Fourier-Stieltjes transforms of positive measures are continuous positive definite

For fixed $x\in G$ the map $\gamma \mapsto \gamma (x)$ is continuous on $\widehat{G}$ by [F3], hence Borel measurable, and $\mathrm{\mid }\gamma (x)\mathrm{\mid }=1$ by [F1]; since $\mu$ is finite, this bounded measurable function is $\mu$-integrable by [F2]. Thus $\varphi (x)$ is a well-defined complex number with $\mathrm{\mid }\varphi (x)\mathrm{\mid }\le \mu (\widehat{G})$ for every $x$, and $\varphi (0)={\int }_{\widehat{G}}\gamma (0)d\mu ={\int }_{\widehat{G}}1d\mu =\mu (\widehat{G})$.

Let $n\ge 0$, $x_1,\dots ,x_n\in G$ and $c_1,\dots ,c_n\in \mathbb{C}$. Each function $\gamma \mapsto c_j\overline{c_k}\gamma (x_j-x_k)$ is $\mu$-integrable by [F1] and [F2], so [F3] and $\gamma (x_j-x_k)=\gamma (x_j)\overline{\gamma (x_k)}$ give ${\sum }_{j=1}^n{\sum }_{k=1}^n{c}_j\overline{c_k}\varphi (x_j-x_k)={\int }_{\widehat{G}}{\sum }_{j=1}^n{\sum }_{k=1}^n{c}_j\overline{c_k}\gamma (x_j)\overline{\gamma (x_k)}d\mu (\gamma )={\int }_{\widehat{G}}\mid {\sum }_{j=1}^n{c}_j\gamma (x_j){\mid }^{2}d\mu (\gamma )\ge 0,$ the last inequality because the integrand is a nonnegative measurable function. For $n=0$ the sum is $0$. Hence $\varphi$ is positive definite.

Suppose first that $\mu (\widehat{G})=0$; then $\varphi (x)=0$ for every $x$ by step 1.1, so $\varphi$ is uniformly continuous. If $\mu (\widehat{G})>0$, let $\epsilon >0$ and use [F2] with the open set $\widehat{G}$ to choose a compact $K\subseteq \widehat{G}$ with $\mu (\widehat{G}\setminus K)<\epsilon \mathrm{/}4$; put $\delta :=\epsilon \mathrm{/}(2\mu (\widehat{G}))$. Consider all pairs $(W,U)$ with $W$ open in $\widehat{G}$, $U$ an open identity neighbourhood in $G$, and $\mathrm{\mid }\eta (x)-1\mathrm{\mid }<\delta \mathrm{/}2$ for $\eta \in W$, $x\in U$. Joint continuity [F3] gives such a pair with each prescribed ${\gamma }_0\in K$ inside $W$. Their first coordinates cover $K$, so compactness gives finitely many pairs $(W_i,U_i)$ covering it. Put $U={\bigcap }_i{U}_i$, or $U=G$ if the finite cover is empty. Then $\mathrm{\mid }\gamma (x)-1\mathrm{\mid }<\delta \mathrm{/}2$ for every $\gamma \in K$ and $x\in U$, without choosing neighborhoods separately for every point of $K$.

For $x\in U$ from step 2.1, $\mathrm{\mid }\gamma (x)-1\mathrm{\mid }\le 2$ by [F1], so [F2] and the linearity and triangle inequality of the integral give $\mathrm{\mid }\varphi (x)-\varphi (0)\mathrm{\mid }\le {\int }_K\mathrm{\mid }\gamma (x)-1\mathrm{\mid }d\mu +{\int }_{\widehat{G}\setminus K}\mathrm{\mid }\gamma (x)-1\mathrm{\mid }d\mu <\frac{\delta }{2}\mu (K)+2\mu (\widehat{G}\setminus K)\le \frac{\epsilon }{4}+\frac{\epsilon }{2}<\epsilon \mathrm{.}$

For arbitrary $h,x\in G$, $\mathrm{\mid }\gamma (h+x)-\gamma (h)\mathrm{\mid }=\mathrm{\mid }\gamma (h)\gamma (x)-\gamma (h)\mathrm{\mid }=\mathrm{\mid }\gamma (x)-1\mathrm{\mid }$ by [F1], and subtraction under the integral together with [F2] gives $\mathrm{\mid }\varphi (h+x)-\varphi (h)\mathrm{\mid }=\mid {\int }_{\widehat{G}}\gamma (h)(\gamma (x)-1)d\mu (\gamma )\mid \le {\int }_{\widehat{G}}\mathrm{\mid }\gamma (x)-1\mathrm{\mid }d\mu (\gamma )<\epsilon$ whenever $x\in U$, where the final inequality repeats the estimate of step 3.1; the bound does not depend on $h$, so $\varphi$ is uniformly continuous. With step 1.2 and step 1.1, $\varphi$ is a continuous positive definite function with $\varphi (0)=\mu (\widehat{G})=\mathrm{\parallel }\mu \mathrm{\parallel }$.