Given a finitely additive positive regular bounded measure $\mu$ on ${\mathbb R}^n$ (i.e. a positive linear functional on $C_b({\mathbb R}^n)$), I wonder what can be said about its Fourier transform. In case of a countably additive measure, Bochner's theorem tells us that the Fourier transform is a continuous positive definite function. This makes use of dominated convergence, which  I assume  fails in general in the finitely additive case. So I wonder how/if such measures can be characterized in terms of their Fourier transforms.

1$\begingroup$ Just a comment: maybe a preliminary question should be, are there any nontrivial examples of finitely but non countably additive measures defined on Lebesgue sets? $\endgroup$ – Piero D'Ancona Sep 30 '13 at 12:40

$\begingroup$ As far as I know, the dual space of $C_b({\mathbb R}^n)$ is precisely the space of all regular bounded finitely additive measures on the Borel sigma algebra of ${\mathbb R}^n$, so there should be many examples. To specify them explicitly seems however to be close to impossible. $\endgroup$ – Gandalf Lechner Sep 30 '13 at 14:47

2$\begingroup$ @Piero: yes, by amenability of $\mathbb{R}$, there are finitely additive measures $m$ such that $m(\mathbb{R}^n)=1$ and $m$ is invariant by translations. (Obviously, such $m$ is zero on bounded subsets and hence cannot be countably additive.) $\endgroup$ – YCor Oct 26 '13 at 22:56
I think there is an error in the statement of the question. $C_b(\mathbb{R}^n)$ is isomorphic, as an ordered Banach space, to $C(\beta \mathbb{R}^n )$ where $\beta \mathbb{R}^n$ is the Stone Cech compactification of $\mathbb{R}^n$. By Riesz's theorem, for example Chapter 7 in Folland's "Real Analysis: modern techniques and their applications", the dual of $C(\beta \mathbb{R}^n)$ is the space of finite $\sigma$additive, signed Radon measures $M(\beta \mathbb{R}^n)$. The positive part of the space of globally defined functionals over $C_b(\mathbb{R}^n)$, by the closed graph theorem, equals the positive cone of $M(\beta \mathbb{R}^n)$. In order to describe the finitely additive measures you need to work with the (positive part of the) dual of $L^{\infty}(\mathbb{R}^n)$.
That said, your question admits a partial answer. The Fourier transform of any finitely additive measure will be a (discontinuous) positive type function over $\mathbb{R}_{disc}^n$. For the converse only the existence holds, i.e.: for any positive type function $\phi$ over $\mathbb{R}_{disc}^n$ there are infinitely many elements in $L^{\infty}(\mathbb{R}^n)^{\ast}_{+}$ whose Fourier transform is $\phi$.
Let $\chi_{\theta}(x) = e^{ \theta x}$ be a character and let $m \in L^{\infty}(\mathbb{R}^n)^{\ast}$. We define the Fourier transform of $m$, as $\mathcal{F}m(\theta) = m(\chi_{\theta})$. As the function $\theta \to \chi_{\theta}$ is not continuous all we can say is that $\mathcal{F}m(\theta) \in C_b(\mathbb{R}_{disc}^n)$ and that $\ \mathcal{F}m(\theta) \_{\infty} \leq \m\_{L^{\infty}(\mathbb{R}^n)^{\ast}}$. Notice that when $m$ is positive $\ m \_{L^{\infty}(\mathbb{R}^n)^{\ast}} = m(\mathbb{1}_{\mathbb{R}^n})$. If $a_1, a_2, ... a_k \in \mathbb{C}$ and $\theta_1, \theta_2, ... \theta_k \in \mathbb{R}^n$ then:
$$ \sum_{i,j}{ \bar{a_i} a_j \mathcal{F}m(\theta^{1}_i \theta_j) } = m( ( \sum_{ i }{ \bar{a_i} \bar{ \chi_{ \theta_i } } } ) ( \sum_{ j }{ a_j \chi_{ \theta_j } } ) ) \geq 0 $$
To see that the injectivity fails let $m \in L^{\infty}(\mathbb{R}^n)^{\ast}_{+}$ be a translation invariant and unital element, which exist because $\mathbb{R}^n$ is amenable. The translation invariance implies that $\mathcal{F}m(\theta) = \chi_{\eta} \mathcal{F}m(\theta)$ for every $\eta$. So $\mathcal{F}m$ can either be identically 0 or be $\mathbb{1}_{\{0\}}$. Since $m$ is unital we are in the second case. As there are infinitely many invariant means on an amenable group, see exercise 14 in Paterson's Amenability, the difference of any two will be in the kernel of the Fourier transform.
The surjectivity can be proved by Bochnner's Theorem in LCA groups applied to the Bohr compactification of $\mathbb{R}^n$. There is a surjective, order preserving map $\pi$ from $L^{\infty}(\mathbb{R})^{\ast}$ to the Borel, $\sigma$ additive measures in the Bohr compactification $M(b \mathbb{R}^n)$ given by dualizing the injective inclusion of $C(b \mathbb{R}^n)$ into $L^{\infty}(\mathbb{R}^n)$ obtained by restricting any continuous function in $b \mathbb{R}^n$ to the open dense subset $\mathbb{R}^n \subset b \mathbb{R}^n$. Bochnner's theorem for $b \mathbb{R}^n$ will give you that any positive type function on $\mathbb{R}_{disc}^n$ is given by the Fourier transform of a positive measure $\mu$ in $b \mathbb{R}^n$. Taking any element in $\pi^{1}[\{\mu\}]$ suffices to see the surjectivity.
I hope this is Helpful for you.

$\begingroup$ You may be interested in contributing to a proposal Spanish language version of math stackexchange; it could use some input from fluent graduate students: area51.stackexchange.com/proposals/64529/… $\endgroup$ – Brian Rushton Feb 1 '14 at 3:14

$\begingroup$ This is an old (1940) result of A.D.Alexandroffsee his Math. Sbornik paper "Additive setfunctions in abstract spaces", which is the first of three seminal papers and is available online. Secondly, the statement about the dual of $L^\infty$ is wrong (I am assuming that this denotes the usual space of equivalence classes under equality a.e. with respect to Lebesgue measure). For the dual one only gets those measures which are absolutely continuous with respect to this measure in the sense that they vanish on Lebesgue null sets. $\endgroup$ – holloway Sep 28 '14 at 8:03

1$\begingroup$ This is a comment on the above answer, not an answer itself (I am not entitled) but which i feel should be made since the first paragraph has two misleading statements. Firstly, while it is true that the dual of the space of bounded, continuous functions can be identified with measures on the StoneČech compactification, it can also, perhaps more naturally, be regarded as a space of measures on the space itself as stated in the OP. (continued in next comment) $\endgroup$ – holloway Sep 28 '14 at 8:03