Finite Total Measure Is Not the Same as Finitely Many Points
A measure space may have infinitely many points but finite total measure, or finitely many points but infinite total measure. The distinction is about the value assigned to the whole space, not its cardinality. Keeping these ideas separate is useful whenever a calculation involves a measurable subset, a complement, or an infinite collection of atoms.
Throughout, \((X,\mathcal{M},\mu)\) is a measure space: \(\mathcal{M}\) is a sigma-algebra of subsets of \(X\), and \(\mu:\mathcal{M}\to[0,\infty]\) is a measure. The measure is allowed to take the value \(+\infty\). A measure space is finite if \(\mu(X)<\infty\), and it has infinite total measure if \(\mu(X)=\infty\). These terms concern \(\mu(X)\), not whether \(X\) is finite or infinite as a set.
In particular, if \(\mu(X)<\infty\), every measurable subset has finite measure: monotonicity gives \(0\leq\mu(E)\leq\mu(X)<\infty\). Conversely, if \(\mu(X)=\infty\), some subsets can still have finite measure, including the empty set. Infinite total measure says only that the whole space has infinite measure.
Finite total measure is also distinct from sigma-finiteness. A measure space is sigma-finite if \(X\) is a countable union of measurable sets of finite measure. A finite measure space is sigma-finite, since \(X\) itself is one such set. But a space of infinite total measure can be sigma-finite too; counting measure on \(\mathbb{N}\) is an example, since it is covered by its singleton sets. This tutorial focuses on total measure and subset calculations, rather than testing sigma-finiteness.
Counting Measure: The Basic Comparison
For counting measure on a set \(X\), every subset is measurable and its measure is its number of points if it is finite, or \(+\infty\) if it is infinite. Thus counting measure on a finite set gives a finite measure space, while counting measure on an infinite set has infinite total measure. A finite subset of an infinite counting-measure space still has finite measure.
Worked Example: Subsets of a Finite Counting-Measure Space
Let \(X=\{a,b,c,d,e\}\), let \(\mathcal{M}\) be the power set of \(X\), and let \(\mu\) be counting measure. Compute the measures of \(E=\{a,c,e\}\) and \(F=\{b,d\}\), and check the total.
Counting measure assigns one unit to each point. The set \(E\) has three points, so \(\mu(E)=3\). The set \(F\) has two points, so \(\mu(F)=2\). The whole space has five points, so \(\mu(X)=5<\infty\); this is a finite measure space. The sets \(E\) and \(F\) are disjoint and their union is \(X\), giving \(\mu(E)+\mu(F)=3+2=5=\mu(X)\). Check: counting the five listed points of \(X\) directly also gives total measure \(5\).
Worked Example: A Finite Subset of an Infinite Counting-Measure Space
Take \(X=\mathbb{N}=\{1,2,3,\ldots\}\) with its power-set sigma-algebra and counting measure. Find the measures of \(A=\{2,5,9\}\), \(B=\{2,4,6,\ldots\}\), and \(X\).
The set \(A\) has exactly three points, so \(\mu(A)=3\). The set \(B\) contains infinitely many even positive integers, so its counting measure is \(\mu(B)=\infty\). The whole space also contains infinitely many points, hence \(\mu(X)=\infty\), and the measure space is not finite. Nevertheless, \(A\) is a measurable subset of finite measure. Check: its three distinct elements each contribute one, so its measure is \(1+1+1=3\); the distinct even integers \(2,4,6,\ldots\) contribute infinitely many units to \(B\).
Weighted Atoms and Finite Total Measure
A weighted atomic measure assigns a nonnegative weight to each point and adds the weights over a subset. On \(\mathbb{N}\), let \(w_n\geq0\) be a sequence and define \(\mu(A)=\sum_{n\in A}w_n\) for every subset \(A\subseteq\mathbb{N}\). A series of nonnegative terms is understood as the supremum of its finite subsums; its value may be infinite. In particular, \(\mu(\mathbb{N})=\sum_{n=1}^{\infty}w_n\). The weights need not all be equal, unlike counting measure.
Proof. The empty set has no terms, so \(\mu(\varnothing)=0\). Let \(A_1,A_2,\ldots\) be pairwise disjoint subsets of \(\mathbb{N}\). Their union is a disjoint collection of indices. Any finite selection of indices in \(\bigcup_j A_j\) meets only finitely many of the sets \(A_j\), and the sum of the selected weights is at most \(\sum_j\mu(A_j)\). Taking the supremum over all such finite selections gives \(\mu(\bigcup_j A_j)\leq\sum_j\mu(A_j)\). For the reverse inequality, fix a finite set \(J\) of positive integers. For each \(j\in J\), choose any finite collection \(F_j\subseteq A_j\). The sets \(F_j\) are disjoint, so their union is a finite subset of \(\bigcup_j A_j\), and \(\sum_{j\in J}\sum_{n\in F_j}w_n\leq\mu(\bigcup_j A_j)\). Taking the supremum over the finite choices \(F_j\) gives \(\sum_{j\in J}\mu(A_j)\leq\mu(\bigcup_j A_j)\). Taking the supremum over finite \(J\) yields \(\sum_j\mu(A_j)\leq\mu(\bigcup_j A_j)\). The two inequalities prove countable additivity, including when either side is infinite. Thus the formula defines a measure. \(\square\)
Worked Example: A Finite Measure on an Infinite Set
On \(X=\mathbb{N}\) with its power-set sigma-algebra, assign weight \(w_n=2^{-n}\) to point \(n\). Compute the total measure, the measure of the even integers \(E=\{2,4,6,\ldots\}\), and the measure of the odd integers \(O=\{1,3,5,\ldots\}\).
The total is a geometric series: \(\mu(X)=\sum_{n=1}^{\infty}2^{-n}=(1/2)/(1-1/2)=1\). Thus the underlying set is infinite, but the measure space is finite. For the even integers, put \(n=2k\): \(\mu(E)=\sum_{k=1}^{\infty}2^{-2k}=\sum_{k=1}^{\infty}(1/4)^k=(1/4)/(1-1/4)=1/3\). For the odd integers, put \(n=2k-1\): \(\mu(O)=\sum_{k=1}^{\infty}2^{-(2k-1)}=2\sum_{k=1}^{\infty}(1/4)^k=2(1/3)=2/3\). The even and odd integers are disjoint and together equal \(X\), and \(1/3+2/3=1\), as required. Check: the two geometric-series calculations give \(1/3+2/3=1=\mu(X)\), independently confirming the total.
Subset Bounds and Complements
Monotonicity gives the basic bound for every measurable subset. In a finite measure space, finite additivity also gives a useful complement identity. The identity must be used with care: when the total measure is infinite, subtracting the measure of a subset from the total is not a valid way to calculate its complement.
Proof. Nonnegativity is part of the definition of a measure. Since \(E\subseteq X\), monotonicity gives \(\mu(E)\leq\mu(X)\). Now suppose \(\mu(X)<\infty\). The sets \(E\) and \(E^c\) are measurable, disjoint, and have union \(X\). Finite additivity therefore gives \(\mu(X)=\mu(E)+\mu(E^c)\). Both terms are finite by the subset bound, so rearranging this equality is valid and yields \(\mu(E^c)=\mu(X)-\mu(E)\). \(\square\)
Worked Example: Complement Calculation in a Finite Atomic Space
On \(X=\{a,b,c\}\) with the power-set sigma-algebra, assign weights \(2\), \(0\), and \(5\) to \(a\), \(b\), and \(c\), respectively. Let \(E=\{a,b\}\). Compute \(\mu(X)\), \(\mu(E)\), and \(\mu(E^c)\) in two ways.
By the atomic weights, \(\mu(X)=2+0+5=7\), and \(\mu(E)=2+0=2\). The complement is \(E^c=\{c\}\), so direct calculation gives \(\mu(E^c)=5\). The complement identity gives the same result: \(\mu(X)-\mu(E)=7-2=5\). Check: the disjoint sets \(E\) and \(E^c\) have measures \(2\) and \(5\), whose sum is \(7=\mu(X)\). The zero-weight point is still in \(E\), but contributes zero to its measure.
Common Pitfall: Treating Infinity Like a Number
For finite total measure, the complement formula involves ordinary subtraction of finite numbers. For infinite total measure, the equation \(\mu(X)=\mu(E)+\mu(E^c)\) still holds by additivity, but it does not permit a useful subtraction rule. In particular, writing \(\mu(E^c)=\infty-\mu(E)\) is not a valid calculation when \(\mu(E)\) is finite, and \(\infty-\infty\) is undefined.
For a concrete correction, use counting measure on \(X=\mathbb{N}\) and \(E=\{2,4,6,\ldots\}\). Both \(E\) and \(E^c=\{1,3,5,\ldots\}\) are infinite, so \(\mu(E)=\mu(E^c)=\infty\), while \(\mu(X)=\infty\). The tempting calculation \(\mu(E^c)=\mu(X)-\mu(E)=\infty-\infty\) has no defined value. The correct method is to identify the complement and count its points; here that gives \(\mu(E^c)=\infty\). Additivity says \(\infty=\infty+\infty\), which is true in the extended nonnegative reals but does not determine either summand from the total.
The key distinctions can be summarized as follows:
| Space and measure | Underlying set | Total measure | Conclusion |
|---|---|---|---|
| Counting measure on a five-point set | Finite | 5 | Finite measure space |
| Counting measure on \(\mathbb{N}\) | Infinite | \(\infty\) | Infinite total measure |
| Weights \(2^{-n}\) on \(\mathbb{N}\) | Infinite | 1 | Finite measure space |
| Any infinite-total space, with a finite subset | May be infinite | \(\infty\) | The subset can still have finite measure |
Check Your Understanding
For each question, distinguish the size of the underlying set from its measure and state whether the total measure is finite.
- Under counting measure on \(\mathbb{N}\), what is the measure of \(\{1,4,10\}\), and what is the total measure of the space?
- For weights \(w_n=3^{-n}\) on \(\mathbb{N}\), compute \(\mu(\mathbb{N})\). Is the measure space finite even though \(\mathbb{N}\) is infinite?
- In a finite measure space with \(\mu(X)=8\) and \(\mu(E)=3\), compute \(\mu(E^c)\) and identify the condition that licenses the subtraction.
- Under counting measure on \(\mathbb{N}\), both the even and odd positive integers are infinite sets. What are their measures?
- Explain why \(\infty-\infty\) cannot be used to find the measure of a complement in an infinite-total measure space.