Measure on $\omega_1$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T11:54:44Z http://mathoverflow.net/feeds/question/81143 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81143/measure-on-omega-1 Measure on $\omega_1$ Nate Ackerman 2011-11-17T05:29:30Z 2011-11-18T03:36:22Z <p>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 $\aleph_1$.)</p> http://mathoverflow.net/questions/81143/measure-on-omega-1/81181#81181 Answer by Gerald Edgar for Measure on $\omega_1$ Gerald Edgar 2011-11-17T15:34:39Z 2011-11-17T15:34:39Z <p>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 $\mu(A) = 0$.</p> http://mathoverflow.net/questions/81143/measure-on-omega-1/81216#81216 Answer by Will Sawin for Measure on $\omega_1$ Will Sawin 2011-11-18T01:45:37Z 2011-11-18T03:36:22Z <p>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.</p> <p>Suppose the set of positive-measure elements is uncountable. Then some countable set must have infinite measure. Proof: Consider the sets <code>$\{x|\mu(x)\}&gt;\epsilon$</code>. For some $\epsilon$, this must be infinite. Choose a countable subset. (Clever ZF-without-choice mojo may enable you to construct a counterexample here).</p> <p>If you're okay with that behavior, then the counting measure is an example.</p> <p>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.)</p> <p>Note: I think Gerald Edgar's measure is the restriction of Michael's measure to the Borel measure space.</p>