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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?

pidgeot
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    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 $A \setminus {x}$. – Nate Eldredge Oct 10 '15 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. – pidgeot Oct 10 '15 at 22:31
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    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. – Nate Eldredge Oct 10 '15 at 22:32
  • I'm not sure how you are thinking "that original argument" will go through? – Nate Eldredge Oct 10 '15 at 22:34
  • Maybe I'm mistaken about the definition, but isn't it enough to prove that $A$ is not an atom by demonstrating a smaller (Borel) subset contained in $A$ that has positive measure? In that case, $A \setminus {x}$ would be such a proper subset of positive measure. – pidgeot Oct 10 '15 at 22:36
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    @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 $A$. – Eric Wofsey Oct 10 '15 at 22:38
  • Oh, right. Thanks for clearing up my misunderstanding. – pidgeot Oct 10 '15 at 22:38

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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.

Eric Wofsey
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  • 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.

     – pidgeot Oct 10 '15 at 22:25
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    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. – Eric Wofsey Oct 10 '15 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). – pidgeot Oct 10 '15 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. – Eric Wofsey Oct 10 '15 at 22:33
  • But such a measure would not be Borel, right? So it doesn't serve as a counterexample to my original approach? – pidgeot Oct 10 '15 at 22:35
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    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. – Eric Wofsey Oct 10 '15 at 22:36
  • Ah, I understand now. I started using a wrong definition for atom. Sorry about that. – pidgeot Oct 10 '15 at 22:39