# What is the formal definition of an "atomless" distribution?

The use of atomless (continuous) distributions is ubiquitous in applied works. While the general idea is somewhat clear to me, I was looking for a formal definition or some useful references on the matter. Any help would be greatly appreciated.

• There are many equivalent definitions. The name may come from the fact that the continuous distributions are precisely those with everywhere continuous distribution functions. Any book on probability and statistics (except for those introductions that avoid as much math as possible) will discuss this: pick your favorite.
– whuber
Jul 30 '15 at 20:00

Given a measure $\mu$ and measurable sets $\mathcal{M} \subseteq \mathcal{P}(X)$, a set $A \in \mathcal{M}$ is said to be an atom in the measureable space $(X,\mathcal{M},\mu)$ if $\mu(A) > 0$ and $\mu(B) \in \{0,\mu(A)\}$ for all $B \subset A, B \in \mathcal{M}.$ The measurable space is atomless, if there is no such $A \in \mathcal{M}$. That is, an atom is a set of positive measure which contains no nontrivial smaller measurable sets. See https://en.wikipedia.org/wiki/Atom_(measure_theory) .
• So, if for instance, the probability space is defined on the interval $[0,1]$ and the probability measure on the lower bound of the interval is $\mu(0)>0$. Then, it is not an atom according to your definition, isn't it?
• @mrb Why not? I assume you meant $\mu(0)$ as shorthand notation for $\mu(\{0\})$. Let $A=\{0\}$ and $B\subset A$. Then, either $B=A$ or $B=\emptyset$. In the first case $\mu(B)=\mu(A)$ and in the second case $\mu(B)=0$. Thus for any $B \subset A$ we have $\mu(B) \in \{0,\mu(A)\}$ and thus $A$ is an atom per the definition given in this answer. Jul 31 '15 at 6:06