A finite Poisson process with atomless intensity is simple

Put $\mathrm{\Lambda }=\nu ({\mathbb{R}}^d)$. If $\mathrm{\Lambda }=0$, [F2] says the process is the zero measure, which is simple by [F1]. Suppose $\mathrm{\Lambda }>0$ and set $Q=\nu \mathrm{/}\mathrm{\Lambda }$. Atomlessness of $\nu$ implies $Q(\{x\})=0$ for every $x\in {\mathbb{R}}^d$.

For $m\in {\mathbb{N}}_0$ and $k\in {\mathbb{Z}}^d$, put $C_{m,k}=2^{-m}(k+[0,1{)}^d)$. The set $H=\{\xi :\xi ({\mathbb{R}}^d)<\mathrm{\infty }\}\cap {\bigcup }_{m\ge 0}{\bigcap }_{k\in {\mathbb{Z}}^d}\{\xi :\xi (C_{m,k})\le 1\}$ is evaluation-measurable, since all cubes are Borel and the unions and intersections are countable. It is exactly the set of finite simple configurations: a repeated location forces a count at least two in its cube at every scale; conversely, a finite set of distinct locations has positive minimum pairwise distance when it has at least two points, so cubes of sufficiently small diameter contain at most one point. Counts zero and one satisfy the condition at every scale.

Fix $n\ge 2$ and a pair $i<j$ of coordinate indices. The diagonal is product-measurable by the preceding basis argument. Under $Q^{\otimes n}$, the event $x_i=x_j$ has measure $\int Q(\{x\})Q(dx)=0$ by [F3] and step 1.1, since every singleton has zero $Q$-mass by step 1.1. The other $n-2$ coordinates contribute total mass one. There are only finitely many pairs among $n$ coordinates, so with $Q^{\otimes n}$-probability one no two coordinates coincide.

In the mixed-binomial construction [F2], the image configuration belongs to $H$ with probability one on each component: step 2.1 treats $n\ge 2$, and counts $n=0,1$ are simple directly. Summing over the component probabilities gives mass one to $H$. Since $H$ is measurable by step 1.2 and every Poisson process with intensity $\nu$ has this push-forward law by [F2], $\mathbb{P}\{\eta \in H\}=1$. Thus $\eta$ is simple almost surely.