Questions on atoms of a measure

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.

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.