Measure on $\omega_1$ - MathOverflow

Measure on $\omega_1$

2011-11-17T05:29:30Z

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 $\aleph_1$.)

Answer by Gerald Edgar for Measure on $\omega_1$

2011-11-17T15:34:39Z

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 $\mu(A) = 0$.

Answer by Will Sawin for Measure on $\omega_1$

2011-11-18T01:45:37Z

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.