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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.

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I adjusted the tags. It is a good question. The fact that it happens to be well-studied is no reason to avoid asking it! –  Gerald Edgar May 12 '10 at 18:32
    
Thanks for adjusting the tags! –  tortortor May 12 '10 at 23:35
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1 Answer

up vote 3 down vote accepted

Your correspondence is equivalent to the existence of a real-valued 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, 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.)

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But of course the OP is asking about real-valued measurable cardinals. Even if their existence may be equiconsistent with 2-valued measurable cardinals, still real-valued measurable cardinals need not be "large" ... In fact the most interesting universe is one where $\mathbb{R}$ itself admits a real-valued measure on its power set. (Ulam showed that $\aleph_1$ is not real-valued measurable.) –  Gerald Edgar May 12 '10 at 18:29
    
Yes, I agree, and I was writing the explanation in the meantime. –  Joel David Hamkins May 12 '10 at 18:41
    
Thank you very much for the fast answer. If I understand correctly: What I called a "state" can be obtained from any real valued measure. Such a measure may exist for P(X) (with X continuum) but only in violation of the Continuum Hypothesis etc. But even if we assume the existence of such measures, the correspondence to the functions (in my sense) on X can no longer hold, as all such measures must assign zero to singletons? –  tortortor May 12 '10 at 23:31
    
Yes, that's it. I should have said that the existence of a counterexample to your correspondence is equivalent to the existence of a real-valued measurable cardinal. –  Joel David Hamkins May 12 '10 at 23:45
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