12.2 Caratheodory Extension Theorem
The Caratheodory extension theorem is a fundamental result which makes possible the
consideration of measures on infinite products among other things. The idea is that if a
finite measure defined only on an algebra is trying to be a measure, then in fact it can be
extended to a measure.
Definition 12.2.1 Let ℰ be an algebra of sets of Ω and let μ0 be a finite measure on ℰ.
This means μ0 is finitely additive and if Ei,E are sets of ℰ with the Ei disjoint
In this definition, μ0 is trying to be a measure and acts like one whenever possible.
Under these conditions, μ0 can be extended uniquely to a complete measure, μ, defined
on a σ algebra of sets containing ℰ such that μ agrees with μ0 on ℰ. The following is the
Theorem 12.2.2 Let μ0 be a measure on an algebra of sets, ℰ, which satisfies
< ∞. Then there exists a complete measure space
for all E ∈ℰ. Also if ν is any such measure which agrees with μ0 on ℰ, then ν = μ on
, the σ algebra generated by ℰ.
Proof: Define an outer measure as follows.
Claim 1: μ is an outer measure.
Proof of Claim 1: Let S ⊆∪i=1∞Si and let Si ⊆∪j=1∞Eij, where
Since ε is arbitrary, this shows μ is an outer measure as claimed.
By the Caratheodory procedure, there exists a unique σ algebra, S, consisting of the
μ measurable sets such that
is a complete measure space. It remains to show μ extends μ0.
Claim 2: If S is the σ algebra of μ measurable sets, S⊇ℰ and μ = μ0 on
Proof of Claim 2: First observe that if A ∈ℰ, then μ
since A = ∪i=1∞Ei ∩ A. Therefore, μ = μ0 on ℰ.
Consider the assertion that ℰ⊆S. Let A ∈ℰ and let S ⊆ Ω be any set. There exist
such that ∪i=1∞Ei ⊇ S
Since ε is arbitrary, this shows A ∈S.
This has proved the existence part of the theorem. To verify uniqueness,
Then G is given to contain ℰ and is obviously closed with respect to countable disjoint
unions and complements. Therefore by Lemma 10.12.3, G⊇ σ
and this proves the
The following lemma is also very significant.
Lemma 12.2.3 Let M be a metric space with the closed balls compact and suppose
μ is a measure defined on the Borel sets of M which is finite on compact sets.
Then there exists a unique Radon measure, μ which equals μ on the Borel sets. In
particular μ must be both inner and outer regular on all Borel sets.
Proof: Define a positive linear functional, Λ
be the Radon
measure which comes from the Riesz representation theorem for positive linear
functionals. Thus for all f ∈ C0
If V is an open set, let
be a sequence of continuous functions in
increasing and converges to
pointwise. Then applying the monotone convergence
and so the two measures coincide on all open sets. Every compact set is a countable
intersection of open sets and so the two measures coincide on all compact sets. Now
be a ball of radius
and let E
be a Borel set contained in this
ball. Then by regularity of μ
there exist sets F,G
such that G
is a countable
intersection of open sets and F
is a countable union of compact sets such that
F ⊆ E ⊆ G
and so μ
If E is an arbitrary Borel set, then
and letting n →∞, this yields μ