σ-finite measure

Inmathematics,a positive (orsigned)measureμdefined on aσ-algebraΣ of subsets of asetXis called a finite measure ifμ(X) is a finitereal number(rather than ∞). A setAin Σ is of finite measure ifμ(A) < ∞.The measureμis calledσ-finiteifXis acountableunionof measurable sets each with finite measure. A set in a measure space is said to haveσ-finite measureif it is a countable union of measurable sets with finite measure. A measure being σ-finite is a weaker condition than being finite, i.e. all finite measures are σ-finite but there are (many) σ-finite measures that are not finite.

A different but related notion that should not be confused with σ-finiteness iss-finiteness.

Definition

Let$(X,{\mathcal {A}})$be ameasurable spaceand$\mu$ameasureon it.

The measure$\mu$is called a σ-finite measure, if it satisfies one of the four following equivalent criteria:

1. the set$X$can be covered with at mostcountably manymeasurable setswith finite measure. This means that there are sets$A_{1},A_{2},\ldots \in {\mathcal {A}}$with$\mu \left(A_{n}\right)<\infty$for all$n\in \mathbb {N}$that satisfy$\bigcup _{n\in \mathbb {N} }A_{n}=X$.
2. the set$X$can be covered with at most countably many measurabledisjoint setswith finite measure. This means that there are sets$B_{1},B_{2},\ldots \in {\mathcal {A}}$with$\mu \left(B_{n}\right)<\infty$for all$n\in \mathbb {N}$and$B_{i}\cap B_{j}=\varnothing$for$i\neq j$that satisfy$\bigcup _{n\in \mathbb {N} }B_{n}=X$.
3. the set$X$can be covered with a monotone sequence of measurable sets with finite measure. This means that there are sets$C_{1},C_{2},\ldots \in {\mathcal {A}}$with$C_{1}\subset C_{2}\subset \cdots$and$\mu \left(C_{n}\right)<\infty$for all$n\in \mathbb {N}$that satisfy$\bigcup _{n\in \mathbb {N} }C_{n}=X$.
4. there exists a strictly positivemeasurable function$f$whose integral is finite.This means that$f(x)>0$for all$x\in X$and$\int f(x)\mu (\mathrm {d} x)<\infty$.

If$\mu$is a$\sigma$-finite measure, themeasure space$(X,{\mathcal {A}},\mu )$is called a$\sigma$-finite measure space.

Examples

Lebesgue measure

For example,Lebesgue measureon thereal numbersis not finite, but it is σ-finite. Indeed, consider theintervals[k,k+ 1)for allintegersk;there are countably many such intervals, each has measure 1, and their union is the entire real line.

Counting measure

Alternatively, consider the real numbers with thecounting measure;the measure of any finite set is the number of elements in the set, and the measure of any infinite set is infinity. This measure is notσ-finite, because every set with finite measure contains only finitely many points, and it would take uncountably many such sets to cover the entire real line. But, the set of natural numbers$\mathbb {N}$with thecounting measureisσ-finite.

Locally compact groups

Locally compact groupswhich areσ-compactare σ-finite under theHaar measure.For example, allconnected,locally compact groupsGare σ-compact. To see this, letVbe a relatively compact, symmetric (that isV=V⁻¹) open neighborhood of the identity. Then

$H=\bigcup _{n\in \mathbb {N} }V^{n}$

is an open subgroup ofG.ThereforeHis also closed since its complement is a union of open sets and by connectivity ofG,must beGitself. Thus all connectedLie groupsare σ-finite under Haar measure.

Nonexamples

Any non-trivial measure taking only the two values 0 and$\infty$is clearly non σ-finite. One example in$\mathbb {R}$is: for all$A\subset \mathbb {R}$,$\mu (A)=\infty$if and only if A is not empty; another one is: for all$A\subset \mathbb {R}$,$\mu (A)=\infty$if and only if A is uncountable, 0 otherwise. Incidentally, both are translation-invariant.

Properties

The class of σ-finite measures has some very convenient properties; σ-finiteness can be compared in this respect toseparabilityof topological spaces. Some theorems in analysis require σ-finiteness as a hypothesis. Usually, both theRadon–Nikodym theoremandFubini's theoremare stated under an assumption of σ-finiteness on the measures involved. However, as shown byIrving Segal,they require only a weaker condition, namelylocalisability.

Though measures which are notσ-finite are sometimes regarded as pathological, they do in fact occur quite naturally. For instance, ifXis ametric spaceofHausdorff dimensionr,then all lower-dimensionalHausdorff measuresare non-σ-finite if considered as measures onX.

Equivalence to a probability measure

Any σ-finite measureμon a spaceXisequivalentto aprobability measureonX:letV_n,n∈N,be a covering ofXby pairwise disjoint measurable sets of finiteμ-measure, and letw_n,n∈N,be a sequence of positive numbers (weights) such that

$\sum _{n=1}^{\infty }w_{n}=1.$

The measureνdefined by

$\nu (A)=\sum _{n=1}^{\infty }w_{n}{\frac {\mu \left(A\cap V_{n}\right)}{\mu \left(V_{n}\right)}}$

is then a probability measure onXwith precisely the samenull setsasμ.

Related concepts

Moderate measures

ABorel measure(in the sense of alocally finite measureon the Borel$\sigma$-algebra)$\mu$is called amoderate measureiff there are at most countably many open sets$A_{1},A_{2},\ldots$with$\mu \left(A_{i}\right)<\infty$for all$i$and$\bigcup _{i=1}^{\infty }A_{i}=X$.

Every moderate measure is a$\sigma$-finite measure, the converse is not true.

Decomposable measures

A measure is called adecomposable measurethere are disjoint measurable sets$\left(A_{i}\right)_{i\in I}$with$\mu \left(A_{i}\right)<\infty$for all$i\in I$and$\bigcup _{i\in I}A_{i}=X$.For decomposable measures, there is no restriction on the number of measurable sets with finite measure.

Every$\sigma$-finite measure is a decomposable measure, the converse is not true.

s-finite measures

A measure$\mu$is called as-finite measureif it is the sum of at most countably manyfinite measures.

Every σ-finite measure is s-finite, the converse is not true. For a proof and counterexample sees-finite measure#Relation to σ-finite measures.

See also

• Finite measure
• Sigma additivity