1.4 Outer Measures

10 CHAPTER 1. MEASURE

1.3.6. (‘Almost everywhere’ and null sets) If (X, A,µ) is a measure space, then a set in A is called a null set (or µ-null) if its measure is 0. Clearly a countable union of null sets is a null set, by the subadditivity condition in Theorem 1.3.4. If a statement about points x ∈ X is true except for points x in some null set, then we say that the statement is true almost everywhere (a.e.), or µ-almost everywhere (µ-a.e.), or for almost every x, or for µ-almost every x. We say that A is complete (with respect to µ), if any subset of a µ-null set is in A (in which case it is also a null set). We also say that the measure space is complete, in this case.

Theorem 1.3.7 If (X, A,µ) is a measure space, deﬁne A¯ to be the set of unions of a set in A and a subset of a µ-null set. Then A¯ is a σ-algebra on X, and there is a unique extension µ¯ of µ to a complete measure on A¯.

∞ ∞ Proof: If (An)n=1, (Bn)n=1 are sets in A with µ(Bn) = 0 for all n, and if Fn ⊂ Bn for ∞ ∞ ∞ ∞ ∞ all n, then ∪n=1 (An ∪ Fn)=(∪n=1 An) ∪ (∪n=1 Fn), and ∪n=1 Fn ⊂ B, where B = ∪n=1 Bn. ∞ ¯ Note that B is µ-null, by a comment above the theorem. So ∪n=1 (An ∪ Fn) ∈ A. If A, B ∈A with µ(B)= 0, and F ⊂ B, then (A ∪ F )c =(A ∪ B)c ∪ (B \ F ) ∈ A¯. Deﬁneµ ¯(A ∪ F )= µ(A) for A, F as above. This is well deﬁned, since if A ∪ F = A′ ∪ F ′ for another A′, B′ ∈ A with µ(B′)=0, F ′ ⊂ B′, then A ⊂ A′ ∪ F ′ ⊂ A′ ∪ B′, so that µ(A) ≤ µ(A′ ∪ B′) ≤ µ(A′)+ µ(B′)= µ(A′). Similarly, µ(A′) ≤ µ(A), so µ(A)= µ(A′). Let An, Fn as above, and suppose further that the (An) are mutually disjoint. We have

∞ ∞ ∞ ∞ ∞ ∞ µ¯(∪n=1 (An ∪ Fn))=µ ¯((∪n=1 An) ∪ (∪n=1 Fn)) = µ(∪n=1 An)= µ(An)= µ¯(An ∪ Fn). n=1 n=1 X X Soµ ¯ is a measure. To see that it is complete, choose C ∈ A¯ which isµ ¯-null, and D ⊂ C. Write C = A ∪ F with A ∈ A, F ⊂ B ∈ A, with µ(B) = 0. Since C isµ ¯-null, µ(A) = 0. Thus D is a subset of the µ-null set A ∪ B, so D ∈ A¯.

We callµ ¯ the completion of µ, and we call A¯ the completion of A with respect to µ.

1.4 Outer measures

A common way of constructing useful measures such as Lebesgue measure is via outer measure and Carath´eodory’s construction. An outer measure on a set X is a function µ∗ : P(X) → [0, ∞] such that

(i) µ∗(∅)=0.

(ii) A ⊆ B ⊆ X ⇒ µ∗(A) ≤ µ∗(B).

∗ ∞ ∞ ∗ (iii) (Countable subadditivity) µ ( k=1 Ak) ≤ k=1 µ (Ak), for any subsets A1, A2, ... of X. S P 1.4. OUTER MEASURES 11

The following is a very general way to construct outer measures:

Proposition 1.4.1 Let X be any set, let E be a collection of subsets of X containing ∅ and a countable cover of X, and let ρ : E → [0, ∞] is any function such that ρ(∅)=0. Define ∗ ∗ ∞ ∞ ∗ µ : P(X) → [0, ∞] by µ (A) = inf{ k=1 ρ(Ek) : Ek ∈ E, A ⊂ k=1 Ek}. Then µ is an outer measure. P S Proof: First, µ∗ is well defined since E contains ∅ and a countable cover of X. Since ∗ ∗ ρ (∅) = 0, we have that µ (∅) = 0 (let Ei = ∅ for i = 1, 2,... , and thus 0 ≤ µ (∅) ≤ ∞ ∞ ∅ i=1 ρ (Ei)= i=1 ρ ( ) = 0). For S1 ⊂ S2 ⊂ X, any countable cover of S2 by elements of ∗ ∗ ∞ E is also a countable cover of S1. Thus µ (S1) ≤ µ (S2). Finally, for {Sj} =1 ⊂ P (X), and P P j given ε> 0, let {Ej,k : k =1, 2,... } be a countable cover for Sj by elements of E such that ∞ ε ρ (E ) ≤ µ∗ (S )+ . j,k j 2j k=1 X ∞ Then the collection {Ej,k : j, k =1, 2,... } is a countable cover of j=1 Sj by elements of E, and ∞ ∞ ∞ ∞ ε S ρ (E ) ≤ µ∗ (S )+ = µ∗ (S )+ ε. j,k j 2j j j=1 k=1 j=1 j=1 X X X X ∗ ∞ ∞ ∗ ∗ ∞ Thus µ j=1 Sj ≤ j=1 µ (Sj)+ ε. Therefore, as ε → 0 we obtain µ j=1 Sj ≤ ∞ µ∗ S . µ∗ j=1 ( Sj) Hence Pis an outer measure. S P1.4.2. (The σ-algebra and measure associated with an outer measure) If µ∗ is an outer measure on a set X, we say a set A in X is µ∗-measurable if µ∗(E) ≥ µ∗(E ∩ A)+ µ∗(E ∩ Ac), ∀E ⊂ X. This is the same as saying that µ∗(E)= µ∗(E ∩ A)+ µ∗(E ∩ Ac), ∀E ⊂ X, since by (iii) above we have µ∗(E) = µ∗((E ∩ A) ∪ (E ∩ Ac)) ≤ µ∗(E ∩ A)+ µ∗(E ∩ Ac). The last centered condition is called Carathéodory’s condition, and we set Cµ∗ to be the set of subsets A of X which satisfy Carathéodory’s condition. This is a σ-algebra, as we shall ∗ see next, and the restriction of µ to Cµ∗ is a complete measure. The completeness is a consequence of:

Proposition 1.4.3 If µ∗ is an outer measure on a set X, then any subset A of X with ∗ µ (A)=0, is in Cµ∗ . Proof: If µ∗(A) = 0 then µ∗(E ∩ A)+ µ∗(E ∩ Ac) ≤ µ∗(A)+ µ∗(E ∩ Ac) ≤ µ∗(E).

∗ Theorem 1.4.4 (Carath´eodory) If µ is an outer measure on a set X, then Cµ∗ is a σ- ∗ algebra, and the restriction of µ to Cµ∗ is a complete measure. 12 CHAPTER 1. MEASURE

Proof: The completeness of µ∗ was stated above. For any E ⊂ X we have that

µ∗ (E)= µ∗ (E ∩ ∅)+ µ∗ (E ∩ X)= µ∗ (E ∩ ∅)+ µ∗ (E ∩ ∅c) and thus ∅ ∈ Cµ∗ . Now let A ∈ Cµ∗ . Then for any E ⊂ X we have that

µ∗ (E)= µ∗ (E ∩ A)+ µ∗ (E ∩ Ac)= µ∗ (E ∩ Ac)+ µ∗ (E ∩ A) , and it follows that Cµ∗ is closed under complements. Next, suppose that A1, A2 ∈ Cµ∗ . We want to show that A1 ∪ A2 satisﬁes Caratheodory’s condition. We note that A1 ∪ A2 = c ∗ A1 ∪ (A2 ∩ A1) is a disjoint union and thus by using the countable subadditivity of µ and the fact that A1 and A2 satisfy Caratheodory’s condition, we have that

∗ ∗ c µ (E ∩ (A1 ∪ A2)) + µ (E ∩ (A1 ∪ A2) )= ∗ c ∗ c =µ (E ∩ (A1 ∪ (A2 ∩ A1))) + µ (E ∩ (A1 ∪ A2) ) ∗ c ∗ c =µ ((E ∩ A1) ∪ (E ∩ (A2 ∩ A1))) + µ (E ∩ (A1 ∪ A2) ) ∗ ∗ c ∗ c ≤µ (E ∩ A1)+ µ (E ∩ (A2 ∩ A1)) + µ (E ∩ (A1 ∪ A2) ) ∗ ∗ c ∗ c c =µ (E ∩ A1)+ µ (E ∩ A2 ∩ A1)+ µ (E ∩ A1 ∩ A2) ∗ ∗ c ∗ c c =µ (E ∩ A1)+ µ (E ∩ A1 ∩ A2)+ µ (E ∩ A1 ∩ A2) ∗ ∗ c =µ (E ∩ A1)+ µ (E ∩ A1) =µ∗(E).

Thus, A1 ∪ A2 satisﬁes Caratheodory’s condition and so A1 ∪ A2 ∈ Cµ∗ . That is, Cµ∗ is closed under ﬁnite unions. Hence, Cµ∗ is an algebra on X. To show that Cµ∗ is a σ-algebra it is enough to show that Cµ∗ is closed under countable ∞ ∗ disjoint unions. So let {Ai}i=1 be a countable collection of disjoint sets in Cµ . Let Bn = n ∞ i=1 Ai and B = j=1 Aj. Then the measurability of An shows that for any E ⊂ X

S S ∗ ∗ ∗ c µ (E ∩ Bn)=µ (E ∩ Bn ∩ An)+ µ (E ∩ Bn ∩ An) ∗ ∗ =µ (E ∩ An)+ µ (E ∩ Bn−1).

∗ n ∗ ∗ By induction, we then have that µ (E ∩Bn)= j=1 µ (E ∩Aj). Now since Cµ is an algebra we have that Bn ∈ Cµ∗ . This, together with the last equality, shows that P ∗ ∗ ∗ c µ (E)=µ (E ∩ Bn)+ µ (E ∩ Bn) n ∗ ∗ c = µ (E ∩ Aj)+ µ (E ∩ Bn) j=1 Xn ∗ ∗ c ≥ µ (E ∩ Aj)+ µ (E ∩ B ) j=1 X 1.5. LEBESGUE MEASURE 13

∗ using the facts that Bn ⊂ B and the monotonicity of µ . Letting n →∞, we now have

∞ ∗ ∗ ∗ c µ (E) ≥ µ (E ∩ Aj)+ µ (E ∩ B ) j=1 X ∞ ∗ ∗ c ≥µ (E ∩ Aj) + µ (E ∩ B ) =1 ! ∗ [ ∞ ∗ c =µ (E ∩ (∪j=1Aj)) + µ (E ∩ B ) =µ∗(E ∩ B)+ µ∗(E ∩ Bc) ≥µ∗(E).

Since the left and right side of the inequality are equal, it follows that we have equalities ∗ ∗ ∗ c throughout. Thus µ (E)= µ (E ∩ B)+ µ (E ∩ B ). Hence, B ∈ Cµ∗ . That is, Cµ∗ is closed under countable disjoint unions, and therefore is a σ-algebra. ∗ ∞ ∗ ∗ c From the above work we have that µ (E)= j=1 µ (E ∩ Aj)+ µ (E ∩ B ) holds for all ∞ E ⊂ X. Thus if we let E = B = ∪ =1Aj we have j P ∞ ∞ ∞ ∗ ∗ ∗ c ∗ ∗ µ (B)= µ (B ∩ Aj)+ µ (B ∩ B )= µ (B ∩ Aj)= µ (Aj). j=1 j=1 j=1 X X X ∗ ∗ ∗ We conclude that µ is countably additive on Cµ . Therefore, µ |Cµ∗ is a complete measure.

Thus, putting together the last two results, we get a very general way to construct measures: Begin with a collection E of subsets of a set X, which covers X and contains ∅, and a function ρ : E → [0, ∞] such that ρ(∅) = 0. Proposition 1.4.1 deﬁnes an outer measure µ∗ built from ρ, and from Carath´eodory’s theorem we get a complete measure space ∗ (X, Cµ∗ ,µ) where µ is the restriction of µ to Cµ∗ .

1.5 Lebesgue measure

The text shows in Section 1.5 that the Borel measures on R are the Lebesgue-Stieljes measures. These are measures constructed essentially by the procedure discussed in the last paragraph, beginning with the collection E of half-open intervals (a, b] in R, together with ∅, and ρ((a, b]) = F (b) − F (a), for a ﬁxed increasing right continuous function F : R → R. We will discuss this much later, and instead consider just the case F (x) = x, which gives Lebesgue measure. More generally, we use this idea to construct Lebesgue measure on Rn. n Let E consist of the ∅ and the ‘open intervals’ k=1 (ak, bk)=(a1, b1)×(a2, b2)×··· (an, bn) n (one could use ‘half-open intervals’ k=1 [ak, bk) = [a1, b1)×[a2, b2)×··· [an, bn) here instead, Q n but this yields the same thing). Deﬁne ρ of this ‘interval’ to be its usual volume =1 (bk−ak). Q k Here ak < bk are real numbers. The associated outer measure from Proposition 1.4.1 is called Q 14 CHAPTER 1. MEASURE

∗ Lebesgue outer measure and is written as λ . The σ-algebra Cλ∗ coming from Carathéodory’s theorem is written as L or L(Rn) or Ln, and is called the Lebesgue σ-algebra. The restriction of λ∗ to L(Rn) is called Lebesgue measure. n n By bounded n-box, we will mean a subset of k=1 [ak, bk] that contains k=1 (ak, bk). Here ak < bk are numbers in R. Unbounded n-boxes are defined similarly. Q Q Proposition 1.5.1 B(Rn) ⊂ L(Rn), and the Lebesgue measure of any n-box in Rn is its usual ‘n-dimensional volume’. Proof: We will assume that n = 1 and leave the cases n ≥ 2 as a homework exercise. To prove the statement about the measure of an interval (= 1-box), first suppose the interval is compact, thus [a, b]. Since [a, b] ⊂ (a − ǫ, b + ǫ) for any ǫ> 0, it is clear that λ∗([a, b]) < b − a +2ǫ. Thus λ([a, b]) ≤ b − a. Conversely, if Ik are open intervals whose ADD By Homework sheet 1 question 3c, to see that B(Rn) ⊂L(Rn), it is enough to prove that n every ‘half-open interval’ [a1, b1) × [a2, b2) ×··· [an, bn) is in L(R ). We do this in the case that n = 1 again. Indeed by Proposition 1.2.9, it suffices to prove that [a, ∞) ∈ L(Rn). If E ⊂ R, then ADD