[proofplan] We first treat the diagonal coefficient $\mu_x(\Delta)=(E(\Delta)x,x)_H$. Since $E(\Delta)$ is an [orthogonal projection](/theorems/437), this coefficient is the squared norm of $E(\Delta)x$, so it is positive; strong countable additivity of the projection-valued measure then gives countable additivity after pairing with $x$. For general $x,y$, we use the complex polarization identity, with the convention that the [Hilbert space](/page/Hilbert%20Space) [inner product](/page/Inner%20Product) is linear in the first argument, to write $\mu_{x,y}$ as a fixed finite linear combination of diagonal positive measures. This representation gives countable additivity and finite total variation. [/proofplan]

[step:Prove positivity and total mass for diagonal coefficients] Fix $x\in H$. Define \begin{align*} \mu_x:\mathcal B(K)\to [0,\infty),\qquad \Delta\mapsto (E(\Delta)x,x)_H. \end{align*} For each $\Delta\in\mathcal B(K)$, the operator $E(\Delta)$ is an orthogonal projection, hence $E(\Delta)^*=E(\Delta)$ and $E(\Delta)^2=E(\Delta)$. Therefore \begin{align*} \mu_x(\Delta)=(E(\Delta)x,x)_H=(E(\Delta)^2x,x)_H=(E(\Delta)x,E(\Delta)x)_H=\|E(\Delta)x\|_H^2\ge 0. \end{align*} Thus $\mu_x$ is positive as a set function. Since $E(K)=I$, where $I\in\mathcal L(H)$ is the identity operator on $H$, we have \begin{align*} \mu_x(K)=(E(K)x,x)_H=(x,x)_H=\|x\|_H^2. \end{align*} In particular, the total mass of $\mu_x$ is finite. [/step]

[step:Use strong additivity to prove countable additivity on diagonal coefficients]Let $(\Delta_n)_{n=1}^{\infty}$ be a pairwise disjoint sequence of Borel subsets of $K$, and define \begin{align*} \Delta:=\bigcup_{n=1}^{\infty}\Delta_n. \end{align*} For each $N\in\mathbb N$, define the partial union \begin{align*} \Delta^{(N)}:=\bigcup_{n=1}^{N}\Delta_n. \end{align*} Finite additivity of the projection-valued measure gives \begin{align*} E(\Delta^{(N)})x=\sum_{n=1}^{N}E(\Delta_n)x. \end{align*} By strong countable additivity in the definition of a projection-valued measure, the sequence $E(\Delta^{(N)})x$ converges in $H$ to $E(\Delta)x$. Since the map \begin{align*} \ell_x:H\to\mathbb C,\qquad z\mapsto (z,x)_H \end{align*} is a bounded linear functional, applying $\ell_x$ to this norm convergence gives \begin{align*} \mu_x(\Delta)=\lim_{N\to\infty}(E(\Delta^{(N)})x,x)_H. \end{align*} Using finite additivity once more, \begin{align*} (E(\Delta^{(N)})x,x)_H=\sum_{n=1}^{N}(E(\Delta_n)x,x)_H=\sum_{n=1}^{N}\mu_x(\Delta_n). \end{align*} Hence \begin{align*} \mu_x(\Delta)=\lim_{N\to\infty}\sum_{n=1}^{N}\mu_x(\Delta_n)=\sum_{n=1}^{\infty}\mu_x(\Delta_n). \end{align*} Therefore $\mu_x$ is countably additive. Together with positivity and $\mu_x(K)<\infty$, this proves that $\mu_x$ is a finite positive measure.[/step]

[guided]We must show that the diagonal set function is not merely finitely additive, but countably additive. Let $(\Delta_n)_{n=1}^{\infty}$ be pairwise disjoint Borel subsets of $K$, and set \begin{align*} \Delta:=\bigcup_{n=1}^{\infty}\Delta_n. \end{align*} For each $N\in\mathbb N$, define \begin{align*} \Delta^{(N)}:=\bigcup_{n=1}^{N}\Delta_n. \end{align*} The sets $\Delta_1,\dots,\Delta_N$ are disjoint, so finite additivity of the projection-valued measure gives \begin{align*} E(\Delta^{(N)})=\sum_{n=1}^{N}E(\Delta_n) \end{align*} as an identity in $\mathcal L(H)$. Applying both sides to the fixed vector $x\in H$ gives \begin{align*} E(\Delta^{(N)})x=\sum_{n=1}^{N}E(\Delta_n)x. \end{align*} The countable additivity axiom for a projection-valued measure is strong operator additivity: for each vector in $H$, the projections of the finite partial unions converge in norm to the projection of the countable union. Thus \begin{align*} E(\Delta^{(N)})x\to E(\Delta)x \end{align*} in the norm of $H$ as $N\to\infty$. Now we pass from vector convergence to scalar convergence. Define \begin{align*} \ell_x:H\to\mathbb C,\qquad z\mapsto (z,x)_H. \end{align*} This is a bounded linear functional because the inner product is linear in the first argument and the [Cauchy-Schwarz inequality](/theorems/432) gives $|\ell_x(z)|\le \|z\|_H\|x\|_H$ for all $z\in H$. Therefore norm convergence of $E(\Delta^{(N)})x$ implies \begin{align*} (E(\Delta^{(N)})x,x)_H\to (E(\Delta)x,x)_H. \end{align*} By the definition of $\mu_x$, this says \begin{align*} \mu_x(\Delta)=\lim_{N\to\infty}\mu_x(\Delta^{(N)}). \end{align*} For each finite $N$, finite additivity gives \begin{align*} \mu_x(\Delta^{(N)})=\sum_{n=1}^{N}\mu_x(\Delta_n). \end{align*} Combining the two identities, \begin{align*} \mu_x(\Delta)=\lim_{N\to\infty}\sum_{n=1}^{N}\mu_x(\Delta_n)=\sum_{n=1}^{\infty}\mu_x(\Delta_n). \end{align*} This is precisely countable additivity for the positive set function $\mu_x$.[/guided]

[step:Express general matrix coefficients by polarization] Fix $x,y\in H$. For each $\Delta\in\mathcal B(K)$, apply the complex polarization identity to the vectors $E(\Delta)x$ and $y$. Since $E(\Delta)$ is self-adjoint and idempotent, for every $z\in H$ we have \begin{align*} (E(\Delta)z,z)_H=\mu_z(\Delta). \end{align*} With the convention that $(\cdot,\cdot)_H$ is linear in the first argument, the polarization identity gives \begin{align*} \mu_{x,y}(\Delta)=\frac{1}{4}\mu_{x+y}(\Delta)-\frac{1}{4}\mu_{x-y}(\Delta)+\frac{i}{4}\mu_{x+iy}(\Delta)-\frac{i}{4}\mu_{x-iy}(\Delta). \end{align*} Thus $\mu_{x,y}$ is a finite linear combination of the finite positive measures $\mu_{x+y}$, $\mu_{x-y}$, $\mu_{x+iy}$, and $\mu_{x-iy}$. [/step]

[step:Conclude countable additivity and finite total variation for all coefficients] Because each diagonal set function appearing in the polarization formula is a countably additive finite positive measure, their finite linear combination $\mu_{x,y}$ is countably additive. It remains to verify finiteness as a complex measure. For every finite Borel partition $(A_j)_{j=1}^{m}$ of $K$, the polarization formula and the triangle inequality imply \begin{align*} \sum_{j=1}^{m}|\mu_{x,y}(A_j)| \le \frac{1}{4}\sum_{j=1}^{m}\mu_{x+y}(A_j)+\frac{1}{4}\sum_{j=1}^{m}\mu_{x-y}(A_j)+\frac{1}{4}\sum_{j=1}^{m}\mu_{x+iy}(A_j)+\frac{1}{4}\sum_{j=1}^{m}\mu_{x-iy}(A_j). \end{align*} Since each $\mu_z$ is a positive measure, the right-hand side equals \begin{align*} \frac{1}{4}\mu_{x+y}(K)+\frac{1}{4}\mu_{x-y}(K)+\frac{1}{4}\mu_{x+iy}(K)+\frac{1}{4}\mu_{x-iy}(K). \end{align*} Using the diagonal mass formula already proved, this is \begin{align*} \frac{1}{4}\|x+y\|_H^2+\frac{1}{4}\|x-y\|_H^2+\frac{1}{4}\|x+iy\|_H^2+\frac{1}{4}\|x-iy\|_H^2<\infty. \end{align*} Taking the supremum over all finite Borel partitions of $K$ gives \begin{align*} |\mu_{x,y}|(K)<\infty, \end{align*} where $|\mu_{x,y}|$ denotes the total variation measure of $\mu_{x,y}$. Hence $\mu_{x,y}$ is a finite complex measure on $K$. The diagonal case also gives $\mu_x(K)=\|x\|_H^2$, completing the proof. [/step]