10.1 Outer Measures
What are some examples of measure spaces? In this chapter, a general procedure is
discussed called the method of outer measures. It is due to Caratheodory (1918). This
approach shows how to obtain measure spaces starting with an outer measure.
This will then be used to construct measures determined by positive linear
Definition 10.1.1 Let Ω be a nonempty set and let μ : P(Ω) → [0,∞] satisfy
Such a function is called an outer measure. For E ⊆ Ω, E is μ measurable if for all
S ⊆ Ω,
To help in remembering 10.1.1, think of a measurable set, E, as a
process which divides a given set into two pieces, the part in E and the
part not in E as in 10.1.1. In the Bible, there are four incidents recorded
in which a process of division resulted in more stuff than was originally
Measurable sets are exactly those for which no such miracle occurs. You might think of
the measurable sets as the nonmiraculous sets. The idea is to show that they form a σ
algebra on which the outer measure, μ is a measure.
First here is a definition and a lemma.
Definition 10.1.2 (μ⌊S)(A) ≡ μ(S ∩A) for all A ⊆ Ω. Thus μ⌊S is the name of
a new outer measure, called μ restricted to S.
The next lemma indicates that the property of measurability is not lost by
considering this restricted measure.
Lemma 10.1.3 If A is μ measurable, then A is μ⌊S measurable.
Proof: Suppose A is μ measurable. It is desired to to show that for all T ⊆ Ω,
Thus it is desired to show
But 10.1.2 holds because A is μ measurable. Apply Definition 10.1.1 to S ∩T instead of
If A is μ⌊S measurable, it does not follow that A is μ measurable. Indeed, if you
believe in the existence of non measurable sets, you could let A = S for such a μ non
measurable set and verify that S is μ⌊S measurable.
The next theorem is the main result on outer measures. It is a very general result
which applies whenever one has an outer measure on the power set of any set.
This theorem will be referred to as Caratheodory’s procedure in the rest of the
Theorem 10.1.4 The collection of μ measurable sets, S, forms a σ algebra
Fn ⊆ Fn+1 ⊆
, then if F
= ∪n=1∞Fn and Fn ∈S, it follows that
Fn ⊇ Fn+1 ⊇
, and if F
= ∩n=1∞Fn for Fn ∈S then if μ
) < ∞,
Also, (S,μ)is complete. By this it is meant that if F ∈S and if E ⊆ Ω with
μ(E ∖ F) + μ(F ∖ E) = 0, then E ∈S.
Proof: First note that ∅ and Ω are obviously in S. Now suppose A,B ∈S. I will show
A ∖ B ≡ A ∩ BC is in S. To do so, consider the following picture.
Since μ is subadditive,
Now using A,B ∈S,
It follows equality holds in the above. Now observe using the picture if you like
Therefore, since S
is arbitrary, this shows A ∖ B ∈S
Since Ω ∈S, this shows that A ∈S if and only if AC ∈S. Now if A,B ∈S,
A ∪ B = (AC ∩ BC)C = (AC ∖ B)C ∈S. By induction, if A1,
, then so is
. If A,B ∈S
, with A ∩ B
By induction, if Ai ∩ Aj = ∅ and Ai ∈S, μ(∪i=1nAi) = ∑
Now let A = ∪i=1∞Ai where Ai ∩ Aj = ∅ for i≠j.
Since this holds for all n, you can take the limit as n →∞ and conclude,
which establishes 10.1.3. Part 10.1.4 follows from part 10.1.3 just as in the proof of
Theorem 9.1.5 on Page 581. That is, letting F0 ≡∅, use part 10.1.3 to write
In order to establish 10.1.5, let the Fn be as given there. Then from what was just
Now I don’t know whether F ∈S and so all that can be said is that
but this implies
But since F ⊆ Fn,
and this establishes 10.1.5. Note that it was assumed μ
subtracted from both sides.
It remains to show S is closed under countable unions. Recall that if A ∈S, then
AC ∈S and S is closed under finite unions. Let Ai ∈S, A = ∪i=1∞Ai, Bn = ∪i=1nAi.
By Lemma 10.1.3 Bn
) measurable and so is BnC
. I want to show
) ≥ μ
) + μ
). If μ
) = ∞
, there is nothing to prove. Assume μ
) < ∞
Then apply Parts 10.1.5
to the outer measure, μ⌊S
and let n →∞
and this yields
Therefore A ∈S and this proves Parts 10.1.3, 10.1.4, and 10.1.5. It remains to prove
the last assertion about the measure being complete.
Let F ∈S and let μ(E ∖ F) + μ(F ∖ E) = 0. Consider the following picture.
Then referring to this picture and using F ∈S,
) = μ
) + μ
) and so E ∈S
. This shows that
is complete and
proves the theorem.
Completeness usually occurs in the following form. E ⊆ F ∈S and μ
= 0. Then