Stationarity and intensity of a periodized empirical field

By [F1], $R_{L,\omega }$ is the push-forward of normalized Lebesgue probability on ${\mathrm{\Lambda }}_L$, so it is a probability measure. For a bounded measurable local $f$ and $y\in {\mathbb{R}}^d$, $R_{L,\omega }(f\circ T_y)=L^{-d}{\int }_{{\mathrm{\Lambda }}_L}f(T_{x+y}{\omega }^{(L)})dx\mathrm{.}$ The integrand is $L{\mathbb{Z}}^d$-periodic in $x$. The translated cube ${\mathrm{\Lambda }}_L+y$ is another fundamental region for this lattice; partitioning it and ${\mathrm{\Lambda }}_L$ into their finitely many translated intersections and using translation invariance of Lebesgue measure gives equal integrals. Hence $R_{L,\omega }(f\circ T_y)=R_{L,\omega }(f)$ for every $y$, so $R_{L,\omega }$ is stationary.

For bounded Borel $A$, Tonelli's theorem and the periodic continuation give $R_{L,\omega }(N_A)=L^{-d}{\int }_{{\mathrm{\Lambda }}_L}{\sum }_{q\in \omega }{\sum }_{k\in {\mathbb{Z}}^d}{\mathbf{1}}_A(q+Lk+x)dx\mathrm{.}$ For each fixed $q\in \omega$, the sets ${\mathrm{\Lambda }}_L+Lk$ partition ${\mathbb{R}}^d$ up to null boundaries. Translating variables in each integral therefore makes the inner sum of integrals equal to $\mathrm{\mid }A\mathrm{\mid }$. Summing over the $N$ representatives gives $R_{L,\omega }(N_A)=N\mathrm{\mid }A\mathrm{\mid }\mathrm{/}{L}^d$. Taking $A=[0,1{)}^d$ proves $\rho (R_{L,\omega })=N\mathrm{/}{L}^d$ and finite intensity.

The barycenter ${\bar{C}}_L$ is a probability mixture of the stationary laws $R_{L,\omega }$, so it is stationary. By step 2.1 and Tonelli, $\rho ({\bar{C}}_L)=E_{C_L}[R_{L,\omega }(N_{[0,1{)}^d})]=E_{C_L}N\mathrm{/}{L}^d<\mathrm{\infty }\mathrm{.}$ Thus it belongs to the finite-intensity stationary domain of ${\tau }_L$.