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.)


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.

  • 2
    $\begingroup$ 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! $\endgroup$ May 12, 2010 at 18:32

1 Answer 1


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.)

  • $\begingroup$ 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.) $\endgroup$ May 12, 2010 at 18:29
  • $\begingroup$ Yes, I agree, and I was writing the explanation in the meantime. $\endgroup$ May 12, 2010 at 18:41
  • $\begingroup$ 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? $\endgroup$
    – tortortor
    May 12, 2010 at 23:31
  • $\begingroup$ 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. $\endgroup$ May 12, 2010 at 23:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.