# Questions on atoms of a measure

1. An atom of any probability measure $\mu$ on $(\mathbb{R}, \mathcal{B})$ is a singleton $\{x\}$ such that $\mu({x}) > 0$.

In Wikipedia:

Given a measurable space $(X,Σ)$ and a measure $\mu$ on that space, a set A in Σ is called an atom if $\mu (A) >0$ and for any measurable subset B of A with $\mu(A) > \mu (B)$, one has $\mu(B) = 0$.

I was wondering if Chung's and Wikipedia's definitions agree with each other? If yes, does Chung's definition mean that in the special probability measure space $(\mathbb{R}, \mathcal{B}, \mu)$, there is no atom, as defined as in Wikipedia, that is not a singleton?

2. Chung says in his book that the number of atoms of any $\sigma$-finite measure is countable.

1. I was wondering if suppose there are uncountably many atoms, how will it contradicts $\sigma$-finite measure?
2. Is this conclusion also true for general measure space $(X,Σ, \mu)$ instead of just for probability measure space $(\mathbb{R}, \mathcal{B}, \mu)$?

Thanks and regards!

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To give a trivial example, take a measure space with all atoms singletons, and add in one point for each atom, and add it to any measurable set containing that atom. Then all the atoms become sets of two points. –  Zhen Lin May 24 '11 at 22:39

The wikipedia definition is more general.

Here is a trivial example: on $X = \mathbb{R}$, take the $\sigma$-algebra $\{ \varnothing, \mathbb{R} \}$ and the measure $\mu(\emptyset) = 0$, $\mu(\mathbb{R}) = 1$. Then $\mathbb{R}$ is an atom which is not a singleton set.

The above example is totally contrived. I believe that if you have a regular Borel measure on a topological space, the atoms will all be points. This certainly holds for $(\mathbb{R},\mathcal{B},\mu)$ (meaning in this case that there are no atoms at all, since all points have Lebesgue measure zero): if you have a subset $A$ with $\mu(A) = \delta > 0$, partition the real line into a countable union of half open intervals $I_n$ of length less than $\delta$. Since $A = \coprod_{n=1}^{\infty} A \cap I_n$, $\sum_{n=1}^{\infty} \mu(A \cap I_n) = \delta$, so there exists $N$ with $0 < \mu(A \cap I_n) < \delta = \mu(A)$.

Finally, suppose you have a measure with uncountably many atoms in Chung's sense, i.e., points of positive measure, and let $\{X_n\}_{n=1}^{\infty}$ be a covering by countably many measurable subsets. Since there are uncountably many atoms, there exist at least one $n$ such that $X_n$ contains uncountably many atoms, so it has infinite measure. Therefore the measure is not $\sigma$-finite.

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I think that the condition needed both for those relations and the equivalence are that the 1-element sets are measurable. –  N. S. May 24 '11 at 22:57
@user9176: it sounds plausible at least. (I don't make any strong claims about anything when it comes to measure theory!) I'm about to sign off for a while, so it would be helpful if you could work this out and include it in your answer. –  Pete L. Clark May 24 '11 at 22:59
Thanks! I was wondering why a measurable subset that contains uncountably many atoms has infinite measure? –  Tim May 25 '11 at 1:01
This follows from the fact that an uncountable sum of positive real numbers is necessarily infinite, which came up for instance on Math Overflow recently. –  Pete L. Clark May 25 '11 at 1:24
Peter, this is not clear; the atoms can contain each other. –  Bombyx mori Nov 13 '12 at 5:00

Let $(\Omega,\Sigma)$ be a measurable space. An atom of $\Sigma$ is a set $B\in\Sigma$ such that for all $A\subseteq B$ either $A=\emptyset$ or $B=A$. A measurable space is atomic if every element lies in some atom. The $\sigma$-algebra $\Sigma$ is countably generated if there is a countable family of measurable sets such that $\Sigma$ is the smallest $\sigma$-algebra containing all of them. For example $(\mathbb{R},\mathcal{B})$ is countably generated since $\mathcal{B}$ is generated by the open intervals with rational endpoints. The atoms of $\mathcal{B}$ are the singletons.

Proposition: If $\Sigma$ is countably generated, then $(X,\Sigma)$ is atomic.

Proof: If there is a countable family generating $\Sigma$, there is also a countable family closed under complementation that generates $\Sigma$. If $\mathcal{C}$ is such a family, we get all atoms of $\Sigma$ as the intersection of all elements of $\mathcal{C}$ that contain a given point.

Now if $(\Omega,\Sigma,\mu)$ is a probability space, we call $B\in\Sigma$ a $\mu$-atom if $\mu(B)>0$ and for all $A\in\Sigma$ such that $A\subseteq B$, either $\mu(A)=0$ or $\mu(A)=\mu(B)$. The probability space is atomless if it contains no $\mu$-atom.

Lemma: If $(\Omega,\Sigma,\mu)$ is a probability space such that $\Sigma$ is countably generated and $\mu$ takes on only the values $0$ and $1$, then there exists an atom $A\in\Sigma$ such that $\mu(A)=1$.

Proof: Let $\mathcal{C}$ be a countable family closed under complementation that generates $\Sigma$. For each element of $\mathcal{C}$, either itself or its complement has probability one $1$. The intersection of all elements in $\Sigma$ with probability $1$ is an atom with probability $1$.

Proposition: If $(\Omega,\Sigma,\mu)$ is a probability space with $\Sigma$ countably generated, then it is atomless if and only if every atom in $\Sigma$ has probability $0$.

Proof: Clearly, in an atomless probability space, every atom must have probability $0$. Supppose now that $A$ is a $\mu$-atom. Let $A\cap\Sigma=\{A\cap S:S\in\Sigma\}$ be the trace $\sigma$-algebra. It is countably generated too. Then $(A,A\cap\Sigma,1/\mu(A)\cdot\mu)$ is a probability space such that the probability takes on only the values $0$ and $1$. So by the lemma, there is an atom $B$ such that $1/\mu(A)⋅\mu(B)=1$. But $B$ is also an atom of $\Sigma$ and $\mu(B)>0$.

So it follows that a probability measure on $(\mathbb{R},\mathcal{B})$ is atomless if and only if it puts probability $0$ on all singletons, which justifies the definition in the book of Kai Lai Chung.

Finally, an example of a probability space in which each atom has probability $0$ but such that the space is not atomless. Let $\Omega$ be any uncountable set, let $\Sigma$ consists of those subsets of $\Omega$ that are either countable or have an uncountable complement. Let $\mu(A)=0$ if $A$ is countable and $\mu(A)=1$ if its complement is countable. Every set with countable complement is an $\mu$-atom, but the atoms of $\Sigma$ are the singletons which all have probability $0$. Note that $\Sigma$ is not countably generated.

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in lemma's proof, why cant the intersection of all elements in sigma with probability 1 have probability 0? –  Theta33 Feb 11 '13 at 6:19
Does this work? Suppose the intersection has probability 1, then its complement has probability 0. If we choose only the countable set, the probability of complement of intersection , which is union of compelments has probability 1. Then there exist one element with probability 1. Then its complement has probability 0. Contradiction with the choise of elements in intersection –  Theta33 Feb 11 '13 at 7:01
@Theta30 I'm not completely sure I understand your argument. But if $(A_n)$ is a sequence of events that all have probability $1$, then we know $\mu(\bigcap_n A_n)=\mu((\bigcap_n A_n^C)^C)=1-\mu(\bigcup_n A_n^C)\geq 1-\sum_n \mu(A_n)=1$. –  Michael Greinecker Feb 11 '13 at 14:25

If the measure is positive, consider $A_n := \{ x | \mu( \{ x \}) \geq \frac{1}{n} \}$.

Any atom is in some $A_n$, thus in the union. If the number of atoms is not countable, then some $A_n$ must be uncountable thus infinite. But this contradicts $\mu$ is $\sigma$-finite.

If $\mu$ is not positive, write it as $\mu=\mu_+-\mu_-+i\mu_{i+}-i\mu_{i-}$ and do the same for each of the four measures.

Edti: And this works in the general case, as long as $\{x\} \in \Sigma$ for all $x$.

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