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 functionals.
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,
 (10.1.1) 
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 present.^{1} 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.
Proof: Suppose A is μ measurable. It is desired to to show that for all T ⊆ Ω,

Thus it is desired to show
 (10.1.2) 
But 10.1.2 holds because A is μ measurable. Apply Definition 10.1.1 to S ∩T instead of S.
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 book.
Theorem 10.1.4 The collection of μ measurable sets, S, forms a σ algebra and
 (10.1.3) 
If
 (10.1.4) 
If
 (10.1.5) 
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 ∩ B^{C} is in S. To do so, consider the following picture.
Since μ is subadditive,

Now using A,B ∈S,

and therefore,
Since Ω ∈S, this shows that A ∈S if and only if A^{C} ∈S. Now if A,B ∈S, A ∪ B = (A^{C} ∩ B^{C})^{C} = (A^{C} ∖ B)^{C} ∈S. By induction, if A_{1},

By induction, if A_{i} ∩ A_{j} = ∅ and A_{i} ∈S, μ(∪_{i=1}^{n}A_{i}) = ∑ _{i=1}^{n}μ(A_{i}).
Now let A = ∪_{i=1}^{∞}A_{i} where A_{i} ∩ A_{j} = ∅ 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 F_{0} ≡∅, use part 10.1.3 to write
In order to establish 10.1.5, let the F_{n} be as given there. Then from what was just shown,

Then, since

Now I don’t know whether F ∈S and so all that can be said is that

but this implies

Hence

which implies

But since F ⊆ F_{n},

and this establishes 10.1.5. Note that it was assumed μ
It remains to show S is closed under countable unions. Recall that if A ∈S, then A^{C} ∈S and S is closed under finite unions. Let A_{i} ∈S, A = ∪_{i=1}^{∞}A_{i}, B_{n} = ∪_{i=1}^{n}A_{i}. Then
By Lemma 10.1.3 B_{n} is (μ⌊S) measurable and so is B_{n}^{C}. I want to show μ(S) ≥ μ(S ∖A) + μ(S ∩A). If μ(S) = ∞, there is nothing to prove. Assume μ(S) < ∞. Then apply Parts 10.1.5 and 10.1.4 to the outer measure, μ⌊S in 10.1.6 and let n →∞. Thus

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,
Completeness usually occurs in the following form. E ⊆ F ∈S and μ