Let $L$ be the poset (ordered by set inclusion) that is the power set of some set $X$.

A *state* is a function $s:L \rightarrow [0,1]$ satisfying

i) for {$p_1,p_2,...$}, $p_i \in L$ a pairwise orthogonal (i.e. $p_i \leq p_j'$ where $a'$ is the complement of $a$) countable sequence, $\bigvee_i p_i$ exists, and $s(\bigvee_i p_i) = \sum_i s(p_i)$.

ii) s(X) = 1.

also consider functions $f:X \rightarrow [0,1]$.

Then for $X$ **countably infinite**, every state $s$ is in one to one correspondance to a function $f$ by the following argument (skip the next three paragraphs if you are OK with this):

As $L$ is atomistic, we have for arbitrary $p \in L$, $p = \bigvee_i a_i$ ($= \bigcup_i a_i$ as the join of the poset corresponds to the union of subsets) where $a_i \in L$ are atoms (which are pairwise orthogonal).

The atoms $a_i$ are in one to one correspondence to the elements of $X$ such that, given $f:X \rightarrow [0,1]$, we can define $f$ on the set of atoms via $f(a) = f(x_a)$ where $x_a$ is the element of $X$ associated to the atom $a$.

Then for a given state $s$ and arbitrary $p \in L$, $s(p) = s(\bigvee_i a_i) = \sum_i s(a_i)$. So $s$ is determined by its values on the atoms and we can associate a state to a function $f$ by setting for all atoms $a$, $s_f(a) = f(a)$. This is bijective. (End of argument.)

Questions:

1) Now for $X = R^n$ (or uncountable) is there a similar correspondence?

2) Or do I need to adapt condition i) ?

What I fail to show in the uncountable case, is whether condition i) is strong enough to ensure that any state on the power set of such an $X$ is uniquely determined by its values on the atoms.

I hope this question is worthy of a response, it is my first one and I hesitated for the last 4 days.