A translated lattice exposes recentering, atomic pair measures, and one-occurrence deletion

For every Borel $B$, $u\mapsto {\eta }_u(B)={\sum }_{k\in \mathbb{Z}}{\mathbf{1}}_B(k+u)$ is measurable as a countable sum of Borel functions; every compact set meets only finitely many points of each ${\eta }_u$, so this is a locally finite random counting measure. For fixed $x$, put $t=x-\lfloor x\rfloor \in [0,1)$. The map $u\mapsto \{u-x\}$ is a translation on each of $[0,t)$ and $[t,1)$, with the two image intervals partitioning $[0,1)$; by [F2] it preserves uniform probability. Reindexing $k\in \mathbb{Z}$ gives ${\theta }_x{\eta }_u={\eta }_{\{u-x\}}$, so the process is stationary. Since exactly one $k+u$ lies in $Q=[0,1)$ for every $u\in [0,1)$, ${\eta }_U(Q)=1$ and $\rho =1$.

For every compact $C\subseteq \mathbb{R}$, choose an integer $m\ge 1$ with $C\subseteq [-m,m]$. If $u\in [0,1)$ and $k+u\in C$, then $-m-1\le k\le m$, so ${\eta }_u(C)\le 2m+2$ uniformly in $u$. Consequently $\mathbb{E}[{\eta }_U(C{)}^2]\le (2m+2{)}^2<\mathrm{\infty }$ and $\mathbb{E}[(2{\eta }_U)(C{)}^2]\le 4(2m+2{)}^2<\mathrm{\infty }$. Both lattice processes satisfy the local square-integrability premise in the second-order and K definitions.

For every configuration event $A$, the unit-cube Palm construction in [F3] gives $P^0(A)=\mathbb{E}{\int }_Q{\mathbf{1}}_A({\theta }_x{\eta }_U){\eta }_U(dx)\mathrm{.}$ There is exactly one occurrence $x=U$ in $Q$, and ${\theta }_U{\eta }_U={\sum }_{k\in \mathbb{Z}}{\delta }_k=:\mathrm{\Lambda }$. Therefore $P^0={\delta }_{\mathrm{\Lambda }}$. In particular, using the uncentered configuration in the indicator ${\mathbf{1}}_{\{\mu (\{0\})>0\}}$ would give $0$ on the left Campbell sum almost surely, whereas the Palm-side value is $1$; this is why recentering cannot be dropped.

For each selected lattice occurrence $X_i=k+U\in Q$, the displacements to all other occurrences are $\mathbb{Z}\setminus \{0\}$. Hence [F4] gives ${\alpha }_{\mathrm{r}\mathrm{e}\mathrm{d}}^{(2)}(D)=\mathrm{\#}(D\cap (\mathbb{Z}\setminus \{0\}))$ for bounded Borel $D$. In particular ${\alpha }_{\mathrm{r}\mathrm{e}\mathrm{d}}^{(2)}(\{1\})=1$ while $\mathrm{\mid }\{1\}\mathrm{\mid }=0$, so no measurable density $g$ can satisfy [F5].

Now put ${\tilde{\eta }}_U=2{\eta }_U$, so every lattice site has two occurrence indices and $\tilde{\rho }=2$. A selected occurrence has one other occurrence at the same site, so ${\alpha }_{\mathrm{r}\mathrm{e}\mathrm{d}}^{(2)}(\{0\})=2$. Equivalently, $P^0$ is concentrated on $2\mathrm{\Lambda }$, and $\mu (\{0\})=2$, so [F6] gives $2=\tilde{\rho }{\mathbb{E}}^0[(\mu -{\delta }_0)(\{0\})]=2\cdot 1$. Thus $K(0)=1\mathrm{/}2$; deleting both origin occurrences would incorrectly give zero.

The simple lattice has an atomic reduced measure and no pair-correlation density, while the doubled lattice verifies that factorial pairs distinguish occurrence indices and Palm deletion removes exactly one occurrence. The origin-centered convention and the closed-ball endpoint $r=0$ are both used explicitly. AC and Countable Choice [A1] are inherited assumptions only.