Proof [Theorem 7.17, Fol99]. Let $f \in C_{0}(X; \complex)$, then

so $\norm{I_\mu}_{C_0(X; \complex)}\le \norm{\mu}_{\text{var}}$.

On the other hand, let $\seqf{A_j}\subset \cb_{X}$ such that $X = \bigsqcup_{j = 1}^{n}A_{j}$. Let $\eps > 0$, then by Proposition 17.1.3 applied to $|\mu|$, there exists compact sets $\seqf{K_j}$ such that for each $1 \le j \le n$, $K_{j} \subset A_{j}$ and $|\mu(K_{j}) - \mu(A_{j})| < \eps$. By outer regularity and Urysohn’s lemma, there exists $\seqf{\phi_j}\subset C_{c}(X; [0, 1])$ with disjoint support such that for each $1 \le j \le n$, $\norm{\phi_j - \one_{K_j}}_{L^1(|\mu|)}< \eps$. Let $\phi = \sum_{j = 1}^{n} \ol{\sgn(\mu(K_j))}\phi_{j}$, then $\norm{\phi}_{u} \le 1$ and

so

As such a $\phi$ exists for all $\eps > 0$, $\norm{I_\mu}_{C_0(X; \complex)}\ge \sum_{j = 1}^{n} |\mu(A_{j})|$. Since this holds for all such partitions, $\norm{I_\mu}_{C_0(X; \complex)}\ge |\mu|(X)$. Therefore the map $\mu \mapsto I_{\mu}$ is isometric.

Finally, let $I \in C_{0}(X; \complex)^{*}$, then there exists bounded linear functionals $I_{r}, I_{i} \in C_{0}(X; \real)^{*}$ such that for any $f \in C_{0}(X; \real)$,

By Lemma 17.3.1, there exists bounded positive linear functionals $I_{r}^{+}, I_{r}^{-}, I_{i}^{+}, I_{i}^{-}$ such that for any $f \in C_{0}(X; \real)$,

Thus by the Riesz Representation Theorem, there exists finite Radon measures $\mu_{r}^{+}, \mu_{r}^{-}, \mu_{i}^{+}, \mu_{i}^{-} \in M_{R}(X; \complex)$ such that for any $f \in C_{0}(X; \real)$,

Let $\mu = \mu_{r}^{+} - \mu_{r}^{-} + i\mu_{i}^{+} - i\mu_{i}^{-}$, then $I = I_{\mu}$, and the map $\mu \mapsto I_{\mu}$ is surjective.$\square$