Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The Question

Let $X$ be a set. Let $\mathcal{F}\subseteq P(X)$ be a $\sigma$-algebra. (Or, if it makes a difference, let $X$ be a topological space and $\mathcal{F}$ the Borel sets.) When can we guarantee the existence of and how can we construct a non-atomic probability measure $\mu$ on $(X,\mathcal{F})$? (In addition to being a measure, we ask that $\mu(X) = 1$ and that if $F\in\mathcal{F}$ and $\mu(F) \neq 0$, there exists a proper measurable subset $E\subsetneq F$ such that $0 < \mu(E) < \mu(F)$.)

Some remarks

The result of Sierpinski guarantees that $\mu$ actually takes on a continuum of different values. So necessarily $\mathcal{F}$ needs to be uncountable and hence $P(X)$ has to be uncountable. This puts a lower bound on how many elements there can be in $X$. (And trivially one sees that if $X$ is finite any probability measure must have atoms.)

I expect, however, that there may be other necessary conditions for the existence of a non-atomic probability measure.

On the sufficient side, the only result I am familiar with that explicitly constructs a measure is the various constructions of the Lebesgue measure. This construction makes use of the local structure of Euclidean space and hence also works for, say, topological manifolds. (Okay, there's also the ultrafilter construction for additive measures, but while the constructed measures are probability, they have a lot of atoms.)

share|improve this question
    
I think the intention of this question on MO by Michael Greinecker was essentially the same as yours. –  t.b. Mar 8 '12 at 16:25
    
Huh, I searched both MO and Math.SE before I posted this question, and somehow I missed that question. I'll have to read that one in more detail to understand what was actually said by the various comments/answers. –  Willie Wong Mar 8 '12 at 16:38
2  
ince that thread isn't particulary illuminating, let me just mention that it seems unknown whether it is decidable in ZFC whether there is purely non-atomic probability measure on $(X,P(X))$ if $|X| \geq |\mathbb{R}|$ (see Fremlin's book, volume 4I, Ch. 438), so you might want to specify more clearly what kind of spaces you're interested in. On uncountable separable complete metric spaces there always is a non-atomic Borel probability measure, for example. –  t.b. Mar 8 '12 at 16:55
    
Correct me if I am wrong: so Maharam's theorem essentially implies that the answer is iff our sigma algebra is isomorphic to the measurable sets on the product of $\gamma$ copies of $[0,1]$, where $\gamma$ is an ordinal? this I guess is just a more refined version of the construction on a manifold? –  Willie Wong Mar 8 '12 at 16:56
1  
This is a bit simplistic, since Maharam's theorem doesn't talk about measure spaces but rather about measure algebras (what you obtain from the $\sigma$-algebra by identifying sets that coincide up to a null set). Those can be obtained up to isomorphism from combining scaled versions of products $[0,1]^\kappa$ where $\kappa$ is a cardinal. I don't think this tells us so much about the $\sigma$-algebras themselves. The product measure on an arbitary product of probability spaces indeed is not much more than a (somewhat subtle) refinement of the usual product measure construction. –  t.b. Mar 8 '12 at 17:05

1 Answer 1

Following Willie Wong, the answer to his question depends on the cardinality of $X$. In the case when $card(X)=c$ his question is not uniquely solvable
in the theory $(ZF)\&(DC)$ for $X=[0,1]$ and $\cal{F}=P[0,1]$, where
$(ZF)$ denotes the Zermelo-Fraenkel set theory and $(DC)$ denotes the axiom of Dependent Choices.

Indeed, on the one hand, in the consistent theory $ZF \&DC \& AD$, where $AD$ denotes an Axiom of Determinacy, the answer to his question is yes, because Mycielski and Swierczkowski well known result asserts that every subset of the real axis is Lebesgue measurable. Hence such a measure is exactly Lebesgue measure in $[0.1]$.

On the other hand, in the consistent theory $ZF\& DC \& AC \& \omega_1=2^{\omega}$ by Ulam's well known result on the powerset of $\omega_1$ (correspondingly, of $2^{\omega}$) we can not define a probability measure which vanishes on singletons.

Since both theories are consistent extensions of the theory $ZF\& DC$ we deduce that Willie Wong's question is not solvable within the theory $ZF\&DC$ for $X=[0,1]$ and $\cal{F}=P[0,1]$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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