The Kolmogorov extension theorem

Jordan Bell
June 21, 2014

1 σ-algebras and semirings

If X is a nonempty set, an algebra of sets on X is a collection 𝒜 of subsets of X such that if {Ai}⊂𝒜 is finite then ⋃iAi∈𝒜, and if A∈𝒜 then X∖A∈𝒜. An algebra 𝒜 is called a σ-algebra if {Ai}⊂𝒜 being countable implies that ⋃iAi∈𝒜.

If X is a set and 𝒢 is a collection of subsets of X, we denote by σ⁢(𝒢) the smallest σ-algebra containing 𝒢, and we say that σ⁢(𝒢) is the σ-algebra generated by G.

Later we will also use the following notion. If X is a nonempty set, a semiring of sets on X is a collection 𝒮 of subsets of X such that (i) ∅∈𝒮, (ii) if A,B∈𝒮 then A∩B∈𝒮, and (iii) if A,B∈𝒮 then there are pairwise disjoint S1,…,Sn∈𝒮 such that A∖B=⋃i=1nSi; we do not demand that this union itself belong to 𝒮. We remark that a semiring on X need not include X.

If 𝒮 is a semiring of sets and μ0:𝒮→[0,∞], we say that μ0 is finitely additive if {Si}⊂𝒮 being finite, pairwise disjoint, and satisfying ⋃iSi∈𝒮 implies that μ0⁢(⋃iSi)=∑iμ0⁢(Si), and countably additive if {Si}⊂𝒮 being countable, pairwise disjoint, and satisfying ⋃iSi∈𝒮 implies that μ0⁢(⋃iSi)=∑iμ0⁢(Si). If 𝒢 is a collection of subsets of X, the algebra generated by G is the smallest algebra containing 𝒢. We shall use the following lemma in the proof of Lemma 10.

Lemma 1.

If S is a semiring on a set X and X∈S, then the algebra generated by S is equal to the collection of finite unions of members of S.

For a bounded countably additive function, the Carathéodory extension theorem states the following.11 1 René L. Schilling, Measures, Integrals and Martingales, p. 37, Theorem 6.1. If we had not specified that μ0:𝒮→[0,1] but rather had talked about μ0:𝒮→[0,∞], then the Carathéodory extension theorem shows that there is some extension of μ0 to σ⁢(𝒮), but this extension need not be unique.

Theorem 2 (Carathéodory extension theorem).

Suppose that X is a nonempty set, that S is a semiring on X, and that μ0:S→[0,1] is countably additive. Then there is one and only one measure on σ⁢(S) whose restriction to S is equal to μ0.

2 Product σ-algebras

Suppose that X is a set, that {(Yi,ℳi):i∈I} is a family of measurable spaces, and that fi:X→Yi are functions. The smallest σ-algebra on X such that each fi is measurable is called the σ-algebra generated by {fi:i∈I}. This is analogous to the initial topology induced by a family of functions on a set. Calling this σ-algebra ℳ and supposing that σ⁢(𝒢i)=ℳi for each i∈I, we check then that

ℳ=σ⁢({fi-1⁢(A):i∈I,A∈𝒢i}). (1)

Suppose that {(Xi,ℳi):i∈I} is a family of measurable spaces. Let

X=∏i∈IXi,

the cartesian product of the sets Xi, and let πi:X→Xi be the projection maps. The product σ-algebra on X is the σ-algebra ℳ generated by {πi:i∈I}, and is denoted

ℳ=⊗i∈Iℳi.

This is analogous to the product topology on a cartesian product of topological spaces, which has the initial topology induced by the family of projection maps.

For H⊂I, we define

XH=∏i∈HXi.

Thus, XI=X and X∅={∅}, and for G⊂H,

XH=XG×XH∖G.

For H⊂I, let

ℳH=⊗i∈Hℳi,

the product σ-algebra on XH. Thus, ℳI=ℳ and ℳ∅={∅,{∅}}, and for G⊂H we have

ℳH=ℳG⊗ℳH∖G.

For G⊂H, we define PH,G:XH→XG to be the projection map: an element of XH is a function x on H such that x⁢(i)∈Xi for all i∈H, and PH,G⁢(x) is the restriction of x to G.

Lemma 3.

For G⊂H, PH,G:(XH,MH)→(XG,MG) is measurable.

If F is a finite subset of I and A∈ℳF, we call A×XI∖F∈ℳ an F-cylinder set. Cylinder sets for the product σ-algebra are analogous to the usual basic open sets for the product topology.

Lemma 4.

The collection of all cylinder sets is an algebra of sets on ∏i∈IXi, and this collection generates the product σ-algebra ⊗I∈IMi.

The product σ-algebra can in fact be generated by a smaller collection of sets. (The following collection of sets is not a minimal collection of sets that generates the product σ-algebra, but it is smaller than the collection of all cylinder sets and it has the structure of a semiring, which will turn out to be useful.) An intersection of finitely many sets of the form A×ℳI∖{t}, A∈ℳt, is called a product cylinder.

Lemma 5.

The collection of all product cylinders is a semiring of sets on ∏i∈IXi, and this collection generates the product σ-algebra ⊗i∈IMi.

3 Borel σ-algebras

If (X,τ) is a topogical space, the Borel σ-algebra on X is σ⁢(τ), and is denoted ℬX. A member of ℬX is called a Borel set.

Lemma 6.

If X is a topological space and G is a countable subbasis for the topology of X, then

ℬX=σ⁢(𝒢).

A separable metrizable space is second-countable, so we can apply the following theorem to such spaces.

Theorem 7.

Suppose that Xi, i∈N, are second-countable topological spaces and let X=∏i∈NXi, with the product topology. Then

ℬX=⊗i∈ℕℬXi.
Proof.

For each i∈ℕ, let 𝒢i be a countable subbasis for the topology of Xi. Because 𝒢i is a subbasis for the topology of Xi for each i, we check that 𝒢 is a subbasis for the product topology of X, where

𝒢={πi-1⁢(A):i∈ℕ,A∈𝒢i}.

Because each 𝒢i is countable and ℕ is countable, 𝒢 is countable. Hence by Lemma 6,

ℬX=σ⁢(𝒢).

On the other hand, for each i∈ℕ we have by Lemma 6 that ℬXi=σ⁢(𝒢i), and so by (1),

⊗i∈ℕℬXi=σ⁢(𝒢).

∎

4 Product measures

If {(Xi,ℳi,μi):1≤i≤n} are σ-finite measure spaces, let X=∏i=1nXi and ℳ=⊗i=1nℳi. It is a fact that there is a unique measure μ on ℳ such that for Ai∈ℳi,

μ⁢(∏i=1nAi)=∏i=1nμi⁢(Ai),

and μ is a σ-finite measure. We write μ=∏i=1nμi and call μ the product measure.22 2 Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, second ed., pp. 64–65, and p. 31, Theorem 1.14.

5 Compact classes

If X is a set and 𝒞 is a collection of subsets of X, we say that 𝒞 is a compact class if every countable subset of 𝒞 with the finite intersection property has nonempty intersection. We remind ourselves that a collection ℰ of sets is said to have the finite intersection property if for any finite subset ℱ of ℰ we have ⋂A∈ℱA≠∅. Usually one speaks about a collection of sets having the finite intersection property in the following setting: A topological space Y is compact if and only if every collection of closed sets that has the finite intersection property has nonempty intersection.

We will employ the following lemma in the proof of the Kolmogorov extension theorem.33 3 V. I. Bogachev, Measure Theory, volume I, p. 50, Proposition 1.12.4.

Lemma 8.

Suppose that C0 is a compact class of subsets of a set X and let C be the collection of countable intersections of finite unions of members of C0. C is the smallest collection of subsets of X containing C0 that is closed under finite unions and countable intersections, and C is itself a compact class.

We state the following result that gives conditions under which a finitely additive functions on an algebra of sets is in fact countably additive,44 4 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 378, Theorem 10.13. and then use it to prove an analogous result for semirings.

Lemma 9.

Suppose that A is an algebra of sets on a set X, and that μ0:A→[0,∞) is finitely additive and μ0⁢(X)<∞. If there is a compact class C⊂A such that

μ0⁢(A)=sup⁡{μ0⁢(C):C∈𝒞 and C⊂A},A∈𝒜,

then μ0 is countably additive.

The following lemma gives conditions under which a finitely additive function on a semiring of sets is in fact countably additive.55 5 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 521, Lemma 15.25.

Lemma 10.

Suppose that S is a semiring of sets on X with X∈S and that μ0:S→[0,∞) is finitely additive and μ0⁢(X)<∞. If there is a compact class C⊂S such that

μ0⁢(A)=sup⁡{μ0⁢(C):C∈𝒞 and C⊂A},A∈𝒮,

then μ0 is a countably additive.

Proof.

Let 𝒞u be the collection of finite unions of members of 𝒞. 𝒞u is a subset of the compact class produced in Lemma 8, hence is itself a compact class. Let 𝒜 be the collection of finite unions of members of 𝒮, which by Lemma 1 is the algebra generated by 𝒮. Because 𝒞⊂𝒮⊂𝒜 and 𝒜 is closed under finite unions, it follows that 𝒞u⊂𝒜.

Because 𝒮 is a semiring, it is a fact that if A1,…,An,A∈𝒮, then there are pairwise disjoint S1,…,Sm∈𝒮 such that A∖⋃i=1nAi=⋃i=1mSi.66 6 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 134, Lemma 4.7. Thus, if A1,…,An∈𝒮, defining Ei=Ai∖⋃j=1i-1Aj, with E1=A1∖∅=A1, the sets E1,…,En are pairwise disjoint, and for each i there are pairwise disjoint Si,1,…,Si,ai∈𝒮 such that Ei=⋃j=1aiSi,j. Then the sets Si,j, 1≤i≤n, 1≤j≤ai are pairwise disjoint and their union is equal to ⋃i=1nAi. This shows that any element of 𝒜 can be written as a union of pairwise disjoint elements of 𝒮.

Furthermore, because 𝒮 is a semiring, if A1,…,AN∈𝒮, there are pairwise disjoint S1,…,Sk∈𝒮 such that for each 1≤i≤k there is some 1≤n≤N such that Si∈An, and for each 1≤n≤N, there is a subset F⊂{1,…,k} such that An=⋃i∈FSi.77 7 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 134, Lemma 4.8.

Let E∈𝒜 and suppose that E=⋃n=1NAn, where A1,…,AN∈𝒮 are pairwise disjoint, and that E=⋃m=1MBm, where B1,…,BM∈𝒮 are pairwise disjoint. There are pairwise disjoint S1,…,Sk∈𝒮 such that for each 1≤i≤k there is some 1≤n≤N or 1≤m≤M such that, respectively, Si∈An or Si∈Bm, and for each 1≤n≤N there is some subset F⊂{1,…,k} such that An=⋃i∈FSi, and for each 1≤m≤M there is some subset F⊂{1,…,k} such that Bm=⋃i∈FSi. It follows that E=⋃i=1kSi, and because μ0 is finitely additive,

∑n=1Nμ0⁢(An)=∑i=1kμ0⁢(Si)=∑m=1Mμ0⁢(Bm).

Therefore, for E∈𝒜 it makes sense to define

μ⁢(E)=∑n=1nμ0⁢(Ai),

where A1,…,An are pairwise disjoint elements of 𝒮 whose union is equal to E. Also, μ⁢(X)=μ0⁢(X)<∞.

We shall now show that the function μ:𝒜→[0,∞) is finitely additive. If E1,…,EN∈𝒜 are pairwise disjoint, for each n there are pairwise disjoint An,1,…,An,an∈𝒮 such that En=⋃j=1anAn,j, and there are pairwise disjoint S1,…,Sk∈𝒮 such that for each 1≤i≤k, there is some 1≤n≤N and some 1≤j≤an such that Si∈An,j, and for each 1≤n≤N and each 1≤j≤an there is some subset F⊂{1,…,k} such that An,j=⋃i∈FSi. It follows that ⋃n=1NEn=⋃i=1kSi, and

μ⁢(⋃n=1NEn)=μ⁢(⋃i=1kSi)=∑i=1kμ0⁢(Si)=∑n=1N∑j=1anμ0⁢(An,j)=∑n=1Nμ⁢(En),

showing that μ is finitely additive.

For E=⋃i=1nAi∈𝒜 with pairwise disjoint A1,…,An∈𝒮, let ϵ>0, and for each 1≤i≤n let Ci∈𝒞 with μ0⁢(Ci)>μ0⁢(Ai)+ϵn and Ci⊂Ai. Then C=⋃i=1nCi∈𝒞u. As A1,…,An are pairwise disjoint and Ci⊂Ai, C1,…,Cn are pairwise disjoint, so because μ is finitely additive on 𝒜,

μ⁢(C)=∑i=1nμ⁢(Ci)=∑i=1nμ0⁢(Ci)>∑i=1n(μ0⁢(Ai)+ϵn)=μ⁢(E)+ϵ.

Lemma 9 tells us now that μ:𝒜→[0,∞) is countably additive, and therefore μ0, its restriction to the semiring 𝒮, is countably additive. ∎

6 Kolmogorov consistent families

Suppose that {(Xi,ℳi):i∈I} is a family of measurable spaces. The collection D of all finite subsets of I, ordered by set inclusion, is a directed set. Suppose that for each F∈D, μF is a probability measure on ℳF; we defined the notation ℳF in §2 and we use that here. We say that the family of measures {μF:F∈D} is Kolmogorov consistent if whenever F,G∈D with F⊂G, it happens that PG,F*⁢μG=μF, where f*⁢μ denotes the pushforward of μ by f, i.e. f*⁢μ=μ∘f-1. It makes sense to talk about PG,F*⁢μG because PG,F:(XG,ℳG)→(XF,ℳF) is measurable, as stated in Lemma 3.

We are now prepared to prove the Kolmogorov extension theorem.88 8 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 522, Theorem 15.26.

Theorem 11 (Kolmogorov extension theorem).

Suppose that {(Xi,Mi):i∈I} is a family of measurable spaces and suppose that for each F∈D, μF is a probability measure on MF. If the family of probability measures {μF:F∈D} is Kolmogorov consistent and if for each i∈I there is a compact class Ci⊂Mi satisfying

μi⁢(A)=sup⁡{μi⁢(C):C∈𝒞i and C⊂A},A∈ℳi, (2)

then there is a unique probability measure on MI such that for each F∈D, the pushforward of μ by the projection map PI,F:XI→XF is equal to μF.

Proof.

Define

𝒞0={C×XI∖{i}:i∈I,C∈𝒞i}.

We shall show that 𝒞0 is a compact class. Suppose

{Cn×XI∖{in}:n∈ℕ,Cn∈𝒞in}⊂𝒞0

has empty intersection. For each i∈I, let

Qi=⋂in=iCn,

and if there are no such in, then Qi=Xi. Then

⋂n∈ℕCn×XI∖{in}=∏i∈IQi.

Because this intersection is equal to ∅, one of the factors in the product is equal to ∅. (For some purposes one wants to keep track of where the axiom of choice is used, so we mention that concluding that some factor of any empty cartesian product is itself empty is equivalent to the axiom of choice). No Xi is empty, so this empty Qi must be of the form ⋂in=iCn for which at least one in is equal to i. But if in=i then Cn∈𝒞i, and because 𝒞i is a compact class, ⋂in=iCn=∅ implies that there are finitely many a1,…,aN such that ⋂n=1NCan=∅, and this yields ⋂n=1NCan×XI∖{ian}=∅. We have thus proved that if an intersection of countably many members of 𝒞0 is empty then some intersection of finitely many of these is empty, showing that 𝒞0 is a compact class. Let 𝒞1 be the smallest collection of subsets of XI containing 𝒞0 that is closed under finite unions and countable intersections, and by Lemma 8 we know that 𝒞1 is a compact class; we use the notation 𝒞1 because presently we will use a subset of 𝒞1.

Let 𝒜 be the collection of all cylinder sets of the product σ-algebra ℳI. Explicitly,

𝒜={A×XI∖F:F∈D,A∈ℳF}.

Suppose F,G∈D, F⊂G, A∈ℳF,B∈ℳG, and that A×XI∖F=B×XI∖G. It follows that B=A×XG∖F, and then using PG,F*⁢μG=μF we get

μG⁢(B)=μG⁢(A×XG∖F)=μG⁢(PG,F-1⁢(A))=μF⁢(A).

Therefore it makes sense to define μ0:𝒜→[0,1] as follows: for A×XI∖F∈𝒜,

μ0⁢(A×XI∖F)=μF⁢(A).

Let F1,…,Fn∈D and A1∈ℳF1,…,An∈ℳFn, and suppose that A1×XI∖F1,…,An×XI∖Fn∈𝒢 are pairwise disjoint. With F=⋃j=1nFj∈D,

⋃j=1nAj×XI∖Fj=(⋃j=1nAj×XF∖Fj)×XI∖F,

which is an F-cylinder set. Then,

μ0⁢(⋃j=1nAj×XI∖Fj) = μF⁢(⋃j=1nAj×XF∖Fj)
= ∑j=1nμF⁢(Aj×XF∖Fj)
= ∑j=1nμ0⁢(Aj×XI∖Fj),

showing that μ0:𝒜→[0,1] is finitely additive.

Let 𝒢 be the collection of all product cylinder sets of the product σ-algebra ℳI. Explicitly,

𝒢={⋂i∈FAi×XI∖{i}:F∈D, and Ai∈ℳi for i∈F}.

It is apparent that 𝒞0⊂𝒢. Let 𝒞 be the intersection of 𝒞1 and 𝒢. A subset of a compact class is a compact class, so 𝒞 is a compact class, and 𝒞0⊂𝒞.

Suppose that E∈𝒢: there is some F∈D and Ai∈ℳi for each i∈F such that

E=⋂i∈FAi×XI∖{i}=(∏i∈FAi)×XI∖F.

Take n=|F|, and let ϵ>0. Then, for each i∈F, by (2) there is some Ci∈𝒞i such that Ci⊂Ai and μi⁢(Ai)<μi⁢(Ci)+ϵn, and we set

C=⋂i∈FCi×XI∖{i}=(∏i∈FCi)×XI∖F,

which is a finite intersection of members of 𝒞0 and hence belongs to 𝒞1, and which visibly belongs to 𝒢, and hence belongs to 𝒞. We have

E∖C = (⋃i∈F(Ai∖Ci)×∏j∈F∖{i}Aj)×XI∖F
⊂ ⋃i∈F(Ai∖Ci)×XI∖{i}.

Both E∖C and the above union are cylinder sets so it makes sense to apply μ0 to them, and because μ0 is finitely additive,

μ0⁢(E∖C) ≤ ∑i∈Fμ0⁢((Ai∖Ci)×XI∖{i})
= ∑i∈Fμi⁢(Ai∖Ci)
= ∑i∈Fμi⁢(Ai)-μi⁢(Ci)
< ∑i∈Fϵn
= ϵ.

Hence μ0⁢(E)-μ0⁢(C)=μ0⁢(E∖C)<ϵ, i.e. μ0⁢(E)<μ0⁢(C)+ϵ. Thus, we have proved that for each ϵ>0, there is some C∈𝒞 such that C⊂E and μ0⁢(E)<μ0⁢(C)+ϵ, which means that

μ0⁢(E)=sup⁡{μ0⁢(C):C∈𝒞 and C⊂E}.

Using the restriction of μ0:𝒜→[0,1] to the semiring 𝒢 and the compact class 𝒞, the conditions of Lemma 10 are satisfied, and therefore the restriction of μ0 to 𝒢 is countably additive.

By Lemma 5, ℳI=σ⁢(𝒢). Because the restriction of μ0 to the semiring 𝒢 is countably additive, we can apply the Carathéodory extension theorem, which tells us that there is a unique measure μ on σ⁢(𝒢)=ℳI whose restriction to 𝒢 is equal to the restriction of μ0 to 𝒢. Check that the restriction of μ to 𝒜 is equal to μ0. For F∈D and A∈ℳF,

PI,F*⁢μ⁢(A)=μ⁢(PI,F-1⁢(A))=μ⁢(A×XI∖F)=μ0⁢(A×XI∖F)=μF⁢(A),

showing that PI,F*⁢μ=μF. Certainly μ⁢(XI)=1, namely, μ is a probability measure. If ν is a probability measure on XI whose pushforward by PI,F is equal to μF for each F∈D, then check that the restriction of ν to 𝒢 is equal to the restriction of μ0 to 𝒢, and then by the assertion of uniqueness in Carathéodory’s theorem, ν=μ, completing the proof. ∎

If X is a Hausdorff space, we say that a Borel measure μ on X is tight if for every A∈ℬX,

μ⁢(A)=sup⁡{μ⁢(K):K is compact and K⊂A}.

A Polish space is a topological space that is homeomorphic to a complete separable metric space, and it is a fact that a finite Borel measure on a Polish space is tight.99 9 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 438, Theorem 12.7. In particular, any Borel probability measure on a Polish space is tight. We use this in the proof of the following version of the Kolmogorov extension theorem, which applies for instance to the case where Xi=ℝ for each i∈I, with I any index set.

Corollary 12.

Suppose that {Xi:i∈I} is a family of Polish spaces and suppose that for each F∈D, μF is a Borel probability measure on XF. If the family of measures {μF:F∈D} is Kolmogorov consistent, then there is a unique probability measure on MI=⊗i∈IBXi such that for each F∈D, the pushforward of μ by the projection map PI,F:XI→XF is equal to μF.

Proof.

For each i∈I, let 𝒞i be the collection of all compact subsets of Xi. In any topological space, check that a collection of compact sets is a compact class. The fact that μi is a Borel probability measure on a Polish space then implies that it is tight, which we can write as

μi⁢(A)=sup⁡{μi⁢(C):K∈𝒞i and K⊂A},A∈ℳi.

Therefore the conditions of Theorem 11 are satisfied, so the claim follows. ∎

If the index set I in the above corollary is countable, then by Theorem 7 the product σ-algebra ⊗i∈IℬXi is equal to the Borel σ-algebra of the product ∏t∈TXi, so that the probability measure μ on the product σ-algebra is in this case a Borel measure.