Earlier in Theorem 7.8.3 an outer measure on P
Definition 7.9.1 Let Ω be a nonempty set and let μ : P(Ω) → [0,∞] be an outer measure. For E ⊆ Ω, E is μ measurable if for all S ⊆ Ω,
 (7.10) 
To help in remembering 7.10, 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 7.10. In the Bible, there are several incidents recorded in which a process of division resulted in more stuff than was originally present.^{1} Measurable sets are exactly those which are incapable of such a miracle. 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 7.9.2 (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
 (7.11) 
But 7.11 holds because A is μ measurable. Apply Definition 7.9.1 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 which shows that starting with an outer measure you can obtain a measure.
Theorem 7.9.4 Let Ω be a set and let μ be an outer measure on P
 (7.12) 
If
 (7.13) 
If
 (7.14) 
This measure space is also complete which means that if μ
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.
It is required to show that

First consider S ∖

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,
 (7.15) 
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 7.12.
Consider part 7.13. Without loss of generality μ

and so if μ

Therefore, letting

which also equals

it follows from part 7.12 just shown that
In order to establish 7.14, let the F_{n} be as given there. Then, since

The problem is, I don’t know F ∈S and so it is not clear that μ

which implies

But since F ⊆ F_{n},

and this establishes 7.14. 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 7.9.3 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 7.14 and 7.13 to the outer measure μ⌊S in 7.16 and let n →∞. Thus

and this yields μ(S) = (μ⌊S)(A) + (μ⌊S)(A^{C}) = μ(S ∩ A) + μ(S ∖ A).
Therefore A ∈S and this proves Parts 7.12, 7.13, and 7.14.
It only remains to verify the assertion about completeness. Letting G and F be as described above, let S ⊆ Ω. I need to verify

However,
Corollary 7.9.5 Completeness is the same as saying that if
Proof: If the new condition holds, then suppose G ⊆ F where μ
Now suppose the earlier version of completeness and let

where μ

and all have measure zero. It follows E ∖

The measure μ which results from the outer measure of Theorem 7.8.3 is called the Lebesgue Stieltjes measure associated with the integrator function F. Its properties will be discussed more in the next section.
class=”left” align=”middle”(ℝ)7.10. ONE DIMENSIONAL LEBESGUE STIELTJES MEASURE