Counting Measure On Finite And Infinite Sets

Counting Sets as a Measure

Counting measure turns the question “How many elements does this set contain?” into a measure. For a finite set, the answer is its ordinary number of elements. For an infinite set, the measure is $+\infty$, whether the set is countable or uncountable. This convention gives values in the extended nonnegative real numbers and makes counting measure a useful basic example of a measure that can take an infinite value.

Let $X$ be any set, and let $\mathcal{P}(X)$ denote the sigma-algebra of all subsets of $X$. The measure space $(X,\mathcal{P}(X),\mu_c)$ is complete: its only null set is the empty set, and every subset of $X$ is measurable. Its total measure is finite when $X$ is finite and infinite when $X$ is infinite. No sigma-finiteness assumption is needed below.

Definition: The counting measure $\mu_c$ on $(X,\mathcal{P}(X))$ is defined for every $A\subseteq X$ by $$ \mu_c(A)= \begin{cases} \#A, & \text{if }A\text{ is finite},\\ +\infty, & \text{if }A\text{ is infinite}. \end{cases} $$ Here $\#A$ is the number of elements of a finite set $A$. In particular, $\mu_c(\varnothing)=0$. The value $+\infty$ is an extended-real measure value, not an infinite cardinal number.

Because the measurable sets here are all subsets of $X$, every set considered below is measurable. On a measurable space with a smaller sigma-algebra $\mathcal{M}\subseteq\mathcal{P}(X)$, the same formula can be restricted to sets $A\in\mathcal{M}$. The computations and additivity argument still apply to measurable sets in that space.

Computing Finite and Infinite Values

For finite sets, counting measure is literal counting. It is the number of distinct elements, so repetitions in a list do not add to the count. For infinite sets, the definition deliberately identifies all infinite sizes: a countably infinite set and an uncountable set both have counting measure $+\infty$. This is different from recording cardinality, which distinguishes different kinds of infinite sets.

Worked Example: A Finite Set and Its Subsets

Take $X=\{a,b,c,d,e\}$ and $A=\{a,c,e\}$. Since $A$ has exactly three distinct elements, the definition gives $\mu_c(A)=3$. For $B=\{b,d\}$, it gives $\mu_c(B)=2$, and for $A\cap B=\varnothing$, it gives $\mu_c(A\cap B)=0$.

To check the counts directly, list the elements of $A$: $a,c,e$, so $\#A=3$. List those of $B$: $b,d$, so $\#B=2$. The sets have no common element, so their intersection has zero elements. Thus the values agree with the defining formula in all three cases.

Worked Example: Counting a Disjoint Union with Finite Pieces

Let $X=\{1,2,3,4,5,6\}$, and define $A_1=\{1,2\}$, $A_2=\varnothing$, $A_3=\{3,4,5\}$, and $A_n=\varnothing$ for every $n\ge4$. These sets are pairwise disjoint. Their union is $\{1,2,3,4,5\}$, which has five elements, so its counting measure is $5$.

Term by term, the measures are $\mu_c(A_1)=2$, $\mu_c(A_2)=0$, $\mu_c(A_3)=3$, and $\mu_c(A_n)=0$ for $n\ge4$. Hence $$ \sum_{n=1}^{\infty}\mu_c(A_n)=2+0+3+0+\cdots=5. $$ The sum matches the measure of the union. Checking by listing the union gives the five elements $1,2,3,4,5$, confirming the calculation.

For an infinite set, the result is not a finite count. In particular, a sum with infinitely many nonzero terms can have value $+\infty$, just as the measure of the corresponding union can be $+\infty$. The next result explains why these values agree for every countable disjoint union.

Countable Additivity

Theorem: Let $X$ be any set and equip it with the complete measure space $(X,\mathcal{P}(X),\mu_c)$. If $(A_n)_{n\ge1}$ is a sequence of pairwise disjoint subsets of $X$, then $$ \mu_c\!\left(\bigcup_{n=1}^{\infty}A_n\right) =\sum_{n=1}^{\infty}\mu_c(A_n). $$ The series is understood as a sum of nonnegative extended-real values, so it may equal $+\infty$.

Proof. Write $U=\bigcup_{n=1}^{\infty}A_n$. If $U$ is finite, its elements are partitioned among the disjoint sets $A_n$. Only finitely many $A_n$ can then be nonempty: each nonempty set contains at least one distinct element of $U$, and $U$ has only finitely many elements. Each nonempty $A_n$ is finite, and counting the elements in these disjoint pieces counts every element of $U$ exactly once. Therefore $\#U=\sum_n\#A_n$, with all but finitely many terms zero. By the definition of $\mu_c$, this is the required equality.

Now suppose $U$ is infinite. If some $A_j$ is infinite, then $\mu_c(A_j)=+\infty$. The series on the right is therefore $+\infty$, and the left side is $\mu_c(U)=+\infty$. If every $A_n$ is finite, there must be infinitely many nonempty sets $A_n$; otherwise $U$ would be a finite union of finite sets and hence finite. For any positive integer $k$, choose $k$ distinct indices with nonempty sets. Each selected set has measure at least $1$, so a partial sum containing those terms is at least $k$. Since this holds for every $k$, the series on the right is unbounded and equals $+\infty$, again matching $\mu_c(U)$. These cases cover all possibilities for $U$ and the sets $A_n$, proving the equality. $\square$

Worked Example: The Natural Numbers as a Union of Singletons

Let $X=\mathbb{N}=\{1,2,3,\ldots\}$, with its power-set sigma-algebra and counting measure. For each $n\ge1$, set $A_n=\{n\}$. The sets are pairwise disjoint, and their union is $\mathbb{N}$. Each singleton has measure $\mu_c(A_n)=1$, so $$ \sum_{n=1}^{\infty}\mu_c(A_n)=\sum_{n=1}^{\infty}1=+\infty. $$ The union is infinite, so its measure is also $\mu_c(\mathbb{N})=+\infty$.

Check: the first $N$ sets have union $\{1,\ldots,N\}$, containing exactly $N$ elements, and the first $N$ terms of the sum add to $N$. As $N$ increases without bound, these finite counts are unbounded, so the full sum is $+\infty$, consistent with the measure of the union.

Infinite Sets and a Common Mistake

Worked Example: Counting an Uncountable Set

Consider $X=[0,1]$ with the power-set sigma-algebra and counting measure. The set is infinite: it contains the distinct points $1,1/2,1/3,\ldots$. Therefore the definition gives $\mu_c([0,1])=+\infty$. This conclusion does not require determining the precise cardinality of $[0,1]$; every infinite set receives the same counting-measure value.

Check: the finite subsets $\{1,1/2,\ldots,1/N\}$ have counting measure $N$ for each positive integer $N$. These sizes have no finite bound, in agreement with the assigned value $+\infty$ for the whole interval.

Common mistake: treating an infinite cardinality as a measure value. A wrong calculation might write $\mu_c([0,1])=\# [0,1]=\mathfrak{c}$, where $\mathfrak{c}$ denotes the cardinality of the continuum. That is a cardinality, not a value in the extended nonnegative real numbers used by this measure. The correct application of the definition is simply $\mu_c([0,1])=+\infty$. The same measure value applies to $\mathbb{N}$, even though these sets have different cardinalities.

Countable additivity is useful because it allows a set to be analyzed through disjoint pieces without losing track of its measure. For counting measure, the verification has a direct interpretation: disjoint pieces do not count any element twice, and every element in the union belongs to one of the pieces. If the union is infinite, either an individual piece is already infinite or there are infinitely many nonempty finite pieces; in both cases the sum of the measures is infinite.

Check Your Understanding

Use the definition and the disjoint-union theorem to answer each question. State whether the set in question is finite or infinite.

For $A=\{p,q,r,s\}$, what is $\mu_c(A)$?
Let $A_n=\{2n\}$ for $n\ge1$. Compute $\sum_{n=1}^{\infty}\mu_c(A_n)$ and identify the union.
If $B_1=\{x,y\}$, $B_2=\{z\}$, and $B_n=\varnothing$ for $n\ge3$, what is the measure of their union and the sum of their measures?
What is the counting measure of any uncountable subset of $X$? Explain why its cardinality is not the measure value.
For a disjoint sequence of finite sets with infinitely many nonempty terms, explain why the sum of their counting measures is $+\infty$.