Earlier in Theorem 19.8.3 an outer measure on P
Definition 19.9.1 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,
 (19.11) 
To help in remembering 19.11, 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 19.11. 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 19.9.2 (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
 (19.12) 
But 19.12 holds because A is μ measurable. Apply Definition 19.9.1 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 19.9.4 Let Ω be a set and let μ be an outer measure on P
 (19.13) 
If
 (19.14) 
If
 (19.15) 
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.

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,
 (19.16) 
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 19.13.
Consider part 19.14. Without loss of generality μ

and so if μ

Therefore, letting

which also equals

it follows from part 19.13 just shown that
In order to establish 19.15, 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 19.15. 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 19.9.3 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 19.15 and 19.14 to the outer measure μ⌊S in 19.17 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 19.13, 19.14, and 19.15.
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 19.9.5 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 19.8.3 is called the Lebesgue Stieltjes measure associated with the integrator function F. Its properties will be discussed more in the next section.
Here is a general result. If you have a measure μ, then by Proposition 19.6.2, defined there is an outer measure which agrees with μ on the σ algebra of measurable sets ℱ. What of the measure determined by ? Denoted still by . Is = μ on ℱ? Is ℱ a subset of the measurable sets, those which satisfy the criterion of being measurable? Suppose E ∈ℱ. Is it the case that

As usual, there is nothing to show if

because everything is measurable. Then
Proof: If S is a set,
Suppose now that
class=”left” align=”middle”(ℝ)19.10. ONE DIMENSIONAL LEBESGUE STIELTJES MEASURE