What Sigma-Finiteness Means
Some measure spaces have infinite total measure but can still be handled by examining them in countably many finite-measure regions. Sigma-finiteness formalizes this idea. The regions need not be disjoint or have any special geometric form; the key requirement is that they are measurable, each has finite measure, and together they cover the whole space.
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 space is not assumed to be finite or complete, and no particular underlying measure is assumed. A set \(E\in\mathcal{M}\) has finite measure when \(\mu(E)<\infty\); this condition concerns its measure, not its number of elements.
A finite-measure cover is not required to be disjoint, nested, or strictly increasing. If a cover has only finitely many sets, it can be regarded as a sequence by adjoining empty sets; the empty set is measurable and has measure zero. Sigma-finiteness also does not say that \(\mu(X)<\infty\): a countable cover by finite-measure sets may have total measure infinite.
Three Equivalent Forms
For many purposes, the cover in the definition is not the most convenient form. It can always be replaced by nested sets, or by disjoint sets. The following equivalence records precisely what those replacements guarantee.
- \((X,\mathcal{M},\mu)\) is sigma-finite.
- There are sets \(F_n\in\mathcal{M}\) with \(F_n\subseteq F_{n+1}\), \(\mu(F_n)<\infty\) for every \(n\), and \(X=\bigcup_{n=1}^{\infty}F_n\).
- There are pairwise disjoint sets \(D_n\in\mathcal{M}\) with \(\mu(D_n)<\infty\) for every \(n\), and \(X=\bigcup_{n=1}^{\infty}D_n\).
Proof. Assume first that (1) holds, and choose a cover \((E_n)\) as in the definition. Define \(F_n=\bigcup_{k=1}^{n}E_k\). Each \(F_n\) is measurable because \(\mathcal{M}\) is closed under finite unions, and \(F_n\subseteq F_{n+1}\). By finite subadditivity of a measure, \(\mu(F_n)\le\sum_{k=1}^{n}\mu(E_k)<\infty\). Also, \(\bigcup_nF_n=\bigcup_nE_n=X\). Thus (2) holds.
Now assume (2). Set \(D_1=F_1\), and, for \(n\ge2\), set \(D_n=F_n\setminus F_{n-1}\). These sets are measurable because \(\mathcal{M}\) is closed under differences of measurable sets. If \(i<j\), then \(D_i\subseteq F_i\subseteq F_{j-1}\), whereas \(D_j\subseteq F_{j-1}^{c}\); hence \(D_i\cap D_j=\varnothing\). Also, \(D_n\subseteq F_n\), so monotonicity gives \(\mu(D_n)\le\mu(F_n)<\infty\). For each \(N\), the construction gives \(\bigcup_{n=1}^{N}D_n=F_N\); taking the union over \(N\) yields \(\bigcup_nD_n=X\). Thus (3) holds.
Finally, if (3) holds, the sequence \((D_n)\) itself is a countable measurable cover by finite-measure sets, so (1) holds by definition. These arguments also cover \(X=\varnothing\): taking every set in the relevant sequence to be empty satisfies each condition. Therefore all three statements are equivalent. \(\square\)
The conversion from a cover to a nested sequence uses only finite unions. The further conversion to disjoint pieces assigns each point to the first set in the original cover that contains it. In particular, disjointification does not require the original cover to be nested.
Worked Constructions
Worked Example: Exhausting the Real Line
Let \((\mathbb{R},\mathcal{L},m)\) be the real line with Lebesgue measure. Use intervals to give a finite-measure cover, an increasing exhaustion, and disjoint finite-measure pieces.
For each integer \(n\ge1\), let \(E_n=[-n,n]\). These sets are measurable, \(m(E_n)=2n<\infty\), and every real number belongs to some \(E_n\), so \(\bigcup_nE_n=\mathbb{R}\). This is a finite-measure cover and is already increasing. Its measure at stage \(n\) is \(2n\); for example, substituting \(n=1,2,3\) gives \(m(E_1)=2\), \(m(E_2)=4\), and \(m(E_3)=6\).
Disjointify this cover by taking \(D_1=[-1,1]\) and, for \(n\ge2\),
These are exactly \(E_n\setminus E_{n-1}\). Their two intervals each have length \(1\), so \(m(D_n)=1+1=2\) for \(n\ge2\); also \(m(D_1)=2\). The pieces are pairwise disjoint by the theorem, and their union is \(\mathbb{R}\). As a check, for any \(N\ge2\), the first \(N\) pieces fill \([-N,N]\), whose measure is \(2N\); their measures sum to \(2+(N-1)\cdot2=2N\), the same value.
Worked Example: Disjointifying a Non-Nested Cover
On \(([0,1],\mathcal{L}|_{[0,1]},m)\), where \(\mathcal{L}|_{[0,1]}\) denotes the Lebesgue-measurable subsets of \([0,1]\), consider \(E_1=[0,3/4]\) and \(E_2=[1/4,1]\). Disjointify this cover and compute the measures of the resulting pieces.
The two sets are measurable and cover \([0,1]\): the first contains every point from \(0\) through \(3/4\), and the second contains every point from \(1/4\) through \(1\). They are not nested, since \(0\in E_1\setminus E_2\) and \(1\in E_2\setminus E_1\). Apply the construction \(D_1=E_1\), \(D_2=E_2\setminus E_1\). It gives \(D_1=[0,3/4]\) and \(D_2=(3/4,1]\). The interval lengths yield \(m(D_1)=3/4\) and \(m(D_2)=1-3/4=1/4\). For \(n\ge3\), take \(D_n=\varnothing\), with \(m(D_n)=0\).
The pieces are disjoint: \(D_1\) ends at \(3/4\), while \(D_2\) excludes \(3/4\); the empty pieces meet neither. Their union is \([0,1]\), and their measures add to \(3/4+1/4=1=m([0,1])\). The nested finite-measure exhaustion obtained directly from the cover is \(F_1=E_1\), \(F_n=[0,1]\) for \(n\ge2\); the measures are \(3/4\) at the first stage and \(1\) at every later stage. This verifies the construction even though the original cover was not nested.
Worked Example: Square Exhaustions in the Plane
Let \((\mathbb{R}^2,\mathcal{L},m)\) denote the plane with Lebesgue measure. Use centered squares to construct disjoint finite-measure pieces and compute the measure of the first \(N\) pieces.
For \(n\ge1\), let \(E_n=[-n,n]^2\). The squares increase and cover \(\mathbb{R}^2\). Each has area \(m(E_n)=(2n)(2n)=4n^2<\infty\). Define \(D_1=E_1\) and \(D_n=E_n\setminus E_{n-1}\) for \(n\ge2\). The theorem ensures these sets are measurable, pairwise disjoint, and cover the plane.
For \(n\ge2\), the inner square \(E_{n-1}\) lies inside the finite-measure square \(E_n\), and \(E_n\) is the disjoint union of \(E_{n-1}\) and \(D_n\). Finite additivity therefore gives \(m(D_n)=m(E_n)-m(E_{n-1})=4n^2-4(n-1)^2=8n-4\). For instance, at \(n=2\), this is \(8\cdot2-4=12\), agreeing with the outer area \(16\) minus the inner area \(4\). Also, \(m(D_1)=4\).
For \(N\ge2\), summing the pieces gives \(m(D_1)+\sum_{n=2}^{N}m(D_n)=4+\sum_{n=2}^{N}(8n-4)=4N^2\). To check the last equality, the sum from \(2\) to \(N\) is \(8\bigl(N(N+1)/2-1\bigr)-4(N-1)=4N^2-4\), and adding \(4\) gives \(4N^2=m(E_N)\). Thus the first \(N\) disjoint pieces fill exactly the square \(E_N\), with the corresponding measure.
A Common Mistake: Finite Pieces Do Not Give Finite Total Measure
A frequent error is to conclude that a countable union of finite-measure sets must itself have finite measure. Countable subadditivity gives \(\mu(X)\le\sum_n\mu(E_n)\), but the series on the right may be infinite. Sigma-finiteness requires each individual \(\mu(E_n)\) to be finite; it does not require the sum of those measures to be finite.
For the real-line cover above, the disjoint pieces all have measure \(2\). A wrong calculation would be: “Each shell has measure \(2\), so the union has measure \(2\).” This omits all but one piece. Since the pieces are disjoint and cover \(\mathbb{R}\), countable additivity gives \(m(\mathbb{R})=\sum_{n=1}^{\infty}m(D_n)=\sum_{n=1}^{\infty}2=\infty\). The correct conclusion is that \(\mathbb{R}\) is sigma-finite, even though its total Lebesgue measure is infinite. The calculation also shows why a cover by finite-measure sets is different from a single finite-measure set covering the whole space.
Another point to keep in view is that “finite-measure” does not mean “bounded” or “finite as a set.” The definition uses only membership in \(\mathcal{M}\) and the numerical condition \(\mu(E_n)<\infty\). No topology, geometry, or completeness assumption is part of sigma-finiteness. The increasing and disjoint forms are useful because they organize the same coverage in different ways, not because those forms add extra conditions to the definition.
Check Your Understanding
For each question, distinguish the cover condition from any claim about the total measure of the space.
- Explain why \(E_n=[-n,n]\) is a finite-measure cover of \(\mathbb{R}\) under Lebesgue measure, and compute \(m(E_4)\).
- Given a finite-measure cover \((E_n)\), write the formulas for a nested exhaustion \(F_n\) and the disjoint pieces \(D_n\).
- In the cover \(E_1=[0,3/4]\), \(E_2=[1/4,1]\), why does \(E_2\setminus E_1\) exclude the endpoint \(3/4\)?
- What is the precise error in concluding that a countable union of finite-measure sets has finite measure?
- Does a sigma-finite measure space have to be complete? Does it have to have finite total measure?