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The following is an overly fancy way of asking a question suggested in Borel Measures: Atoms vs. Point Masses

Let $\phi$ be a property that topological spaces can have (such as "compact", "$T_1$", etc.) and let $M(\phi)$ be the statement "There exists a topological space $X$ with property $\phi$, and a nontrivial, countably additive, 0-1 valued Borel measure on $X$ which is not a point mass." (For the purposes of this question, a measure $\mu$ is a point mass if there exists $a \in X$ such that $\mu(A) = 1$ iff $a \in A$.)

For various values of $\phi$, what is the consistency strength of $M(\phi)$?

A few examples:

  • $M(\text{compact Hausdorff})$ is a theorem of ZFC (let $X = [0,\omega_1] = \omega_1+1$ with the order topology, and consider the Dieudonné measure.)

  • $M(\text{finite discrete})$ is inconsistent with ZFC.

  • $M(\text{separable metrizable})$ is also inconsistent with ZFC, as I show in this answer.

  • $M(\text{discrete})$ is implied by "there exists a measurable cardinal", and if I understand the Wikipedia article correctly, they are actually equivalent. It's known that this has greater consistency strength than ZFC but is not known to be inconsistent.

I would be particularly interested in knowing about $M(\text{metrizable})$ and $M(\text{completely metrizable})$.

Community
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Nate Eldredge
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  • I'm not entire sure if that's what you mean, but a real-valued measurable cardinal seems to be a relevant notion here. It's really a measurable cardinal in disguise. But if the continuum is real-valued measurable then there is a measure extending the Lebesgue measure which measures every subset. I suppose that by extending the co-null filter you can obtain a $0$-$1$ measure. – Asaf Karagila Oct 24 '14 at 06:45

1 Answers1

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Suppose $X$ is a metric space with a two valued diffused Borel measure. Let $\mathcal{U} = \bigcup \{\mathcal{U}_n : n < \omega\}$ be a basis for $X$ where each $\mathcal{U}_n$ is a disjoint family of open sets - See here.

Let $\langle x_i : i < \kappa \rangle$ list $X$. For each $i$, let $B_i$ be a open ball around $x_i$ whose measure is zero. Let $\theta \leq \kappa$ be least such that $\bigcup \{B_i : i < \theta\}$ is not null.

For each $n$, consider the family $\mathcal{F}_n = \{U \in \mathcal{U}_n :(\exists i < \theta)(U \subseteq B_i)\}$. Each $\mathcal{F}_n$ is a disjoint family of open sets and if $\mathcal{F} = \bigcup \{\mathcal{F}_n : n < \omega\}$, then $\bigcup \mathcal{F}$ is not null. Hence for some $n$, $\bigcup \mathcal{F}_n$ is not null. So we can define a total two valued measure on $|\mathcal{F_n}| \leq \kappa$ hence also on $\kappa$.

hot_queen
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  • I'm afraid I don't see how you conclude that $\theta$ supports a total 0-1 valued measure. – Nate Eldredge Oct 24 '14 at 17:13
  • Oops! I forgot to make $B_i$'s disjoint. I will edit my answer. – hot_queen Oct 24 '14 at 18:06
  • Yeah, all that is used in your original argument is that each point has a null neighborhood. But that holds for many non-metric spaces. Take the Dieudonné measure on $\omega_1 + 1$ and restrict it to $\omega_1$ - now every point of $\omega_1$ has a null neighborhood (for instance $i$ has $[0,i+1)$.) But ZFC proves that $\aleph_1$ is not measurable. – Nate Eldredge Oct 24 '14 at 18:20
  • Actually for that matter, every point of $\omega_1$ is $G_\delta$. – Nate Eldredge Oct 24 '14 at 18:33
  • @Nate: One more reason to ditch choice, and work in models where $\aleph_1$ is in fact measurable! :-) – Asaf Karagila Oct 25 '14 at 04:02
  • Sorry, I still don't quite follow your last sentence - how do you propose to define a total measure on $\theta$? – Nate Eldredge Oct 25 '14 at 06:29
  • Let $\mathcal{F}_n = {U_i : i < \lambda}$ (where $\theta \leq \lambda \leq \kappa$). For $A \subseteq \lambda$, put $\mu(A) = m(\bigcup {U_i : i \in A})$. So $\kappa$ (not $\theta$) is above a measurable. – hot_queen Oct 25 '14 at 13:03
  • @NateEldredge Just wondering if your read the updated answer. – hot_queen Apr 30 '15 at 23:57
  • So having completely forgot about this question/answer for a year, I "independently" reconstructed it here. :-/ – Nate Eldredge Mar 05 '16 at 00:35