22.3.2 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 22.3.7 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 main result.
Theorem 22.3.8 Let μ0 be a measure on an algebra of sets, ℰ, which
< ∞. 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 σ
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 sets
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, Let
Then G is given to contain ℰ and is obviously closed with respect to countable disjoint
unions and complements. Therefore by Lemma 7.4.2, G⊇ σ
and this proves the
The following lemma is also very significant. Actually Lemmas 7.5.3 and 7.5.4 are of even
more use, but one can use the following to get useful information in some cases. In particular,
in the proof of the Kolmogorov extension theorem, one can consider the special case that
Mt = ℝ or ℝnt in that theorem.
Lemma 22.3.9 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 ∈ Cc
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 let
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