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

Throughout, $\mathbb{R}$ carries the absolute-value metric $d_{\mathbb{R}}$, and measurable means measurable with respect to $\mathcal{B}(X)$ and the Borel $\sigma$-algebra of the real line.

Claim 1.

Step 1: $\mu(X)=\nu(X)$. The constant function with value $1$ on $X$ is Lipschitz with constant $0$ and bounded, so the hypothesis and claim 6(a) of Borel Measurability and Bounded Integration on a Metric Space give

$$\mu(X)=\int_X 1\,d\mu=\int_X 1\,d\nu=\nu(X),$$

a real number because $\mu$ and $\nu$ are finite.

Step 2: $\mu(F)=\nu(F)$ for every closed $F\subseteq X$. If $F$ is empty then $\mu(F)=0=\nu(F)$ by Measure, Measure Space, and Probability Measure. So let $F$ be nonempty and closed in the topology of open subsets of $(X,d)$. For $k\in\mathbb{N}$ let $\lambda_k=\iota(k)$, where $\iota$ is the canonical map from $\mathbb{N}$ to $\mathbb{R}$, and define $g_k:X\to\mathbb{R}$ by

$$g_k(x)=\max\bigl\{0,\;1-\lambda_k\operatorname{dist}_d(x,F)\bigr\},$$

with $\operatorname{dist}_d$ the distance to a set. Then $0\le g_k(x)\le1$ for every $x$, so each $g_k$ is bounded.

Each $g_k$ is Lipschitz with constant $\lambda_k$. Indeed, for real $s,t$ one has $|\max\{0,s\}-\max\{0,t\}|\le|s-t|$: if $s\le0$ and $t\le0$ the left-hand side is $0$; if $0\le s$ and $0\le t$ it equals $|s-t|$; and if $t\le0\le s$ it equals $s$, while $s\le s-t=|s-t|$ because $-t\ge0$ (and symmetrically if $s\le0\le t$). Applying this with $s=1-\lambda_k\operatorname{dist}_d(x,F)$ and $t=1-\lambda_k\operatorname{dist}_d(y,F)$ and using claim 4 of The Distance to a Set is Nonexpansive,

$$|g_k(x)-g_k(y)|\le\lambda_k\bigl|\operatorname{dist}_d(x,F)-\operatorname{dist}_d(y,F)\bigr|\le\lambda_k\,d(x,y).$$

By hypothesis, therefore, $\int_X g_k\,d\mu=\int_X g_k\,d\nu$ for every $k\in\mathbb{N}$.

Next, $(g_k(x))_{k\in\mathbb{N}}$ converges to $\mathbf{1}_F(x)$ for every $x\in X$, where $\mathbf{1}_F$ is the indicator function of $F$. If $x\in F$, then $\operatorname{dist}_d(x,F)=0$ by claims 1 and 2 of The Distance to a Set is Nonexpansive applied with $a=x$, so $g_k(x)=1=\mathbf{1}_F(x)$ for every $k$. If $x\notin F$, then $x$ lies in the open set $X\setminus F$, so by Open Subset of a Metric Space there is a real $r>0$ with $B_d(x,r)\subseteq X\setminus F$; consequently $r\le d(x,a)$ for every $a\in F$, so $r$ is a lower bound of $\{d(x,a):a\in F\}$ and $r\le\operatorname{dist}_d(x,F)$. By claim 2 of The Archimedean Property of the Real Numbers there is $K\in\mathbb{N}$ with $1<\lambda_K\,r$, and for every $k\ge K$ we get $1<\lambda_K\,r\le\lambda_k\operatorname{dist}_d(x,F)$, using claim 6 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field; hence $1-\lambda_k\operatorname{dist}_d(x,F)<0$ and $g_k(x)=0=\mathbf{1}_F(x)$.

Each $g_k$ is continuous on $X$, hence measurable by claim 3 of Borel Measurability and Bounded Integration on a Metric Space; $|g_k(x)|\le1$ for every $x$ and every $k$; and the constant function with value $1$ is integrable with respect to $\mu$ and with respect to $\nu$ by claim 6(a) of that lemma. The hypotheses of Dominated Convergence Theorem, namely convergence at every point and domination at every point by a fixed integrable function, are thus satisfied, so $\bigl(\int_X g_k\,d\mu\bigr)_{k}$ converges to $\int_X\mathbf{1}_F\,d\mu$ and $\bigl(\int_X g_k\,d\nu\bigr)_{k}$ converges to $\int_X\mathbf{1}_F\,d\nu$.

The set $F$ is closed, hence lies in $\mathcal{B}(X)$ by claim 1 of Borel Measurability and Bounded Integration on a Metric Space, and $\mathbf{1}_F$ is a nonnegative simple function whose integral in the sense of Simple Function and Its Integral is $\rho(F)$ for a measure $\rho$; by the agreement recorded in Lebesgue Integral of a Nonnegative Measurable Function together with claim 6(c) of Borel Measurability and Bounded Integration on a Metric Space, $\int_X\mathbf{1}_F\,d\mu=\mu(F)$ and $\int_X\mathbf{1}_F\,d\nu=\nu(F)$. The two sequences of integrals are equal term by term, so their limits coincide by the uniqueness of limits in the metric space $(\mathbb{R},d_{\mathbb{R}})$ (Uniqueness of Limits in a Metric Space); that is, $\mu(F)=\nu(F)$.

Step 3. By claim 1 of Borel Measurability and Bounded Integration on a Metric Space, the family $\mathcal{C}$ of closed subsets of $X$ is a $\pi$-system whose generated $\sigma$-algebra is $\mathcal{B}(X)$. By Steps 1 and 2, $\mu$ and $\nu$ agree on $\mathcal{C}$ and satisfy $\mu(X)=\nu(X)<\infty$. Claim 1 of Uniqueness of Finite Measures on a Generating Pi-System and the Density of the Exponential Law therefore gives $\mu(B)=\nu(B)$ for every $B\in\mathcal{B}(X)$.

Claim 2. Let $f:X\to\mathbb{R}$ be bounded and Lipschitz. By A Lipschitz Map is Uniformly Continuous it is continuous on $X$, and it is bounded, so by Weak Convergence of Finite Borel Measures on a Metric Space the sequence $\bigl(\int_X f\,d\mu_n\bigr)_{n\in\mathbb{N}}$ converges both to $\int_X f\,d\mu$ and to $\int_X f\,d\nu$. By the uniqueness of limits in $(\mathbb{R},d_{\mathbb{R}})$ (Uniqueness of Limits in a Metric Space), $\int_X f\,d\mu=\int_X f\,d\nu$. Since $f$ was an arbitrary bounded Lipschitz function, claim 1 gives $\mu=\nu$.