# Regular Borel measures and the measure of a singleton

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, $$\mu(\bigcap^\infty_{j=1} E_j)=\beta$$.

• I guess "measure" here means "signed measure"? Is there an obvious reduction to positive measures? Anyway, I would suppose the contrary; then by regularity every point has an open neighborhood of measure $< 2/3 + \epsilon$. By compactness you can find a finite cover by such neighborhoods. It seems like one ought to be able to obtain a contradiction eventually. – Nate Eldredge Oct 5 '18 at 4:14
• No, "measure" here means "positive measure". – André Porto Oct 5 '18 at 4:21
• Okay, it just sounded weird to talk about the variation of a positive measure. – Nate Eldredge Oct 5 '18 at 4:22

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.

• Wow! That's a very nice argument! – André Porto Oct 5 '18 at 5:08
• Just as an useful comment, the same argument is true for Lindelöff spaces. When we consider the covering, we get a countable subcovering $(V_n)_{n\in\mathbb N}$. Then, since $$\mu\left(\bigcup_{k=1}^nV_k\right)\to\mu(X)=1,$$ there exists the least $n_0$ such that $\mu(\bigcup_{k=1}^{n_0}V_k)>2/3$. Then, we know that $\mu(\bigcup_{k=1}^{n_0-1}V_k)<1/3$, but this implies that $\mu(\bigcup_{k=1}^{n_0}V_k)<1/3+\mu(V_{n_0})< 2/3$, a contradiction. – André Porto Oct 26 '18 at 0:40

Here's a variation on Nate's argument that needs neither compactness of $$S$$ nor regularity of $$\mu$$; instead one has to know that the range of an atomless measure is an interval (in fact, it would be enough to know that it is dense in an interval). Now, let $$A_1, A_2,\dots$$ be the finite or infinite sequence of atoms of $$\mu$$; put $$\mu_i=\mu(A_i)$$ and $$A= \bigcup_i A_i$$. If $$\mu(A)<1/3$$, then $$\mu(S\setminus A)>2/3$$, and since $$\mu$$ is atomless on $$S\setminus A$$, it takes the forbidden value $$1/2$$: contradiction. Therefore $$\mu(A)>2/3$$; thus we have $$\sum_i \mu_i>2/3$$. Let $$N$$ be the maximal integer such that $$\sum_{i=1}^N \mu_i <1/3$$. Then $$\sum_{i=1}^{N+1} \mu_i > 2/3$$ and, consequently, $$\mu_{N+1}>1/3$$ and therefore even $$\mu_{N+1}>2/3$$. Thus $$A_{N+1}$$ is the atom you're looking for. (Actually, to answer the original query it is left to argue that atoms are singletons; for this additional assumptions come in handy ...)

• Just one question: how do you know that the family $A_1, A_2, ...$ is countable? – André Porto Oct 20 '18 at 20:53
• I think you meant to say: if $\mu(A)<\frac{2}{3}$, then ... and so $\mu$ takes a forbidden value $\frac{1}{3}$. @AndréPorto Let $A_1$ be an atom with the largest mass, let $A_2$ be an atom in $S\backslash A_1$ with the largest mass, and so on. Since $\mu(S)=1$ we have $\mu(A_n)\to 0$, from where $\mu$ is atomless on $S\backslash A$. – erz Oct 20 '18 at 21:25
• If we assume that $\mu$ is inner regular and that $S$ a $T_2$ (no need of compactness) we conclude that any atom has a singleton in which $\mu$ is concentrated. Fix an atom $A$ and the family $F$ of compact subsets of $A$ with measure $\mu(A)$. Since $\mu$ is inner regular, $F$ is non-empty. Since $A$ is an atom, the family $F$ has the F.I.P.. Let $L=\cap F\neq\emptyset$. If we prove that $\mu(L)=\mu(A)$, then $L$ is the least element of $F$ and, as such, using inner regularity again, we prove that $L$ has no proper non-empty measurable subset. Since $S$ is $T_2$, $L$ must be singleton. – André Porto Oct 20 '18 at 21:42
• Actually, it lacks to prove that $\mu(L)=\mu(A)$ in the argument of the last comment. – André Porto Oct 20 '18 at 21:46
• @erz, is there such a thing as an atom with largest mass? I ask that because unions of atoms are not atoms. – André Porto Oct 20 '18 at 21:49

The atom given by Dirk's answer may be obtained in a much more simple way by just using the item 2) exposed in the question.

Fix a borelian set $$U$$ such that $$\mu(U)=\beta=\inf\{\mu(A): \mu(A)\geq1/3\}.$$ It follows, by the assumption on $$\mu$$, that $$\mu(U) = \beta\geq 2/3.$$

Let us see that $$U$$ is an atom of $$\mu$$. If some borel set $$E\subset U$$ satisfies $$0<\mu(E)<\mu(U)=\beta,$$ then it also satisfies $$0<\mu(U\setminus E)<\mu(U)=\beta.$$ Since $$\mu(E),\ \mu(U\setminus E)<\beta$$, it follows by the minimality of $$\beta$$ that $$\mu(E),\ \mu(U\setminus E)<1/3$$, so $$\mu(U)=\mu(E) + \mu(U\setminus E)<2/3\leq\beta,$$ a contradiction.