Radon measures on $[0, \omega_1]$ Denote by $\omega_1$ the first uncountable ordinal and consider the compact Hausdorff space $[0, \omega_1]$ with its order topology. Is every Radon measure (= regular Borel measure) on $[0, \omega_1]$ necessarily a countable convergent sum of Dirac measures on the ordinals $0 \leq \alpha \leq \omega_1$, i.e. of the form $\sum_{0 \leq i < \alpha} a_i \delta_i + b \delta_{\omega_1}$ where $\alpha < \omega_1$, $\sum_{0 \leq i < \alpha} |a_i| < \infty$ and $b \in \mathbb{R}$?
In [Rao, Rao, "Borel $\sigma$-algebra of $[0, \omega_1]$" (1971)] it is shown that every (positive finite) Borel measure on $[0, \omega_1]$ is not non-atomic and of the form $c \mu + \tau$ where $\mu$ is the famous Dieudonné measure (a non-regular Borel measure) and $\tau$ is concentrated on a countable subset of $[0, \omega_1]$. Hence, the Radon measures are precisely those with $c = 0$, i.e. concentrated on a countable set. So I would expect the answer to the above question to be affirmative, unless I overlook something. Recall also that every non-atomic measure has the intermediate value property (due to Sierpinski), see here.
 A: Edit: Below I assume that "measure" means "positive measure". Looking back at the question, it seems you are probably  talking about real measures. The result for real measures follows, since any real measure is the difference of two finite positive measures.
If I'm  not mistaken the answer to the question is yes (at least for finite $\mu$; recall that Radon measures are finite):


Suppose $\mu$ is a finite regular Borel measure on $[0,\omega_1]$. There exist $\alpha<\omega_1$, $a_j\ge 0$ for $0\le j\le\alpha$ and $c\ge0$ such that $\mu=\sum_{j\le\alpha}a_j\delta_j+c\delta_{\omega_1}$.


Proof: Let $E=\{j\in[0,\omega_1]:\mu(\{j\})>0\}$. Countable additivity shows that $E$ is countable. So there exists $\alpha<\omega_1$ with $$E\subset[0,\alpha]\cup\{\omega_1\}.$$Define $a_j=\mu(\{j\})$ for $j\le\alpha$ and $c=\mu(\{\omega_1\})$, and let $$\mu_1=\sum_{j\le\alpha}a_j\delta_j+c\delta_{\omega_1},$$ $$\nu=\mu-\mu_1.$$We need to show $\nu=0$. Note first that $\mu_1\le\mu$ (because $\mu_1(A)=\mu(A\cap E)$ [hint: $A\cap E$ is countable]), so $\nu\ge0$; hence it's enough to show that $\nu([0,\omega_1])=0$.
But $\nu(\{j\})=0$ for every $j\in[0,\omega_1]$, so countable additivity shows that$$\nu([0,\alpha])=0\quad(\alpha<\omega_1).$$If $K$ is a compact subset of $[0,\omega_1)$ there exists $\alpha<\omega_1$ with $K\subset[0,\alpha)$; hence $$\nu(K)=0\quad(\text{compact } K\subset[0,\omega_1)).$$So regularity shows that $\nu([0,\omega_1))=0$, and since $\nu(\{\omega_1\})=0$ we have $\nu([0,\omega_1])=0$.
Note. This is closely related to the result of Rao & Rao mentioned in the question. Indeed, we could have used Rao & Rao to obtain $\mu_1$ and $\nu$ above (I didn't do that because the construction is very simple a priori, and the version of Rao & Rao in the original version of the question was obviously false.) And in fact the result above suffices to reduce Rao & Rao to a special case:
Say $\mu$ is a (finite positive) Borel measure on $[0,\omega_1]$. Define $\mu_1$ and $\nu$ as above. Then $\mu_1$ is concentrated on a countable set, so we "only" have to show that $\nu$ is a multiple of the Dieudonne measure. So Rao & Rao is equivalent to this:


Lemma (Rao & Rao). If $\nu$ is a Borel probability measure on $[0,\omega_1]$ and $\nu(\{j\})=0$ for every $j\in[0,\omega_1]$ then $\nu$ is the Dieudonne measure.


I'm totally stuck on that, in fact it seems implausible. Edit: There's a proof here.
