# Must atoms of a Borel measure space be singletons?

It's been a while since I've done any real analysis, so I'd appreciate some guidance.

Suppose we're working on the real line, with some Borel measure induced by a non-decreasing, right-continuous function $F$. Clearly all the points of discontinuity of $F$ are atoms (of which there may only be countably many). So if we had a non-singleton atom $A$, it would have to be uncountable. I wanted to conclude the argument by considering the set $A \setminus \{x\}$ for any $x \in A$, but since the Borel $\sigma$-algebra isn't complete, there's no reason why I should expect that to be a measurable set. Is there a better way to see why this result might be true, or is it false?

Edit: I think I figured it out. Suppose $A$ is an atomic with positive measure $\epsilon$. Then if we partition the real line into half-open intervals of measure less than $\epsilon$, then the intersection of $A$ with one of these intervals should be a proper subset of $A$, with positive measure.

Edit: I think that might not work in general? Can we even partition the real line into countably many intervals of measure $< \epsilon$ for any $\epsilon > 0$? I suppose it must work for finite measure spaces?

• If $A,B$ are Borel then so is $A \setminus B = (A^c \cup B)^c$. Singletons are Borel because they are closed. So if $A$ is Borel then so is $A \setminus \{x\}$. Oct 10, 2015 at 22:30
• Yup, I realized that just a few minutes ago. I abandoned that idea because I didn't want to assume singletons are Borel sets, but they obviously are. Am now wondering why that original argument wouldn't go through. After all, $A \setminus \{x\} \subset A$ and has positive measure. Oct 10, 2015 at 22:31
• Really, the essential property being used here is that $\mathbb{R}$ is first countable. On a topological space which is not first countable, a Borel measure can have an atom which is not a singleton. The Dieudonne measure on $\omega_1$ is a standard example. Oct 10, 2015 at 22:32
• I'm not sure how you are thinking "that original argument" will go through? Oct 10, 2015 at 22:34
• @pidgeot: Being an atom means there is no smaller subset of smaller positive measure. If $\{x\}$ has measure zero, then $A\setminus \{x\}$ has the same measure as $A$. Oct 10, 2015 at 22:38

## 1 Answer

Suppose $A$ is an atom of some Borel measure $\mu$. For simplicity, let us assume $A\subseteq[0,1)$ and $\mu(A)=1$ (it is easy to generalize the argument). For each integer $n$ and each integer $k$ such that $0\leq k<2^n$, let $I_{n,k}=[k/2^n,(k+1)/2^n)$. Since $A$ is an atom, $\mu(A\cap I_{n,k})$ must be either $0$ or $1$. Since these sets (for fixed $n$) partition $A$, we conclude that exactly one of them has measure $1$; that is, there is a unique $k_n$ such that $\mu(A\cap I_{n,k_n})=1$. It is now easy to see that $I_{n,k_n}\subset I_{m,k_m}$ for $n>m$. It follows that $\bigcap_n I_{n,k_n}$ consists of a single point $x$ and that $\mu(A\cap\{x\})=\inf_n \mu(A\cap I_{n,k_n})=1$. That is, $x\in A$, $\{x\}$ is an atom, and $A$ differs from $\{x\}$ by a set of measure zero.

• Thanks, that clears things up. It looks as though the proof I gave in my edit is similar to yours, so I'm glad. Just one question though. Would my argument to consider $A \setminus \{x\}$ have worked? I dismissed it earlier because I didn't expect $\{x\}$ to necessarily be a Borel set (since Borel $\sigma$-algebra is not complete), but then it occurred to me that the Borel $\sigma$-algebra still contains all the singletons. Oct 10, 2015 at 22:25
• It does not work to just consider sets of the form $A\setminus\{x\}$, since $A$ might be uncountable, and we only know the measure is countably additive. Oct 10, 2015 at 22:27
• I'm not sure I see the problem. Even if $A$ is uncountable, $A \setminus \{x\} \subset A$, and $A \setminus \{x\}$ should have positive measure (since we can WLOG choose a point with nonzero mass, since we know there are only countably many atoms). Oct 10, 2015 at 22:29
• Consider the following measure, defined on the $\sigma$-algebra of all subsets of $\mathbb{R}$ which are either countable or cocountable: $\mu(A)=1$ if $A$ is cocountable, and $\mu(A)=0$ if $A$ is countable. Then $\mathbb{R}$ is an atom for $\mu$, but every singleton has measure zero. This example shows that you need to use more about Borel sets than just the fact that points are Borel. Oct 10, 2015 at 22:33
• Well, by my answer there is no Borel counterexample. But it is a counterexample to your approach, because the only property of the $\sigma$-algebra that your approach is using is the fact that it contains all singletons. Oct 10, 2015 at 22:36