## Finite measure on the power set

Let $X$ be an uncountable set, and let $\Omega$ be the power set of $X$, viewed as a $\sigma$-algebra. Does there exist a positive $\sigma$-additive measure of finite total mass on $(X, \Omega)$ such that each point of $X$ has measure zero?

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 Maybe I'm missing something, but couldn't you just take a free ultrafilter on $X$ and let the measure of $S\subseteq X$ equal $1$ is $S$ belongs to the ultrafilter, and $0$ otherwise? – Philip Brooker Jul 31 at 5:55 @Philip: a measure on a $\sigma$-algebra is usually required to be countably additive, not just finitely additive. – Trevor Wilson Jul 31 at 6:03 @Trevor Wilson: ah, yes, of course. That makes the problem much more interesting! Thanks. – Philip Brooker Jul 31 at 9:15

I assume you mean a $\sigma$-additive measure. This is Ulam's measure problem. A positive answer is closely tied up to the existence of real-valued measurable cardinals, so it is equiconsistent with the existence of a measurable cardinal, which is a large cardinal assumption significantly beyond the usual axioms of set theory.

You can see a quick write up of the argument here. A good reference is the beginning of David Fremlin, "Real-valued measurable cardinals", in Set Theory of the reals, Haim Judah, ed., Israel Mathematical Conference Proceedings 6, Bar-Ilan University (1993), 151–304, that I also mention in the notes linked to above.

In short (this is expanded in the notes): If $(X,\mathcal P(X),\lambda)$ is such a measure space, we may as well assume (by concentrating on an appropriate subset, which may be of smaller size than $X$, and renormalizing) that $\lambda$ is a probability measure. Its additivity is the smallest cardinal $\kappa$ such that the measure of the disjoint union of some collection of $\kappa$ many disjoint subsets of $Y$ is not the sum of the measures of the sets in the union. (So the additivity is at least $\aleph_1$, and it is well-defined, since we are assuming that $\lambda(X)>0$.)

Then we can in fact assume $X=\kappa$ (identifying cardinals with sets of ordinals). If $\lambda$ is non-atomic (meaning, for any $E\subseteq\kappa$, if $\lambda(E)>0$ then there is $F\subset E$ with $0<\lambda(F)<\lambda(E)$), then $\lambda$ is (atomlessly) real valued measurable. On the one hand, these cardinals are not too large: $\kappa\le|\mathbb R|$. On the other, $\kappa$ must be weakly inaccessible, and in fact limit of weakly inaccessibles that themselves are limit of weakly inaccessibles, etc. This is very very large.

The other possibility is that $\lambda$ is atomic. Then, after further renormalization, $\lambda$ can be identified with the characteristic function of a non-principal $\kappa$-complete ultrafilter, that is, $\kappa$ is measurable.

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Ack! This is the second time in two years that I've accidentally bumped into some serious set theory / foundations in my research. @Andres - Thanks for putting a name to this question and pointing me to some literature. And yes, I did mean $\sigma$-additive. Is $W = E$ in your third paragraph, line 6? – Xander Faber Jul 31 at 7:00
I don't have any sense of whether or not a given set $X$ can be put in bijection with a measurable cardinal (again identifying cardinals with sets of ordinals). So for example, is it known when $X = \mathbb{R}$? I assume this case was the original motivation for the question. – Xander Faber Jul 31 at 7:17
Xander, measurable cardinals are much bigger than $\mathbb R$. It is the so-called real valued measurables that could possibly be $\leq|\mathbb R|$ and that are connected to measures on $\mathcal P(\mathbb R)$. – Stefan Geschke Jul 31 at 8:02
I should add to my comment that the reason Andres mentions measurable cardinals is that their equiconsistency with real valued measurables shows that you cannot construct a $\sigma$-additive measure on $\mathcal P(\mathbb R)$ without the help of some strong additional axioms. – Stefan Geschke Jul 31 at 8:08
Xander: Yes, $W$ was a typo for $E$. Fixed now. Thanks. It may perhaps be worth pointing out two remarks: 1. If there is a real-valued measurable $\kappa$, then any $X$ with $|X|\ge\kappa$ admits such a measure: Simply concentrate it on subsets of $Y$, where $Y$ is a subset of $X$ of size $\kappa$; and on subsets of $Y$ simply assign a measure via a bijection with $\kappa$. 2. In fact, if there is an atomless real-valued measurable, and $X=\mathbb R$, we can find a measure on all subsets of $X$ that extends Lebesgue measure. – Andres Caicedo Jul 31 at 14:50
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