Proof. Since $\OM$ is LCH, it is a completely regular space so the map $\om\mto \ha{\om}$ is a homeomorphism of $\OM$ into $\OM(A)$. Let $\rho\in\OM(A)$ then $\rho$ is a linear functional on $C_{\infty}(\OM)$ such that $\rho\geq 0$ if $x\geq 0$ as a function. Hence there exists a positive Radon measure $\mu$ on $\OM$ such that $\rho(x)=\integral{\OM}{}x(\om)d\mu(\om)$ and $\mu(\OM)=1$. By multiplicativity of $\rho$

Hence for each $x\in C_{\infty}(\OM)$, $E_{x}\define\curl{\om\in\OM:x(\om)\neq \rho(x)}$ is a $\mu$-nullset. If $\supp{\mu}$ contains $\om_{1}\neq \om_{2}$ then $\exists x\in C_{\infty}(\OM)$ with $x(\om_{1})\neq x(\om_{2})$ since $C_{\infty}(\OM)$ separates points since $\OM$ is completely regular. By continuity $\exists U_{i}\ni \om_{i}$ open neighborhoods where $x|_{U_1}\no\equiv x|_{U_2}$ hence on $U_{1}\cup U_{2}$ we have $\abs{x(\om)-\rho(x)}^{2}>0$. Since $\mu$ is a positive Radon measure, $\mu(U_{i})>0$ thus

We conclude $\mu$ is concentrated at a single point $\om\in\OM$ so $\rho(x)=x(\om)$ and $\rho=\ha{\om}$.$\square$