Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $E$ be a normed $\mathbb R$-vector space, $\lambda$ denote the Lebesgue measure on $\mathcal B(\mathbb R)$, $\mu$ be a measure on $E$ with $\mu\{0\}=0$ and $$\int1\wedge\|x\|_E\:\mu({\rm d}x)<\infty\tag1$$ and $\pi$ be a Poisson random measure$^1$ on $[0,\infty)\times E$ with respect to $(\Omega,\mathcal A,\operatorname P)$.
We can show that there is a $\operatorname P$-null set $N$ such that $\pi_\omega$ is a counting measure, $$\left|\left\{x\in E:\pi_\omega\{(t,x)\}\ne0\right\}\right|\le1\tag2$$ and $$A_t(\omega):=\int_{[0,\:t]\times E}\pi_\omega({\rm d}(s,x))\|x\|_E<\infty\tag3$$ and hence $$X_t(\omega):=\int_{[0,\:t]\times E}\pi_\omega({\rm d}(s,x))x=\sum_{\substack{s\in[0,\:t]\\\Delta X_s(\omega)\ne0}}\Delta X_s(\omega)\tag4$$ for all $t\ge0$.
How can we show that $$\operatorname E\left[e^{{\rm i}\langle X_1,\:x'\rangle}\right]=e^{f(x')},$$ where $$f(x'):=\int e^{{\rm i}\langle x,\:x'\rangle}-1\:\lambda({\rm d}x),$$ for all $x'\in E'$? (Feel free to consider the case $E=\mathbb R^d$.)
i.e. $\pi$ is a transition kernel with source $(\Omega,\mathcal A)$ and target $(E,\mathcal B(E))$. Moreover, denoting $\pi_\omega:=\pi(\omega,\;\cdot\;)$ for $\omega\in\Omega$ and $\pi(B):=\pi(\;\cdot\;,B)$ for $B\in\mathcal B([0,\infty))\otimes\mathcal B(E)$,
- $\pi(B)$ is Poisson distributed with intensity $(\lambda\otimes\mu)(B)$ for all $B\in\mathcal B([0,\infty))\otimes\mathcal B(E)$, where a random variable $N$ is said to be Poisson distributed with intensity $0$, if $N=0$ almost surely, and Poisson distributed with intensity $\infty$, if $N=\infty$ almost surely;
- If $\mathcal F\subseteq\mathcal B([0,\infty))\otimes\mathcal B(E)$ is finite, then $(\pi(B))_{B\in\mathcal F}$ is independent.