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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?

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    $\begingroup$ 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 $A \setminus \{x\}$. $\endgroup$ Oct 10, 2015 at 22:30
  • $\begingroup$ 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. $\endgroup$
    – pidgeot
    Oct 10, 2015 at 22:31
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    $\begingroup$ 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. $\endgroup$ Oct 10, 2015 at 22:32
  • $\begingroup$ I'm not sure how you are thinking "that original argument" will go through? $\endgroup$ Oct 10, 2015 at 22:34
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    $\begingroup$ @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 $A$. $\endgroup$ Oct 10, 2015 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.

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  • $\begingroup$ 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. $\endgroup$
    – pidgeot
    Oct 10, 2015 at 22:25
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    $\begingroup$ 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. $\endgroup$ Oct 10, 2015 at 22:27
  • $\begingroup$ 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). $\endgroup$
    – pidgeot
    Oct 10, 2015 at 22:29
  • $\begingroup$ 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. $\endgroup$ Oct 10, 2015 at 22:33
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    $\begingroup$ 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. $\endgroup$ Oct 10, 2015 at 22:36

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