# Atom (measure theory)

In mathematics, more precisely in measure theory, an atom is a measurable set which has positive measure and contains no set of smaller but positive measure. A measure which has no atoms is called non-atomic or atomless.

## Definition

Given a measurable space $(X, \Sigma)$ and a measure $\mu$ on that space, a set $A$ in $\Sigma$ is called an atom if

$\mu (A) >0\,$

and for any measurable subset $B$ of $A$ with

$\mu(A) > \mu (B) \,$

one has $\mu(B)=0.$

## Examples

• Consider the set X={1, 2, ..., 9, 10} and let the sigma-algebra $\Sigma$ be the power set of X. Define the measure $\mu$ of a set to be its cardinality, that is, the number of elements in the set. Then, each of the singletons {i}, for i=1,2, ..., 9, 10 is an atom.
• Consider the Lebesgue measure on the real line. This measure has no atoms.

## Non-atomic measures

A measure which has no atoms is called non-atomic. In other words, a measure is non-atomic if for any measurable set $A$ with $\mu (A) >0$ there exists a measurable subset B of A such that

$\mu(A) > \mu (B) > 0. \,$

A non-atomic measure with at least one positive value has an infinite number of distinct values, as starting with a set A with $\mu (A) >0$ one can construct a decreasing sequence of measurable sets

$A=A_1\supset A_2 \supset A_3 \supset \cdots$

such that

$\mu(A)=\mu(A_1) > \mu(A_2) > \mu(A_3) > \cdots > 0.$

This may not be true for measures having atoms; see the first example above.

It turns out that non-atomic measures actually have a continuum of values. It can be proved that if μ is a non-atomic measure and A is a measurable set with $\mu (A) >0,$ then for any real number b satisfying

$\mu (A) \geq b \geq0\,$

there exists a measurable subset B of A such that

$\mu(B)=b.\,$

This theorem is due to Wacław Sierpiński.[1][2] It is reminiscent of the intermediate value theorem for continuous functions.

Sketch of proof of Sierpiński's theorem on non-atomic measures. A slightly stronger statement, which however makes the proof easier, is that if $(X,\Sigma, \mu)$ is a non-atomic measure space and $\mu(X)=c$, there exists a function $S:[0, c]\to\Sigma$ that is monotone with respect to inclusion, and a right-inverse to $\mu:\Sigma\to[0,\,c]$. That is, there exists a one-parameter family of measurable sets S(t) such that for all $0\leq t \leq t'\leq c$

$S(t)\subset S(t'),$
$\mu\left (S(t)\right)=t.$

The proof easily follows from Zorn's lemma applied to the set of all monotone partial sections to $\mu$ :

$\Gamma:=\{S:D\to\Sigma\; :\; D\subset[0,\,c],\, S\; \mathrm{ monotone }, \forall t\in D\; (\mu\left (S(t)\right)=t)\},$

ordered by inclusion of graphs, $\mathrm{graph}(S)\subset \mathrm{graph}(S').$ It's then standard to show that every chain in $\Gamma$ has an upper bound in $\Gamma$, and that any maximal element of $\Gamma$ has domain $[0,c],$ proving the claim.