Let $G$ be a locally compact group and we put $L^{1}(G)= \ \text {The space of all complex functions which are integrable with respect to the Haar measure of } \ G.$
Let $\Gamma$ is the dual group (or character group) of the locally compact abelian group $G$ (with addition as group operation), and we define the map(Fourier transform) from $L^{1}(G)$ to $C(\Gamma)$ as $$Fg(y)= \hat{g}(y):=\int_{G} <-x, y> g(x) dx; (y\in \Gamma),$$ where the symbol $<x, y>$ denotes the value of the character $y$ at the point $x$.
We consider, the algebras of all Fourier transforms, $A(\Gamma) = \{ f:\Gamma \to \mathbb C : \exists g \in L^{1}(G) \ \text {such that} \ f(y)= \hat{g}(y)\}$
and the algebra of all Fourier-Stieltjes transforms on $\Gamma$,$B(\Gamma) = \{f:\Gamma \to \mathbb C : \exists \ \text {a bounded complex Borel measure } \ \mu \ \text{on} \ G \ \text {such that} \ f(y) = \hat {\mu(y)} \}.$ ($f(y)= \hat {\mu}(y)= \int_{G} <-x, y> d\mu(x); (y\in \Gamma )$)
My questions are: (1) In what sense we can think of $B(\Gamma)$ as a generalization of $A(\Gamma)$ ?
(2) If $\Gamma$ is compact; why, $A(\Gamma) = B(\Gamma)$ ?
Thanks,