Proof of Lipschitz Test Functions Determine a Finite Borel Measure, and Uniqueness of Weak Limits

Throughout, $\rho(B)=\int_{S}\mathbf{1}_{B}\,d\rho$ for $\rho\in\{\mu,\nu\}$ and $B\in\mathcal{B}(S)$ (Basic Properties of the Integral of Nonnegative Measurable Maps (ii), Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vi)), and closed and open sets are Borel (Borel Sets of a Metric Space: Open and Closed Sets, Points, Open Sets of Subspaces, and Subspaces (ii), Borel Sets of a Metric Space: Open and Closed Sets, Points, Open Sets of Subspaces, and Subspaces (i)). A bounded Lipschitz function $S\to\mathbb{R}$ lies in $C_{b}(S)$ (Bounded Continuous Functions on a Metric Space: Measurability, Integrals, Operations, Lipschitz Functions and Approximation of Closed Sets (vi), Bounded Continuous Functions on a Metric Space (i)), so it is Borel measurable and integrable against every finite Borel measure on $S$ (Bounded Continuous Functions on a Metric Space: Measurability, Integrals, Operations, Lipschitz Functions and Approximation of Closed Sets (i), Bounded Continuous Functions on a Metric Space: Measurability, Integrals, Operations, Lipschitz Functions and Approximation of Closed Sets (ii)). Sequences of functions given by explicit formulas, such as the $f_{k}$ of Step 1 (built from the distance, the real number $k$ of Maps Given by Iterated Operations: Partial Iterates, Images of Natural Numbers, Powers and Operations over a Finite Set (ii) and the maximum of The Functions f ↦ f⁺ and f ↦ f⁻ on the Real-Valued Functions on a Set, and a Predicative Form of g = f⁺ and g = f⁻ (i)), are maps on $\mathbb{N}$ by Maps from Functional Predicative Relations, and Families of Sections (ii) (with The Graph Reading of a Formula Tree Is Predicative and Agrees with Its Reading, So the Reading Is Predicative (i), The Graph Reading of a Formula Tree Is Predicative and Agrees with Its Reading, So the Reading Is Predicative (iii) and Predicative Formulas for Suprema and Infima, and Maps Defined by Suprema and Infima (iii)).

Step 1 (closed sets; under the hypothesis of (i)). Let $F\subseteq S$ be closed; we show $\mu(F)=\nu(F)$. If $F=\emptyset$ both sides vanish. Otherwise, for $k\in\mathbb{N}$ let

$$f_{k}(x)=\max\bigl\{0,\ 1-k\operatorname{dist}(x,F)\bigr\}\qquad(x\in S),$$

with $\operatorname{dist}(x,F)$ the distance from $x$ to $F$. By Bounded Continuous Functions on a Metric Space: Measurability, Integrals, Operations, Lipschitz Functions and Approximation of Closed Sets (vii), each $f_{k}$ is Lipschitz with values in $[0,1]$, hence lies in $C_{b}(S)$ (Bounded Continuous Functions on a Metric Space: Measurability, Integrals, Operations, Lipschitz Functions and Approximation of Closed Sets (vi)), and $f_{k}\to\mathbf{1}_{F}$ pointwise.

By hypothesis $\int_{S}f_{k}\,d\mu=\int_{S}f_{k}\,d\nu$ for every $k$. The constant function $1$ lies in $C_{b}(S)$ (Bounded Continuous Functions on a Metric Space: Measurability, Integrals, Operations, Lipschitz Functions and Approximation of Closed Sets (iii)), so it is integrable for $\mu$ and for $\nu$, and it dominates every $f_{k}$. Dominated convergence ($F$ is Borel, as noted at the start) therefore gives $\mathbf{1}_{F}\in\mathcal{L}^{1}(\mu)\cap\mathcal{L}^{1}(\nu)$ (Dominated Convergence Theorem (i)) and

$$\int_{S}f_{k}\,d\mu\to\int_{S}\mathbf{1}_{F}\,d\mu=\mu(F),\qquad\int_{S}f_{k}\,d\nu\to\int_{S}\mathbf{1}_{F}\,d\nu=\nu(F)$$

(Dominated Convergence Theorem (iii)). The two sequences coincide, so $\mu(F)=\nu(F)$ by uniqueness of limits (Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (ii)).

Step 2 ((i)). Let $B\in\mathcal{B}(S)$. We first show $\mu(B)\le\nu(B)$. Suppose instead that $\mu(B)>\nu(B)$, and let $\varepsilon=\mu(B)-\nu(B)>0$, a real number as both measures are finite. By regularity of $\mu$ (Closed–Open Regularity of a Finite Borel Measure on a Metric Space (i)) there are a closed set $F$ and an open set $V$ with $F\subseteq B\subseteq V$ and $\mu(V\setminus F)<\varepsilon$. By monotonicity (Basic Properties of a Measure (iii)), by $\mu(V)=\mu(F)+\mu(V\setminus F)$ (Basic Properties of a Measure (iv)) and by Step 1,

$$\mu(B)\le\mu(V)=\mu(F)+\mu(V\setminus F)<\nu(F)+\varepsilon\le\nu(B)+\varepsilon=\mu(B),$$

a contradiction. Exchanging the roles of $\mu$ and $\nu$, which the hypothesis allows, gives $\nu(B)\le\mu(B)$, now using the regularity of $\nu$. Hence $\mu(B)=\nu(B)$ for every $B\in\mathcal{B}(S)$, that is, $\mu=\nu$.

Step 3 ((ii)). A Lipschitz $f:S\to\mathbb{R}$ with values in $[0,1]$ lies in $C_{b}(S)$, so by Weak Convergence of Finite Borel Measures on a Metric Space (i) the sequence $\bigl(\int_{S}f\,d\mu_{n}\bigr)$ converges both to $\int_{S}f\,d\mu$ and to $\int_{S}f\,d\nu$; these are equal by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (ii). Hence $\mu=\nu$ by (i).