Riesz-Markov Representation Theorem on a Compact Metric Space

Let $(K,d)$ be a metric space with $K$ nonempty, let $\mathcal{T}_d$ be the collection of subsets of $K$ that are open in $(K,d)$, which is a topology on $K$ by Metric Open Sets Form a Topology, and assume that $K$ is compact in $(K,\mathcal{T}_d)$. Let $\mathcal{B}(K)$ be the Borel $\sigma$-algebra of $(K,d)$.

Let $\mathbb{R}$ denote the real numbers, with the addition, multiplication, identities, additive inverses, multiplicative inverses and order of their ordered field structure; for $s,t\in\mathbb{R}$ write $s-t$ for $s+(-t)$, write $s<t$ to mean that $s\le t$ and $s\ne t$, write $s^{-1}$ for the multiplicative inverse of $s$ when $s\ne 0$, let $|s|$ be the absolute value of $s$, and let $d_{\mathbb{R}}$ be given by $d_{\mathbb{R}}(s,t)=|s-t|$, which is a metric on $\mathbb{R}$ by The Absolute Value Metric on the Real Line.

Let $\mathcal{C}$ be the set of all maps $f:K\to\mathbb{R}$ that are continuous on $K$ as maps from $(K,d)$ to $(\mathbb{R},d_{\mathbb{R}})$. By claims 1 and 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space, $\mathcal{C}$ contains every constant map on $K$, and contains $f+g$ and $cf$ whenever $f,g\in\mathcal{C}$ and $c\in\mathbb{R}$, these maps being defined pointwise as in that theorem. Write $\mathbf{1}$ for the constant map on $K$ with value $1$.

Let $\Lambda$ be a map from $\mathcal{C}$ to $\mathbb{R}$ such that, for all $f,g\in\mathcal{C}$ and every $c\in\mathbb{R}$,

(i) (additivity) $\Lambda(f+g)=\Lambda(f)+\Lambda(g)$;

(ii) (homogeneity) $\Lambda(cf)=c\,\Lambda(f)$;

(iii) (positivity) if $0\le f(x)$ for every $x\in K$, then $0\le\Lambda(f)$.

Then the following hold.

1. (Existence of a representing measure) There is a Borel measure $\mu$ on $(K,d)$ with $\mu(K)=\Lambda(\mathbf{1})$, so that $\mu$ is finite, and such that every $f\in\mathcal{C}$ is integrable with respect to $\mu$ and satisfies

2. (Uniqueness of the representing measure) Let $\mu$ and $\nu$ be finite Borel measures on $(K,d)$ such that every $f\in\mathcal{C}$ is integrable with respect to each of them and

Then $\mu(B)=\nu(B)$ for every $B\in\mathcal{B}(K)$.