Finite Measure and Sigma-Finiteness
A measure space need not have finite total measure for its measurable sets to be manageable. The key condition is that the space can be covered by countably many measurable sets, each having finite measure. Such a cover allows arguments to be carried out on finite-measure pieces. When the pieces are chosen to increase, they form a finite-measure exhaustion.
Throughout, a measure space means \((X,\mathcal{M},\mu)\), where \(\mu\) is a measure taking values in \([0,\infty]\). No completeness assumption is needed for the definitions or results below. In particular, the measure space may be complete or incomplete.
Every finite measure space is sigma-finite: the single set \(X\) is already a finite-measure cover. Sigma-finiteness is more general. It permits \(\mu(X)=\infty\), provided \(X\) can be built up from countably many pieces of finite measure. An exhaustion is an actual cover of \(X\), not merely a cover outside a null set.
From a Cover to an Increasing Exhaustion
Proof. If \((E_n)\) is a finite-measure exhaustion, then its members are measurable, have finite measure, and cover \(X\). Thus the space is sigma-finite. Conversely, suppose that \((A_n)\) is a measurable cover of \(X\) with \(\mu(A_n)<\infty\) for every \(n\). Define \(E_n=\bigcup_{k=1}^{n}A_k\). Each \(E_n\) is measurable because a sigma-algebra is closed under finite unions, and \(E_n\subseteq E_{n+1}\). Finite subadditivity of a measure gives \(\mu(E_n)\le\sum_{k=1}^{n}\mu(A_k)<\infty\). Finally, \(\bigcup_{n=1}^{\infty}E_n=\bigcup_{n=1}^{\infty}A_n=X\). Hence \((E_n)\) is a finite-measure exhaustion. If \(X\) is empty, the same construction gives \(E_n=\varnothing\) for every \(n\), which also satisfies the definition. \(\square\)
This equivalence explains why it is enough to begin with any countable finite-measure cover: taking successive finite unions makes the cover increasing without losing finiteness of measure. The measures \(\mu(E_n)\) need not be bounded by one common finite constant. They are required to be finite one at a time.
Worked Example: Exhausting Euclidean Space for Lebesgue Measure
Consider \((\mathbb{R}^d,\mathcal{L},m)\), where \(d\ge1\), \(\mathcal{L}\) is the Lebesgue sigma-algebra, and \(m\) is Lebesgue measure. For each integer \(n\ge1\), set \(E_n=[-n,n]^d\). These are measurable and increasing, and every point of \(\mathbb{R}^d\) belongs to \(E_n\) for all sufficiently large \(n\). The volume of this cube is the product of its \(d\) side lengths, each \(2n\), so
Thus \((E_n)\) is a finite-measure exhaustion, and Lebesgue measure on \(\mathbb{R}^d\) is sigma-finite. For \(d=1\), the first three pieces are \([-1,1]\), \([-2,2]\), and \([-3,3]\), with measures \(2\), \(4\), and \(6\), respectively; the formula gives \((2n)^1=2n\) in each case. The total space has infinite measure: since the increasing pieces have measures \(2n\), no finite number can bound the measure of all of \(\mathbb{R}\). Equivalently, \(\mathbb{R}\) contains intervals of arbitrarily large finite length.
Check: when \(n=2\), \(E_2=[-2,2]^d\) has side length \(4\), so its volume is \(4^d=(2\cdot2)^d\), as the general formula requires.
Counting Measure and Finite Measure Spaces
Worked Example: Exhaustion for a Countable Set
Let \(S\) be a countably infinite set, let \(\mathcal{M}=\mathcal{P}(S)\), and let \(\mu\) be counting measure, so \(\mu(A)\) is the number of elements of \(A\) when \(A\) is finite and is \(+\infty\) when \(A\) is infinite. Enumerate \(S=\{s_1,s_2,\ldots\}\) without repetition and set \(E_n=\{s_1,\ldots,s_n\}\). Each \(E_n\) is measurable, \(E_n\subseteq E_{n+1}\), and \(\bigcup_nE_n=S\). Its measure is exactly its number of elements:
For example, \(E_1=\{s_1\}\) has measure \(1\), \(E_2=\{s_1,s_2\}\) has measure \(2\), and \(E_3=\{s_1,s_2,s_3\}\) has measure \(3\). This proves sigma-finiteness. The space is not finite, because \(\mu(S)=+\infty\). If \(S\) instead has exactly \(q\) elements, use \(E_n=\{s_1,\ldots,s_{\min(n,q)}\}\); then \(\mu(E_n)=\min(n,q)\) and the sequence covers \(S\). If \(S=\varnothing\), take \(E_n=\varnothing\), whose measure is \(0\).
Check: for the infinite-set construction, the measure of the third exhaustion set is \(\mu(\{s_1,s_2,s_3\})=3\), matching the formula \(n=3\).
Worked Example: A Finite Measure Space
Let \(X=\{a,b,c,d\}\), let \(\mathcal{M}=\mathcal{P}(X)\), and assign masses \(1/8,1/4,1/2,1/8\) to \(a,b,c,d\), respectively. For \(A\subseteq X\), define \(\mu(A)\) to be the sum of the masses of its members. This is a measure: the empty set has measure \(0\), and for disjoint subsets the sum of the masses in their union is the sum of the two separate sums. The total measure is
Therefore \((X,\mathcal{M},\mu)\) is finite. Set \(E_n=X\) for every \(n\ge1\). This sequence is increasing, covers \(X\), and has \(\mu(E_n)=1<\infty\) for each \(n\); it is a finite-measure exhaustion. This illustrates that an exhaustion need not consist of strictly growing sets.
Check: the exhaustion set \(E_2\) is still \(X\), so direct substitution gives \(\mu(E_2)=1\), the same as \(\mu(X)\).
Restricting a Sigma-Finite Measure
Proof. Choose a finite-measure exhaustion \((E_n)\) of \(X\), which exists by the theorem. For each \(n\), let \(F_n=A\cap E_n\). Then \(F_n\in\mathcal{M}|_A\), \(F_n\subseteq F_{n+1}\), and \(\bigcup_nF_n=A\cap\bigcup_nE_n=A\). Since \(F_n\subseteq E_n\), monotonicity of the measure gives \(\mu|_A(F_n)=\mu(F_n)\le\mu(E_n)<\infty\). Thus the sequence \((F_n)\) is a finite-measure exhaustion of \(A\). The argument includes \(A=\varnothing\), for which every \(F_n=\varnothing\). \(\square\)
This restriction property is useful when an argument only concerns a measurable part of a larger space: sigma-finiteness is retained on that part. It does not assert that every measurable subset has finite measure; it asserts that the subset can itself be exhausted by countably many finite-measure pieces.
A Useful Nonexample and a Common Pitfall
Worked Example: Uncountable Counting Measure Is Not Sigma-Finite
Let \(S\) be uncountable, let \(\mathcal{M}=\mathcal{P}(S)\), and let \(\mu\) be counting measure. A set has finite counting measure exactly when it is finite. If \(S\) had a countable cover \(S=\bigcup_{n=1}^{\infty}A_n\) with every \(\mu(A_n)<\infty\), every \(A_n\) would be finite. A countable union of finite sets is countable: list the elements of each finite set in a finite list, then arrange those lists in countably many rows; the resulting collection has at most countably many entries. This would make \(S\) countable, contrary to its assumed uncountability. Therefore this counting-measure space is not sigma-finite.
Check: the candidate finite-measure pieces cannot cover \(S\), because each contains only finitely many points and countably many finite sets contain at most countably many points altogether.
A common mistake is to treat sigma-finiteness as if it required a finite total measure, or to require a uniform bound on the measures of all exhaustion sets. Neither condition is part of the definition. In the Lebesgue example on \(\mathbb{R}\), the exhaustion \([-n,n]\) has measure \(2n\): substituting \(n=1,2,3\) gives \(2,4,6\), so the measures grow without a finite uniform bound, while every individual measure remains finite.
The condition that each piece have finite measure matters. For example, the constant sequence \(E_n=\mathbb{R}\) covers \(\mathbb{R}\) and is increasing, but \(m(E_n)=m(\mathbb{R})=+\infty\), so it is not a finite-measure exhaustion. The intervals \([-n,n]\), by contrast, satisfy the finite-measure requirement separately at every \(n\).
Check Your Understanding
For each question, identify the measure space and verify which part of the definition is being used.
- Show that \([-n,n]\) is a finite-measure exhaustion of \(\mathbb{R}\) for Lebesgue measure, and compute the measure of its \(n\)-th set.
- If \((A_n)\) is a countable cover with \(\mu(A_n)<\infty\), explain why \(E_n=\bigcup_{k=1}^n A_k\) also has finite measure.
- Give a finite-measure exhaustion for a finite set equipped with counting measure, including the case of the empty set.
- Explain why counting measure on an uncountable set is not sigma-finite.
- Suppose \((X,\mathcal{M},\mu)\) is sigma-finite and \(A\in\mathcal{M}\). Describe an exhaustion of \(A\) using an exhaustion of \(X\).