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):




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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.)




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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.



Correspondence between functions on a set and "states" on its power set
Your correspondence is equivalent to the existence of a [real-valued measurable cardinal](http://en.wikipedia.org/wiki/Measurable_cardinal), a large cardinal concept equiconsistent with the existence of a measurable cardinal. 


First, note that if $\kappa$ is a measurable cardinal, then there is a 2-valued measure $\mu$ on $P(\kappa)$ which is not only countably-additive but $\kappa$-additive, in the sense the measure of the union of fewer than $\kappa$ many disjoint sets is the sum of the measures. For this measure, every set gets measure either 0 or 1, and so there are no disjoint sets of positive measure, and also every singleton gets measure 0. So it is an instance of a violation of your correspondence. 


More generally, if $\kappa$ is a [real-valued measurable cardinal](http://en.wikipedia.org/wiki/Measurable_cardinal), then there is a real-valued $\kappa$-additive measure $\mu$ on $P(\kappa)$ giving measure 0 to singletons. In particular, such a measure would be countably additive, and it would not correspond to function in your sense. 


Conversely, suppose that there were a countably additive real-valued measure $\mu$ on $P(X)$ for some set $X$. If this measure does not correspond to a function, let's subtract from it the sum measure on singletons, to arrive without loss of generality at a measure that gives measure zero to singletons, but positive measure to the whole space. In this case, let $\kappa$ be the additivity of $\mu$, the largest cardinal such that the $\mu$ measure of any less-than-$\kappa$ sized disjoint union is equal to the sum of the measures individually. In this case, there is a set $Y\subset X$ of positive measure and a $\kappa$ partition of $Y=\cup\_{\alpha\lt\kappa} Y\_\alpha$ such that each $Y\_\alpha$ has measure $0$. We may now define a $\kappa$-additive measure on $P(\kappa)$ by $\mu\_0(I)=\Sigma\_{\alpha\in I}\mu(Y\_\alpha)$. Thus, $\kappa$ is a real-valued measurable cardinal. 


So your question is equivalent to the existence of a real-valued measurable cardinal. Such a hypothesis is equiconsistent with the existence of a measurable cardinal. 


The particular case when the set $X$ has size continuum $c$ corresponds to the situation where $c$ is a real-valued measurable cardinal. This implies a strong failure of the Continuum Hypothesis, since in this case $c$ would be weakly inaccessible. It is equivalent to the existence of a countably-additive extension of Lebesgue measure measuring all sets. (Such an extension cannot be translation invariant by Vitali.)


