Weak* convergence of measures on Boolean algebras

The Dieudonné-Grothendieck theorem asserts that given a compact Hausdorff space $K$ and a uniformly bounded family $\mathcal{K}\subset C(K)^*$ (the dual to the Banach space of continuous real-valued functions on $K$), $\mathcal{K}$ is relatively weakly compact if and only if for every sequence $(O_n)_{n\in\mathbb{N}}$ of pairwise disjoint open subsets of $K$ we have $\lim_{n\to\infty}\sup_{\mu\in\mathcal{K}}|\mu(O_n)|=0$. In the case of a Boolean algebra $\mathcal{A}$ and a uniformly bounded sequence $(\mu_n)_{n\in\mathbb{N}}$ of bounded finitely additive measures on $\mathcal{A}$, by the Eberlein-Šmulian theorem and regularity of measures, we can rephrase the theorem as follows: $(\mu_n)$ is not weakly convergent if and only if there exist an antichain $(A_k)_{k\in\mathbb{N}}$ in $\mathcal{A}$, a subsequence $(\mu_{n_k})_{k\in\mathbb{N}}$ and $\varepsilon>0$ such that $|\mu_{n_k}(A_k)|>\varepsilon$ for every $k\in\mathbb{N}$. (By antichain I mean that for any $n\neq k$ we have $A_n\cdot A_k=0$.)

I am interested in the following issue. Is there an analogous characterization for non-weak*-convergence? I.e. can we characterize sequences of bounded finitely additive measures on a Boolean algebra which are not weak* convergent in terms of existence of some (finite?) antichains in this algebra? Maybe with some additional assumption on $(\mu_n)$ or $(A_k)$?

(Recall that a sequence $(\mu_n)$ on $\mathcal{A}$ is weak* convergent to $\mu$ if $\int_{St(\mathcal{A})}fd\mu_n\to\int_{St(\mathcal{A})}fd\mu$ for every $f\in C(St(\mathcal{A}))$, where $St(\mathcal{A})$ denotes the Stone space of $\mathcal{A}$.)

• Consider the sequence $\{\delta_n\colon n\in \mathbb{N}\}$ in $C(\beta \mathbb N)^*$, which is conspicuously not weak*-convergent and neither is any subsequence. What antichain would you associate with this sequence? – Tomek Kania Jun 27 '17 at 16:37
• Tomek, actually for this sequence you have a nice antichain about which I (possibly) ask, namely $(\{n\})_{n\in\mathbb{N}}$ ($\subset\wp(\mathbb{N})=\mathcal{A}$), which gives $\delta_n(\{n\})=1$. Btw, $C(\beta\omega)$ has the Grothendieck property, so for every not weak*-convergent sequence you get a nice antichain from the Dieudonné-Grothendieck theorem. – Damian Sobota Jun 27 '17 at 18:01
• Note everyone that I am not asking precisely for the exact copy of the D-G theorem for weak* topology, but about its variation -- possibly quite far but still somehow similar, also involving some combinatorics or descriptive set theory (Borel hierarchy to some low level). – Damian Sobota Jun 27 '17 at 18:02
• Right, I see the point now. Cheers. – Tomek Kania Jun 27 '17 at 18:04