Background: for a discrete abelian group $G$, a character of $G$ is a homomorphism $\chi:G\to \mathbf S^1$, $\mathbf S^1$ being the circle group $\{z\in \mathbb C:|z|=1\}$ with ordinary multiplication. The set of characters of $G$ is denoted $\widehat{G}$, and the trivial character $\operatorname{id}_{\widehat{G}}$ is the one where $\chi(g)=1$ for all $g\in G$. With the group operation of pointwise multiplication and the topology of pointwise convergence, $\widehat{G}$ is a compact abelian group. Given a Borel measure $\sigma$ on $\widehat{G}$, its Fourier transform $\hat{\sigma}:G\to \mathbb C$ is given by $\hat{\sigma}(g):=\int \chi(g)\, d\sigma(\chi)$.
A sequence of elements $g_n\in G$ is called Hartman uniformly distributed (Hartman-u.d.) if for every nontrivial character $\chi\in\widehat{G}$, $\lim_{N\to\infty} \frac{1}{N}\sum_{n=1}^N \chi(g_n)=0$. The following conditions are well known to be equivalent.
(i) $(g_n)_{n\in \mathbb N}$ is Hartman-u.d.
(ii) For every positive finite Borel measure $\sigma$ on $\widehat{G}$, $\lim_{N\to\infty} \frac{1}{N}\sum_{n=1}^N \hat{\sigma}(g_n) = \sigma(\{\operatorname{id}_{\widehat{G}}\})$.
The usual proof of (i)$\implies$(ii) uses the dominated convergence theorem: when $(g_n)$ is Hartman-u.d., we have $$ \lim_{N\to\infty} \frac{1}{N}\sum_{n=1}^N \hat{\sigma}(g_n) = \lim_{N\to\infty} \frac{1}{N}\sum_{n=1}^N \int \chi(g_n)\, d\sigma(\chi)=\int \lim_{N\to\infty} \frac{1}{N}\sum_{n=1}^N \chi(g_n)\, d\sigma(\chi)=\int 1_{\operatorname{id}_{\widehat{G}}}\, d\sigma. $$
The proof of (ii) $\implies$ (i) is merely the observation that a given character $\chi$ is the Fourier transform of the Dirac mass $\delta_{\{\chi\}}$.
We write $\ell^\infty(G)$ for the space of bounded $\mathbb C$-valued functions on $G$ with the supremum norm. A mean on $\ell^\infty(G)$ is a linear functional such that $m(f)\geq 0$ for every bounded nonnegative $f$ on $G$, and $m(1_G)=1$.
Question: Given a discrete abelian group $G$, is it true that for every mean $m$ on $\ell^\infty(G)$, the following are equivalent?
(i') $m(\chi)=0$ for every nontrivial $\chi\in\widehat{G}$.
(ii') $m(\hat{\sigma})=\sigma(\{\operatorname{id}_{\widehat{G}}\})$ for every positive finite Borel measure $\sigma$ on $\widehat{G}$.
As above, (ii')$\implies$(i') is straightforward. But the natural generalization of the proof that (i)$\implies$(ii) might not work: since $m$ is not necessarily countably additive, it may be that $m(\hat{\sigma})\neq \int m(\chi)\, d\sigma(\chi)$.
I don't know the answer for any infinite abelian group $G$; the special case where $G=\mathbb Z$ is already interesting.
