When Is a Measure Space Infinite?
A measure can assign a finite value to some sets and an infinite value to others. In the previous tutorial, a measure was called finite when the whole space had finite measure. The corresponding distinction here concerns the total mass: an infinite measure space has infinite measure on the whole space. This does not mean that every measurable subset has infinite measure.
Throughout, let \((X,\mathcal{F},\mu)\) be a measure space. As in the definition of a measure, values of \(\mu\) lie in the extended nonnegative real numbers \([0,\infty]\), so \(+\infty\) is an allowed value. The measure remains countably additive; an infinite total is not a failure of the measure axioms.
This definition is about one particular measurable set, namely \(X\). A measure space may have infinite total measure even though it contains many sets of finite measure. Conversely, an infinite measure need not have any nonempty sets of finite measure. The examples below illustrate both possibilities.
Worked Example: Harmonic Weights on the Natural Numbers
On \((\mathbb{N},\mathcal{P}(\mathbb{N}))\), assign weight \(w_n=1/n\) to the point \(n\) and set \(\mu(A)=\sum_{n\in A}1/n\). The weighted atomic measure theorem from “Examples of Measures” shows that this defines a measure. Its total mass is infinite. To verify divergence, for each \(k\geq 1\) consider the integers \(n\) with \(2^{k-1}<n\leq 2^k\). There are \(2^{k-1}\) such integers, and each satisfies \(1/n\geq 1/2^k\). Therefore $$ \sum_{n=2^{k-1}+1}^{2^k}\frac{1}{n} \geq 2^{k-1}\frac{1}{2^k} =\frac12. $$ The sums over these disjoint blocks are each at least \(1/2\), so the partial sums of the harmonic series are unbounded. Thus \(\mu(\mathbb{N})=\infty\). However, every finite set \(A\) has finite measure: if \(A\) has \(m\) elements, then \(\mu(A)\) is a sum of \(m\) finite numbers. Infinite total mass and finite-measure subsets occur in the same space.
A Finite Piece Has an Infinite Complement
There is a useful consequence of the definition that does not require any further assumptions about the space. In an infinite measure space, if a measurable set has finite measure, the part of the space outside it must carry infinite measure. Otherwise, the whole space would be the disjoint union of two finite-measure sets and would itself have finite measure.
Proof. Since \(\mathcal{F}\) is a sigma-algebra, \(X\setminus A\in\mathcal{F}\). The sets \(A\) and \(X\setminus A\) are disjoint and have union \(X\). Finite additivity, which follows from countable additivity, gives $$ \mu(X)=\mu(A)+\mu(X\setminus A). $$ Suppose, for contradiction, that \(\mu(X\setminus A)<\infty\). The hypothesis also gives \(\mu(A)<\infty\), so their sum is finite. The displayed equality would then imply \(\mu(X)<\infty\), contradicting \(\mu(X)=\infty\). Hence \(\mu(X\setminus A)=\infty\). \(\square\)
The finiteness assumption on \(A\) matters. If \(\mu(A)=\infty\), this theorem says nothing about the measure of its complement: that complement may have finite measure, infinite measure, or measure zero.
Worked Example: A Finite Initial Segment and Its Tail
For the harmonic-weight measure above, let \(A=\{1,2,\ldots,N\}\). Its measure is the finite sum $$ \mu(A)=\sum_{n=1}^{N}\frac{1}{n}<\infty. $$ Since \(\mu(\mathbb{N})=\infty\), the theorem gives \(\mu(\mathbb{N}\setminus A)=\infty\). Directly, this complement is \(\{N+1,N+2,\ldots\}\); its sum diverges because removing finitely many terms from a divergent series leaves a divergent tail. Thus no finite initial segment contains all but a finite amount of the total mass.
Increasing Exhaustions Reveal Infinite Total Mass
Often the whole space is easier to understand by approximating it from within. An increasing sequence of measurable sets \((E_n)\) is an exhaustion of \(X\) if \(E_n\subseteq E_{n+1}\) for each \(n\) and \(\bigcup_{n=1}^{\infty}E_n=X\). Continuity from below, established in “Definition of a Measure,” says that the measures of such a sequence converge to the measure of its union. In an infinite measure space, an exhaustion therefore has measures that grow without bound.
Proof. By continuity from below, $$ \lim_{n\to\infty}\mu(E_n) =\mu\left(\bigcup_{n=1}^{\infty}E_n\right) =\mu(X) =\infty. $$ The sequence \(\mu(E_n)\) is nondecreasing because \(E_n\subseteq E_{n+1}\) and measures are monotone. Convergence to \(+\infty\) means that for each finite \(M\geq 0\), some \(N\) satisfies \(\mu(E_N)>M\). Monotonicity then gives \(\mu(E_n)\geq\mu(E_N)>M\) for every \(n\geq N\), as required. \(\square\)
This is a consequence of continuity from below, not a new assumption about how quickly the measures grow. An exhaustion may grow slowly, and its individual terms may all have finite measure, but their measures cannot stay bounded if the union has infinite measure.
Worked Example: Exhausting the Real Line by Bounded Intervals
Let \(\mu\) be Lebesgue measure on the Borel subsets of \(\mathbb{R}\), and take \(E_n=[-n,n]\). These sets are increasing and their union is \(\mathbb{R}\). Each has finite measure \(\mu(E_n)=2n\), while $$ \lim_{n\to\infty}\mu(E_n) =\lim_{n\to\infty}2n =\infty. $$ The exhaustion theorem is visible directly: given a finite \(M\geq0\), choose an integer \(N>M/2\). For every \(n\geq N\), \(\mu(E_n)=2n\geq2N>M\). Thus finite intervals approximate the whole space, but their lengths must grow without bound.
How Infinite Mass Appears in a Countable Partition
A measurable partition of \(X\) is a sequence of pairwise disjoint measurable sets whose union is \(X\). Countable additivity describes the total mass through the masses of its pieces. For an infinite measure space, this gives a simple alternative: either at least one piece already has infinite measure, or every piece has finite measure and the sum of those finite values diverges.
Proof. By countable additivity on pairwise disjoint measurable sets, $$ \mu(X)=\sum_{n=1}^{\infty}\mu(A_n). $$ The left side is \(\infty\), so the series on the right has value \(\infty\). If any term \(\mu(A_n)\) is infinite, the first alternative holds. If no term is infinite, then every term is finite, while the series still has value \(\infty\); equivalently, its nondecreasing partial sums are unbounded. This is the second alternative. \(\square\)
This result does not imply that one particular piece must have infinite measure. An infinite total can be distributed across countably many pieces, with each piece finite, provided their measures have a divergent sum. The distinction is important whenever the space is divided into convenient regions.
Worked Example: Unit Intervals Partition the Real Line
The half-open intervals \(A_k=[k,k+1)\), indexed by \(k\in\mathbb{Z}\), are pairwise disjoint and have union \(\mathbb{R}\). Each interval has Lebesgue measure \(1\). To apply the countable-partition theorem using positive integer indices, enumerate the intervals as \(A_1=[0,1)\), \(A_2=[1,2)\), \(A_3=[-1,0)\), \(A_4=[2,3)\), \(A_5=[-2,-1)\), and continue outward. Every piece still has measure \(1\). The sum over the first \(N\) pieces is therefore $$ \sum_{j=1}^{N}\mu(A_j) =\sum_{j=1}^{N}1 =N, $$ which tends to infinity. All pieces have finite measure, but their total mass is infinite.
Infinite Measures and Sigma-Finiteness
Infinite total measure should not be confused with failure of sigma-finiteness. Recall that a measure is sigma-finite if \(X\) is a countable union of measurable sets of finite measure. The real line with Lebesgue measure is infinite but sigma-finite, since \(\mathbb{R}=\bigcup_{n=1}^{\infty}[-n,n]\). By contrast, some infinite measures do not admit any such cover.
Worked Example: An Infinite Measure with No Nonempty Finite Pieces
Let \(X\) be any nonempty set, take \(\mathcal{F}=\mathcal{P}(X)\), and define \(\mu(\varnothing)=0\) and \(\mu(A)=\infty\) for every nonempty \(A\subseteq X\). This is a measure. To check countable additivity, consider pairwise disjoint sets \(A_n\). If their union is empty, every \(A_n\) is empty and both sides of the additivity equation equal zero. If their union is nonempty, at least one \(A_n\) is nonempty, so \(\sum_n\mu(A_n)=\infty\); the union also has measure \(\infty\). Thus countable additivity holds. The whole space has infinite measure, and its only finite-measure subset is the empty set. A countable union of empty sets cannot cover the nonempty space, so this measure is not sigma-finite.
This example is deliberately extreme, but it clarifies the distinction. Being infinite concerns the total mass \(\mu(X)\); sigma-finiteness concerns whether the space can be covered by finite-measure pieces. One property does not determine the other: an infinite measure can be sigma-finite or not.
A Common Pitfall: Do Not Subtract Infinite Measures
For finite measures, the disjoint decomposition \(X=A\cup(X\setminus A)\) gives a familiar complement identity. When \(\mu(X)=\infty\), writing \(\mu(X\setminus A)=\mu(X)-\mu(A)\) is not generally valid. If both terms on the right are infinite, the expression \(\infty-\infty\) is undefined. The finite-complement theorem above avoids subtraction: it uses disjoint additivity and a contradiction, and it requires the set \(A\) to have finite measure.
A related caution concerns continuity from above. The earlier continuity-from-above theorem requires the first set in a decreasing sequence to have finite measure. Without that hypothesis, the conclusion can fail. For Lebesgue measure on \(\mathbb{R}\), let \(F_n=[n,\infty)\). Then \(F_n\) decreases and \(\bigcap_{n=1}^{\infty}F_n=\varnothing\), but each \(\mu(F_n)=\infty\). The measures therefore do not converge to the measure of the intersection, which is zero. The finite-measure hypothesis in continuity from above cannot be dropped.
Infinite measure spaces are therefore understood through countable additivity, finite-measure subsets, and useful approximations such as exhaustions—not by informal cancellation of infinite quantities. An exhaustion can show how total mass grows, while a partition can show whether the mass is concentrated in a piece or spread across infinitely many finite pieces.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- What condition on \(\mu(X)\) makes a measure space an infinite measure space?
- If \(\mu(X)=\infty\) and \(\mu(A)<\infty\), what must be true of \(\mu(X\setminus A)\), and why?
- What happens to the measures along an increasing measurable exhaustion of an infinite measure space?
- For a countable measurable partition of an infinite measure space, what are the two possible ways the infinite total can arise?
- Why can continuity from above fail for decreasing sets in an infinite measure space?