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Consider the Dieudonné measure on $[0, \omega_1)$ which is defined on the Borel $\sigma$-algebra of $[0, \omega_1)$ by $\mu(A) = 1$ if $A$ contains a closed and unbounded set and $\mu(A) = 0$ otherwise. (The Borel $\sigma$-algebra is the collection of all subsets $A$ of $[0, \omega_1)$ such that either $A$ or its complement $A^c$ contains a closed and unbounded set.) It is known that $\mu$ is purely atomic, $\mu$ has empty support and $\mu$ vanishes on all singletons (in fact $\mu$ vanishes on all compact subsets of $[0, \omega_1)$ - this turns $\mu$ into a standard example of a non-regular Borel measure). Hence, singletons are not atoms.

A natural question would be: Is there an explicit characterization of all the atoms of $\mu$?

I think a possible issue is that if $C \subseteq [0, \omega_1)$ is a closed unbounded set then $\mu(C) > 0$ and if $C$ contains a further closed unbounded set $C' \subsetneq C$ then $\mu(C') > 0$. So this reduction process can be performed ad infinitum (but for how long?). For instance, if $C$ is the set of all countable limit ordinals (which is closed and unbounded) then removing the first countable ordinal $\omega_0$ from $C$ gives a set $C'$ that is also an unbounded closed set in $[0, \omega_1)$.

yada
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  • Yes, you can remove the elements of $C$, one by one, from the smallest and onwards. And this can go through any countably many steps. Now, why do you think there are atoms? Also, what happens when you take $S$ such that $S$ and its complement are stationary? Is it just that those are provably not Borel, or does something else break? (I honestly know nothing about the Borel sets of $\omega_1$...) – Asaf Karagila Aug 05 '19 at 11:13
  • @AsafKaragila It was proven in [Rao, Rao, "Borel $\sigma$-algebra on $[0, \Omega]$" (1971)] that on $[0, \omega_1]$ (and $[0, \omega_1)$) there are no non-atomic measures; hence $\mu$ has atoms and is moreover purely atomic. I don't know anything about stationary sets, but as you mentioned here, these sets are not Borel measurable (provable in ZFC). – yada Aug 05 '19 at 11:42
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    You seem to be ignoring the definition. $A$ is an atom if $\mu(A)>0$ and there does not exist $B\subset A$ with $0<\mu(B)<\mu(A)$. For the Dieudonne measure $\mu(E)$ is $0$ or $1$ for every $E$, hence any set of positive measure is an atom. – David C. Ullrich Aug 05 '19 at 16:41
  • Say $B$ is the Borel algebra and $M$ is the algebra of $E$ such that $E$ or $E^c$ contains an unbounded closed set. You say $B=M$. Are you sure about that? I know $B\subset M$... – David C. Ullrich Aug 05 '19 at 16:47
  • Why is the paragraph "I think a possible issue is..." an "issue"? – David C. Ullrich Aug 05 '19 at 16:58
  • @AsafKaragila Say $B$ and $M$ are as in my comment above. If $C$ is closed then $C$ is either unbounded or not, hence $C\in M$. The intersection of countably many clubs is club, hence $M$ is a $\sigma$-algebra, hence $B\subset M$. If $S$ and $S^c$ are both stationary then $S\notin M$, so yes, it follows that $S\notin B$. – David C. Ullrich Aug 05 '19 at 18:58
  • @David: Thanks. So you seem to suggest that the Borel sets are generated by clubs and singletons maybe? – Asaf Karagila Aug 05 '19 at 19:30
  • @AsafKaragila I wasn't aware of suggesting that. But come to think of it yes, that's more or less obvious: The Borel sets are certainly generated by the closed sets, and a bounded closed set is a countable union of singletons, while an unbounded closed set is unbounded... – David C. Ullrich Aug 05 '19 at 20:58
  • @David: Yes, but are there any other Borel sets? (Other than clubs mod countable and their complements...) – Asaf Karagila Aug 05 '19 at 21:18
  • @AsafKaragila I don't really know anything about this stuff except what I learned here a few days ago thinking about a problem about the Dieudonne measure. The answer "must" be yes I suppose probably – David C. Ullrich Aug 05 '19 at 22:37
  • @AsafKaragila What are Borel sets: We know that $B\subset M$. It's stated above that $B=M$; I don't believe that but haven't been able to show otherwise. – David C. Ullrich Aug 05 '19 at 22:48
  • @AsafKaragila Hmm, maybe it's true - it's also asserted that $B=M$ here: https://math.stackexchange.com/questions/268794/non-measurable-subset-of-omega-1?rq=1 – David C. Ullrich Aug 05 '19 at 22:54
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    @DavidC.Ullrich You are right that every set of full measure ($=1$) is an atom here. I was thinking about an atom like something being minimal as a set in a certain sense. But such a "minimality" condition need not be well-defined, i.e. "minimal" sets need not exist. So, in general, an atom can be contained in another atom. – yada Aug 06 '19 at 06:19
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    @DavidC.Ullrich For a proof of $B = M$ on $[0, \omega_1)$ see [Fremlin, "Measure Theory", 4A3J] and for $B = M$ on $[0, \omega_1]$ see [Rao, Rao, "Borel $\sigma$-algebra on $[0, \Omega$]"]: In both cases $X = [0, \omega_1)$ or $X = [0, \omega_1]$, a set $A \subseteq X$ is a Borel set if and only if (either) $A$ or $A^c$ contains a closed unbounded set. [Aliprantis, "Infinite Dimensional Analysis", Example 12.9] refers to a set $A$ as "big" if it contains a closed uncountable set and to "small" if $A^c$ is big. The Borel $\sigma$-algebra is then the collection of all the big and small sets. – yada Aug 06 '19 at 06:28
  • For those who have access, the paper of Rao and Rao can be found here. The monumental measure theory treatise of Fremlin can be found here. – yada Aug 06 '19 at 06:37
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    @yadaddy Thanks. Just before seeing your comment I came up with my own possibly correct proof that $B=M$. Say $K$ is an unbounded closed set; let $\omega_1\setminus K=\bigcup (\alpha_j, \beta_j)$ and assume $\gamma_j\in(\alpha_j,\beta_j)$ for all $j$. It's enough to show that $K\cup{\gamma_j}$ is closed, since a big set is a countable union of such sets; I think it's fairly clear that this set is in fact closed. – David C. Ullrich Aug 06 '19 at 13:05
  • @yadaddy This all started, from my pov, when I heard that Rao&Rao prove that any Borel measure on $\omega_1$ is the sum of a measure concentrated on a countable set and a multiple of the Dieudonne measure. That's immediate from the result proved here; hence the question: Are there more big results in Rao&Rao? – David C. Ullrich Aug 06 '19 at 13:19
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    @yadaddy About atoms being minimal: Say $A$ is the algebra of measurable sets modulo null sets. Then an atom is a minimal element of $A$ of positive measure... – David C. Ullrich Aug 06 '19 at 13:21
  • @AsafKaragila Yes, $B=M$. Seems to me this is the best answer you're going to get to the question of what other Borel sets there are... – David C. Ullrich Aug 06 '19 at 16:25
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    @DavidC.Ullrich Rao&Rao's paper is quite short. A corollary of their results is: a real function on $[0,\omega_1]$ is Borel measurable if and only if it is constant on a closed unbounded set. It is also well known that a function on $[0, \omega_1]$ is continuous if and only if it is eventually constant - the Stone-Cech and the one-point compactification of $[0, \omega_1)$ coincide. – yada Aug 07 '19 at 05:59
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    Fremlin also proved that the Baire $\sigma$-algebra on $[0, \omega_1)$ or on $[0, \omega_1]$ (generated by the continuous functions) is just the countable-cocountable $\sigma$-algebra. – yada Aug 07 '19 at 06:29
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    @yadaddy Say a function is ann R&R function if it's constant on some club. At first blush the idea that a function is R&R if and only if it's Borel measurable sounded like nonsense. But it's actually pretty clear. If $f$ is R&R then it's Borel, just because any subset of a small set is small and hence Borel. Otoh if $E$ is a Borel set then $\chi_E$ is R&R, hence similarly for Borel simple functions, and the class of R&R functions is closed under "complements" ($f\mapsto 1-f$) and sequential pointwise limits, qed. Fascinating, thanks... – David C. Ullrich Aug 07 '19 at 11:06

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