Preliminaries. Let $X$ be a set and let $\mathcal A$ be a Boolean algebra of subsets of $X$ (i.e., $\mathcal A\subset 2^X$ such that $\mathcal A$ contains the empty set and is closed under finite unions and complements). Let $\mu$ be a finitely additive probability measure on $(X,\mathcal A)$ with the following property, sometimes called convexity: for each $A\in\mathcal A$ and each $r\in[0,1]$, there is $B\in\mathcal A$ such that $B \subset A$ and $\mu(B) = r \mu(A)$.
Problem. Does there exist a chain $(A_t)_{t\in[0,1]}$ such that
(i) for each $t\in[0,1]$, $A_t\in\mathcal A$,
(ii) for all $s,t\in[0,1]$ with $s < t$, $A_s\subset A_t$, and
(iii) $\{\mu(A_t)\colon t\in[0,1]\} = [0,1]$.
Comments. It is easy to see that for any maximal chain $(A_t)_{t\in[0,1]}$, $\{\mu(A_t)\colon t\in[0,1]\}$ is dense in $[0,1]$. However, since $\mathcal A$ is not assumed to be complete, one cannot easily conclude that this dense subset of all of $[0,1]$. Here, we say that $\mathcal A$ is complete if each subset of $\mathcal A$ has a supremum.
