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Let $\mathcal{O}$ be the $\sigma$-algebra on $\omega_1$ generated by its countable subsets. Is there a ($\sigma$-additive) probability measure on $\mathcal{O}$ that is not concentrated on a countable set? (I am trying to construct a real random variable whose support has size $\aleph_1$.)

edited Nov 17 2011 at 5:42

asked Nov 17 2011 at 5:29

Nate Ackerman
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A set is in the sigma-algebra if it is countable or has a countable complement. Put measure 0 on the countable sets and measure 1 on the sets with countable complement. – Michael Greinecker Nov 17 2011 at 5:44

Thanks for your comment -- I realize that I need to refine my question substantially. (I also see that there need be no probability measure on $\mathbb{R}$ whose support has size $\aleph_1$, because any closed set of reals satisfies the continuum hypothesis.) – Nate Ackerman Nov 17 2011 at 6:09

Every uncountable Borel set has the size of the contiuum. If you just want such a measure for some $\sigma$a-gebra on $\mathbb{R}$, you can do the following: If the continuum hypothesis holds, you do what I described above. If not, you can do the same with the $\sigma$-algebra of those subsets whose size is at most $\aleph_1$ or whose complement has at most size $\aleph_1$. – Michael Greinecker Nov 17 2011 at 6:25

Nate, I think the $\sigma$-algebra you're thinking of is substantially larger than the one you describe. Do you mean to restrict the Borel algebra to a set of reals with size $\aleph_1$? – François G. Dorais♦ Nov 17 2011 at 6:36

David Roberts · Answer 1 · 2011-11-18 03:36:22Z UTC

The $\sigma$-algebra you describe includes all the sets of one element. Due to the countable additivity of measure, we conclude that the measure of any countable set is determined by the sum of the measure of its elements.

Suppose the set of positive-measure elements is uncountable. Then some countable set must have infinite measure. Proof: Consider the sets $\{x|\mu(x)\}>\epsilon$ . For some $\epsilon$, this must be infinite. Choose a countable subset. (Clever ZF-without-choice mojo may enable you to construct a counterexample here).

If you're okay with that behavior, then the counting measure is an example.

If you're not okay with that behavior, then all measures will look like the sum of Michael Greinecker's measure and a measure of countable support. (Let $S$ be the set of positive-measure elements and $T$ another set, then $\mu(T)=\mu(S\cap T)+\mu(T-S)$, the first being the sum of its elements and the second $0$ on all countable sets, and therefore equal on all co-countable sets.)

Note: I think Gerald Edgar's measure is the restriction of Michael's measure to the Borel measure space.

Gerald Edgar · Answer 2 · 2011-11-17 15:34:39Z UTC

Another nice example is related to this. Let $\Omega$ be the set of all countable ordinals, with its order topology. I may write $\Omega = [0,\omega_1)$. Note that the last point is missing, but any countable subset has a supremum in $\Omega$. Topologically, $\Omega$ is not compact, but is locally compact and pseudo-compact: Indeed, even stronger, any continuous function $f : \Omega \to \mathbb R$ is eventually constant. The linear functional $\Lambda$ that assigns to each continuous function $f$ this eventual value, is what we want. As usual, there is a measure $\mu$ so that $\Lambda(f) = \int_\Omega f\,d\mu$ for all $f$. This is a good example of a measure with "empty support". For every point $t \in \Omega$, there is a neighborhood $A$ with $\mu(A) = 0$.

Measure on $\omega_1$

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