Spatial summaries and Palm laws

Choose $x_0\in \mathrm{int}W$ and $r>0$ with the open ball $B(x_0,r)\subseteq W$. For $0<t<1$ put $W_t=x_0+(1-t)(W-x_0)$. If $w\in W$ and $\mathrm{\parallel }z\mathrm{\parallel }<tr$, then $x_0+z\mathrm{/}t\in B(x_0,r)\subseteq W$ and $(1-t)w+t(x_0+z\mathrm{/}t)=(1-t)w+t{x}_0+z\in W$ by convexity; thus every point of $W_t$ has an open ball of radius $tr$ contained in $W$, so $W_t\subseteq \mathrm{int}W$.

If $0<t^{\mathrm{\prime }}<t<1$, put $w^{\mathrm{\prime }}=\frac{1-t}{1-t^{\mathrm{\prime }}}w+\frac{t-t^{\mathrm{\prime }}}{1-t^{\mathrm{\prime }}}{x}_0$. The two coefficients are nonnegative and sum to one, so $w^{\mathrm{\prime }}\in W$ by convexity, and $x_0+(1-t)(w-x_0)=x_0+(1-t^{\mathrm{\prime }})(w^{\mathrm{\prime }}-x_0)\mathrm{.}$ Thus $W_t\subseteq W_{t^{\mathrm{\prime }}}$. Moreover, every $y\in \mathrm{int}W$ lies in some $W_t$: for $y\mathrm{\ne }{x}_0$, choose $s>0$ small enough that $w=x_0+(1+s)(y-x_0)\in W$, and then $y=x_0+(1-t)(w-x_0)$ for $t=s\mathrm{/}(1+s)$; $x_0$ lies in every $W_t$. Hence for any sequence $t_n\downarrow 0$, $W_{t_n}\uparrow \mathrm{int}W$.

By [F2] and [F3], $\mathrm{\mid }{W}_{t_n}\mathrm{\mid }=(1-t_n{)}^d\mathrm{\mid }W\mathrm{\mid }$ and the sets increase to $\mathrm{int}W$, so [F3] gives $\mathrm{\mid }\mathrm{int}W\mathrm{\mid }=\mathrm{\mid }W\mathrm{\mid }$. Thus the compact boundary $\mathrm{\partial }W=W\setminus \mathrm{int}W$ is a null set. Since $W$ contains a nonempty open ball and is bounded, [F1] gives $0<\mathrm{\mid }W\mathrm{\mid }<\mathrm{\infty }$.

Given $ϵ>0$, choose $t>0$ so that $\mathrm{\mid }W\mathrm{\mid }-\mathrm{\mid }{W}_t\mathrm{\mid }=(1-(1-t{)}^d)\mathrm{\mid }W\mathrm{\mid }<ϵ$. Choose a grid of half-open cubes of side $\delta <tr\mathrm{/}\sqrt{d}$. Each cube that intersects $W_t$ is contained in $W$, because its points lie within distance at most $\sqrt{d}\delta <tr$ of a point whose radius-$tr$ ball lies in $W$. Only finitely many grid cubes intersect bounded $W_t$; their union $R$ is disjoint, finite, contains $W_t$, and is contained in $W$. Therefore $\mathrm{\mid }W\setminus R\mathrm{\mid }\le \mathrm{\mid }W\setminus W_t\mathrm{\mid }<ϵ$.

Fix $ϵ>0$ and choose $R\subseteq W$ as in step 4.1, a finite union of disjoint rectangles. Translation invariance and the symmetric-difference inclusion give $\mathrm{\mid }(W+h)\mathrm{△}W\mathrm{\mid }\le 2\mathrm{\mid }W\setminus R\mathrm{\mid }+\mathrm{\mid }(R+h)\mathrm{△}R\mathrm{\mid }\mathrm{.}$ For each half-open rectangle, its overlap with a translate by $h$ has volume tending to the rectangle's volume as $h\to 0$, by the box-volume formula; the same limit holds for their finite union. Thus ${\mathrm{lim\hspace{0.17em}sup}}_{h\to 0}\mathrm{\mid }(W+h)\mathrm{△}W\mathrm{\mid }\le 2ϵ$, and arbitrariness of $ϵ$ gives convergence to zero.

For $a>0$, [F2] gives $\mathrm{\mid }aW\mathrm{\mid }=a^d\mathrm{\mid }W\mathrm{\mid }$ and $\mathrm{\mid }(aW+x)\mathrm{△}aW\mathrm{\mid }=a^d\mathrm{\mid }(W+x\mathrm{/}a)\mathrm{△}W\mathrm{\mid }$. Dividing and applying step 5.1 with $h=x\mathrm{/}a\to 0$ proves the asserted boundary ratio.

Steps 3.1, 4.1, 5.1, and 6.1 prove positive finite volume, the finite inner rectangle approximation, dilation scaling, and the translated-boundary limit. Countable Choice is used only through [F1]–[F2]; no AC or DC is used.

Fix a coordinate $i$ and either face $F$ of $[0,1{]}^d$ in that coordinate. For each integer $n\ge 1$, translate the lower face by $j\mathrm{/}(n+1)e_i$, or the upper face by $-j\mathrm{/}(n+1)e_i$, for $j=1,\dots ,n$. These $n$ translates are pairwise disjoint compact subsets of $[0,1{]}^d$, each has measure $m(F)$ by translation invariance, and $m([0,1{]}^d)<\mathrm{\infty }$ by local finiteness. Hence $nm(F)\le m([0,1{]}^d)$ for every $n$, which forces $m(F)=0$. The boundary of $[0,1{]}^d$ is a finite union of such faces.

The symmetric difference of $Q$ and $(0,1{]}^d$ is contained in that boundary, so [F1] and step 1.1 give $m(Q)=m((0,1{]}^d)=c$.

If $c=0$, the integer translates $Q+k$, $k\in {\mathbb{Z}}^d$, are a countable disjoint partition of ${\mathbb{R}}^d$. Translation invariance gives $m(Q+k)=0$ for every $k$, and countable additivity gives $m({\mathbb{R}}^d)=0$. Monotonicity then gives $m(B)=0$ for every Borel $B$.

If $c>0$, $m\mathrm{/}c$ is a translation-invariant Borel measure and, by step 2.1, $(m\mathrm{/}c)((0,1{]}^d)=1$. Applying [F2] gives $m\mathrm{/}c={\lambda }_d$, hence $m(B)=c{\lambda }_d(B)$ for every Borel $B$.

The cases $c=0$ and $c>0$ exhaust the nonnegative value $m(Q)$, proving the statement. The proof uses Countable Choice only through [F2]; no stronger choice principle is used.

First suppose $\mathbb{E}{X}_1=0$ and $\mathrm{\mid }{X}_n\mathrm{\mid }\le 1$ almost surely for each $n$. Put $A_n=n^{-1}{\sum }_{j=1}^n{X}_j$ and $L={\mathrm{lim\hspace{0.17em}sup}}_n{A}_n$. Then $-1\le L\le 1$, and for every fixed $p\ge 1$ the averages $m^{-1}{\sum }_{j=p}^{p+m-1}{X}_j$ have the same limsup $L$: their difference from $A_m$ is at most $2(p-1)\mathrm{/}m$, which tends to zero.

Fix $0<ϵ<1\mathrm{/}4$. For each $p\ge 1$ let $T_p$ be the least positive integer $m$ for which $m^{-1}{\sum }_{j=p}^{p+m-1}{X}_j>L-ϵ$. Such an $m$ exists by the definition of limsup; [F1] and shift invariance show that the $T_p$ all have the same distribution. Choose an integer $k$ so that $\mathbb{P}(T_1>k)<ϵ$, and set $X_p^{\ast }=X_p+{\mathbf{21}}_{\{T_p>k\}}$.

For every $p$, $X_p^{\ast }\ge -1$ and $\mathbb{E}{X}_p^{\ast }=2\mathbb{P}(T_1>k)<2ϵ$. Define $T_p^{\ast }$ as the first $m\ge 1$ for which the average of $X_p^{\ast },\dots ,X_{p+m-1}^{\ast }$ exceeds $L-ϵ$. If $T_p\le k$, the same average of the larger sequence $X_j^{\ast }$ at length $T_p$ exceeds $L-ϵ$; if $T_p>k$, then $X_p^{\ast }=X_p+2\ge 1>L-ϵ$. Thus $1\le T_p^{\ast }\le k$ in either case.

Starting from $M_0=0$, recursively set $M_{q+1}=M_q+T_{M_q+1}^{\ast }$. Each increment is between $1$ and $k$, and the average of $X_j^{\ast }$ on each block $(M_q,M_{q+1}]$ exceeds $L-ϵ$. Therefore $S_{M_q}^{\ast }:={\sum }_{j=1}^{M_q}{X}_j^{\ast }>(L-ϵ)M_q$.

For $n\ge k$, take the last $M_q\le n$; then $0\le n-M_q<k$. The remaining terms are at least $-1$, so $S_n^{\ast }>S_{M_q}^{\ast }-k>(L-ϵ)M_q-k\ge (L-ϵ)n-2k$, using $L-ϵ\le 1$. Hence $n^{-1}{S}_n^{\ast }>L-ϵ-2k\mathrm{/}n$.

Stationarity gives $\mathbb{E}[n^{-1}{S}_n^{\ast }]=\mathbb{E}{X}_1^{\ast }<2ϵ$. Taking expectations in step 5.1 and then letting $n\to \mathrm{\infty }$ gives $\mathbb{E}L\le 3ϵ$. Since this holds for every such $ϵ$, $\mathbb{E}L\le 0$. Applying the same argument to $(-X_n)$ shows $\mathbb{E}[{\mathrm{lim\hspace{0.17em}inf}}_n{A}_n]\ge 0$. The bounded nonnegative difference ${\mathrm{lim\hspace{0.17em}sup}}_n{A}_n-{\mathrm{lim\hspace{0.17em}inf}}_n{A}_n$ therefore has expectation at most zero, so it vanishes almost surely. Thus $A_n$ converges almost surely; [F2] gives convergence in $L^1$.

If instead $\mathrm{\mid }{X}_n\mathrm{\mid }\le C<\mathrm{\infty }$ without centering, the case $C=0$ is immediate; for $C>0$ apply the preceding result to $(X_n-\mathbb{E}{X}_1)\mathrm{/}(2C)$, whose terms are centered and have absolute value at most $1$. Rescaling and adding $\mathbb{E}{X}_1$ proves the bounded case.

Suppose $X_n\ge 0$ and is integrable but not necessarily bounded. For $r\ge 1$ set $X_{n,r}=X_n\wedge r$ and $A_{n,r}=n^{-1}{\sum }_{j=1}^n{X}_{j,r}$. Define the limit canonically by $Y_r:={\mathrm{lim\hspace{0.17em}sup}}_{n\to \mathrm{\infty }}{A}_{n,r}$; it is measurable by [F1]. By step 7.1, the averages converge almost surely and in $L^1$ to this integrable $Y_r$. On the common probability-one event for these countably many convergences, $Y_r\le Y_{r+1}$; also $Y_r\ge 0$ and $\mathbb{E}{Y}_r=\mathbb{E}{X}_{1,r}$.

Let $Y={\lim }_{r\to \mathrm{\infty }}{Y}_r$. By [F3], $Y$ is integrable and $\mathbb{E}Y=\mathbb{E}{X}_1$. Given $ϵ>0$, choose $r$ with $\mathbb{E}(X_1-X_{1,r})<ϵ$; then also $\mathbb{E}(Y-Y_r)<ϵ$. For this fixed $r$, stationarity and nonnegativity give $\mathbb{E}\mathrm{\mid }{A}_n-A_{n,r}\mathrm{\mid }=\mathbb{E}(X_1-X_{1,r})<ϵ$, while $\mathbb{E}\mathrm{\mid }{A}_{n,r}-Y_r\mathrm{\mid }\to 0$. The triangle inequality proves ${\mathrm{lim\hspace{0.17em}sup}}_n\mathbb{E}\mathrm{\mid }{A}_n-Y\mathrm{\mid }\le 2ϵ$, and arbitrariness of $ϵ$ gives $L^1$ convergence.

For a signed integrable stationary sequence, apply the nonnegative case in steps 8.1–9.1 separately to $X_n^{+}=\max (X_n,0)$ and $X_n^{-}=\max (-X_n,0)$. Both are stationary nonnegative integrable sequences, and subtracting their $L^1$ limits gives an integrable $L^1$ limit for the averages of $X_n$. The entire argument used no choice principle.

Fix a Borel set $B\subseteq {\mathbb{R}}^d$. On parameter space ${\mathbb{R}}^d\times {\mathcal{N}}_{\mathrm{l}\mathrm{f}}({\mathbb{R}}^d)$ define the evaluation kernel $K((x,\mu ),C)=\mu (C)$. Its evaluations are measurable by [F1]; the fixed compact exhaustion $[-n,n{]}^d$ has finite $K((x,\mu ),\cdot )$-mass for every $(x,\mu )$ by local finiteness. The function $((x,\mu ),y)\mapsto {\mathbf{1}}_B(y-x)$ is product measurable because subtraction is continuous and $B$ is Borel.

Applying [F2] to this kernel gives measurability of $(x,\mu )⟼\int {\mathbf{1}}_B(y-x)\mu (dy)=\mu (B+x)=({\theta }_x\mu )(B)\mathrm{.}$ Since this holds for every Borel $B$ and evaluations generate the target sigma-algebra [F1], $(x,\mu )\mapsto {\theta }_x\mu$ is measurable.

The argument applies to every Borel evaluation and uses no selection; the AC and Countable Choice tags record only the carried supplier conventions [A1].

First assume $\lambda :=\nu ({\mathbb{R}}^d)<\mathrm{\infty }$. If $\lambda =0$, then $\eta ({\mathbb{R}}^d)=0$ almost surely by [F1], so both identities are zero. If $\lambda >0$, put $Q_{\nu }=\nu \mathrm{/}\lambda$ and $p_n=e^{-\lambda }{\lambda }^n\mathrm{/}n!$. For each $n$, let $P_n$ be the law of ${\sum }_{j=1}^n{\delta }_{X_j}$ when $(X_1,\dots ,X_n)$ has product law $Q_{\nu }^{\otimes n}$, with $P_0$ the point mass at the zero measure. The countable weighted sum $P^{\ast }={\sum }_{n\ge 0}{p}_n{P}_n$ is a probability measure by the normalized Poisson masses [F3] and the countable weighted-sum result [F7]; evaluation maps of the finite Dirac sum are measurable by [F2].

Now let $\nu$ be sigma-finite and locally finite. Partition ${\mathbb{R}}^d$ into the explicit bounded Borel cells $A_1=[-1,1{)}^d$ and $A_n=[-n,n{)}^d\setminus [-(n-1),n-1{)}^d$ for $n\ge 2$. Let ${\eta }_i$ be the restriction of $\eta$ to $A_i$ and ${\eta }_{-i}$ its restriction to $A_i^c$. Each ${\eta }_i$ is a finite-intensity Poisson process. The two random configurations ${\eta }_i$ and ${\eta }_{-i}$ are independent: finite evaluation cylinders refine to disjoint Borel pieces in $A_i$ and $A_i^c$, whose counts are independent by [F1], and the generating rectangle pi-system extends this factorization to their evaluation sigma-algebras by [F4].

For disjoint Borel sets $B_1,\dots ,B_m$, add their complement as $B_0$ and put $q_j=Q_{\nu }(B_j)$. On the $n$th component the vector of counts has multinomial mass $n!{\prod }_{j=0}^m{q}_j^{k_j}\mathrm{/}{k}_j!$ when ${\sum }_j{k}_j=n$. Mixing by $p_n$ and summing $k_0$ gives $P^{\ast }(N(B_j)=k_j,1\le j\le m)={\prod }_{j=1}^m{e}^{-\nu (B_j)}\frac{\nu (B_j{)}^{k_j}}{k_j!},$ using the exponential series; hence $P^{\ast }$ has the defining independent Poisson counts [F1]. Any other Poisson process with finite $\nu$ has the same finite-dimensional evaluation laws: refine any finite family of Borel sets into its finitely many disjoint Boolean atoms, apply [F1] to their counts, and recover the original counts by summation. Evaluation cylinders form a generating pi-system for the configuration sigma-algebra [F2], so probability-measure uniqueness [F4] shows that $\eta$ has law $P^{\ast }$.

For any nonnegative product-measurable $f$, represent $\eta$ by the mixture law from steps 1.1–1.2. Pathwise the integral against its finite counting measure is a sum over occurrences, so Tonelli [F6] gives $\mathbb{E}\int f(x,\eta )\eta (dx)={\sum }_{n\ge 1}{p}_n{\int }_{({\mathbb{R}}^d{)}^n}{\sum }_{i=1}^{n}f(x_i,{\sum }_{j=1}^n{\delta }_{x_j})Q_{\nu }^{\otimes n}(d\mathbf{x})\mathrm{.}$ Coordinate-permutation invariance of $Q_{\nu }^{\otimes n}$ makes the $n$ summands equal. Reindexing with $m=n-1$ and $n{p}_n=\lambda p_{n-1}$ changes this to $\lambda {\sum }_{m\ge 0}{p}_m{\int }_{({\mathbb{R}}^d{)}^{m+1}}f(x_1,{\delta }_{x_1}+{\sum }_{j=2}^{m+1}{\delta }_{x_j})Q_{\nu }^{\otimes (m+1)}(d\mathbf{x})\mathrm{.}$ Integrating first in $x_1$ and using $\nu =\lambda Q_{\nu }$, the inner mixture over $m$ remaining points is $\mathbb{E}[f(x,\eta +{\delta }_x)]$ by steps 1.1–1.2. This proves the add-one identity for finite total intensity.

Fix $i$. Independence makes the joint law of $({\eta }_i,{\eta }_{-i})$ the product of their laws, as it agrees with that product on measurable rectangles [F4]. For each fixed background configuration $\mu$, the function $f_{\mu }(x,\zeta )=f(x,\zeta +\mu )$ is product-measurable by the addition evaluation formula in [F5]. Apply the finite-intensity add-one identity of step 3.1 to ${\eta }_i$ and $f_{\mu }$, then integrate over $\mu ={\eta }_{-i}$; this gives $\mathbb{E}\int f(x,\eta ){\eta }_i(dx)={\int }_{A_i}\mathbb{E}[f(x,\eta +{\delta }_x)]\nu (dx)\mathrm{.}$ Summing over $i$ and using [F6] proves the add-one identity for $\nu$.

Let $g$ be the nonnegative integrand in the deletion formula, and put $D=\{(x,\mu ):\mu (\{x\})\ge 1\}$. Define $H(x,\mu )=g(x,\mu -{\delta }_x)$ on $D$ and $H=0$ off $D$; [F5] makes $H$ measurable. Apply the add-one identity of step 4.1 to $H$. At every occurrence counted by $\eta (dx)$, $(x,\eta )\in D$ and $H(x,\eta )=g(x,\eta -{\delta }_x)$; for every $x$, $(x,\eta +{\delta }_x)\in D$ and $H(x,\eta +{\delta }_x)=g(x,\eta )$. This gives the deletion identity. Since $\eta -{\delta }_x$ removes one unit mass, all other co-located occurrences remain.

Steps 1.1–4.1 establish both forms, including the zero-intensity boundary case and multiplicity convention. AC and Countable Choice [A1]–[A2] are inherited supplier assumptions; the local Mecke derivation itself uses no selection, and no DC is used.

By [F1], $\nu (B)=\mathbb{E}[\eta (B)]$ is a Borel measure, and $\nu (Q)=\rho$.

Let $B$ be bounded Borel. Choose an integer $n$ with $B\subseteq [-n,n{)}^d$. This cube is a finite disjoint union of translates $Q+k$, $k\in \{-n,\dots ,n-1{\}}^d$. Stationarity [F2] gives $\nu (Q+k)=\mathbb{E}[\eta (Q+k)]=\rho$ for every such $k$. Monotonicity and finite additivity therefore give $\nu (B)\le (2n{)}^d\rho <\mathrm{\infty }$. Hence $\nu$ is locally finite.

For any Borel $B$ and translation $x$, the definition of recentering gives $\nu (B+x)=\mathbb{E}[\eta (B+x)]=\mathbb{E}[({\theta }_x\eta )(B)]=\mathbb{E}[\eta (B)]=\nu (B),$ where the third equality is stationarity [F2]. Thus $\nu$ is translation invariant.

Apply [F3] to the locally finite translation-invariant measure $\nu$. Since $\nu (Q)=\rho$ by step 1.1, it follows that $\nu (B)=\rho {\lambda }_d(B)$ for every Borel $B$. This is the claimed mean-measure formula.

For each configuration event $A$ and Borel set $B\subseteq {\mathbb{R}}^d$, define $C_A(B):=\mathbb{E}\int {\mathbf{1}}_B(x){\mathbf{1}}_A({\theta }_x\eta )\eta (dx)\mathrm{.}$ The integrand is measurable by [F3], so this is defined. For disjoint $B_j$, the indicators of their finite unions increase to the indicator of their union; monotone convergence [F5] and countable additivity of each $\eta$ give $C_A({\bigcup }_j{B}_j)={\sum }_j{C}_A(B_j)$. Also $0\le C_A(B)\le \mathbb{E}\eta (B)=\rho {\lambda }_d(B)$ by [F2], so $C_A$ is locally finite.

For any $y\in {\mathbb{R}}^d$, recentering has the flow property ${\theta }_{x+y}={\theta }_x{\theta }_y$. Changing the location variable in the integral gives $C_A(B+y)=\mathbb{E}\int {\mathbf{1}}_B(u){\mathbf{1}}_A({\theta }_u({\theta }_y\eta ))({\theta }_y\eta )(du)=C_A(B),$ where the last equality uses stationarity of $\eta$. Thus $C_A$ is translation invariant.

By [F4] there is a constant $c_A=C_A(Q)$ such that $C_A(B)=c_A{\lambda }_d(B)$ for every Borel $B$. For disjoint configuration events $A_j$, monotone convergence in the indicator sum gives $c_{{\cup }_j{A}_j}={\sum }_j{c}_{A_j}$. Also $c_{\mathcal{N}}=\mathbb{E}\eta (Q)=\rho$ and $c_{\mathrm{\varnothing }}=0$. Hence $P^0(A):=c_A\mathrm{/}\rho$ is a probability measure on the configuration sigma-algebra.

For Borel $B$ and configuration event $A$, the definition gives $\mathbb{E}\int {\mathbf{1}}_B(x){\mathbf{1}}_A({\theta }_x\eta )\eta (dx)=C_A(B)=\rho {\lambda }_d(B)P^0(A)\mathrm{.}$ Define the Campbell measure $\mathrm{\Pi }(E)=\mathbb{E}\int {\mathbf{1}}_E(x,{\theta }_x\eta )\eta (dx)$ on the product space. Joint measurability [F3] and monotone convergence [F5] make $\mathrm{\Pi }$ a measure. It is sigma-finite since $\mathrm{\Pi }(Q_n\times \mathcal{N})=\mathbb{E}\eta (Q_n)=\rho \mathrm{\mid }{Q}_n\mathrm{\mid }<\mathrm{\infty }$ for the increasing bounded cubes $Q_n=[-n,n{)}^d$. The product measure $\rho ({\lambda }_d\otimes P^0)$ is sigma-finite by [F6] and agrees with $\mathrm{\Pi }$ on rectangles by the displayed identity; uniqueness on the generating rectangle pi-system therefore gives equality of these measures. Integrating any nonnegative product-measurable $h$ against them proves the refined Campbell identity.

If $P_1^0$ is another probability law satisfying the refined identity, use $h(x,\mu )={\mathbf{1}}_Q(x){\mathbf{1}}_A(\mu )$ to obtain $\rho P_1^0(A)=C_A(Q)=\rho P^0(A)$ for every configuration event $A$; since $\rho >0$, $P_1^0=P^0$.

Let $A_0=\{\mu :\mu (\{0\})=0\}$. For every counted occurrence $x$ of $\eta$, the recentered configuration ${\theta }_x\eta$ has an occurrence at $0$, so the left side of the refined identity with $h(x,\mu )={\mathbf{1}}_Q(x){\mathbf{1}}_{A_0}(\mu )$ is zero. Its right side is $\rho P^0(A_0)$ because $\mathrm{\mid }Q\mathrm{\mid }=1$. Hence $P^0(A_0)=0$: the full Palm law retains at least the distinguished occurrence at the origin, even when other occurrences coincide there.

The constructed law satisfies the defining identity by step 4.1, is unique by step 5.1, and has the origin-atom property by step 5.2. AC and Countable Choice [A1]–[A2] are inherited supplier assumptions; the construction itself uses no choice beyond the Lebesgue-measure supplier.