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Say $\omega_1$ is the first uncountable ordinal, and let $X=[0,\omega_1]$ (with the order topology). Elsewhere it is said that Rao & Rao proved a result that is easily seen to be equivalent to this:

Lemma (Rao & Rao). If $\nu$ is a Borel probability measure on $X$ and $\nu(\{j\})=0$ for every $j\in X$ then $\nu$ is the Dieudonne measure.

Which is the same as saying that $\nu(K)=1$ for every uncountable compact $K$.

I find this hard to believe. Stuck at home right now with no access to the paper; anyone have a hint?

A brief reminder re the Dieudonne measure:

Say $M_1$ is the class of all $E\subset X$ such that $E\cup\{\omega_1\}$ contains an uncountable compact set; let $M_0=\{X\setminus E:E\in M_1\}$ and $M=M_1\cup M_0$. Then $M$ is a $\sigma$-algebra containing every Borel set (hint: it's clear that $M$ contains every compact set); noting that $M_1\cap M_0=\emptyset$ we define the Dieudonne measure $\lambda$ on $M$ by $$\lambda(E)=\begin{cases}1,&(E\in M_1),\\0,&(E\in M_0).\end{cases}$$See for example Exercise 18 in Chapter 2 of Rudin Real and Complex Analysis...

David C. Ullrich
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  • To make this more plausible, note that if $U$ is the set of successor ordinals, then $\nu$ restricts to a measure on the entire power set $P(U)$ (since $U$ is discrete), which forces $\nu$ to be $0$ on $U$ (hopefully this sounds plausible, though it is nontrivial to prove). – Eric Wofsey Aug 01 '19 at 18:04
  • @EricWofsey Right, every subset of $U$ is open, hence Borel. But supposing for the sake of argument that indeed $\nu(U)=0$, I don't see why that makes the lemma more plausible... (oh: that says that $\nu$ agrees with the Dieudonne measure on $U$; was that your point or is there more to it?) – David C. Ullrich Aug 01 '19 at 18:18
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    Well, the point is that the complement of $U$ (the limit ordinals) is in some sense the archetypical closed unbounded set. So if it is forced to have measure $1$, it is plausible that by some similar argument every other closed unbounded set is forced to have measure $1$. – Eric Wofsey Aug 01 '19 at 18:19

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The key ingredient is the following theorem of Ulam:

Theorem: Let $\mu$ be a finite measure defined on the entire power set $P(\omega_1)$ which vanishes on singletons. Then $\mu=0$.

In particular, we can apply this to your setup as follows.

Corollary: Let $\nu$ be a Borel probability measure on $\omega_1$ which vanishes on singletons, and suppose $U\subset\omega_1$ can be written as a disjoint union of bounded open sets. Then $\nu(U)=0$.

Proof of Corollary from Theorem: Suppose $U=\bigcup_{\alpha<\omega_1} U_\alpha$ where the $U_\alpha$ are disjoint, bounded, and open. Define a measure $\mu$ on $P(\omega_1)$ by $\mu(A)=\nu(\bigcup_{\alpha\in A} U_\alpha)$ (this is well-defined since such a union is always open and a measure since the $U_\alpha$ are disjoint). Then $\mu$ vanishes on singletons since each $U_\alpha$ is bounded. Thus by the Theorem, $\mu=0$, and in particular $\nu(U)=\mu(\omega_1)=0$.

Now to prove your Lemma, let $K\subset\omega_1$ be any closed unbounded set and let $U=\omega_1\setminus K$. Enumerating the elements of $K$ in order as $(c_\alpha)_{\alpha<\omega_1}$, then we can partition $U$ into the bounded open sets $[0,c_0)$ and $(c_\alpha,c_{\alpha+1})$ as $\alpha$ ranges over all of $\omega_1$. Thus by the Corollary, $\nu(U)=0$ and so $\nu(K)=1$.

Finally, here is a proof of the Theorem. Suppose $\mu$ is nonzero. For each $\alpha<\omega_1$, let $f_\alpha:\alpha\to\omega$ be an injection. For $\beta<\omega_1$ and $n<\omega$, let $A_{\beta,n}=\{\alpha:f_\alpha(\beta)=n\}$. Note that for fixed $\beta$, $\bigcup_n A_{\beta,n}=\omega_1\setminus(\beta+1)$ (since $f_\alpha(\beta)$ is defined as long as $\alpha>\beta$), which has full measure since $\mu$ vanishes on countable sets. Thus for each $\beta$ there is some $n$ such that $A_{\beta,n}$ has positive measure. Since there are uncountably many $\beta$'s and only countably many $n$'s, then must be some fixed $n$ such that $A_{\beta,n}$ has positive measure for uncountably many different $\beta$. But for fixed $n$, the sets $A_{\beta,n}$ are disjoint since the functions $f_\alpha$ are injective. Since a finite measure cannot have an uncountable family of disjoint sets of positive measure, this is a contradiction.

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Eric Wofsey
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  • Cool. (Note how I work around the injunction to avoid comments that just say "thanks"). – David C. Ullrich Aug 01 '19 at 19:08
  • Let $B$ be the Borel sets of $\omega_1+1.$ Suppose $(i)...m:B\to [0,\infty]$ where $m(b)<\infty$ when $b\in B$ with $\sup b<\omega_1$ and $(ii)... m(b)\le m(c)$ when $b,c\in B$ with $b\subset c$ and $(iii)... m({\omega_1})=0.$ Now let $f(x)=m(x+1)$ for $x\in \omega_1.$ Then $f$ is increasing from $\omega_1$ to $[0,\infty).$ We can easily show that any monotonic $f:\omega_1\to \Bbb R$ is bounded and eventually constant. So if $m$ is also finitely additive then $\exists x\in \omega_1,(,m(,(\omega_1+1)\setminus x)=0,).$ – DanielWainfleet Aug 01 '19 at 20:34
  • @DanielWainfleet: Your final sentence does not follow (and in fact is false, as the Dieudonne measure is a counterexample). – Eric Wofsey Aug 01 '19 at 20:45
  • Sorry. U R right....... – DanielWainfleet Aug 01 '19 at 21:16