Measures On Finite Sigma-Algebras

Atoms Record the Structure of a Finite Sigma-Algebra

A finite sigma-algebra may have only finitely many measurable sets even when its underlying space is infinite. Its structure is captured by a finite partition: the smallest nonempty measurable pieces from which every measurable set is assembled. These pieces are called atoms. Once they have been identified, calculating any measurable-set measure becomes a finite sum, with $+\infty$ allowed.

Let $(X,\mathcal{M},\mu)$ be a measure space in which $\mathcal{M}$ is a finite sigma-algebra. No assumption is made that $X$ or $\mu(X)$ is finite, or that the measure space is complete or sigma-finite. All sets considered below belong to $\mathcal{M}$.

Definition: An atom of $\mathcal{M}$ is a nonempty set $A\in\mathcal{M}$ such that the only measurable subsets of $A$ are $\varnothing$ and $A$. Equivalently, there is no $E\in\mathcal{M}$ with $\varnothing\subsetneq E\subsetneq A$.

The definition is relative to $\mathcal{M}$: an atom need not be a singleton in $X$. A finite measurable partition, as considered in the earlier material on finite measurable partitions, gives a sigma-algebra whose atoms are precisely its partition blocks. For a general finite sigma-algebra, the atoms can be found directly from its measurable sets.

The Atoms Form a Partition

Proposition: If $\mathcal{M}$ is a finite sigma-algebra on $X$, then its atoms are pairwise disjoint, their union is $X$, and every $E\in\mathcal{M}$ is a union of atoms.

Proof. Since $X\in\mathcal{M}$ and $X$ is a finite set of measurable sets in the sense that $\mathcal{M}$ has finitely many members, consider any $x\in X$. The collection $\{E\in\mathcal{M}:x\in E\}$ is nonempty because it contains $X$, and it is finite. Its intersection $A_x$ belongs to $\mathcal{M}$, because finite intersections of members of a sigma-algebra are measurable. Also $x\in A_x$, so $A_x\ne\varnothing$.

If $B\in\mathcal{M}$ and $B\subseteq A_x$, then either $x\in B$ or $x\notin B$. In the first case, $B$ is one of the sets intersected to form $A_x$, hence $A_x\subseteq B$, so $B=A_x$. In the second case, $B\cap A_x=\varnothing$: otherwise $y\in B\cap A_x$ for some $y$, but this does not alone force $x\in B$. Instead use the minimal-set characterization directly: $A_x$ is the intersection of all measurable sets containing $x$. If a measurable $B\subseteq A_x$ contains any $y\in A_x$, then $A_x\subseteq B$ follows because $y$ belongs to every measurable set containing $x$ only if the corresponding membership classes agree. More explicitly, define $y\sim x$ when $x$ and $y$ belong to exactly the same members of $\mathcal{M}$. The class $[x]$ is measurable: it is the intersection of the sets in $\mathcal{M}$ containing $x$ and the complements of those not containing $x$, a finite intersection. Any measurable subset $B\subseteq[x]$ containing $y\in[x]$ must contain $x$, since $x,y$ have identical membership in every member of $\mathcal{M}$; thus $B=[x]$. This proves that $[x]$ is an atom.

Conversely, if $A$ is any atom and $x\in A$, then $A\subseteq[x]$: for each $E\in\mathcal{M}$ containing $x$, the intersection $A\cap E$ is a nonempty measurable subset of $A$, so it equals $A$; and if $E$ does not contain $x$, it cannot contain any point of $A$, since $A\cap E$ would again be a nonempty proper measurable subset of $A$. Thus $A=[x]$. Therefore the atoms are exactly the equivalence classes $[x]$. Equivalence classes are disjoint or equal and cover $X$. Finally, if $E\in\mathcal{M}$ and $x\in E$, every point equivalent to $x$ also belongs to $E$. Hence $E$ is a union of the equivalence classes it meets, which are atoms. $\square$

The proof also shows how to identify atoms: points are grouped together precisely when no measurable set separates them. In particular, a generating family need not itself be the atom partition; intersections of its members and complements may be needed to split the space into the smallest measurable pieces.

Measures Are Sums of Atom Weights

Theorem: Let $\mathcal{M}$ be a finite sigma-algebra on $X$, with atoms $A_1,\dots,A_r$. For any measure $\mu$ on $(X,\mathcal{M})$ and any $E\in\mathcal{M}$, $$ \mu(E)=\sum_{\{i:A_i\subseteq E\}}\mu(A_i). $$ Conversely, any assignment of weights $w_i\in[0,\infty]$ to the atoms defines a measure on $\mathcal{M}$ by $\mu(E)=\sum_{\{i:A_i\subseteq E\}}w_i$. The empty sum is $0$, and a finite sum containing $+\infty$ is $+\infty$.

Proof. By the proposition, each $E\in\mathcal{M}$ is the union of exactly the atoms it contains. These atoms are pairwise disjoint. Finite additivity of a measure, which follows from countable additivity by appending empty sets, therefore gives the displayed formula. This direction includes $E=\varnothing$, whose union has no atoms and whose measure is $0$.

For the converse, define $\mu$ by the displayed finite sum. It is nonnegative and $\mu(\varnothing)=0$. Let $E_1,E_2,\dots\in\mathcal{M}$ be pairwise disjoint, and set $E=\bigcup_{n=1}^{\infty}E_n$, which belongs to $\mathcal{M}$. Each atom $A_i$ contained in $E$ must be contained in some $E_n$: since $A_i\subseteq E$, the measurable sets $A_i\cap E_n$ have union $A_i$; if every intersection were empty, their union would be empty, and if any intersection were nonempty, atomicity makes that intersection all of $A_i$. There can be at most one such $n$, because the $E_n$ are disjoint and $A_i\ne\varnothing$. Thus the atoms contained in $E$ are partitioned according to the unique $E_n$ that contains them. Consequently, summing their weights gives $$ \mu(E)=\sum_{n=1}^{\infty}\mu(E_n). $$ This equality holds also when a weight is infinite: both sides are $+\infty$ if any included atom has infinite weight, and otherwise the sums are finite sums of nonnegative weights, with only finitely many atoms in total. Hence $\mu$ is countably additive and is a measure. $\square$

This characterization says that specifying a measure on a finite sigma-algebra is exactly the same as specifying one nonnegative extended-real weight for each atom. It does not say that the measure is finite: a single atom may have infinite measure.

Worked Example: All Sets in a Three-Atom Sigma-Algebra

Let $X=\{a,b,c,d,e,f\}$, and let $\mathcal{M}$ consist of all unions of $A_1=\{a,b\}$, $A_2=\{c,d\}$, and $A_3=\{e,f\}$. Assign atom measures $\mu(A_1)=2$, $\mu(A_2)=0$, and $\mu(A_3)=5$. Calculate the measure of every measurable set.

The three blocks are nonempty, disjoint, and cover $X$. Every measurable set is one of their eight possible unions. Summing the weights of its blocks gives:

Measurable set	Calculation	Measure
$\varnothing$	empty sum	$0$
$A_1$	$2$	$2$
$A_2$	$0$	$0$
$A_3$	$5$	$5$
$A_1\cup A_2$	$2+0$	$2$
$A_1\cup A_3$	$2+5$	$7$
$A_2\cup A_3$	$0+5$	$5$
$X$	$2+0+5$	$7$

Check: The disjoint sets $A_1\cup A_2$ and $A_3$ have measures $2$ and $5$; their union is $X$, and $2+5=7=\mu(X)$. The zero-weight atom is still a nonempty measurable set: its measure is $0$, not because it is absent from the partition, but because its assigned weight is zero.

Worked Example: Atoms from Overlapping Generators

On $X=\{1,2,3,4,5,6\}$, let $\mathcal{M}=\sigma(S,T)$, where $S=\{1,2,3\}$ and $T=\{3,4,5\}$. Find the atoms and, if their respective weights are $4,1,0,7$, calculate $\mu(S)$, $\mu(T)$, and $\mu(S\triangle T)$. Here $S\triangle T$ denotes the points in exactly one of $S,T$.

Membership in $S$ and $T$ gives the four intersections $$ S\cap T=\{3\},\quad S\cap T^c=\{1,2\},\quad S^c\cap T=\{4,5\},\quad S^c\cap T^c=\{6\}. $$ All are nonempty and disjoint, and their union is $X$. They are the atoms: each is a membership-pattern class, and the sets generated by $S,T$ are exactly unions of these four classes. In the order displayed, the weights are $4,1,0,7$. Thus $S$ consists of the first and second atoms, so $\mu(S)=4+1=5$. The set $T$ consists of the first and third atoms, so $\mu(T)=4+0=4$. The symmetric difference consists of the second and third atoms, giving $\mu(S\triangle T)=1+0=1$.

Check: The four atom weights sum to $4+1+0+7=12$, so $\mu(X)=12$. Also $S\cap T=\{3\}$ has measure $4$; the disjoint decomposition $S\cup T=(S\triangle T)\cup(S\cap T)$ gives measure $1+4=5+4-4=5$, agreeing with $\mu(S\cup T)$ computed from its atoms $4+1+0=5$.

Finite Sigma-Algebras Can Carry Infinite Measures

Worked Example: An Infinite Underlying Space and Infinite Atom Weight

Let $X=\mathbb{R}$, $A=[0,1]$, and $\mathcal{M}=\{\varnothing,A,A^c,\mathbb{R}\}$. Assign weights $w(A)=3$ and $w(A^c)=+\infty$. Calculate the measure of every set in $\mathcal{M}$.

The two atoms are $A$ and $A^c$, both nonempty measurable sets. The atom-weight formula gives $\mu(\varnothing)=0$, $\mu(A)=3$, $\mu(A^c)=+\infty$, and $\mu(\mathbb{R})=3+(+\infty)=+\infty$. The theorem guarantees countable additivity for this assignment, even though the measure is not finite. The sigma-algebra has four members, despite the underlying space being uncountable.

Check: The disjoint union $A\cup A^c=\mathbb{R}$ has measure $3+(+\infty)=+\infty$, exactly the assigned value of $\mathbb{R}$. Also $\mu(\mathbb{R})$ is not the number of measurable sets or the size of the underlying space; it is determined by the two atom weights.

A Common Mistake: Treating Generators as Atoms

For the overlapping sets $S=\{1,2,3\}$ and $T=\{3,4,5\}$ above, a tempting but incorrect calculation is to call $S$ and $T$ the atoms. They overlap at $3$, so they are not even a partition. Assigning them weights $4$ and $1$ and then adding those weights to find $\mu(S\cup T)$ would give $4+1=5$, but it counts the overlap as though the sets were disjoint and ignores the rest of the membership classes.

The correction is to split the space by membership in each generator before assigning or summing atom weights. The four disjoint classes are $\{3\}$, $\{1,2\}$, $\{4,5\}$, and $\{6\}$. With weights $4,1,0,7$, respectively, $S\cup T$ contains the first three atoms, so its measure is $4+1+0=5$; the remaining atom has weight $7$, and $5+7=12=\mu(X)$. The atom partition, not the original overlapping family, is the correct basis for the calculation.

Check Your Understanding

Use the definition of atom and the finite-sum formula to justify each answer.

If a finite sigma-algebra has atoms $A_1,A_2,A_3$ with measures $2,0,6$, what are the measures of $A_1\cup A_2$ and $A_2\cup A_3$?
On $X=\{1,2,3,4\}$, take $S=\{1,2\}$ and $T=\{2,3\}$. Find the atoms of $\sigma(S,T)$, including the membership class that contains $4$.
Can a finite sigma-algebra on an infinite set have an atom that is not a singleton? Give an example and identify its atoms.
Suppose one atom of a finite sigma-algebra has infinite measure. What is the measure of any measurable set containing that atom?
Explain why assigning a nonnegative extended-real weight to each atom always defines a measure, including when some weights equal $+\infty$.

Measurable set	Calculation	Measure
\(\varnothing\)	empty sum	\(0\)
\(A_1\)	\(2\)	\(2\)
\(A_2\)	\(0\)	\(0\)
\(A_3\)	\(5\)	\(5\)
\(A_1\cup A_2\)	\(2+0\)	\(2\)
\(A_1\cup A_3\)	\(2+5\)	\(7\)
\(A_2\cup A_3\)	\(0+5\)	\(5\)
\(X\)	\(2+0+5\)	\(7\)