Finite-window homogeneous Poisson construction and ordered event-time density

Put $\lambda =\theta T$ and let $Q=\mathrm{\ell }\mathrm{/}T$, a probability measure on $S$ by [A1] and [F2]. For each $n\ge 0$, use the finite product probability $Q^{\otimes n}$ on $S^n$; for $n=0$, $S^0$ is the singleton empty tuple with unit mass. Form the countable disjoint union $\mathrm{\Omega }={\coprod }_{n\ge 0}(\{n\}\times S^n)$ with the sigma-algebra whose sections in each component are product-measurable. Put $p_n=e^{-\lambda }{\lambda }^n\mathrm{/}n!$ and define $P(E)={\sum }_{n\ge 0}{p}_n{Q}^{\otimes n}(E_n)$. [F1] gives ${\sum }_n{p}_n=1$; [F4] makes $P$ a probability measure. On component $n$, set $\eta ={\sum }_{i=1}^n{\delta }_{x_i}$. Every evaluation $\eta (B)={\sum }_{i=1}^n{1}_B(x_i)$ is measurable, and each realization has at most $n$ atoms, so it is locally finite. The space $S$ is locally compact, second countable, and Hausdorff by the displayed neighborhood and base checks; hence [F8] makes $\eta$ a point process.

Let $B_1,\dots ,B_r$ be disjoint Borel subsets of $S$, put $B_0=S\setminus {\bigcup }_{j=1}^r{B}_j$, and set $q_j=Q(B_j)$. Conditional on component $n$, a count vector $(k_0,\dots ,k_r)$ with sum $n$ occurs for $n!\mathrm{/}{\prod }_{j=0}^r{k}_j!$ coordinate assignments; each assignment has probability ${\prod }_{j=0}^r{q}_j^{k_j}$ by [F3]. Multiplying by $p_n$ and summing over $k_0\ge 0$ gives, for fixed $k_1,\dots ,k_r\in {\mathbb{N}}_0$, $P\{\eta (B_j)=k_j,1\le j\le r\}={\sum }_{k_0=0}^{\mathrm{\infty }}{e}^{-\lambda }\frac{(\lambda q_0{)}^{k_0}}{k_0!}{\prod }_{j=1}^r\frac{(\lambda q_j{)}^{k_j}}{k_j!}=e^{-\lambda }{\prod }_{j=1}^r\frac{(\lambda q_j{)}^{k_j}}{k_j!}{\sum }_{k_0=0}^{\mathrm{\infty }}\frac{(\lambda q_0{)}^{k_0}}{k_0!}={\prod }_{j=1}^r\left[e^{-\lambda q_j}\frac{(\lambda q_j{)}^{k_j}}{k_j!}\right]\mathrm{.}$ Here $0^0=1$ for zero-mass cells, and $q_0=1-{\sum }_{j=1}^r{q}_j$. The exponential series evaluates the sum, and $\lambda q_j=\theta \mathrm{\ell }(B_j)$. Thus the joint count mass factors into the independent Poisson masses with means ${\nu }_{\theta }(B_j)$; [F5] makes $\eta$ a Poisson point process of intensity ${\nu }_{\theta }$.

Conditional on component $n$, the coordinate vector has density $T^{-n}$ on $S^n$. For each pair $i<j$, the diagonal $\{x_i=x_j\}$ is Borel; its section in coordinate $x_j$ is a singleton and has Lebesgue measure zero by [F2], so [F7] gives product measure zero. The finite union of pairwise diagonals is null. Coordinate permutations preserve the product-uniform law: they preserve the rectangle formula in [F3], and uniqueness there extends that equality from rectangles to the product sigma-algebra. Thus the $n!$ ordering regions have equal measure and partition the cube outside the diagonals. More precisely, for every Borel $C\subseteq {\mathrm{\Delta }}_n$, the preimage of $C$ under the Borel sorting map, off the diagonals, is the disjoint union of the $n!$ coordinate permutations of $C$. Each has product-uniform probability $T^{-n}{\mathrm{\ell }}^n(C)$, so sorting gives conditional density $n!\mathrm{/}{T}^n$ on the ordered simplex. Multiplying by $p_n=e^{-\theta T}(\theta T{)}^n\mathrm{/}n!$ gives joint density $e^{-\theta T}{\theta }^n$. The endpoint set where any coordinate equals $T$ has product measure zero because $Q(\{T\})=0$ by [F2] and [F3]; thus the same density applies to the stated simplex with $t_n\le T$. For $n=0$, the construction gives empty-pattern mass $p_0=e^{-\theta T}$. Each component has tie probability zero, and their countable mixture also has tie probability zero. This proves the event-time density and simplicity claims.