1.4 Outer measures 10 CHAPTER 1. MEASURE

Transcription

1 10 CHAPTER 1. MEASURE ( 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 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 If (X, A, µ is a measure space, define Ā to be the set of unions of a set in A and a subset of a µ-null set. Then Ā is a σ-algebra on X, and there is a unique extension µ of µ to a complete measure on Ā. Proof: If (A n n=1, (B n n=1 are sets in A with µ(b n = 0 for all n, and if F n B n for all n, then n=1 (A n F n = ( n=1 A n ( n=1 F n, and n=1 F n B, where B = n=1 B n. Note that B is µ-null, by a comment above the theorem. So n=1 (A n F n Ā. If A, B A with µ(b = 0, and F B, then (A F c = (A B c (B \ F Ā. Define µ(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 A n, F n as above, and suppose further that the (A n are mutually disjoint. We have µ( n=1 (A n F n = µ(( n=1 A n ( n=1 F n = µ( n=1 A n = µ(a n = n=1 n=1 µ(a n F n. So µ is a measure. To see that it is complete, choose C Ā which is µ-null, and D C. Write C = A F with A A, F B A, with µ(b = 0. Since C is µ-null, µ(a = 0. Thus D is a subset of the µ-null set A B, so D Ā. We call µ the completion of µ, and we call Ā 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 Carathéodory 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 A k k=1 µ (A k, for any subsets A 1, A 2,... of X.

2 1.4. OUTER MEASURES 11 The following is a very general way to construct outer measures: Proposition 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(e k : E k E, A k=1 E k}. Then µ is an outer measure. Proof: First, µ is well defined since E contains and a countable cover of X. Since ρ ( = 0, we have that µ ( = 0 (let E i = for i = 1, 2,..., and thus 0 µ ( i=1 ρ (E i = i=1 ρ ( = 0. For S 1 S 2 X, any countable cover of S 2 by elements of E is also a countable cover of S 1. Thus µ (S 1 µ (S 2. Finally, for {S j } P (X, and given ε > 0, let {E j,k : k = 1, 2,... } be a countable cover for S j by elements of E such that ρ (E j,k µ (S j + ε 2. j k=1 Then the collection {E j,k : j, k = 1, 2,... } is a countable cover of S j by elements of E, and ρ (E j,k (µ (S j + ε = µ (S 2 j j + ε. k=1 ( Thus µ S j ( µ (S j + ε. Therefore, as ε 0 we obtain µ S j µ (S j. Hence µ is an outer measure (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 This is the same as saying that µ (E µ (E A + µ (E A c, E X. µ (E = µ (E A + µ (E A c, E X, since by (iii above we have µ (E = µ ((E A (E A c µ (E A + µ (E A c. The last centered condition is called Carathéodory s condition, and we set C µ 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 C µ is a complete measure. The completeness is a consequence of: Proposition If µ is an outer measure on a set X, then any subset A of X with µ (A = 0, is in C µ. Proof: If µ (A = 0 then µ (E A + µ (E A c µ (A + µ (E A c µ (E. Theorem (Carathéodory If µ is an outer measure on a set X, then C µ is a σ- algebra, and the restriction of µ to C µ is a complete measure.

3 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 C µ. Now let A C µ. Then for any E X we have that µ (E = µ (E A + µ (E A c = µ (E A c + µ (E A, and it follows that C µ is closed under complements. Next, suppose that A 1, A 2 C µ. We want to show that A 1 A 2 satisfies Caratheodory s condition. We note that A 1 A 2 = A 1 (A 2 A c 1 is a disjoint union and thus by using the countable subadditivity of µ and the fact that A 1 and A 2 satisfy Caratheodory s condition, we have that µ (E (A 1 A 2 + µ (E (A 1 A 2 c = =µ (E (A 1 (A 2 A c 1 + µ (E (A 1 A 2 c =µ ((E A 1 (E (A 2 A c 1 + µ (E (A 1 A 2 c µ (E A 1 + µ (E (A 2 A c 1 + µ (E (A 1 A 2 c =µ (E A 1 + µ (E A 2 A c 1 + µ (E A c 1 Ac 2 =µ (E A 1 + µ (E A c 1 A 2 + µ (E A c 1 Ac 2 =µ (E A 1 + µ (E A c 1 =µ (E. Thus, A 1 A 2 satisfies Caratheodory s condition and so A 1 A 2 C µ. That is, C µ is closed under finite unions. Hence, C µ is an algebra on X. To show that C µ is a σ-algebra it is enough to show that C µ is closed under countable disjoint unions. So let {A i } i=1 be a countable collection of disjoint sets in C µ. Let B n = n i=1 A i and B = A j. Then the measurability of A n shows that for any E X µ (E B n =µ (E B n A n + µ (E B n A c n =µ (E A n + µ (E B n 1. By induction, we then have that µ (E B n = n µ (E A j. Now since C µ is an algebra we have that B n C µ. This, together with the last equality, shows that µ (E =µ (E B n + µ (E Bn c n = µ (E A j + µ (E Bn c n µ (E A j + µ (E B c

4 1.5. LEBESGUE MEASURE 13 using the facts that B n B and the monotonicity of µ. Letting n, we now have µ (E µ (E A j + µ (E B c ( µ (E A j + µ (E B c =1 =µ (E ( A j + µ (E B c =µ (E B + µ (E B c µ (E. Since the left and right side of the inequality are equal, it follows that we have equalities throughout. Thus µ (E = µ (E B + µ (E B c. Hence, B C µ. That is, C µ is closed under countable disjoint unions, and therefore is a σ-algebra. From the above work we have that µ (E = µ (E A j + µ (E B c holds for all E X. Thus if we let E = B = A j we have µ (B = µ (B A j + µ (B B c = µ (B A j = µ (A j. We conclude that µ is countably additive on C µ. 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 defines an outer measure µ built from ρ, and from Carathéodory s theorem we get a complete measure space (X, C µ, µ where µ is the restriction of µ to C µ. 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 R n. Let E consist of the and the open intervals n k=1 (a k, b k = (a 1, b 1 (a 2, b 2 (a n, b n (one could use half-open intervals n k=1 [a k, b k = [a 1, b 1 [a 2, b 2 [a n, b n here instead, but this yields the same thing. Define ρ of this interval to be its usual volume n k=1 (b k a k. Here a k < b k are real numbers. The associated outer measure from Proposition is called

5 14 CHAPTER 1. MEASURE Lebesgue outer measure and is written as λ. The σ-algebra C λ coming from Carathéodory s theorem is written as L or L(R n or L n, and is called the Lebesgue σ-algebra. The restriction of λ to L(R n is called Lebesgue measure. By bounded n-box, we will mean a subset of n k=1 [a k, b k ] that contains n k=1 (a k, b k. Here a k < b k are numbers in R. Unbounded n-boxes are defined similarly. Proposition B(R n L(R n, and the Lebesgue measure of any n-box in R n 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 I k are open intervals whose ADD By Homework sheet 1 question 3c, to see that B(R n L(R n, it is enough to prove that every half-open interval [a 1, b 1 [a 2, b 2 [a n, b n is in L(R n. We do this in the case that n = 1 again. Indeed by Proposition 1.2.9, it suffices to prove that [a, L(R n. If E R, then ADD