Let $\mu$ be a regular Borel measure on $\mathbb Z$ and we put, $$\|\mu\|:= |\mu| (\mathbb Z)= \text {total variation of} \ \mu . $$ and define
$$M(\mathbb Z):= \{\mu: P(\mathbb Z)\to \mathbb C : \mu \ \text{is regular and } \ \|\mu\|< \infty \}$$ If $\mu \in M(\mathbb Z)$, the function $\hat{\mu}$ defined on $\mathbb T$ by $$\hat{\mu}(e^{i\theta})= \int_{\mathbb Z} e^{-in \theta} d\mu(n); (e^{i\theta} \in \mathbb T)$$ is called the Fourier-Stieltjes transform of $\mu$ and we define
$$B(\mathbb T) := \{f:\mathbb T \to \mathbb C : \exists \ \mu \in M(\mathbb Z) \ \text {such that} \ f(y) = \hat{\mu}(y) \}.$$
My naive Questions are: (1) What are members of $M(\mathbb Z)$ and the what are its corresponding members in $B(\mathbb T)$ ?
My attempt: Example: Fix $n_{0}\in \mathbb Z$ and consider Dirac measure $\delta_{n_{0}}: P(\mathbb Z) \to \{0,1 \}$ then $\delta_{n_{0}}\in M(\mathbb Z)$; now let us try to compute corresponding member of $B(\mathbb T)$; namely, $\hat{\delta_{n_{0}}}(e^{i\theta})= \int_{\mathbb Z} e^{-in\theta} d\delta_{n_{0}}(n)$; but, now I don't know how to interpret this last integral ? (I want to see few more specific examples of different types)
(2) How to prove, $B(\mathbb T)= A(\mathbb T):= \{f:\mathbb T \to \mathbb C : \exists \ g\in \ell^{1}(\mathbb Z) \ \text {such that} \ f(e^{i\theta})= \hat{g}(e^{i\theta})\} $
My attempt:
Let $f\in A(\mathbb T)$. Then there is a $g\in \ell^{1}(\mathbb Z)$ such that $\hat{g}(e^{i\theta})=f(e^{i\theta})$, that is, $\sum_{n\in \mathbb Z} g(n)e^{-in\theta} = f(e^{i\theta})$. I must show, $f\in B(\mathbb T)$, that is to find $\mu \in M(\mathbb Z)$ such that $f(e^{i\theta})=\hat{\mu} (e^{i \theta})= \int_{\mathbb Z} e^{-in\theta} d\mu (n);$ the hypothesis suggests me that, I should take, $\mu:P(\mathbb Z)\to \mathbb C$, in terms of $g$, but I don’t know precisly how to take $\mu$;
On the other hand, let $f\in B(\mathbb T)$. Then there exists $\mu \in M(\mathbb Z)$ such that $\hat{\mu}(e^{i\theta})= f(e^{i\theta})$. I must show, $f\in A(\mathbb T)$, that is, to find $g\in \ell^{1}(\mathbb Z)$ such that, $\hat{g}(e^{i\theta})= f(e^{i\theta})$; so the hypothesis suggests that, take a $g$ such that, $\hat{g}(e^{i\theta})= \hat{\mu}(e^{i\theta})$; by some thing like uniqueness theorem, $\mu = g$ (I am not sure here, this is bit vague!) and my confusion is here $\mu$ is a set function define on $P(\mathbb Z) $ while $g$ is function define on $\mathbb Z$ ?
Definitions: Let $\mathbb Z$ be a topological space with the discrete topology and take $B=P(\mathbb Z)$= the power set of $\mathbb Z$, as the Borel sets on $\mathbb Z$ and a measure on $\mathbb Z$ is a set function $\mu$, defined for all Borel sets of $\mathbb Z$, which is countably additive and for which the $\mu (E)$ is finite if the closure of $E$ is compact. For each measure $\mu$ on $\mathbb Z$ there is associated set function $|\mu|$, the total variation of $\mu$ is defined by , $|\mu|(E)= \sup \sum |\mu(E_{i})|$, the supremum being taken over all finite collections of pairwise disjoints Borel sets $E_{i}$ whose union is $E$. We note that $|\mu|$ is also a measure on $\mathbb Z$. If $|\mu|(E)=\sup |\mu|(K)= \inf |\mu|(v)$ for every Borel set $E$, where $K$ ranges over all compact subset of $E$ and $V$ ranges over all open supersets of $E$, then $\mu$ is called a regular.