Measure theory class notes - 11 August 2010, class 5
1
Carathéodory extension theorem - proof outline
We have shown that proving the Carathéodory extension theorem for finite measures suffices to
prove it for measures which are σ-finite on the field. Consider our (Ω, F , µ). Now we have to prove
the theorem for the case when µ(Ω) < ∞. If µ(Ω) = 0, then the zero measure (which assigns a
measure of 0 to all sets in σ(F )) is an extension, and monotonicity of µ implies that this is the
only extension. Otherwise, if 0 < µ(Ω) < ∞, we can scale µ by µ(Ω) and so assume µ(Ω) = 1. We
will prove:
Theorem (Carathéodory extension theorem for probability measures). Let Ω be a set and F a
field on it. Let µ be a probability measure on F . Then there is a unique probability measure µ̃
on σ(F ) such that µ and µ̃ agree on F .
We first describe the outline of the proof:
Step 1: Let
Fσ = {A : ∃{An }∞
n=1 , each An ∈ F , An ↑ A as n → ∞}
Define µ̃ : Fσ → [0, ∞]R̄ as follows: if An ↑ A as n → ∞, then µ̃(A) = limn→∞ µ(An ). We will
show the following:
• µ̃ is well-defined and extends µ.
• µ̃ is monotonic: If A, B ∈ Fσ and A ⊆ B, then µ̃(A) ≤ µ̃(B).
• µ̃ is strongly finitely additive: If A, B ∈ Fσ , then A ∪ B, A ∩ B ∈ Fσ and µ̃(A) + µ̃(B) =
µ̃(A ∪ B) + µ̃(A ∩ B).
• µ̃ respects increasing limits: If {An }∞
n=1 is an increasing sequence of sets from Fσ whose
union is A, then A ∈ Fσ and limn→∞ µ̃(An ) = µ̃(A).
If at all we want to assign a measure to a set in Fσ , it has better be the one given by µ̃, as we
have shown earlier that measures respect increasing unions.
Step 2: Define µ∗ : 2Ω → [0, ∞] by
µ∗ (B) = inf{µ̃(A) : A ∈ Fσ , A ⊇ B}
The inf is well-defined since Ω ∈ Fσ . µ∗ is called the outer measure. Note that it is defined for all
subsets of Ω. We will show the following:
• µ∗ extends µ̃.
• If B1 ⊆ B2 , then µ∗ (B1 ) ≤ µ∗ (B2 )
• µ∗ (B1 ) + µ∗ (B2 ) ≥ µ∗ (B1 ∪ B2 ) + µ∗ (B1 ∩ B2 )
∗
∗
• If {Bn }∞
n=1 increases to B as n → ∞, then µ (Bn ) ↑ µ (B) as n → ∞
If at all we want to assign a measure to a set B ⊆ Ω, because of monotonicity, it must be at most
µ∗ (B). Also, it must be at least 1 − µ∗ (Ω \ B) (since the measure assigned to Ω \ B must be at
Measure theory class notes - 11 August 2010, class 5
2
most its outer measure too). Let us define the measure for the sets for which the upper bound and
lower bound agree.
Step 3: Let
C = {B ⊆ Ω : µ∗ (B) + µ∗ (Ω \ B) = 1}
We will show that
• C is a σ-field.
• µ∗ is a probability measure on C .
• F ⊆ C.
Clearly σ(F ) ⊆ C , but in general equality will not hold. This is not an issue: µ∗ restricted to
σ(F ) is of course a probability measure on σ(F ) as required.
Carathéodory extension theorem - proof details
We now execute the steps 1, 2, and 3, independently of each other.
Details of step 3
We will show that C is a σ-field. Since µ∗ extends µ̃ which extends µ, we have µ∗ (∅) = 0 and
µ∗ (Ω) = 1. So ∅, Ω ∈ C . It is clear that C is closed under complements.
If D ⊆ Ω, then from step 2 we know that
µ∗ (D) + µ∗ (Ω \ D) ≥ µ∗ (D ∪ (Ω \ D)) + µ∗ (D ∩ (Ω \ D)) = 1
Let A, B ∈ C . We will show that A ∪ B and A ∩ B also belong to C . We have
∗
µ∗ (A) + µ∗ (Ω \ A) = 1
(1)
µ∗ (B) + µ∗ (Ω \ B) = 1
(2)
∗
µ (A ∪ B) + µ (Ω \ (A ∪ B)) ≥ 1
(3)
µ∗ (A ∩ B) + µ∗ (Ω \ (A ∩ B)) ≥ 1
∗
∗
(4)
∗
∗
µ (A) + µ (B) ≥ µ (A ∪ B) + µ (A ∩ B)
µ∗ (Ω \ A) + µ∗ (Ω \ B) ≥ µ∗ (Ω \ (A ∩ B)) + µ∗ (Ω \ (A ∪ B))
(5)
(6)
Adding (5) and (6) and using (1) and (2), we get
µ∗ (A ∪ B) + µ∗ (A ∩ B) + µ∗ (Ω \ (A ∩ B)) + µ∗ (Ω \ (A ∪ B)) ≤ 2
(7)
If either of (3) or (4) was a strict inequality, this would not be possible. So both (3) or (4) are
equalities. This shows that A ∪ B and A ∩ B belong to C . C is a field. Adding (3) and (4), we
get that equality holds in (7). Since (7) is also obtained by adding (5) and (6), equality must hold
in (5) and (6) too. (5) implies that µ∗ is finitely additive on C .
Measure theory class notes - 11 August 2010, class 5
Let {Cn }n∈N be a countable collection of sets in C . We need to show that B :=
all n ∈ N, let
[
Bn =
Cm
3
S
n∈N
Cn ∈ C . For
1≤m≤n
{Bn }∞
n=1
increases to B. Since each Bn ∈ C (C is a field),
µ∗ (Bn ) + µ∗ (Ω \ Bn ) = 1
Using the monotonicity of µ∗ from step 2,
µ∗ (Bn ) + µ∗ (Ω \ B) ≤ 1
From step 2 we know that µ∗ respects increasing unions, so taking the limit as n → ∞,
µ∗ (B) + µ∗ (Ω \ B) ≤ 1
We know that
µ∗ (B) + µ∗ (Ω \ B) ≥ 1
(this holds for any set B, in fact) So equality holds, and B ∈ C . C is a σ-field.
We will show that µ∗ is countably additive on C : Take {Bn }∞
n=1 , a countable collection of disjoint
∗
sets in C whose union is B. We have shown that µ is finitely additive on C , so for all k ∈ N,
!
k
k
X
[
∗
µ∗ (Bn )
µ
Bn =
n=1
n=1
Taking the limit as k → ∞ and using that µ∗ respects increasing unions, we get
∗
µ (B) =
∞
X
µ∗ (Bn )
n=1
This shows that µ∗ is countably additive.
Since µ∗ extends µ̃ which extends µ, it is clear that F ⊆ C .
Details of step 1
∞
Let {An }∞
n=1 and {Bn }n=1 be two increasing sequences of sets from F , each having union A. To
show that µ̃ is well-defined, we need to show that
lim µ(An ) = lim µ(Bn )
n→∞
n→∞
Fix n ∈ N. {An ∩ Bm }∞
m=1 is an increasing sequence of sets from F , increasing to An ∩ A = An .
All these sets are in F , so
lim µ(An ∩ Bm ) = µ(An )
m→∞
Since An ∩ Bm ⊆ Bm , so
lim µ(Bm ) ≥ µ(An )
m→∞
Measure theory class notes - 11 August 2010, class 5
4
Taking the limit as n → ∞,
lim µ(Bm ) ≥ lim µ(An )
m→∞
n→∞
By a symmetric argument,
lim µ(An ) ≥ lim µ(Bm )
n→∞
m→∞
Equality holds, and µ̃ is well-defined. Clearly µ̃ extends µ, since for A ∈ F we can take An = A
for all n, and so µ̃(A) = µ(A).
∞
We will show that µ̃ is monotonic. Let A, B ∈ Fσ , A ⊆ B. Let {An }∞
n=1 and {Bn }n=1 be increasing
sequences of sets in F whose union is A and B respectively. {An ∩ Bn }∞
n=1 is also an increasing
sequence, and increases to A ∩ B = A.
µ̃(A) = lim µ(An ∩ Bn ) ≤ lim µ(Bn ) = µ̃(B)
n→∞
n→∞
∞
We now show strong finite additivity. Let A, B ∈ Fσ . Let {An }∞
n=1 and {Bn }n=1 be increasing
∞
sequences of sets in F whose union is A and B respectively. Then {An ∪ Bn }n=1 and {An ∩ Bn }∞
n=1
are increasing sequences of sets in F whose union is A ∪ B and A ∩ B respectively. So A ∪ B and
A ∩ B belong to Fσ . By finite additivity on F , we have
µ(An ) + µ(Bn ) = µ(An \ Bn ) + µ(An ∩ Bn ) + µ(Bn \ An ) + µ(Bn ∩ An )
= µ(An \ Bn ) + µ(Bn \ An ) + µ(An ∩ Bn ) + µ(An ∩ Bn )
= µ(An ∪ Bn ) + µ(An ∩ Bn )
Taking the limit as n → ∞, we get
µ̃(A) + µ̃(B) = µ̃(A ∪ B) + µ̃(A ∩ B)
We now show that µ̃ respects increasing unions. Let {An }∞
n=1 be an increasing sequence of sets in
∞
Fσ with union A. For each n, let {An,m }m=1 be an increasing sequence of sets in F whose union
is An . For each m ∈ N, let
[
Bm =
Ai,j
1≤i≤m,1≤j≤m
{Bm }∞
m=1 is an increasing sequence of sets in F whose union is A. So A ∈ Fσ and µ̃(A) =
limm→∞ µ(Bm )
Since An ⊆ A, µ̃(An ) ≤ µ̃(A). Taking limits,
lim µ̃(An ) ≤ µ̃(A)
n→∞
Bn ⊆ An , so µ̃(Bn ) ≤ µ̃(An ). Taking limits as n → ∞,
µ̃(A) = lim µ(Bn ) = lim µ̃(Bn ) ≤ lim µ̃(An )
n→∞
n→∞
So
µ̃(A) = lim µ̃(An )
n→∞
This completes step 1.
n→∞