Earlier in Theorem 5.7.1 an outer measure on P
Definition 5.8.1 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,
 (5.4) 
To help in remembering 5.4, 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 5.4. 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 5.8.2 (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
 (5.5) 
But 5.5 holds because A is μ measurable. Apply Definition 5.8.1 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 5.8.4 Let Ω be a set and let μ be an outer measure on P
 (5.6) 
If
 (5.7) 
If
 (5.8) 
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,
 (5.9) 
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 5.6.
Consider part 5.7. Without loss of generality μ

and so if μ

Therefore, letting

which also equals

it follows from part 5.6 just shown that
In order to establish 5.8, 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 5.8. 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 5.8.3 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 5.8 and 5.7 to the outer measure μ⌊S in 5.10 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 5.6, 5.7, and 5.8.
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,
The measure m which results from the outer measure of Theorem 5.7.1 is called Lebesgue measure. The following is a general result about completion of a measure space.

Then
Proof: The first part of this follows from Proposition 5.5.2. It only remains to verify that ℱ⊆

due to the fact that μ is a measure. As usual, if

Since ε is arbitrary, this shows that E ∈
Why are these two σ algebras equal if
class=”left” align=”middle”(ℝ)5.9. WHEN IS A MEASURE A BOREL MEASURE?