Nondegenerate representations of C0 have regular PVMs

If $\mathcal{H}=\{0\}$, let $P$ be the zero PVM, $P(B)=0$ for every Borel $B$; then $P(X)=I=0$ and $\int fdP=0=T(f)$ for every $f$, and it is the only PVM on $\mathcal{H}=0$. So assume $\mathcal{H}\mathrm{\ne }\{0\}$ from now on.

Extend $T$ to a unital star homomorphism $T^{+}:C(X^{+})\to \mathcal{B}(\mathcal{H})$ by $T^{+}(f)=T(f-f(\mathrm{\infty })\mathbf{1})+f(\mathrm{\infty })I$, where $f-f(\mathrm{\infty })\mathbf{1}$ is regarded as an element of $C_0(X)$ through the open inclusion $X\subseteq X^{+}$: it is continuous on $X$ and tends to $0$ at $\mathrm{\infty }$ because $f$ does. The map $f\mapsto f-f(\mathrm{\infty })\mathbf{1}$ is linear, so $T^{+}$ is linear and $T^{+}(1)=I$; and $T^{+}$ is multiplicative and conjugation preserving because for $f,g\in C(X^{+})$, writing $f=f_0+c$, $g=g_0+d$ with $c=f(\mathrm{\infty })$, $d=g(\mathrm{\infty })$ and $f_0,g_0\in C_0(X)$, one has $fg=f_0{g}_0+d{f}_0+c{g}_0+cd$ with $f_0{g}_0+d{f}_0+c{g}_0\in C_0(X)$, so $T^{+}(fg)=T(f_0)T(g_0)+dT(f_0)+cT(g_0)+cdI=T^{+}(f)T^{+}(g)$, and $T^{+}(\bar{f})=T(\overline{f_0})+\bar{c}I=T(f_0{)}^{\ast }+\bar{c}I=T^{+}(f{)}^{\ast }$.

By [F2] applied to the nonempty compact Hausdorff space $X^{+}$ and the unital star homomorphism $T^{+}$ there is a unique regular PVM $E^{+}$ on $X^{+}$ with $T^{+}(h)={\int }_{X^{+}}hd{E}^{+}$ for every $h\in C(X^{+})$.

For every $f\in C_0(X)$ one has $T(f)E^{+}(\{\mathrm{\infty }\})=0$: since $f{\mathbf{1}}_{\{\mathrm{\infty }\}}=0$ as a bounded Borel function on $X^{+}$ and the bounded integral is multiplicative, $T(f)E^{+}(\{\mathrm{\infty }\})={\mathrm{\Phi }}_{E^{+}}(f){\mathrm{\Phi }}_{E^{+}}({\mathbf{1}}_{\{\mathrm{\infty }\}})={\mathrm{\Phi }}_{E^{+}}(0)=0$.

Nondegeneracy forces $E^{+}(\{\mathrm{\infty }\})=0$: suppose $E^{+}(\{\mathrm{\infty }\})\mathrm{\ne }0$ and pick $\xi =E^{+}(\{\mathrm{\infty }\})\xi \mathrm{\ne }0$ in its range; then for every $f\in C_0(X)$ and $\eta \in \mathcal{H}$, $⟨\xi ,T(\bar{f})\eta ⟩=⟨T(f)\xi ,\eta ⟩=⟨T(f)E^{+}(\{\mathrm{\infty }\})\xi ,\eta ⟩=0$ by [step 4.1], so $\xi$ is orthogonal to the linear span of $T(C_0(X))\mathcal{H}$, which is dense by nondegeneracy; hence $\xi =0$, a contradiction.

Define $P(B):=E^{+}(B)$ for Borel $B\subseteq X$. This is a PVM on $X$: the Borel sets of the open subspace $X$ are exactly the traces of Borel sets of $X^{+}$, the values are orthogonal projections with $P(\mathrm{\varnothing })=0$, $P(X)=E^{+}(X)=I-E^{+}(\{\mathrm{\infty }\})=I$ by [step 5.1], multiplicativity and countable additivity are inherited from $E^{+}$. For $f\in C_0(X)$, $T(f)=T^{+}(f)={\int }_{X^{+}}fd{E}^{+}={\int }_{X}fdP$, since $f$ vanishes at $\mathrm{\infty }$ and $E^{+}(\{\mathrm{\infty }\})=0$.

$P$ is regular: for each $x$, the finite measure $P_x(B)=E_x^{+}(B)$ on $X$ is the restriction of the regular Borel measure $E_x^{+}$ on $X^{+}$; inner regularity holds because each compact subset of $X$ in the subspace topology is compact in $X^{+}$, and outer regularity holds because open subsets of $X$ are open in $X^{+}$.

Uniqueness: if $P^{\mathrm{\prime }}$ is any regular PVM on $X$ with $T(f)={\int }_{X}fd{P}^{\mathrm{\prime }}$ for all $f\in C_0(X)$, let $\widetilde{P^{\mathrm{\prime }}}$ be its extension by zero at infinity, $\widetilde{P^{\mathrm{\prime }}}(B):=P^{\mathrm{\prime }}(B\cap X)$ for Borel $B\subseteq X^{+}$. This is a regular PVM on $X^{+}$: values are orthogonal projections, $\widetilde{P^{\mathrm{\prime }}}(X^{+})=P^{\mathrm{\prime }}(X)=I$ and $\widetilde{P^{\mathrm{\prime }}}(\{\mathrm{\infty }\})=0$, countable additivity and multiplicativity are inherited from $P^{\mathrm{\prime }}$, and its finite scalar measures are inner regular on all Borel sets, including those containing $\mathrm{\infty }$, by compact approximation inside $X$. Outer regularity follows by applying inner regularity to complements in the compact space $X^{+}$; thus the extension is regular. For $h\in C(X^{+})$ write $h=h_0+c$ with $h_0\in C_0(X)$ and $c=h(\mathrm{\infty })$; then $\int hd\widetilde{P^{\mathrm{\prime }}}=\int h_{0}d{P}^{\mathrm{\prime }}+c\widetilde{P^{\mathrm{\prime }}}(X^{+})=T(h_0)+cI=T^{+}(h)$, so $\widetilde{P^{\mathrm{\prime }}}$ represents $T^{+}$ and the uniqueness in [F2] gives $\widetilde{P^{\mathrm{\prime }}}=E^{+}$ and hence $P^{\mathrm{\prime }}=P$.

Thus for nonzero $\mathcal{H}$ there is exactly one regular PVM $P$ on $X$ with $P(X)=I$ and $T(f)={\int }_{X}fdP$, namely the restriction of $E^{+}$; for $\mathcal{H}=\{0\}$ the zero PVM is the unique one by [step 1.1]. Both cases together prove the claim.