Continuous functional calculus produces a regular PVM

Contractivity of $\pi$: if $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$ then $1-\bar{f}f\ge 0$ pointwise, so $1-\bar{f}f=\bar{g}g$ with $g=\sqrt{1-\bar{f}f}$ continuous and hence $I-\pi (f{)}^{\ast }\pi (f)=\pi (g{)}^{\ast }\pi (g)$; for every $x$ this gives $\mathrm{\parallel }x{\mathrm{\parallel }}^2-\mathrm{\parallel }\pi (f)x{\mathrm{\parallel }}^2=\mathrm{\parallel }\pi (g)x{\mathrm{\parallel }}^2\ge 0$, so $\mathrm{\parallel }\pi (f)x\mathrm{\parallel }\le \mathrm{\parallel }x\mathrm{\parallel }$ and $\mathrm{\parallel }\pi (f)\mathrm{\parallel }\le 1$; for general $f\mathrm{\ne }0$ apply this to $f\mathrm{/}\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$.

For all $x,y\in H$ the map $L_{x,y}(f):=⟨\pi (f)x,y⟩$ is a bounded complex linear functional on $C(K;\mathbb{C})$ with $\mathrm{\parallel }{L}_{x,y}\mathrm{\parallel }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$, because $\pi$ is linear and $\mathrm{\mid }⟨\pi (f)x,y⟩\mathrm{\mid }\le \mathrm{\parallel }\pi (f)x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$.

Since $K$ is compact, $C(K;\mathbb{C})=C_0(K;\mathbb{C})$, so the representation theorem applies: for each pair $x,y$ there is a unique finite regular complex Borel measure ${\mu }_{x,y}$ with $⟨\pi (f)x,y⟩=\int fd{\mu }_{x,y}$ for all continuous $f$, and $\mathrm{\mid }{\mu }_{x,y}\mathrm{\mid }(K)=\mathrm{\parallel }{L}_{x,y}\mathrm{\parallel }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$.

Sesquilinearity: for all $x,z,y\in H$, $\lambda \in \mathbb{C}$ and continuous $f$ one has $\int fd{\mu }_{x+z,y}=⟨\pi (f)(x+z),y⟩=\int fd{\mu }_{x,y}+\int fd{\mu }_{z,y}$ and $\int fd{\mu }_{\lambda x,y}=\lambda \int fd{\mu }_{x,y}$, while $\int fd{\mu }_{x,y+z}=\int fd{\mu }_{x,y}+\int fd{\mu }_{x,z}$ and $\int fd{\mu }_{x,\lambda y}=\bar{\lambda }\int fd{\mu }_{x,y}$; both sides in each identity are finite regular complex measures with equal integrals against every continuous $f$, so they coincide by uniqueness in the representation theorem.

For a bounded Borel $h$ the form ${\mathrm{\Lambda }}_h(x,y):=\int hd{\mu }_{x,y}$ is sesquilinear by the previous step, and $\mathrm{\mid }{\mathrm{\Lambda }}_h(x,y)\mathrm{\mid }\le \mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }{\mu }_{x,y}\mathrm{\mid }(K)\le \mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$; hence for each $x$ the map $y\mapsto {\mathrm{\Lambda }}_h(x,y)$ is a bounded conjugate-linear functional and Hilbert Riesz representation gives a unique vector $E(h)x$ with $⟨E(h)x,y⟩=\int hd{\mu }_{x,y}$ for all $y$, where $\mathrm{\parallel }E(h)x\mathrm{\parallel }\le \mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }x\mathrm{\parallel }$; the map $x\mapsto E(h)x$ is linear by sesquilinearity, so $E(h)\in \mathcal{B}(H)$ and $\mathrm{\parallel }E(h)\mathrm{\parallel }\le \mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}$.

For bounded Borel $h_1,h_2$ and scalars $a,b$ one has $E(a{h}_1+b{h}_2)=aE(h_1)+bE(h_2)$ because the defining pairings agree, and $E({\mathbf{1}}_K)=I$ because $⟨E({\mathbf{1}}_K)x,y⟩=\int {\mathbf{1}}_{K}d{\mu }_{x,y}={\mu }_{x,y}(K)=⟨\pi (1)x,y⟩=⟨x,y⟩$; moreover $E(f)=\pi (f)$ for every continuous $f$, since their pairings are equal by the defining property of ${\mu }_{x,y}$.

Conjugate symmetry of the scalar measures: for continuous $f$ one has $\int fd{\mu }_{y,x}=⟨\pi (f)y,x⟩=\overline{⟨\pi (\stackrel{‾}{f})x,y⟩}=\overline{\int \stackrel{‾}{f}d{\mu }_{x,y}}=\int fd\overline{{\mu }_{x,y}}$, so ${\mu }_{y,x}=\overline{{\mu }_{x,y}}$ by uniqueness; consequently for bounded Borel $h$ and all $x,y$, $⟨E(h{)}^{\ast }x,y⟩=\overline{⟨E(h)y,x⟩}=\overline{\int hd{\mu }_{y,x}}=\overline{\int hd\stackrel{‾}{{\mu }_{x,y}}}=\int \bar{h}d{\mu }_{x,y}=⟨E(\bar{h})x,y⟩$, where the third expression inserts ${\mu }_{y,x}=\overline{{\mu }_{x,y}}$ and the fourth uses $\int hd\overline{{\mu }_{x,y}}=\overline{\int \stackrel{‾}{h}d{\mu }_{x,y}}$; hence $E(h{)}^{\ast }=E(\bar{h})$.

Multiplicativity with a continuous factor: fix continuous $f$; for the measures $\nu :={\int }_{\cdot }fd{\mu }_{x,y}$ and $\rho :={\mu }_{x,E(\bar{f})y}$ and every continuous $g$ one computes $\int gd\nu =\int fgd{\mu }_{x,y}=⟨E(fg)x,y⟩=⟨E(f)E(g)x,y⟩=⟨E(g)x,E(\bar{f})y⟩=\int gd\rho$, where the middle identity uses $E(f)=\pi (f)$, the multiplicativity of $\pi$ and $E(g)=\pi (g)$ for continuous $g$; here $\nu$ is regular because $\mathrm{\mid }\nu \mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }{\mu }_{x,y}\mathrm{\mid }$ and $\rho$ is regular by construction, so uniqueness in the representation theorem gives $\nu =\rho$ and hence $\int hd(f{\mu }_{x,y})=\int hd{\mu }_{x,E(\bar{f})y}$ for every bounded Borel $h$; therefore $E(fh)=E(f)E(h)$ for every bounded Borel $h$.

Measure identity for a Borel density: for every bounded Borel $f$ and all $x,y$ the finite complex measures $f{\mu }_{x,y}:B\mapsto {\int }_{B}fd{\mu }_{x,y}$ and ${\mu }_{x,E(\bar{f})y}$ are equal, because for every continuous $g$ one has $\int gd(f{\mu }_{x,y})=\int fgd{\mu }_{x,y}=⟨E(fg)x,y⟩=⟨E(f)E(g)x,y⟩=⟨E(g)x,E(\bar{f})y⟩=\int gd{\mu }_{x,E(\bar{f})y}$, using multiplicativity with a continuous second factor, which follows from the continuous-factor case together with $E(h{)}^{\ast }=E(\bar{h})$.

Full multiplicativity: for bounded Borel $f,h$ and all $x,y$, $⟨E(fh)x,y⟩=\int fhd{\mu }_{x,y}=\int hd(f{\mu }_{x,y})=\int hd{\mu }_{x,E(\bar{f})y}=⟨E(h)x,E(\bar{f})y⟩=⟨E(h)x,E(f{)}^{\ast }y⟩=⟨E(f)E(h)x,y⟩$, so $E(fh)=E(f)E(h)$.

The set function $B\mapsto E(B):=E({\mathbf{1}}_B)$ takes values in orthogonal projections: $E(B{)}^2=E({\mathbf{1}}_B^2)=E({\mathbf{1}}_B)=E(B)$ by multiplicativity and $E(B{)}^{\ast }=E({\mathbf{1}}_B)=E(B)$ by conjugation symmetry; moreover $E(\mathrm{\varnothing })=0$, $E(K)=I$ and $E(B\cap C)=E(B)E(C)$ for all Borel $B,C$.

Strong countable additivity and regularity: for pairwise disjoint Borel sets $B_n$ with union $B$ and $C_N:={\bigcup }_{n\le N}{B}_n$ one has $E(B)-E(C_N)=E({\mathbf{1}}_{B\setminus C_N})$ by linearity, so $\mathrm{\parallel }E(B)x-E(C_N)x{\mathrm{\parallel }}^2=⟨E({\mathbf{1}}_{B\setminus C_N})x,x⟩={\mu }_{x,x}(B\setminus C_N)$; the sets $B\setminus C_N$ decrease to $\mathrm{\varnothing }$ and ${\mu }_{x,x}$ is a finite complex measure, so its values on them tend to $0$, giving $E(B)x={\lim }_{N}E(C_N)x={\sum }_{n}E(B_n)x$ in norm; moreover $E_x(B)=⟨E(B)x,x⟩={\mu }_{x,x}(B)$ is a regular Borel measure, since ${\mu }_{x,x}$ is regular by construction.

Uniqueness: if $E^{\mathrm{\prime }}$ is a regular PVM on the Borel $\sigma$-algebra of $K$ with $\pi (f)=\int fd{E}^{\mathrm{\prime }}$ for every continuous $f$, then for each $x$ the finite regular positive measures $E_x$ and $E_x^{\mathrm{\prime }}$ have $\int fd{E}_x=⟨\pi (f)x,x⟩=\int fd{E}_x^{\mathrm{\prime }}$ for every continuous $f$, so $E_x=E_x^{\mathrm{\prime }}$ by the uniqueness theorem for Radon measures; the polarization formula $⟨E(B)x,y⟩=\frac{1}{4}{\sum }_k{i}^k{E}_{x+i^{k}y}(B)$ for the sesquilinear form $(x,y)\mapsto ⟨E(B)x,y⟩$ then gives $⟨E(B)x,y⟩=⟨E^{\mathrm{\prime }}(B)x,y⟩$ for all $B,x,y$, hence $E(B)=E^{\mathrm{\prime }}(B)$.

Consequently $E(B)=E({\mathbf{1}}_B)$ defines a regular PVM on the Borel $\sigma$-algebra of $K$ with $\int fdE=\pi (f)$ for every continuous $f$, and it is the unique such regular PVM; for bounded Borel $h$ the operator $\int hdE$ of Bounded borel pvm integral coincides with $E(h)$, since both have the pairings $\int hd{\mu }_{x,y}$.