Proof of A Probability Measure on Euclidean Space Is Determined by the Integrals of Lipschitz Functions with Values in the Unit Interval

Each result cited is universally quantified over the data in its own statement. Let $X=(\mathbb{R}^{q},d_{E})$, a metric space by Euclidean Distance is a Metric on $\mathbb{R}^n$. By Probability Measures on Euclidean Space and Random Vectors: Standing Notation §spaces the $\sigma$-algebra $\mathcal{B}(\mathbb{R}^{q})$ is the Borel $\sigma$-algebra of $X$, so claim 1 of Borel Measurability and Bounded Integration on a Metric Space applies to it: every closed subset $F$ of $X$ belongs to $\mathcal{B}(\mathbb{R}^{q})$, and the family $\mathcal{C}$ of closed subsets of $X$ is a $\pi$-system whose generated $\sigma$-algebra is $\mathcal{B}(\mathbb{R}^{q})$. Since $\mu(\mathbb{R}^{q})=1=\nu(\mathbb{R}^{q})$, claim 1 of Uniqueness of Finite Measures on a Generating Pi-System and the Density of the Exponential Law reduces the assertion $\mu=\nu$ to

which we now prove. If $F=\varnothing$ both sides are $0$ (Measure, Measure Space, and Probability Measure). Let $F$ be nonempty and closed, write $\operatorname{dist}(x,F)$ for the distance from $x$ to $F$, the infimum of $\{d_{E}(x,a):a\in F\}$, and for $n\in\mathbb{N}$, read in $\mathbb{R}$ as in The Real Numbers: Standing Notation and Background §numbers, let

the larger of the two numbers, which exists by the trichotomy of the order of Ordered Field.

Step 1: $\phi_{n}$ is Lipschitz with values in $[0,1]$. Since $0\le\operatorname{dist}(x,F)$ (claim 1 of The Distance to a Set is Nonexpansive) and $0\le n$, one has $1-n\operatorname{dist}(x,F)\le1$ (claim 5 of Elementary Arithmetic in an Ordered Field and the compatibility of the order with addition, an axiom of Ordered Field), so $0\le\phi_{n}(x)\le1$. For real numbers $a,b$ one has $|\max\{0,a\}-\max\{0,b\}|\le|a-b|$: if $0\le a$ and $0\le b$ the left side is $|a-b|$; if $a\le0$ and $b\le0$ it is $|0-0|=0\le|a-b|$ (claim 1 of Properties of the Absolute Value in an Ordered Field); and if $b\le0\le a$ it is $|a|=a\le a-b=|a-b|$, by Absolute Value in an Ordered Field, claim 3 of Elementary Arithmetic in an Ordered Field applied to $b\le0$, and $0\le a-b$; the case $a\le0\le b$ is symmetric (claim 2 of Properties of the Absolute Value in an Ordered Field). Hence, for $x,y\in\mathbb{R}^{q}$,

by claim 4 of Properties of the Absolute Value in an Ordered Field (with $|n|=n$ by Absolute Value in an Ordered Field) and claim 4 of The Distance to a Set is Nonexpansive. Thus $\phi_{n}$ is Lipschitz with constant $n$, and by hypothesis

Step 2: $\phi_{n}(x)\to\mathbf{1}_{F}(x)$ for every $x$. Let $x\in F$. Then $\operatorname{dist}(x,F)\le d_{E}(x,x)=0$ by claim 2 of The Distance to a Set is Nonexpansive with $a=x$, so $\operatorname{dist}(x,F)=0$ by claim 1 there and the antisymmetry of $\le$, and $\phi_{n}(x)=\max\{0,1\}=1=\mathbf{1}_{F}(x)$ for every $n$ (using $0\cdot n=0$, claim 1 of Zero Products and Elementary Identities in a Field). Let $x\notin F$. The complement $\mathbb{R}^{q}\setminus F$ is open in $X$ by Closed Subset of a Topological Space, so there is a positive real $r$ such that every $y$ with $d_{E}(x,y)<r$ lies outside $F$; hence $r\le d_{E}(x,a)$ for every $a\in F$, so $r$ is a lower bound of $\{d_{E}(x,a):a\in F\}$ and $r\le\operatorname{dist}(x,F)$, the infimum being the greatest lower bound. By claim 2 of The Archimedean Property of the Real Numbers there is $N\in\mathbb{N}$ with $1<Nr$; for $n\in\mathbb{N}$ with $N\le n$ one has $1<Nr\le nr\le n\operatorname{dist}(x,F)$ (claim 5 of Elementary Arithmetic in an Ordered Field, the natural number order being transported by claim 6 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field, the case $N=n$ being trivial), so $1-n\operatorname{dist}(x,F)<0$ and $\phi_{n}(x)=0=\mathbf{1}_{F}(x)$. In either case the sequence $(\phi_{n}(x))_{n}$ is eventually constant with value $\mathbf{1}_{F}(x)$, hence converges to it.

Step 3: conclusion. Each $\phi_{n}$ is Borel and integrable, by the preamble of the statement, and satisfies $|\phi_{n}|\le1$, and the constant $1$ is integrable with respect to $\mu$ and to $\nu$ by claim 6(a) of Borel Measurability and Bounded Integration on a Metric Space. By Dominated Convergence Theorem, applied once for $\mu$ and once for $\nu$, $\int\phi_{n}\,d\mu\to\int\mathbf{1}_{F}\,d\mu$ and $\int\phi_{n}\,d\nu\to\int\mathbf{1}_{F}\,d\nu$; the integral of the indicator of the Borel set $F$ is the measure of $F$ by Simple Function and Its Integral with Lebesgue Integral of a Nonnegative Measurable Function, $\mathbf{1}_{F}$ being Borel by claim 1 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, and the integral of the nonnegative integrable function $\mathbf{1}_{F}$ delivered by dominated convergence agrees with it by claim 6(c) of Borel Measurability and Bounded Integration on a Metric Space. The two sequences of integrals coincide term by term by step 1, so their limits coincide by Uniqueness of Limits and Boundedness of Convergent Real Sequences: $\mu(F)=\nu(F)$. This holds for every closed $F$, and $\mu=\nu$ follows as explained at the start.