Measuring Sets by Counting Their Points
In the previous tutorial, restriction, pushforward, and addition provided ways to construct measures from existing ones. Counting measure is a particularly direct example: it assigns to each set the number of points it contains, with every infinite set assigned measure infinity. Unlike length, it does not make small intervals small. A set with even one point has positive counting measure, and every infinite measurable set has infinite counting measure.
We begin on the power set \(\mathcal{P}(X)\), so every subset of \(X\) is measurable. The same assignment can also be used on any sigma-algebra \(\mathcal{F}\) on \(X\), by considering only its members. The central point to check is countable additivity: a countable disjoint union should have measure equal to the sum of the measures of its pieces.
This definition deliberately does not distinguish between different infinite cardinalities: every infinite set receives the value \(\infty\). Measure values lie in the extended nonnegative real numbers, rather than in the full range of cardinal numbers.
Why Counting Measure Is a Measure
Proof. The empty set has no elements, so \(\#(\varnothing)=0\). Let \(A_1,A_2,\ldots\) be pairwise disjoint subsets of \(X\), and set \(A=\bigcup_{n=1}^{\infty}A_n\). We verify that $$ \#(A)=\sum_{n=1}^{\infty}\#(A_n). $$ If \(A\) is finite, only finitely many of the disjoint sets \(A_n\) can be nonempty. Each is finite, and every element of \(A\) belongs to exactly one of them. Thus the number of elements in \(A\) is the sum of the numbers of elements in the pieces, as required.
Now suppose \(A\) is infinite. If some \(A_n\) is infinite, then \(\#(A_n)=\infty\), so the right-hand side is \(\infty\). Otherwise all the \(A_n\) are finite. There must then be infinitely many nonempty \(A_n\): if only finitely many were nonempty, their union would be a finite union of finite sets and hence finite, contradicting that \(A\) is infinite. Each nonempty \(A_n\) contributes at least \(1\), so the partial sums of \(\sum_n\#(A_n)\) are unbounded and the sum is \(\infty\). In either case the right-hand side equals \(\#(A)=\infty\). This proves countable additivity. Finally, if \(\mathcal{F}\) is a sigma-algebra, its sets and countable unions of its sets are among the subsets just considered, so the same proof establishes that the restriction is a measure. \(\square\)
The proof covers both ways an infinite union can arise. One piece might already be infinite, or infinitely many finite pieces might accumulate to an infinite union. Omitting the second case would leave a gap in a countable-additivity proof.
Worked Example: Counting Subsets of a Finite Set
Let \(X=\{a,b,c,d,e\}\), with counting measure on \(\mathcal{P}(X)\). For \(A=\{a,c,e\}\), there are exactly three elements, so \(\#(A)=3\). If \(B=\{b,d\}\), then \(A\) and \(B\) are disjoint and \(A\cup B=X\). The additivity calculation is $$ \#(A\cup B)=5=3+2=\#(A)+\#(B). $$ For the empty set, \(\#(\varnothing)=0\), and for the whole space, \(\#(X)=5\). Since every subset of this \(X\) is finite, all the measure values are ordinary nonnegative integers.
Worked Example: Counting Measure on the Natural Numbers
Take \(X=\mathbb{N}=\{1,2,3,\ldots\}\). For the finite set \(F=\{2,5,9,12\}\), we have \(\#(F)=4\). The set \(E=\{2,4,6,\ldots\}\) of even positive integers is infinite, so \(\#(E)=\infty\). Its complement in \(\mathbb{N}\), the odd positive integers, is also infinite and has measure infinity.
The even numbers are the disjoint union of the singleton sets \(\{2k\}\), \(k\in\mathbb{N}\). Each singleton has measure \(1\), giving $$ \#(E)=\sum_{k=1}^{\infty}\#(\{2k\}) =\sum_{k=1}^{\infty}1 =\infty. $$ This is an instance of countable additivity in which infinitely many finite pieces, each of measure \(1\), form an infinite-measure set.
Finite Measure and Sigma-Finiteness
Counting measure makes the distinction between finite and infinite sets especially sharp. A measurable set has finite measure exactly when it has finitely many points. This also determines when the whole space can be covered by countably many sets of finite counting measure.
Proof. Recall that a measure is sigma-finite if its space is a countable union of measurable sets, each having finite measure. Suppose first that counting measure is sigma-finite. Then there are sets \(E_1,E_2,\ldots\) such that \(X=\bigcup_{n=1}^{\infty}E_n\) and \(\#(E_n)<\infty\) for every \(n\). Each \(E_n\) is finite by the definition of counting measure. A countable union of finite sets is at most countable, so \(X\) is at most countable.
Conversely, suppose \(X\) is at most countable. If \(X\) is finite, it is itself a finite-measure set, and the sequence \(E_n=X\) for all \(n\) is a countable cover. If \(X\) is countably infinite, list its elements as \(x_1,x_2,\ldots\). The singleton sets \(E_n=\{x_n\}\) each have measure \(1\), and their union is \(X\). Thus \(X\) is a countable union of finite-measure sets. In both cases the counting measure is sigma-finite. \(\square\)
The assumption that we are using \(\mathcal{P}(X)\) matters for the converse: every singleton is measurable there. If counting measure is restricted to a sigma-algebra, the argument that an at-most-countable space is sigma-finite requires an appropriate measurable cover—for example, measurable singletons. The necessity remains: any sigma-finite cover by finite sets makes \(X\) at most countable.
Worked Example: Counting Measure on the Borel Sets of the Real Line
Every Borel subset of \(\mathbb{R}\) can be assigned its counting measure. A singleton such as \(\{3\}\) is Borel and has measure \(1\). The finite set \(\{-2,0,4\}\) has measure \(3\), while the countably infinite set \(\mathbb{Q}\) has measure \(\infty\). The interval \([0,1]\) is also Borel, but it is uncountable, so its counting measure is \(\infty\).
This measure is not sigma-finite on the Borel sets of \(\mathbb{R}\). Indeed, any Borel set of finite counting measure must be finite. A countable union of finite sets is at most countable and therefore cannot cover \(\mathbb{R}\). This contrasts with Lebesgue measure: a bounded interval has finite Lebesgue measure, but infinite counting measure.
Counting Measure Versus Length
Counting measure and Lebesgue measure answer different questions. Counting measure records how many points lie in a measurable set, collapsing every infinite answer to \(\infty\). Lebesgue measure instead records length on the real line. For example, a singleton has counting measure \(1\) but Lebesgue measure \(0\); an interval of positive length has finite Lebesgue measure but infinite counting measure.
| Set in \(\mathbb{R}\) | Counting measure | Lebesgue measure |
|---|---|---|
| \(\{4\}\) | \(1\) | \(0\) |
| \(\{1,3,8\}\) | \(3\) | \(0\) |
| \([2,5]\) | \(\infty\) | \(3\) |
A common pitfall is to assume that a set with small length must have small counting measure, or that a set of length zero must contain no points. The singleton row already disproves both assumptions. Another is to reason that a countable set must have finite counting measure. Countability says that its points can be listed; counting measure still assigns infinity to any infinite list of distinct points.
Counting measure is also a useful reference point for measures built from individual points. Every singleton has mass \(1\), and finite disjoint unions of singletons have measure equal to the number of points. On a countable space, it is sigma-finite because the space can be covered by its singletons. On an uncountable space, no countable cover by finite sets can reach every point. These properties make counting measure both a basic example of a measure and a clear illustration of how the choice of measure determines what “size” means.
Check Your Understanding
Use the definition and results in this tutorial to answer the following questions.
- What is the counting measure of a set with exactly seven elements? What is the counting measure of an infinite set?
- In the countable-additivity proof, why must infinitely many of the sets be nonempty if their union is infinite and every set in the sequence is finite?
- Is counting measure on \(\mathbb{N}\) sigma-finite? Give a countable cover that verifies your answer.
- Why is counting measure on the Borel sets of \(\mathbb{R}\) not sigma-finite?
- How do counting measure and Lebesgue measure differ on a singleton and on a nondegenerate interval?