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?

$\begingroup$ 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? $\endgroup$ – Philip Brooker Jul 31 '12 at 5:55

$\begingroup$ @Philip: a measure on a $\sigma$algebra is usually required to be countably additive, not just finitely additive. $\endgroup$ – Trevor Wilson Jul 31 '12 at 6:03

$\begingroup$ @Trevor Wilson: ah, yes, of course. That makes the problem much more interesting! Thanks. $\endgroup$ – Philip Brooker Jul 31 '12 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 realvalued 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, "Realvalued measurable cardinals", in Set Theory of the reals, Haim Judah, ed., Israel Mathematical Conference Proceedings 6, BarIlan 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 welldefined, 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 nonatomic (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 nonprincipal $\kappa$complete ultrafilter, that is, $\kappa$ is measurable.

$\begingroup$ 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? $\endgroup$ – Xander Faber Jul 31 '12 at 7:00

$\begingroup$ 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. $\endgroup$ – Xander Faber Jul 31 '12 at 7:17

$\begingroup$ Xander, measurable cardinals are much bigger than $\mathbb R$. It is the socalled real valued measurables that could possibly be $\leq\mathbb R$ and that are connected to measures on $\mathcal P(\mathbb R)$. $\endgroup$ – Stefan Geschke Jul 31 '12 at 8:02

$\begingroup$ 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. $\endgroup$ – Stefan Geschke Jul 31 '12 at 8:08

1$\begingroup$ 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 realvalued 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 realvalued measurable, and $X=\mathbb R$, we can find a measure on all subsets of $X$ that extends Lebesgue measure. $\endgroup$ – Andrés E. Caicedo Jul 31 '12 at 14:50
Just to complement Andres's excellent answer with another reference, you can find a nice summary of the status of this question, as well as further references, in chapter 1.12(x) of Bogachev's monograph "Measure Theory I".
The (very) short summary is that in all concrete cases the answer is no.

1$\begingroup$ Regarding your summary, I would dispute your statement, since as Andres explains, it is known to be (relatively) consistent with the axioms of set theory that the reals $\mathbb{R}$ support such a measure, and this would seem to count as a concrete case. So we don't really seem to know that the answer is no in all concrete cases. $\endgroup$ – Joel David Hamkins Nov 2 '13 at 16:57