Finite-intensity Poisson-process existence and mixed-binomial law

If $\lambda =0$, take a one-point probability space and let $\eta$ be the zero measure. For every Borel $B$, $\nu (B)=0$, so $\eta (B)=0$ has the required Poisson law and every finite vector of counts is independent.

Now suppose $\lambda >0$ and put $Q=\nu \mathrm{/}\lambda$. Let $\mathrm{\Omega }={\coprod }_{n\ge 0}(\{n\}\times S^n)$ with the sigma-algebra whose sections in every $S^n$ are product-measurable. For each $n$, the section map defines a probability measure $Q_n(A)=Q^{\otimes n}(A_n)$ on $\mathrm{\Omega }$, by [F3]. The countable sum $P={\sum }_{n\ge 0}{e}^{-\lambda }{\lambda }^n{Q}_n\mathrm{/}n!$ is a probability measure by [F2] and [F4].

On the component $(n,x_1,\dots ,x_n)$ define $\eta ={\sum }_{i=1}^n{\delta }_{x_i}$. For each Borel $B$, $\eta (B)={\sum }_{i=1}^n{1}_B(x_i)$ is measurable on each component, so $\eta$ is a random locally finite counting measure.

Let $B_1,\dots ,B_m$ be pairwise-disjoint Borel sets and let $B_0=S\setminus {\bigcup }_{j=1}^m{B}_j$. Put $q_j=Q(B_j)$. Conditional on the component $S^n$, independence of the $n$ coordinates gives, for $k_0+\cdots +k_m=n$, $P\{\eta (B_j)=k_j,0\le j\le m\mid n\}=\frac{n!}{{\prod }_{j=0}^m{k}_j!}{\prod }_{j=0}^m{q}_j^{k_j}$, where $0^0=1$. This is the multinomial count of coordinate assignments to the partition $B_0,\dots ,B_m$.

Put $K={\sum }_{j=1}^m{k}_j$. Summing the conditional multinomial masses in step 1.4 over $k_0\ge 0$ gives $P\{\eta (B_j)=k_j,1\le j\le m\}=e^{-\lambda }{\prod }_{j=1}^m\frac{(\lambda q_j{)}^{k_j}}{k_j!}{\sum }_{k_0=0}^{\mathrm{\infty }}\frac{(\lambda q_0{)}^{k_0}}{k_0!}={\prod }_{j=1}^m{e}^{-\lambda q_j}\frac{(\lambda q_j{)}^{k_j}}{k_j!}={\prod }_{j=1}^m{e}^{-\nu (B_j)}\frac{\nu (B_j{)}^{k_j}}{k_j!}$, since the exponential series gives the sum as $e^{\lambda q_0}$ and $1-q_0={\sum }_{j=1}^m{q}_j$. Thus these counts are independent Poisson variables with respective means $\nu (B_j)$, proving existence with the required count laws.

For uniqueness in distribution, take any finite family of Borel sets and partition $S$ into the finitely many atoms generated by that family. Each original count is a sum of counts on these disjoint atoms. For either process, [F1] gives independent Poisson counts on those atoms with the same means $\nu (C)$, so the atom-count vectors have the same product law and every original finite evaluation vector has the same law. The point-process law characterization [F5] then gives equality of the laws.

Steps 1.1-3.1 prove the zero-mass case, construct the mixed-binomial law for positive finite mass, and show that every finite-intensity Poisson process has that law. No choice principle is used; only finite products and a specified countable sum occur.