Finite measure on the power set - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T12:10:30Z http://mathoverflow.net/feeds/question/103583 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103583/finite-measure-on-the-power-set Finite measure on the power set Xander Faber 2012-07-31T05:29:15Z 2012-07-31T14:46:03Z <p>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? </p> http://mathoverflow.net/questions/103583/finite-measure-on-the-power-set/103585#103585 Answer by Andres Caicedo for Finite measure on the power set Andres Caicedo 2012-07-31T06:07:16Z 2012-07-31T14:46:03Z <p>I assume you mean a $\sigma$-additive measure. This is Ulam's <em>measure problem</em>. 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.</p> <p>You can see a quick write up of the argument <a href="http://caicedoteaching.wordpress.com/2009/03/05/580-cardinal-arithmetic-8/" rel="nofollow">here</a>. A good reference is the beginning of David Fremlin, "Real-valued measurable cardinals", in <strong>Set Theory of the reals</strong>, Haim Judah, ed., Israel Mathematical Conference Proceedings 6, Bar-Ilan University (1993), 151–304, that I also mention in the notes linked to above.</p> <p>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 <em>additivity</em> is the smallest cardinal $\kappa$ such that the measure of the disjoint union of some collection of $\kappa$ many disjoint subsets of $Y$ is <em>not</em> 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 $\lambda(X)>0$.)</p> <p>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 $0&lt;\lambda(F)&lt;\lambda(E)$), then $\lambda$ is <em>(atomlessly) real valued measurable</em>. 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.</p> <p>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 <a href="http://en.wikipedia.org/wiki/Measurable_cardinal" rel="nofollow">measurable</a>. </p>