Let $G$ be a locally compact Abelian group (we can think that $G={\mathbb R}$). Let $C_0(G)$ be the space of continuous functions $u:G\to{\mathbb C}$ vanishing at infinity with the usual $\sup$-norm, and $M(G)$ the space of complex Radon measures on $G$, understood as a Banach space dual to the Banach space $C_0(G)$: $$ M(G)=C_0(M)^*. $$ The Banach space $M(G)$ is naturally endowed with the structure of a Banach algebra with the convolution of measures as the multiplication $$ \alpha * \beta,\qquad \alpha,\beta\in M(G), $$ and, moreover, it also has a natural involution: $$ \alpha^\bullet(u)=\overline{\alpha(\overline{\widetilde{u}})},\qquad \alpha\in M(G),\quad u\in C_0(G), $$ where $\overline{z}$ is the usual involition of the complex number $z\in{\mathbb C}$, and $\widetilde{u}$ is the antipode of the function $u\in C_0(G)$: $$ \widetilde{u}(t)=u(t^{-1}),\qquad t\in G. $$

So $M(G)$ is a Banach algebra with involution. And formally we can define states on $M(G)$, as linear continuous functionals $\sigma:M(G)\to{\mathbb C}$ with the properties $$ \sigma(\alpha^\bullet*\alpha)\ge 0, \qquad \sigma(\delta^e)=1,\qquad \alpha\in M(G), $$ where $e\in G$ is the unit in $G$ and $\delta^e$ is the delta-measure supported in $e$.

It is more or less obvious, that each normalized continuous positive-definite function $f:G\to{\mathbb C}$ defines a state on $M(G)$ by the formula $$ \sigma(\alpha)=\int_G f(t)\ \alpha(d t) \tag{1} $$ and by the Bochner theorem, we can represent the action of $\sigma$ as the integral by some positive measure $\mu$ on the Pontryagin dual group $\widehat{G}$: $$ \sigma(\alpha)=\int_{\widehat{G}} {\mathcal F}(\alpha)(\chi)\ \mu(d\chi) \tag{2} $$ where $$ {\mathcal F}(\alpha)(\chi)=\int_G \chi(t)\ \alpha(d t),\qquad \chi\in\widehat{G} $$ is the Fourier transform of the measure $\alpha$.

I would call such states $\sigma$ on $M(G)$ "regular states", since intuitively they are what we expect them to be.

But strange thing, there are some other, "irregular states" on $M(G)$. Namely, the function $$ f(t)=\begin{cases}1, & t=e\\ 0, & t\ne e\end{cases} $$ (where $e\in G$ is the unit of the group $G$) gives another state by the same formula (1), which can be simplified in this case as follows: $$ \sigma(\alpha)=\int_G f(t)\ \alpha(d t)=\alpha(\{e\}). \tag{3} $$ And if $G$ is not discrete, then $f$ cannot be a continuous function on $G$, and in this case the state (3) cannot be represented in the form (2) (i.e., as an integral of the Fourier transform by some measure $\mu\ge 0$ on $\widehat{G}$).

And I deduce from this the there is a qualitative difference between the states on $M(G)$, which can be represented in the form (2) and those that cannot. My question is

what is this qualitative difference between the states (2) and (3) in terms of the properties of $M(G)$ considering it as the Banach algebra with involution?

In other words if we have a state $\sigma:M(G)\to{\mathbb C}$, do we have a possibility to understand if it can be represented in the form (2) if we look only at the properties of $\sigma$ as a functional on $M(G)$? (We can't restore the topology of $G$ from $M(G)$, so our answer can't be like this: $\sigma$ must be generated by some continuous positive-definite function $f$.)

My conjecture is that perhaps the difference between (2) and (3) is that these states relate differently to the monotonicity conditions. Using the Beppo Levi lemma one can show that if a sequence of measures $\alpha_n\in M(G)$ tends to a measure $\alpha\in M(G)$ monotonously with respect to the preorder on $M(G)$ generated by the involution, $$ 0\le\alpha_1\le\alpha_2\le...\le\alpha_n\le...\le\sup_n\alpha_n=\alpha $$ then the values of the state $\sigma$ of the form (2) tend to what we need: $$ \sigma(\alpha_n)\underset{n\to\infty}{\longrightarrow}\sigma(\alpha) \tag{4} $$ But if we consider the state $\sigma$ of the form (3), we cannot use the Bochner theorem, cannot represent $\sigma$ in the form (2), and because of that I am not sure that this state satisfies the condition (4). So a simplified version of my question is the following:

is the state $\sigma$ of the form (3) on $M(G)$ monotone in the sense that it preserves the monotonous convergence of sequences?

I think this must be simple, but I stuck in this, and if somebody could clarify this to me, I would be very grateful.