Borel measures on $\omega_1$. 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...
 A: 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.
