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

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.

$$^1$$ 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 $$\mu(\{x\})\ne0$$.

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 $$n \neq m$$.

• Thank you for your answer. I'll check it out now. But regarding your technical remark: I guess you're saying this, because you think the paths of $X$ are only almost surely càdlàg? In order to rule out this unnecessary problem, I've assumed that all paths are càdlàg. Or did I miss something else? Mar 5 at 15:30
• No, you are right!
– FOE
Mar 5 at 15:32
• Suppose $\pi_{\omega}(D^c) \neq 0$. Then there is at least one $s$ such that $\triangle X_s \neq 0$ and $(s, \triangle X_s) \in D^c$. On the other hand if $s$ is a non-continuity point then there is a $n \in \mathbb{N}$ such that $t_n =s$. However the points $(t_,n, \triangle X_{t_n})$ belongs to $D$. ( This answer is to clarify a deleted question)
– FOE
Mar 5 at 15:47
• Yes, sorry. Figured it out. It's trivial. The sum is $0$ (it is a sum over an empty index set), since for all $s\ge0$ it holds $(s,\Delta X_s(\omega))\in D^c$ iff $\Delta X_s(\omega)=0$. Mar 5 at 15:50
• No, that wouldn't be correct, but it should be "$x\in S$ is an atom of $\mu$ iff $\mu(\{x\})\ne0$". Thanks for mentioning. Mar 5 at 15:52