[proofplan] The proof has four movements. (1) Existence reduces to the positive case via Jordan decomposition of $\varphi$ followed by the Riesz representation theorem applied to each positive piece. (2) The easy direction of the norm identity, $\|\varphi\| \le \|\nu\|_1$, follows from the integral inequality $|\int f\, d\nu| \le \|f\|_\infty\, |\nu|(K)$. (3) The reverse inequality is the technical heart: regularity of $|\nu|$ converts a Borel partition almost realising $|\nu|(K)$ into a topologically tame configuration on which a continuous test function can be built. (4) Uniqueness is then deduced from the norm identity, and the assembly produces the isometric isomorphism. [/proofplan] [step:Reduce existence to the Riesz representation theorem via Jordan decomposition] By the [Jordan Decomposition of Functionals on C(K)](/theorems/2642), part (i), there exist unique $\varphi_1, \varphi_2 \in M_\mathbb{R}(K)$ with $\varphi = \varphi_1 + i\varphi_2$. By part (iv) of the same theorem, there exist unique $\varphi_j^+, \varphi_j^- \in M_+(K)$ ($j = 1, 2$) with $\varphi_j = \varphi_j^+ - \varphi_j^-$. The Riesz Representation Theorem for Positive Functionals on C(K) states: for every $\psi \in M_+(K)$, there exists a unique regular positive Borel measure $\mu_\psi$ on $K$ with \begin{align*} \psi(f) = \int_K f \, d\mu_\psi \quad \text{for all } f \in C(K), \qquad \mu_\psi(K) = \|\psi\| = \psi(\mathbb{1}_K). \end{align*} Apply this theorem to each of $\varphi_1^+, \varphi_1^-, \varphi_2^+, \varphi_2^-$ to obtain regular positive Borel measures $\mu_1^+, \mu_1^-, \mu_2^+, \mu_2^-$ on $K$, each finite. Set \begin{align*} \nu := (\mu_1^+ - \mu_1^-) + i(\mu_2^+ - \mu_2^-). \end{align*} This is a complex Borel measure (linear combination of finite positive measures), and it is regular: each $\mu_j^\pm$ is regular, and finite linear combinations of regular measures are regular (regularity transfers to total variation, and $|\nu| \le \mu_1^+ + \mu_1^- + \mu_2^+ + \mu_2^-$). For any $f \in C(K)$, \begin{align*} \int_K f \, d\nu = \int_K f \, d\mu_1^+ - \int_K f \, d\mu_1^- + i\int_K f \, d\mu_2^+ - i\int_K f \, d\mu_2^- = \varphi_1(f) + i\varphi_2(f) = \varphi(f). \end{align*} If $\varphi \in M_\mathbb{R}(K)$, then $\varphi_2 \equiv 0$ (uniqueness in part (i) of the Jordan decomposition applied to the decomposition $\varphi = \varphi + i \cdot 0$), so $\nu = \mu_1^+ - \mu_1^-$ is a signed measure. [/step] [step:Prove the easy bound $\|\varphi\| \le \|\nu\|_1$] Fix $\nu$ as in Step 1 and let $|\nu|$ denote its total variation, a finite positive Borel measure on $K$ with $|\nu|(K) = \|\nu\|_1$. For $f \in C(K)$, the integral inequality for complex measures gives \begin{align*} |\varphi(f)| = \left|\int_K f \, d\nu\right| \le \int_K |f| \, d|\nu| \le \|f\|_\infty |\nu|(K) = \|f\|_\infty \|\nu\|_1. \end{align*} Taking the supremum over $\|f\|_\infty \le 1$, $\|\varphi\| \le \|\nu\|_1$. [/step] [step:Set up the reverse bound via a finite Borel partition almost realising $|\nu|(K)$] We prove $\|\varphi\| \ge \|\nu\|_1$. Fix $\varepsilon > 0$. By definition of total variation, \begin{align*} \|\nu\|_1 = |\nu|(K) = \sup\left\{\sum_{k=1}^N |\nu(A_k)| : K = \bigsqcup_{k=1}^N A_k,\ A_k \in \mathcal{B}(K)\right\}, \end{align*} the supremum being over finite Borel partitions of $K$. Choose a finite Borel partition $K = \bigsqcup_{k=1}^N A_k$ with \begin{align*} \sum_{k=1}^N |\nu(A_k)| > \|\nu\|_1 - \varepsilon. \end{align*} [/step] [step:Refine the partition to closed sets inside disjoint open neighbourhoods] By regularity of $|\nu|$, for each $k$ there is a closed set $E_k \subseteq A_k$ with $|\nu|(A_k \setminus E_k) < \varepsilon / N$. The closed sets $E_1, \dots, E_N$ are pairwise disjoint (they are subsets of the disjoint $A_k$), so we can separate them by disjoint open sets in the normal compact Hausdorff space $K$. We construct disjoint open sets $U_k$ with $E_k \subseteq U_k$ and $|\nu|(U_k \setminus E_k) < \varepsilon / N$. Since $K$ is normal, for each pair $(j, k)$ with $j \ne k$, $E_j$ and $E_k$ are disjoint closed sets, so there are disjoint open sets separating them. Iterating: by induction on $k$, find an open $U_k$ with $E_k \subseteq U_k$ and $U_k$ disjoint from $E_j$ for $j \ne k$. The intersection of all such $U_k$'s for fixed $k$ remains an open neighbourhood of $E_k$ disjoint from $\bigcup_{j \ne k} E_j$. To also ensure $|\nu|(U_k \setminus E_k) < \varepsilon / N$: by outer regularity of the finite positive measure $|\nu|$, for each $k$ there is an open $\widetilde{U}_k \supseteq E_k$ with $|\nu|(\widetilde{U}_k) < |\nu|(E_k) + \varepsilon/N$, hence $|\nu|(\widetilde{U}_k \setminus E_k) < \varepsilon/N$. Replace $U_k$ by $U_k \cap \widetilde{U}_k$, still an open neighbourhood of $E_k$ disjoint from $\bigcup_{j \ne k} E_j$ and with $|\nu|(U_k \setminus E_k) < \varepsilon / N$. Now make the $U_k$ pairwise disjoint: replace each $U_k$ by $U_k \setminus \bigcup_{j \ne k} \overline{V_j}$, where $V_j$ is an open neighbourhood of $E_j$ with $\overline{V_j}$ disjoint from $E_k$ (such $V_j$ exists by normality applied to $E_j$ and $\{E_i : i \ne j\}$, and we may take $V_j \subseteq U_j$, so the inner shrinking does not lose the bound on $|\nu|(U_k \setminus E_k)$). After this surgery the $U_k$ are pairwise disjoint open neighbourhoods of the $E_k$ with $|\nu|(U_k \setminus E_k) < \varepsilon/N$ each. [/step] [step:Construct a partition-of-unity test function with unimodular phases] For each $k$, choose $\lambda_k \in \mathbb{C}$ with $|\lambda_k| = 1$ and \begin{align*} \lambda_k \nu(E_k) = |\nu(E_k)| \end{align*} (any unit-modulus rotation; if $\nu(E_k) = 0$, take $\lambda_k = 1$). Apply the [Partition of Unity on C(K)](/theorems/2643) to the closed set $E := \bigsqcup_{k=1}^N E_k$ (closed as a finite union of closed sets) covered by the open sets $U_1, \dots, U_N$ (each $E_k \subseteq U_k$ and the $U_k$ are pairwise disjoint, so $E \subseteq \bigcup_k U_k$). This produces $h_1, \dots, h_N \in C(K)$ with $h_k \prec U_k$ (so $0 \le h_k \le 1$ and $\operatorname{supp}(h_k) \subseteq U_k$), $0 \le \sum_k h_k \le 1$ on $K$, and $\sum_k h_k \equiv 1$ on $E$. In particular, since $\operatorname{supp}(h_j) \subseteq U_j$ and $E_k \subseteq U_k$ with $U_j \cap U_k = \varnothing$ for $j \ne k$, we have $h_j \equiv 0$ on $E_k$ for $j \ne k$. Combined with $\sum_j h_j \equiv 1$ on $E_k \subseteq E$, this forces $h_k \equiv 1$ on $E_k$. Define \begin{align*} f: K &\to \mathbb{C} \\ x &\mapsto \sum_{k=1}^N \lambda_k h_k(x). \end{align*} Then $f \in C(K)$. The supports $\{h_k \ne 0\}$ are contained in the disjoint open sets $U_k$, so at each $x \in K$ at most one $h_k(x)$ is non-zero (those $h_k$ with $x \in \overline{U_k} \setminus \bigcup_{j \ne k} U_j$ vanish, and the rest by support disjointness). Therefore \begin{align*} |f(x)| = \left|\sum_k \lambda_k h_k(x)\right| \le \sum_k h_k(x) \le 1, \end{align*} so $\|f\|_\infty \le 1$. [/step] [step:Estimate $\varphi(f)$ to recover $\sum_k |\nu(E_k)|$ up to error $O(\varepsilon)$] We compute \begin{align*} \varphi(f) = \int_K f \, d\nu = \sum_{k=1}^N \lambda_k \int_K h_k \, d\nu. \end{align*} Split each integral into the contribution on $E_k$ and the contribution on $U_k \setminus E_k$ (recall $\operatorname{supp}(h_k) \subseteq U_k = E_k \sqcup (U_k \setminus E_k)$, so the integrand vanishes off $U_k$): \begin{align*} \int_K h_k \, d\nu = \int_{E_k} h_k \, d\nu + \int_{U_k \setminus E_k} h_k \, d\nu = \nu(E_k) + R_k, \end{align*} where we used $h_k \equiv 1$ on $E_k$, and $R_k := \int_{U_k \setminus E_k} h_k \, d\nu$ satisfies \begin{align*} |R_k| \le \int_{U_k \setminus E_k} |h_k| \, d|\nu| \le |\nu|(U_k \setminus E_k) < \varepsilon / N. \end{align*} Therefore \begin{align*} \varphi(f) = \sum_{k=1}^N \lambda_k (\nu(E_k) + R_k) = \sum_{k=1}^N |\nu(E_k)| + \sum_{k=1}^N \lambda_k R_k, \end{align*} using $\lambda_k \nu(E_k) = |\nu(E_k)|$. Bounding the error term: $|\sum_k \lambda_k R_k| \le \sum_k |R_k| < N \cdot \varepsilon/N = \varepsilon$. Now we estimate $\sum_k |\nu(E_k)|$ from below using the original partition. For each $k$, \begin{align*} |\nu(A_k)| \le |\nu(E_k)| + |\nu(A_k \setminus E_k)| \le |\nu(E_k)| + |\nu|(A_k \setminus E_k) < |\nu(E_k)| + \varepsilon/N, \end{align*} so $\sum_k |\nu(E_k)| > \sum_k |\nu(A_k)| - \varepsilon > \|\nu\|_1 - 2\varepsilon$. Putting things together, \begin{align*} |\varphi(f)| \ge \operatorname{Re} \varphi(f) = \sum_{k=1}^N |\nu(E_k)| + \operatorname{Re}\sum_k \lambda_k R_k > (\|\nu\|_1 - 2\varepsilon) - \varepsilon = \|\nu\|_1 - 3\varepsilon. \end{align*} Since $\|f\|_\infty \le 1$, $\|\varphi\| \ge |\varphi(f)| > \|\nu\|_1 - 3\varepsilon$. Letting $\varepsilon \to 0$, $\|\varphi\| \ge \|\nu\|_1$. Combined with Step 2, $\|\varphi\| = \|\nu\|_1$. [guided] We carry out the estimate on $\varphi(f)$ in full, with each object declared and each bound derived. The goal is to show $\|\varphi\| \ge \|\nu\|_1 - 3\varepsilon$ for the test function $f$ built in Step 5; letting $\varepsilon \to 0$ then yields $\|\varphi\| \ge \|\nu\|_1$. **Phase choice.** For each $k \in \{1, \dots, N\}$, pick $\lambda_k \in \mathbb{C}$ with $|\lambda_k| = 1$ and \begin{align*} \lambda_k \nu(E_k) = |\nu(E_k)|. \end{align*} If $\nu(E_k) \ne 0$, write $\nu(E_k) = r_k e^{i\theta_k}$ with $r_k = |\nu(E_k)| > 0$ and set $\lambda_k = e^{-i\theta_k}$; if $\nu(E_k) = 0$, set $\lambda_k = 1$. In either case the displayed identity holds. The purpose of this rotation is to convert the complex number $\nu(E_k)$ into the real non-negative number $|\nu(E_k)|$ when summed against $\lambda_k$, so that the resulting expression $\sum_k |\nu(E_k)|$ matches the partition sum entering the definition of $\|\nu\|_1$. **Test function.** Using the partition of unity $h_1, \dots, h_N$ from Step 5 (with $h_k \prec U_k$, $\sum_k h_k \le 1$ on $K$, $\sum_k h_k \equiv 1$ on $E := \bigsqcup_k E_k$, and $h_k \equiv 1$ on $E_k$), define \begin{align*} f: K &\to \mathbb{C} \\ x &\mapsto \sum_{k=1}^N \lambda_k h_k(x). \end{align*} Pointwise disjointness of the supports of the $h_k$ (forced by pairwise disjointness of the open sets $U_k$) gives $|f(x)| \le \sum_k h_k(x) \le 1$, so $\|f\|_\infty \le 1$. **Splitting the integral.** Since $\varphi$ is represented by $\nu$ (Step 1), \begin{align*} \varphi(f) = \int_K f \, d\nu = \sum_{k=1}^N \lambda_k \int_K h_k \, d\nu. \end{align*} Because $\operatorname{supp}(h_k) \subseteq U_k$ and $U_k = E_k \sqcup (U_k \setminus E_k)$, the integrand vanishes outside $U_k$: \begin{align*} \int_K h_k \, d\nu = \int_{E_k} h_k \, d\nu + \int_{U_k \setminus E_k} h_k \, d\nu. \end{align*} On $E_k$, $h_k \equiv 1$, so $\int_{E_k} h_k \, d\nu = \nu(E_k)$. Define the error \begin{align*} R_k := \int_{U_k \setminus E_k} h_k \, d\nu, \end{align*} so that $\int_K h_k \, d\nu = \nu(E_k) + R_k$. The integral inequality for complex measures, combined with $|h_k| \le 1$, bounds $R_k$ via regularity: \begin{align*} |R_k| \le \int_{U_k \setminus E_k} |h_k| \, d|\nu| \le |\nu|(U_k \setminus E_k) < \frac{\varepsilon}{N}, \end{align*} where the last step uses the bound $|\nu|(U_k \setminus E_k) < \varepsilon/N$ established in Step 4. **Lower bound on $|\varphi(f)|$.** Substituting and using $\lambda_k \nu(E_k) = |\nu(E_k)|$, \begin{align*} \varphi(f) = \sum_{k=1}^N \lambda_k \bigl(\nu(E_k) + R_k\bigr) = \sum_{k=1}^N |\nu(E_k)| + \sum_{k=1}^N \lambda_k R_k. \end{align*} The error sum is bounded by the triangle inequality and $|\lambda_k| = 1$: \begin{align*} \left|\sum_{k=1}^N \lambda_k R_k\right| \le \sum_{k=1}^N |R_k| < N \cdot \frac{\varepsilon}{N} = \varepsilon. \end{align*} Since $\sum_k |\nu(E_k)|$ is real and non-negative, taking real parts, \begin{align*} \operatorname{Re}\varphi(f) = \sum_{k=1}^N |\nu(E_k)| + \operatorname{Re}\sum_{k=1}^N \lambda_k R_k \ge \sum_{k=1}^N |\nu(E_k)| - \varepsilon, \end{align*} using $|\operatorname{Re} z| \le |z| < \varepsilon$ for $z = \sum_k \lambda_k R_k$. **Comparing $\sum_k |\nu(E_k)|$ to $\|\nu\|_1$.** From Step 4, $|\nu|(A_k \setminus E_k) < \varepsilon/N$. The triangle inequality applied to $\nu(A_k) = \nu(E_k) + \nu(A_k \setminus E_k)$ gives \begin{align*} |\nu(A_k)| \le |\nu(E_k)| + |\nu(A_k \setminus E_k)| \le |\nu(E_k)| + |\nu|(A_k \setminus E_k) < |\nu(E_k)| + \frac{\varepsilon}{N}. \end{align*} Summing and using the bound $\sum_k |\nu(A_k)| > \|\nu\|_1 - \varepsilon$ from Step 3, \begin{align*} \sum_{k=1}^N |\nu(E_k)| > \sum_{k=1}^N |\nu(A_k)| - \varepsilon > \|\nu\|_1 - 2\varepsilon. \end{align*} **Conclusion.** Combining the two inequalities, \begin{align*} |\varphi(f)| \ge \operatorname{Re}\varphi(f) \ge \sum_{k=1}^N |\nu(E_k)| - \varepsilon > (\|\nu\|_1 - 2\varepsilon) - \varepsilon = \|\nu\|_1 - 3\varepsilon. \end{align*} Since $\|f\|_\infty \le 1$, the dual-norm definition gives \begin{align*} \|\varphi\| = \sup_{\|g\|_\infty \le 1} |\varphi(g)| \ge |\varphi(f)| > \|\nu\|_1 - 3\varepsilon. \end{align*} The bound holds for every $\varepsilon > 0$, so letting $\varepsilon \to 0$ yields $\|\varphi\| \ge \|\nu\|_1$. Combined with Step 2, $\|\varphi\| = \|\nu\|_1$. The three sources of error — refining $A_k$ to $E_k$ (loss $\varepsilon/N$ per piece, total $\varepsilon$), the integrals $R_k$ over the buffer regions $U_k \setminus E_k$ (loss $\varepsilon/N$ per piece, total $\varepsilon$), and the partition slack $\sum_k |\nu(A_k)| > \|\nu\|_1 - \varepsilon$ — combine to $3\varepsilon$, which is the price paid to upgrade a Borel partition into a configuration that can be tested against continuous functions on $K$. The phase rotation by $\lambda_k$ is what makes the leading term $\sum_k |\nu(E_k)|$ rather than $|\sum_k \nu(E_k)| = |\nu(E)|$: without it, the partition structure of $\|\nu\|_1$ would collapse into a single absolute value and the supremum could not be recovered. [/guided] [/step] [step:Deduce uniqueness of $\nu$ from the norm identity] Suppose $\nu_1, \nu_2$ are regular complex Borel measures on $K$ both representing $\varphi$, i.e. \begin{align*} \varphi(f) = \int_K f \, d\nu_1 = \int_K f \, d\nu_2 \quad \text{for all } f \in C(K). \end{align*} Set $\nu := \nu_1 - \nu_2$, a regular complex Borel measure (the difference of regular complex measures is regular: $|\nu| \le |\nu_1| + |\nu_2|$, and regularity passes through this domination — for any Borel $A$ and $\delta > 0$, find compact $E_j \subseteq A$ with $|\nu_j|(A \setminus E_j) < \delta/2$, then $E := E_1 \cap E_2$ is compact with $|\nu|(A \setminus E) \le |\nu_1|(A \setminus E_1) + |\nu_2|(A \setminus E_2) < \delta$, and similarly for outer regularity). Then $\int_K f \, d\nu = 0$ for all $f \in C(K)$, so the corresponding functional $\varphi_\nu \in M(K)$ is the zero functional, i.e. $\|\varphi_\nu\| = 0$. By the norm identity (Steps 2 and 6 applied to $\varphi_\nu$ and $\nu$), $\|\nu\|_1 = \|\varphi_\nu\| = 0$. Therefore $|\nu|(K) = 0$, hence $\nu = 0$ on $\mathcal{B}(K)$, i.e. $\nu_1 = \nu_2$. [/step] [step:Assemble: the map is an isometric isomorphism] Define \begin{align*} T: \{\text{regular complex Borel measures on } K\} &\to M(K) \\ \nu &\mapsto \varphi_\nu, \qquad \varphi_\nu(f) := \int_K f \, d\nu. \end{align*} $T$ is well-defined and complex-linear (linearity of the integral in the measure argument), with $\|T\nu\| = \|\nu\|_1$ by the norm identity (Steps 2, 6). So $T$ is an isometry. $T$ is surjective: Step 1 produced, for every $\varphi \in M(K)$, a regular complex Borel measure $\nu$ with $T\nu = \varphi$. $T$ is injective by Step 7. Hence $T$ is an isometric isomorphism between regular complex Borel measures on $K$ (with the total-variation norm) and $C(K)^* = M(K)$ (with the dual norm). For $\varphi \in M_\mathbb{R}(K)$, Step 1 noted that $\nu$ is a signed (real) measure. This completes the proof. [/step]