Let $(E,\cal E,\mu)$ be a $\sigma$-finite measure space A poisson random measure with intensity $\mu$ is a map $M:\Omega\times\cal E \rightarrow Z^{+}$ Denote $E^{*}$ the set of integer-valued measures on $\cal E$ and define
$X:E^{*} \times \cal E \rightarrow Z^{+}$
$X_{A}:E^{*} \rightarrow Z^{+}$, $A \in \cal E$
Theorem 7.1.3 There exist a unique probability measure $\mu^{*}$ on $(E^{*},\cal E^{*})$ such that $X$ is Poisson random measure with intensity $\mu$
The proof for uniqueness said
For disjoint sets $A_{1},\ldots,A_{k} \in \cal E$ and $n_{1},\ldots,n_{k} \in Z^{+}$ set $A^{*}=\{m \in E^{*}:m(A_{1})=n_{1},\ldots,m(A_{k})=n_{k}\}$
such of sets $A^{*}$ is a $\pi$-system generating $\cal E^{*}$
How we to verify this????
The attached website is the lecture
http://www.ressources-actuarielles.net/EXT/ISFA/1226.nsf/0/6021110392b6ba43c1256f6a002d5f33/$FILE/AP7.pdf
this doesn't seem right to me. Consider $E = \{ 1 , 2 , 3 \}$ with its counting measure and subsets $\sigma$-algebra, we can consider the sets :
$$A_1 = \{ m, m( \{ 1, 2 \} ) = 2 \}$$ $$A_2 = \{ m, m( \{ 2, 3 \} ) = 2 \}$$
It seems to me that their intersection is not in $A^*$. This is not too hard to work out, and based on the fact that their intersection contains both the measures $m$ and $m'$ :
$$m( 1 ) = m( 2 ) = m( 3 ) = 1$$ $$m'( 1 ) = m'( 3 ) = 0, m'( 2 ) = 2$$
Which proves that the decomposition of $A_1 \cap A_2$ cannot contain conditions on singletons and on the whole space since $m$ and $m'$ disagree on all of them. So the decomposition may only involve doubles, but it is easy to check that the intersection is neither equal to $\{m, m(\{ 1, 2 \}) = 2\}$ or $\{m, m(\{ 2, 3 \}) = 2\}$.