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 22.3.7Let ℰ be an algebra of sets of Ω and let μ_{0}be a finite measureon ℰ. This means μ_{0}is finitely additive and if E_{i},E are sets of ℰ with the E_{i}disjointand
E = ∪∞i=1Ei,
then
∑∞
μ0 (E ) = μ0 (Ei)
i=1
while μ_{0}
(Ω )
< ∞.
In this definition, μ_{0} is trying to be a measure and acts like one whenever possible. Under
these conditions, μ_{0} can be extended uniquely to a complete measure, μ, defined on a σ
algebra of sets containing ℰ such that μ agrees with μ_{0} on ℰ. The following is the main result.
Theorem 22.3.8Let μ_{0}be a measure on an algebra of sets, ℰ, whichsatisfies μ_{0}
(Ω)
< ∞. Then there exists a complete measure space
(Ω,S, μ )
suchthat
μ(E) = μ0(E)
for all E ∈ℰ. Also if ν is any such measure which agrees with μ_{0}on ℰ, then ν = μ on σ
(ℰ)
,the σ algebra generated by ℰ.
Proof:Define an outer measure as follows.
{ }
∑∞ ∞
μ (S ) ≡ inf μ0 (Ei) : S ⊆ ∪ i=1Ei,Ei ∈ ℰ
i=1
Claim 1: μ is an outer measure.
Proof of Claim 1: Let S ⊆∪_{i=1}^{∞}S_{i} and let S_{i}⊆∪_{j=1}^{∞}E_{ij}, where
ε ∑∞
μ (Si)+ -i ≥ μ(Eij).
2 j=1
Then
∑ ∑ ∑ ( ) ∑
μ (S ) ≤ μ(Eij) = μ (Si) + ε- = μ (Si)+ ε.
i j i 2i i
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
(Ω,S, μ)
is a complete measure space. It remains to show μ extends μ_{0}.
Claim 2:If S is the σ algebra of μ measurable sets, S ⊇ℰ and μ = μ_{0} on
ℰ.
Proof of Claim 2: First observe that if A ∈ℰ, then μ
(A)
≤ μ_{0}
(A )
by definition.
Letting
∞∑
μ (A)+ ε > μ0(Ei), ∪∞i=1 Ei⊇A, Ei ∈ ℰ,
i=1
it follows
∞
μ(A)+ ε > ∑ μ0(Ei ∩ A) ≥ μ0(A)
i=1
since A = ∪_{i=1}^{∞}E_{i}∩ A. Therefore, μ = μ_{0} on ℰ.
Consider the assertion that ℰ⊆S. Let A ∈ℰ and let S ⊆ Ω be any set. There exist sets
This has proved the existence part of the theorem. To verify uniqueness, Let
G ≡ {E ∈ σ(ℰ) : μ (E) = ν (E)}.
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
(ℰ)
and this proves the
lemma.
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
M_{t} = ℝ or ℝ^{nt} in that theorem.
Lemma 22.3.9Let M be a metric space with the closed balls compact and supposeμ is a measure defined on the Borel sets of M which is finite on compact sets. Thenthere exists a unique Radon measure, μwhich equals μ on the Borel sets. In particularμ must be both inner and outer regular on all Borel sets.
Proof: Define a positive linear functional, Λ
(f)
= ∫fdμ. Let μ be the Radon measure
which comes from the Riesz representation theorem for positive linear functionals. Thus for
all f ∈ C_{c}
(M )
,
∫ ∫
--
fdμ = fdμ.
If V is an open set, let
{fn}
be a sequence of continuous functions in C_{c}
(M )
which is
increasing and converges to X_{V } pointwise. Then applying the monotone convergence
theorem,
∫ ∫
-- --
XVdμ = μ(V ) = XV dμ = μ(V)
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
B
(a,n)
be a ball of radius n 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 μ