Finite measure on the power set

2012-07-31T05:29:15Z

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?

Answer by Andres Caicedo for Finite measure on the power set

2012-07-31T06:07:16Z

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.

Finite measure on the power set - MathOverflow

Finite measure on the power set

Answer by Andres Caicedo for Finite measure on the power set