I'm studying this paper: http://matwbn.icm.edu.pl/ksiazki/sm/sm73/sm7313.pdf

At the top of page 36, it states the following Proposition:

Let $S$ be a compact and $\mu$ a regular Borel measure on $S$ with total variation 1. If for every partition of $S$ into two Borel subsets the measure of the smaller one is less than $1/3$, then there is an $s_0\in S$ such that $\mu(\{s_0\})>2/3$,.

The author does not prove this Proposition. Maybe it's so obvious, but I simply have no idea of the proof.

I list below some approaches I've been trying with no success:

**1)** I tried to do some kind of argument with ultrafilters. If we consider the family $\mathcal A$ of Borel sets with measure greater than $2/3$ it is a filter contained in the $\sigma$-algebra of borelian sets. Moreover, $\mathcal A$ satisfies the following property:

If $E$ is a Borel set of $S$ then either $E\in \mathcal A$ or $E^c\in \mathcal A$.

Therefore, if we pick an ultrafilter $U$ containing $\mathcal A$ then $U\cap\mathcal B_S=\mathcal A$, where $\mathcal B_S$ denotes the family of borel sets of $S$. The idea to do this was to conclude that $U$ cannot be a free ultrafilter, but I can't see any further argument.

**2)** Define $$\alpha=\sup\{\mu(E): E\in\mathcal B_S \mbox{ and } \mu(E)< 1/3 \},$$
and
$$\beta=\inf\{\mu(E): E\in\mathcal B_S \mbox{ and } \mu(E)\geq 1/3 \}.$$

Let us see that $\beta$ is assumed.

Suppose the contrary, then, for each $n\in \mathbb N$, pick $E_n\in\mathcal B_S$ such that $$ \beta< \mu(E_n)< \beta+\frac{1}{n}. $$ We prove the following

Claim.If $A, B \in \mathcal B_S$ are such that $\mu(A),\ \mu(B)>\beta$, then $ \mu(A\cap B) > \beta. $

Suppose by contradiction that $\mu(A\cap B)\leq \beta$. Since $\beta$ is not assumed, $\mu(A\cap B)< \beta$, and by the minimality of $\beta$, $$ \begin{array}{rl} \mu(A\cap B) < 1/3 & \Rightarrow \mu((A\cap B)^c) > 2/3\\ &\Rightarrow\ \mu((A\cap B)^c) \geq \beta\\ &\Rightarrow\ 1-\mu(A\cap B) \geq \beta\\ &\Rightarrow\ \mu(A\cap B) \leq 1-\beta, \end{array} $$ and consequently, $$ 2\beta < \mu(A)+\mu(B) = \mu(A\cap B) + \mu(A\cup B) \leq 1 - \beta + 1, $$ so it follows that $\beta< 2/3$, which is a contradiction with the assumption on $\mu$.

Usihg the claim, we prove by induction that $$\beta \leq \mu\left(\bigcap^n_{j=1} E_j\right)\leq \beta + 1/n,\ \forall n\in\mathbb N.$$ and consequently, $\mu(\bigcap^\infty_{j=1} E_j)=\beta$.