Poisson Reference Counts and Translation Invariance

Fix one unit cell $C_i$ and a finite Borel partition $B_0,\dots ,B_s$ of it. Put $v_j=\mathrm{\mid }{B}_j\mathrm{\mid }$, so ${\sum }_j{v}_j=1$. Conditional on the cell count $N=n$, the $n$ independent uniform points give category-count probability $n!{\prod }_j{v}_j^{n_j}\mathrm{/}{\prod }_j{n}_j!$ for ${\sum }_j{n}_j=n$. Multiplying by the Poisson count probability $e^{-\lambda }{\lambda }^n\mathrm{/}n!$ gives $e^{-\lambda }{\prod }_j(\lambda v_j{)}^{n_j}\mathrm{/}{n}_j!={\prod }_j{e}^{-\lambda v_j}(\lambda v_j{)}^{n_j}\mathrm{/}{n}_j!$, by [F3]. This factors into the Poisson probabilities of the disjoint cell pieces, including the complement $B_0$; hence their counts are independent Poisson variables with respective means $\lambda v_j$. Zero-volume pieces have count zero almost surely by the same formula.

Each bounded $A_j$ intersects only finitely many unit cells. Apply step 1.1 to the partition formed by the sets $A_j\cap C_i$ and their complement in each such cell. Independence within each cell and independence between cell configurations show that all component counts, indexed by both $j$ and $i$, are independent Poisson variables.

For fixed $j$, sum the component counts over its finitely many cells. The sum of independent Poisson variables of means ${\mu }_1,\dots ,{\mu }_q$ has mass at $n$ equal to $e^{-\sum {\mu }_i}(\sum {\mu }_i{)}^n\mathrm{/}n!$: expand $({\sum }_i{\mu }_i{)}^n$ by [F2], then use the product of the component masses. Thus $N_{A_j}$ is Poisson with mean $\lambda {\sum }_i\mathrm{\mid }{A}_j\cap C_i\mathrm{\mid }=\lambda \mathrm{\mid }{A}_j\mathrm{\mid }$. The groups of components for different $j$ are disjoint independent families, so their sums remain independent.

Put $\theta =\lambda \mathrm{\mid }A\mathrm{\mid }$. The count mass in step 3.1 gives, by nonnegative partial sums and [F5], $E_{Q_{\lambda }}{e}^{a{N}_A}={\sum }_{n=0}^{\mathrm{\infty }}{e}^{an}{e}^{-\theta }\frac{{\theta }^n}{n!}=e^{-\theta }{\sum }_{n=0}^{\mathrm{\infty }}\frac{(\theta e^a{)}^n}{n!}=\exp \text{\hspace{0.17em}⁣}(\theta (e^a-1))\mathrm{.}$ For $\theta =0$, the count is zero almost surely and both sides are one, so no ambiguous zero power is needed. This establishes the exponential-moment interface before entropy estimates use it.

Let $A_1,\dots ,A_r$ be pairwise disjoint and, for each $j$, consider a finite count-cylinder event in the restriction to $A_j$. Refine the finitely many counting sets in each event into their disjoint atoms inside $A_j$. Step 3.1 gives independent Poisson counts on all these atoms across every $j$; the event for each restricted configuration is a measurable function only of its own atom counts. Thus finite cylinder events factor across $j$. The restriction sigma-fields are generated by these cylinders, so repeated pi-lambda extension proves independence of the restrictions themselves.

For a finite family of bounded Borel sets $A_1,\dots ,A_s$, partition their union into the finitely many disjoint Boolean atoms formed by the sets and their complements. Every $N_{A_j}$ is a sum of the counts on these atoms. Translation preserves each atom's volume, so step 3.1 gives the same joint distribution of this whole count vector under $Q_{\lambda }$ and its translate. Thus the laws agree on every finite count cylinder. These cylinders generate the configuration sigma-algebra by [F4], and pi-lambda uniqueness identifies the laws. The one-set count law gives $E_{Q_{\lambda }}{N}_A=\lambda \mathrm{\mid }A\mathrm{\mid }$ for bounded $A$. For arbitrary Borel $A$, increase bounded cubes to all space and apply [F5] to the counting measures and Lebesgue measure to obtain the same extended identity. For disjoint Borel $A_n$, the pointwise counting identity $N_{{\cup }_n{A}_n}={\sum }_n{N}_{A_n}$ and [F5] show countable additivity of ${\mathrm{\Lambda }}_{Q_{\lambda }}$, while the empty set has expectation zero. Thus the defined intensity functional is a measure and equals $\lambda$ times Lebesgue measure.