Proof of Radon Measures and Finite Signed Borel Measures Are Determined by Their Integrals of Compactly Supported Continuous Functions

Throughout, continuous real functions on $X$ are Borel measurable, and for a nonnegative function that is integrable with respect to a measure, its integral as an integrable function and as a nonnegative measurable function agree.

Integrability. Let $f\in C_{c}(X)$ and let $\mu$ be a Radon measure on $X$. The set $K=\operatorname{supp}f$ is compact, so $\mu(K)<\infty$. By The Continuous Functions with Compact Support Form a Real Normed Space under the Uniform Norm (iii), $|f(x)|\le\|f\|_{u}$ for every $x\in X$, and by Continuous Functions with Compact Support: the Zero Function, Closure under Operations, Vanishing off the Support, Existence of the Maximum Modulus, and Uniform Continuity (iii), $f(x)=0$ for $x\notin K$; hence

$$|f(x)|\le\|f\|_{u}\,\mathbf{1}_{K}(x)\qquad(x\in X).$$

By monotonicity of the integral of nonnegative functions,

$$\int_{X}|f|\,d\mu\le\|f\|_{u}\int_{X}\mathbf{1}_{K}\,d\mu=\|f\|_{u}\,\mu(K)<\infty,$$

so $f$ is integrable with respect to $\mu$ by the integrability criterion. If $\nu$ is a finite signed Radon measure on $X$, then $|\nu|$ is a Radon measure by Finite Signed Radon Measure on a Metric Space (i), so $f$ is integrable with respect to $|\nu|$, that is, integrable with respect to $\nu$. This proves (i).

Compact sets. Let $\mu,\mu'$ be as in (ii), and let $K\subseteq X$ be compact. Let $\varepsilon>0$. Since $\mu(K)<\infty$, outer regularity of $\mu$ (Radon Measure on a Metric Space (i)) gives an open $V\supseteq K$ with $\mu(V)<\mu(K)+\varepsilon$. For this $V$, Thickenings, Neighbourhoods with Compact Closure, Urysohn Functions and Partitions of Unity in a Locally Compact Metric Space (iii) gives $f\in C_{c}(X)$ with $0\le f\le1$, $f=1$ on $K$ and $\operatorname{supp}f\subseteq V$; since $f$ vanishes off $\operatorname{supp}f$ (Continuous Functions with Compact Support: the Zero Function, Closure under Operations, Vanishing off the Support, Existence of the Maximum Modulus, and Uniform Continuity (iii)),

$$\mathbf{1}_{K}(x)\le f(x)\le\mathbf{1}_{V}(x)\qquad(x\in X).$$

By monotonicity of the integral and the hypothesis,

$$\mu'(K)=\int_{X}\mathbf{1}_{K}\,d\mu'\le\int_{X}f\,d\mu'=\int_{X}f\,d\mu\le\int_{X}\mathbf{1}_{V}\,d\mu=\mu(V)<\mu(K)+\varepsilon .$$

As $\varepsilon>0$ was arbitrary, $\mu'(K)\le\mu(K)$; exchanging the roles of $\mu$ and $\mu'$ gives $\mu(K)=\mu'(K)$.

Open sets. Let $V\subseteq X$ be open. By the previous step the sets $\{\mu(K):K\subseteq V\text{ compact}\}$ and $\{\mu'(K):K\subseteq V\text{ compact}\}$ coincide, so inner regularity of $\mu$ and of $\mu'$ (Radon Measure on a Metric Space (i)) gives

$$\mu(V)=\sup\{\mu(K):K\subseteq V\text{ compact}\}=\sup\{\mu'(K):K\subseteq V\text{ compact}\}=\mu'(V).$$

Borel sets. Let $B\in\mathcal{B}(X)$. By the previous step and outer regularity of $\mu$ and of $\mu'$,

$$\mu(B)=\inf\{\mu(V):V\supseteq B\text{ open}\}=\inf\{\mu'(V):V\supseteq B\text{ open}\}=\mu'(B).$$

This proves (ii).

Finite signed Borel measures. Let $X$ be separable and let $\nu,\nu'$ be as in (iii). By Finite Borel Measures on a Locally Compact Separable Metric Space Are Radon (iii), $\nu$ and $\nu'$ are finite signed Radon measures, so the integrals in the hypothesis exist by (i). Let

$$\lambda=\nu+(-1)\nu',\qquad\lambda(B)=\nu(B)-\nu'(B)\quad(B\in\mathcal{B}(X)),$$

a finite signed Borel measure by Properties of the Total Variation: a Finite Measure, the Jordan Decomposition, Linear Combinations, the Total Variation Norm, and Integrability (v), applied first to $(-1)\nu'$ and then to the sum. Let $f\in C_{c}(X)$. Then $f$ is Borel measurable and $|f(x)|\le\|f\|_{u}$ for every $x\in X$ by The Continuous Functions with Compact Support Form a Real Normed Space under the Uniform Norm (iii), so $f$ is integrable with respect to $\lambda$ by Properties of the Integral with Respect to a Finite Signed Measure: Linearity, Bounds, Bounded Functions, Linearity in the Measure, Indicators and Finite Measures (iii), with $C=\|f\|_{u}$. By Properties of the Integral with Respect to a Finite Signed Measure: Linearity, Bounds, Bounded Functions, Linearity in the Measure, Indicators and Finite Measures (v), first for the sum of $\nu$ and $(-1)\nu'$ and then for the multiple of $\nu'$ by $c=-1$, and by the hypothesis,

$$\int_{X}f\,d\lambda=\int_{X}f\,d\nu+\int_{X}f\,d\bigl((-1)\nu'\bigr)=\int_{X}f\,d\nu-\int_{X}f\,d\nu'=0 .$$

Let $\lambda^{+}$ and $\lambda^{-}$ be the Jordan decomposition of $\lambda$, formed with a Hahn set for $\lambda$ (one exists by Hahn and Jordan Decompositions of a Finite Signed Measure (i), using the axiom of dependent choice); they are finite Borel measures with $\lambda(B)=\lambda^{+}(B)-\lambda^{-}(B)$ for every $B\in\mathcal{B}(X)$. By Finite Borel Measures on a Locally Compact Separable Metric Space Are Radon (ii), $\lambda^{+}$ and $\lambda^{-}$ are Radon measures, so every $f\in C_{c}(X)$ is integrable with respect to both by (i), and by Integrable Functions and the Integral with Respect to a Finite Signed Measure (ii)

$$\int_{X}f\,d\lambda^{+}-\int_{X}f\,d\lambda^{-}=\int_{X}f\,d\lambda=0\qquad(f\in C_{c}(X)).$$

By (ii), $\lambda^{+}(B)=\lambda^{-}(B)$ for every $B\in\mathcal{B}(X)$. Hence $\lambda(B)=0$, that is, $\nu(B)=\nu'(B)$, for every $B\in\mathcal{B}(X)$, which proves (iii).