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, $\mu(\bigcap^\infty\_{j=1} E\_j)=\beta$.



Regular Borel measures and the measure of a singleton
Let $\mu$ be a (positive, probability) measure satisfying the hypothesis. Note that there cannot be any Borel set $A$ with $1/3 \le \mu(A) \le 2/3$, since then in the partition $S = A \cup A^c$, the smaller set would have measure at least $1/3$. Thus it suffices to show there is an $x\_0$ with $\mu(\{x\_0\}) \ge 1/3$.


Suppose not; then every point has measure less than $1/3$. By regularity, every point thus has an open neighborhood with measure less than $1/3$. By compactness, I can find a finite cover of $S$ by such open sets, $U\_1, \dots, U\_n$. Let $V\_k = U\_1 \cup \dots \cup U\_k$ for $1 \le k \le n$. Let $m = \max\{ k : \mu(V\_k) < 1/3 \}$; in particular $m < n$ (since $V\_n = S$). Then $$1/3 \le \mu(V\_{m+1}) < \mu(V\_m) + 1/3 < 2/3$$ which is a contradiction.


