. However, this shouldn't be important.

This is stated as an exercise in exercise 16 of this lecture notes. So, I guess it's simple, but I've no idea how we should approach this.

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$^1$ Remember that a measure $\mu$ on a measurable space $(S,\mathcal S)$ is called counting measure if

1. $\mu$ is purely atomic, i.e. $$D_\mu:=\{x\in S:x\text{ is an atom of }\mu\}$$ is countable and $\mu\left(D_\mu^c\right)=0$;
2. If $x\in D_\mu$, then $$\mu(\{x\})=1.$$

Moreover, $x\in S$ is an atom of $\mu$ iff $\mu(\{x\})\ne0$.

Answer 1

Technical remark: $\pi_{\omega}$ will be a counting measure in the set which $\lbrace s \geq 0: \triangle X_s \neq 0 \rbrace $ is countable (it has probability one). We will fix $\omega$ is in that set .

First claim : Consider $\lbrace t_n \rbrace $ the countable set which $\triangle X_{t_n} \neq 0 $. Define $D:= \lbrace (t_n,\triangle X_{t_n} ), n \in \mathbb{N} \rbrace $. \ Then $\pi_{\omega}(D^c)$ is clearly zero because each pair $(s,\triangle X_s)$ such that $\triangle X_s \neq 0$ belongs to $D$ (therefore each element in the sum is zero).

Second claim: Suppose $\pi_{\omega}(t_n,\triangle X_{t_n}) >1$ . Then there is at least one $m \in \mathbb{N}, \ m \neq n$ such that: $$ (t_n, \triangle X_{t_n})= (t_m, \triangle X_{t_m}) $$ because $\mathbf{1}_{ \lbrace (t_{n},\triangle X_{t_n}) \rbrace } (s, \triangle X_s) \neq 0$ for at least two $(s, \triangle X_s)$ such that $(\triangle X_s) \neq 0$. This is absurd because $t_n \neq t_m$ if $n \neq m$.