Fourier-Stieltjes transforms determine finite Radon measures

(The transform algebra separates the character space.) Put $\mathcal{A}:=\{\widehat{f}:f\in A\}$. By [F2] the map $f\mapsto \widehat{f}$ is a linear bijection onto $\mathcal{A}$ that carries convolution to pointwise multiplication and $\ast$ to complex conjugation, so $\mathcal{A}$ is a self-adjoint complex algebra of functions on $\widehat{G}$; by [F1] it is contained in $C_0(\widehat{G})$. Viewed on $\mathrm{\Delta }(A^{+})$ through $\chi \mapsto \chi (0,f)$, it is a subalgebra of $C(\mathrm{\Delta }(A^{+}))$ that contains $\widehat{f}$ and vanishes at $q$. If $\gamma \mathrm{\ne }\eta$ in $\widehat{G}$ then $h_{\gamma }^{+}\mathrm{\ne }{h}_{\eta }^{+}$, and since the Gelfand topology is Hausdorff there is $a=(z,f)\in A^{+}$ with $h_{\gamma }^{+}(a)\mathrm{\ne }{h}_{\eta }^{+}(a)$; the two characters agree on constants, so $\widehat{f}(\gamma )\mathrm{\ne }\widehat{f}(\eta )$. Likewise $h_{\gamma }^{+}\mathrm{\ne }q$ gives $f\in A$ with $\widehat{f}(\gamma )\mathrm{\ne }0$. Hence $\mathcal{A}$ separates points of $\widehat{G}$ and has no common zero there, while its extension to $\mathrm{\Delta }(A^{+})$ has the unique common zero $q$.

(The pairing identity.) Let $f\in L^1(G,m_G)$. Then ${\int }_{\widehat{G}}\widehat{f}(\gamma )d\mu (\gamma )={\int }_{\widehat{G}}{\int }_{G}f(x)\overline{\gamma (x)}d{m}_G(x)d\mu (\gamma )\mathrm{.}$ The integrand is absolutely integrable over the product of a $\sigma$-finite essential support of $f$ with $(\widehat{G},\mu )$ by [F4], so Fubini applies and, using $\overline{\gamma (x)}=\gamma (-x)$, ${\int }_{\widehat{G}}\widehat{f}d\mu ={\int }_{G}f(x)({\int }_{\widehat{G}}\gamma (-x)d\mu (\gamma ))d{m}_G(x)={\int }_{G}f(x)\mathrm{\Phi }(-x)d{m}_G(x)=0.$

(Uniform density of the transform algebra.) By step 1.1 the algebra $B:=\{\chi \mapsto \chi (0,f):f\in A\}\subseteq C(\mathrm{\Delta }(A^{+}))$ is self-adjoint, point-separating, and its only common zero is $q$. The vanishing-at-one-point case of [F3] applied to $X=\mathrm{\Delta }(A^{+})$ and $x_0=q$ gives that the uniform closure of $B$ is $\{G\in C(\mathrm{\Delta }(A^{+})):G(q)=0\}$. Under the homeomorphism $\mathrm{\Delta }(A^{+})\cong \widehat{G}\cup \{q\}$ of [F2], this is exactly the space of continuous functions on $\widehat{G}$ vanishing at infinity; hence $\mathcal{A}=\{\widehat{f}:f\in A\}$ is uniformly dense in $C_0(\widehat{G})$.

(Vanishing against all of $C_0(\widehat{G})$.) Let $H\in C_0(\widehat{G})$ and $\epsilon >0$. By step 2.1 choose $f\in A$ with $\mathrm{\parallel }\widehat{f}-H{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon \mathrm{/}(1+\mathrm{\mid }\mu \mathrm{\mid }(\widehat{G}))$. By step 1.2, $\int \widehat{f}d\mu =0$, so $\mid {\int }_{\widehat{G}}Hd\mu \mid \le {\int }_{\widehat{G}}\mathrm{\mid }H-\widehat{f}\mathrm{\mid }d\mathrm{\mid }\mu \mathrm{\mid }+\mid {\int }_{\widehat{G}}\widehat{f}d\mu \mid \le \mathrm{\parallel }\widehat{f}-H{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }\mu \mathrm{\mid }(\widehat{G})<\epsilon \mathrm{.}$ Hence ${\int }_{\widehat{G}}Hd\mu =0$ for every $H\in C_0(\widehat{G})$.

(Conclusion $\mu =0$.) Every $h\in C_c(\widehat{G})$ lies in $C_0(\widehat{G})$, so by step 3.1 $\int hd\mu =0$ for all $h\in C_c(\widehat{G})$. Write $\mu ={\mu }_1-{\mu }_2+i({\mu }_3-{\mu }_4)$ with ${\mu }_1,\dots ,{\mu }_4$ finite positive regular Borel measures, as in [F5]. Then for every real-valued $h\in C_c(\widehat{G})$ one has $\int hd{\mu }_1=\int hd{\mu }_2$ and $\int hd{\mu }_3=\int hd{\mu }_4$; by [F5] applied to the Radon measures ${\mu }_1,{\mu }_2$ and then to ${\mu }_3,{\mu }_4$, all four equalities hold as measures, so $\mu =0$.

(Equivalence.) If finite regular complex Borel measures $\mu ,\nu$ on $\widehat{G}$ have the same inverse transform, then $\sigma :=\mu -\nu$ is again a finite regular complex Borel measure and its inverse transform vanishes identically; step 4.1 gives $\sigma =0$, that is, $\mu =\nu$.