Smooth compact tests determine phase-space probabilities

Fix a bounded Lipschitz function $f:Z\to \mathbb{R}$. Use ${\lambda }_d$ and $m_{{\mathbb{T}}^d}$ from [F5]. Choose the smooth cutoff $\chi$ of [F2] and set $\eta (u)=c\chi (2u)$ on ${\mathbb{R}}^d$, where $c=({\int }_{{\mathbb{R}}^d}\chi (2u)d{\lambda }_d(u){)}^{-1}$; its integral is positive and finite because $\chi =1$ on the unit ball and is supported in the radius-two ball. Thus $\eta$ is a nonnegative smooth probability density supported in the unit ball. For $0<ϵ<1\mathrm{/}4$, define ${\eta }_{ϵ}^{\mathbb{R}}(v)={ϵ}^{-d}\eta (v\mathrm{/}ϵ)$ and periodize it to ${\mathbb{T}}^d$ by ${\eta }_{ϵ}^{\mathbb{T}}(x)={\sum }_{n\in {\mathbb{Z}}^d}{ϵ}^{-d}\eta ((\tilde{x}+n)\mathrm{/}ϵ)$, where $\tilde{x}$ is any representative. The sum is locally finite and independent of the representative; the fundamental-domain formula for $m_{{\mathbb{T}}^d}$ in [F5], followed by the change of variables $u=(\tilde{x}+n)\mathrm{/}ϵ$, gives integral one, and its support satisfies $d_{{\mathbb{T}}^d}(x,0)\le ϵ$. Set $f_{ϵ}(x,v)={\int }_{{\mathbb{T}}^d}{\int }_{{\mathbb{R}}^d}{\eta }_{ϵ}^{\mathbb{T}}(x-y){\eta }_{ϵ}^{\mathbb{R}}(v-w)f(y,w)d{\lambda }_d(w)d{m}_{{\mathbb{T}}^d}(y)$. Differentiating the compactly supported smooth kernels under the integral shows that $f_{ϵ}$ is smooth, and their nonnegativity and unit integrals give $\mathrm{\parallel }{f}_{ϵ}{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$. On their support, $d_{{\mathbb{T}}^d}(x,y)\le ϵ$ and $\mathrm{\mid }v-w\mathrm{\mid }\le ϵ$, so $\mathrm{\parallel }{f}_{ϵ}-f{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2ϵ\mathrm{Lip}(f)$. For each $R>0$, ${\chi }_R(v)f_{ϵ}(x,v)$ is in $C_c^{\mathrm{\infty }}(Z)$, so its integrals agree by the hypothesis. Let $R\to \mathrm{\infty }$ and apply [F3] to each probability measure; then let $ϵ\downarrow 0$ and use uniform convergence. Thus $\mu$ and $\nu$ have equal integrals for every bounded Lipschitz $f$.

If $C\subseteq Z$ is a nonempty closed set, distance to $C$ is 1-Lipschitz by the triangle inequality. Hence $h_n(z)=\max (0,1-n{d}_Z(z,C))$ is bounded Lipschitz and decreases pointwise to ${\mathbf{1}}_C$. Step 1.1 gives equal integrals of $h_n$, and [F3] gives $\mu (C)=\nu (C)$. The empty closed set has mass zero for both. Let $\mathcal{D}:=\{E\in \mathcal{B}(Z):\mu (E)=\nu (E)\}$. Since $\mu$ and $\nu$ are probabilities, $Z\in \mathcal{D}$; equality of total mass makes $\mathcal{D}$ closed under complements in $Z$, and countable additivity makes it closed under countable disjoint unions. Thus $\mathcal{D}$ is a lambda-system. The $d_Z$ sum metric induces the product topology on $Z$, so its closed sets form a pi-system containing $Z$ and generating $\mathcal{B}(Z)$. Apply [F4] to this pi-system and $\mathcal{D}$ to get $\mu =\nu$.