Spatial ergodic limit for counts in dilates of a convex window

Work on the canonical configuration probability space with law $P_{\eta }$, and put $f(\mu )=\mu (Q)$. By [F2], each $T_x$ is measurable; by [F1], stationarity makes it measure preserving and $T_{x+y}=T_x{T}_y$. By [F3], $f\in L^1(P_{\eta })$ and $\mathbb{E}f=\rho$.

For $j=1,\dots ,d$ let $T_j=T_{e_j}$, where $e_j$ is the $j$th coordinate vector. For any $g\in L^1(P_{\eta })$, the sequence $(g\circ T_j^k{)}_{k\ge 0}$ is stationary: shifting a finite vector by $m$ composes it with the measure-preserving map $T_j^m$. By [F4] its averages $A_n^{(j)}g=n^{-1}{\sum }_{k=0}^{n-1}g\circ T_j^k$ converge in $L^1$; denote their limit by $P_{j}g$. The averages are $L^1$ contractions, so limits preserve linearity and satisfy $\mathrm{\parallel }{P}_{j}g-P_{j}h{\mathrm{\parallel }}_1\le \mathrm{\parallel }g-h{\mathrm{\parallel }}_1$. Also $P_{j}g$ is $T_j$-invariant because $\mathrm{\parallel }{A}_n^{(j)}g\circ T_j-A_n^{(j)}g{\mathrm{\parallel }}_1\le 2\mathrm{\parallel }g{\mathrm{\parallel }}_1\mathrm{/}n$. Since the translations commute, $P_j$ preserves invariance under every $T_i$ with $i\mathrm{\ne }j$.

Define $g_0=f$ and $g_j=P_j{g}_{j-1}$. For positive integers $n_1,\dots ,n_d$, let $A_{n_1,\dots ,n_d}=A_{n_d}^{(d)}\cdots A_{n_1}^{(1)}f$. Contractivity and the definition of $P_j$ give inductively $\mathrm{\parallel }{A}_{n_1,\dots ,n_j}-g_j{\mathrm{\parallel }}_1\le \mathrm{\parallel }{A}_{n_1,\dots ,n_{j-1}}-g_{j-1}{\mathrm{\parallel }}_1+\mathrm{\parallel }{A}_{n_j}^{(j)}{g}_{j-1}-g_j{\mathrm{\parallel }}_1\mathrm{.}$ Thus $A_{n_1,\dots ,n_d}\to Z:=g_d$ in $L^1$ whenever every $n_j\to \mathrm{\infty }$. The variable $Z$ is nonnegative, invariant under every integer-coordinate translation, and $\mathbb{E}Z=\mathbb{E}f=\rho$.

The rectangular average with coordinate lengths $n_j$ and starting indices $m_j\in \mathbb{Z}$ is $A_{n_1,\dots ,n_d}f\circ T_m$, where $T_m={\theta }_{(m_1,\dots ,m_d)}$. Since $Z\circ T_m=Z$ almost surely and $T_m$ preserves $P_{\eta }$, its $L^1$ distance from $Z$ equals $\mathrm{\parallel }{A}_{n_1,\dots ,n_d}f-Z{\mathrm{\parallel }}_1$, which tends to zero.

Let $R={\prod }_{j=1}^d[b_j,c_j)$ be a bounded half-open rectangle with positive side lengths. In each coordinate, the complete unit cells contained in $a_{n}R$ form an integer interval with length $n_{j,n}$ satisfying $n_{j,n}\mathrm{/}{a}_n\to c_j-b_j$. Their product is a translated rectangular block of unit cells. By step 3.1, its count divided by ${\prod }_j{n}_{j,n}$ converges in $L^1$ to $Z$. The uncovered part of $a_{n}R$ has volume $a_n^d\mathrm{\mid }R\mathrm{\mid }-{\prod }_j{n}_{j,n}=o(a_n^d)$; by [F3] its expected count has the same order. Therefore $a_n^{-d}\eta (a_{n}R)\to \mathrm{\mid }R\mathrm{\mid }Z$ in $L^1$.

Fix $ϵ>0$ and choose by [F5] a finite disjoint union $R={\bigcup }_{q=1}^N{R}_q$ of half-open rectangles inside $W$ with $\mathrm{\mid }W\setminus R\mathrm{\mid }<ϵ$. Applying step 4.1 to each rectangle gives $a_n^{-d}\eta (a_{n}R)\to \mathrm{\mid }R\mathrm{\mid }Z$ in $L^1$. Also, by [F3], $\mathbb{E}[a_n^{-d}\eta (a_n(W\setminus R))]=\rho \mathrm{\mid }W\setminus R\mathrm{\mid }<\rho ϵ\mathrm{.}$ Since $Z\ge 0$ and $\mathbb{E}Z=\rho$, the triangle inequality yields ${\mathrm{lim\hspace{0.17em}sup}}_n\mathbb{E}\mid a_n^{-d}\eta (a_{n}W)-\mathrm{\mid }W\mathrm{\mid }Z\mid \le 2\rho ϵ\mathrm{.}$ Arbitrariness of $ϵ$ proves $a_n^{-d}\eta (a_{n}W)\to \mathrm{\mid }W\mathrm{\mid }Z$ in $L^1$.

By [F5], $\mathrm{\mid }{a}_{n}W\mathrm{\mid }=a_n^d\mathrm{\mid }W\mathrm{\mid }$ and $\mathrm{\mid }W\mathrm{\mid }>0$. Dividing the convergence in step 5.1 by $\mathrm{\mid }W\mathrm{\mid }$ gives $\eta (a_{n}W)\mathrm{/}\mathrm{\mid }{a}_{n}W\mathrm{\mid }\to Z$ in $L^1$.

For fixed $x\in {\mathbb{R}}^d$, stationarity and [F3] give ${\parallel \frac{\eta (a_{n}W+x)}{\mathrm{\mid }{a}_{n}W\mathrm{\mid }}-\frac{\eta (a_{n}W)}{\mathrm{\mid }{a}_{n}W\mathrm{\mid }}\parallel }_1\le \rho \frac{\mathrm{\mid }(a_{n}W+x)\mathrm{△}{a}_{n}W\mathrm{\mid }}{\mathrm{\mid }{a}_{n}W\mathrm{\mid }}⟶0$ by [F5]. Together with step 6.1 and measure preservation, this implies $Z\circ T_x=Z$ almost surely for every fixed $x$. Hence every rational threshold set $\{Z>q\}$ is in ${\mathcal{I}}_{\eta }$, so $Z$ is ${\mathcal{I}}_{\eta }$-measurable.

Let $A\in {\mathcal{I}}_{\eta }$ and define ${\nu }_A(B)=\mathbb{E}[{\mathbf{1}}_A\eta (B)]$. Monotone convergence makes ${\nu }_A$ a measure; it is locally finite since ${\nu }_A\le \nu =\rho {\lambda }_d$. For any $x$, stationarity and $A$'s invariance modulo null sets give ${\nu }_A(B+x)={\nu }_A(B)$: writing $\zeta ={\theta }_x\eta$, stationarity gives $\mathcal{L}(\zeta )=P_{\eta }$, while ${\mathbf{1}}_A(\eta )={\mathbf{1}}_A({\theta }_{-x}\zeta )={\mathbf{1}}_A(\zeta )$ almost surely by $A\in {\mathcal{I}}_{\eta }$. Thus $\mathbb{E}[{\mathbf{1}}_A\eta (B+x)]=\mathbb{E}[{\mathbf{1}}_A(\eta )({\theta }_x\eta )(B)]=\mathbb{E}[{\mathbf{1}}_A(\zeta )\zeta (B)]=\mathbb{E}[{\mathbf{1}}_A\eta (B)]$. By [F6], ${\nu }_A(B)={\nu }_A(Q)\mathrm{\mid }B\mathrm{\mid }=\mathbb{E}[{\mathbf{1}}_A\eta (Q)]\mathrm{\mid }B\mathrm{\mid }$. Therefore, for each $n$, $\mathbb{E}\left[{\mathbf{1}}_A\frac{\eta (a_{n}W)}{\mathrm{\mid }{a}_{n}W\mathrm{\mid }}\right]=\mathbb{E}[{\mathbf{1}}_A\eta (Q)]\mathrm{.}$ Passing to the $L^1$ limit in step 6.1 shows $\mathbb{E}[{\mathbf{1}}_{A}Z]=\mathbb{E}[{\mathbf{1}}_A\eta (Q)]$ for every $A\in {\mathcal{I}}_{\eta }$. By [F7], $Z$ is a version of $\mathbb{E}[\eta (Q)\mid {\mathcal{I}}_{\eta }]$.

If the translation action is ergodic, every event of ${\mathcal{I}}_{\eta }$ has probability zero or one. If the real ${\mathcal{I}}_{\eta }$-measurable variable $Z$ were not almost surely constant, there would be real $a<b$ with $P_{\eta }(Z\le a)>0$ and $P_{\eta }(Z\ge b)>0$; choosing a rational $q\in (a,b)$ would make $\{Z>q\}$ an invariant event of probability strictly between zero and one, a contradiction. Thus $Z$ is constant almost surely. Its expectation is $\rho$ by step 2.1, so $Z=\rho$ almost surely. The result is only an $L^1$ law of large numbers: it supplies neither a rate nor a CLT. AC and Countable Choice [A1]–[A2] are inherited supplier assumptions; the coordinate-sequence mean argument itself is choice-free.