The Caratheodory Construction of Measures

Chapter 5 The Caratheodory Construction of Measures Recall how our construction of Lebesgue measure in Chapter 2 proceeded from an initial notion of the size of a very restricted class of subsets of R, the intervals. Using this notion, we defined an outer measure by the process (2.1) and proceeded from there to obtain a measure. In general, an outer measure on a set X is a function µ : P(X) [0, ] satisfying the following three conditions: (5.1) µ ( ) = 0, (5.2) A B = µ (A) µ (B), (5.3) A j X countable = µ ( ) A j µ (A j ). j j Recall that P(X) is the collection of all subsets of X. Parallel to (2.1), here is a way to construct outer measures. Let E be some family of subsets of X, such that E and X = j 1 X j for some countable collection X j E. Suppose you have a function (5.4) ρ : E [0, ], ρ( ) = 0. For any S X, set (5.5) µ (S) = inf { ρ(e j ) : E j E, S j 1 j 1 Here {E j } is a countable cover of S by sets in E. E j }. 57

58 5. The Caratheodory Construction of Measures Proposition 5.1. Under the hypotheses above, µ, defined by (5.5), is an outer measure. Proof. Property (5.1) follows from ρ( ) = 0 and (5.2) from the fact that, when A B, any countable cover of B by elements of E is also a cover of A. The proof of (5.3) works the same way as the proof of Proposition 2.1; each A j has a countable cover {E jk : k 1} by elements of E, such that µ (A j ) k ρ(e jk) 2 j ε. Then {E jk : j, k 1} is a countable cover of j A j by elements of E, and we obtain (5.3) in the limit as ε 0. If we have an outer measure µ, then, as suggested when we stated Proposition 2.12, we say that a set A X is µ -measurable if and only if (5.6) µ (Y ) = µ (Y A) + µ (Y \ A), Y X. (By (5.3), always holds, so the condition to check is.) Denote by M the class of µ -measurable subsets of X. The following result is known as Caratheodory s Theorem. Theorem 5.2. If µ is an outer measure on X, then the class M of µ - measurable sets is a σ-algebra, and the restriction of µ to M is a measure. Proof. Clearly M. Also, if A M, then, for all Y X, Y A c = Y \A and Y \ A c = Y A, so M is closed under complements. Next, suppose A j M. We want to show that (5.6) holds with A = A 1 A 2. Since µ is subadditive, it suffices to establish that (5.7) µ (Y ) µ ( Y (A 1 A 2 ) ) + µ ( Y (A 1 A 2 ) c), for all Y X. Note that A 1 A 2 = A 1 (A 2 A c 1 ) is a disjoint union. We see that the right side of (5.7) is (5.8) µ (Y A 1 ) + µ (Y A 2 A c 1) + µ (Y A c 1 A c 2) = µ (Y A 1 ) + µ (Y A c 1 ) = µ (Y ), where the last two identities use A 2 M and A 1 M, respectively. This yields (5.7) and shows that M is an algebra, i.e., M is closed under complements and finite unions. We next check additivity when A 1, A 2 M are disjoint. Indeed, if we set Y = A 1 A 2, A = A 1 in (5.6), we get (5.9) µ (A 1 A 2 ) = µ (A 1 ) + µ (A 2 ), A j M, disjoint.

5. The Caratheodory Construction of Measures 59 Inductively, we have, for A j M, disjoint, (5.10) µ ( N ) A j = N µ (A j ). Taking N and using monotonicity of µ yield µ ( ) A j µ (A j ), j 1 j 1 and since the reverse inequality holds by (5.3), we have for a countable family A j, (5.11) µ ( ) A j = µ (A j ), A j M, disjoint. j 1 j 1 Compare the proof of Proposition 2.7, leading to Theorem 2.8. To finish the proof of Theorem 5.2, we need to show that, if A j M is a countable disjoint family, then A = j 1 A j is µ -measurable. Let B n = j n A j. The measurability of A n implies µ (Y B n ) = µ (Y B n A n ) + µ (Y B n A c n) for any Y X. Inductively, we obtain (5.12) µ (Y B n ) = = µ (Y A n ) + µ (Y B n 1 ), n µ (Y A j ). We know M is an algebra, so B n M, and hence, using (5.12), we have (5.13) µ (Y ) = µ (Y B n ) + µ (Y Bn c ) n µ (Y A j ) + µ (Y A c ), the last inequality holding because B n A. Taking n, we have µ (Y ) j 1 µ (Y A j ) + µ (Y A c ) (5.14) µ ( j 1 ) (Y A j ) + µ (Y A c ) = µ (Y A) + µ (Y A c ) µ (Y ).

60 5. The Caratheodory Construction of Measures Thus A M, and Theorem 5.2 is proved. While the construction (5.4) (5.5) is very general, you need extra structure to relate µ (S) to ρ(s) when S E. The following is a convenient setting. Let A be an algebra of subsets of X, i.e., a nonempty collection of subsets of X, closed under finite unions and under complements, hence under finite intersections. A function µ 0 : A [0, ] is called a premeasure if it satisfies the following two conditions: (5.15) µ 0 ( ) = 0, (5.16) S j A countable, disjoint, j S j = S A = µ 0 (S) = j µ 0 (S j ). As an example, let X = I = [a, b], and let A consist of finite unions of intervals (open, closed, or half-open) in I. If S = N k=1 J k is a disjoint union of intervals, take µ 0 (S) = N k=1 l(j k), so µ 0 is the restriction of Lebesgue (outer) measure to A. In this case, (5.16) can be demonstrated by an argument similar to that needed for Exercise 1 in Chapter 2. If µ 0 is a premeasure on A, it induces an outer measure on X via the construction (5.4) (5.5), i.e., { (5.17) µ (E) = inf A j }. j 0 µ 0 (A j ) : A j A, E j 0 Proposition 5.3. If µ 0 is a premeasure on A and µ is defined by (5.17), then (5.18) S A = µ (S) = µ 0 (S), and every set in A is µ -measurable. Proof. To prove (5.18), first note that µ (S) µ 0 (S) for S A since S covers itself. Suppose S A, and S j 1 A j, A j A. Then let B n = S ( A n \ j<n A j), so {Bn : n 1} is a disjoint family of members of A, whose union is S. Thus, by (5.16), (5.19) µ 0 (S) = j 1 µ 0 (B j ) j 1 µ 0 (A j ). It follows that µ 0 (S) µ (S). This proves (5.18).

5. The Caratheodory Construction of Measures 61 To prove that each A A is µ -measurable, if Y X and ε > 0, there is a sequence {B j : j 1} A with Y j 1 B j and µ 0 (B j ) µ (Y ) + ε. Since µ 0 is additive on A, (5.20) µ (Y ) + ε µ 0 (B j A) + µ 0 (B j A c ) j 1 j 1 µ (Y A) + µ (Y A c ), the latter inequality holding, e.g., because {B j A : j 1} is a cover of Y A by elements of A. Taking ε 0, we obtain (5.6), so any A A is µ -measurable. When we combine the last result with Theorem 5.2, we obtain an extension of the premeasure µ 0 to a measure. Theorem 5.4. Let A P(X) be an algebra, µ 0 a premeasure on A, and M = σ(a) the σ-algebra generated by A. Then there exists a measure µ on M whose restriction to A is µ 0, namely µ M, where µ is given by (5.17). Proof. In fact, the class of µ -measurable sets is a σ-algebra containing A; hence it contains σ(a). There is nothing more to prove. We next examine the extent to which the extension µ of µ 0 to σ(a) is unique. Suppose ν is a measure on M = σ(a), which also agrees with µ 0 on A. If E M and E j 1 A j, A j A, then (5.21) ν(e) j 1 so, by the construction (5.17) of µ, ν(a j ) = j 1 µ 0 (A j ), (5.22) ν(e) µ(e), E σ(a). Also, by the property (3.5), (5.23) B j A, B j B = ν(b) = lim n ν(b n) = lim n µ(b n) = µ(b). Suppose now that E M and µ(e) <. Fix ε > 0. Then we can choose the cover A j in (5.17) such that µ(a j ) < µ(e) + ε. Hence, with B n = A j, j n A j B = j 1 we have µ(b) < µ(e)+ε, so µ(b\e) < ε, and hence, by (5.22), ν(b\e) < ε. Meanwhile, by (5.23), ν(b) = µ(b), so (5.24) µ(e) µ(b) = ν(b) = ν(e) + ν(b \ E) ν(e) + ε. Taking ε 0 gives µ(e) ν(e), as long as µ(e) <. In concert with (5.22), this yields the following.

62 5. The Caratheodory Construction of Measures Proposition 5.5. Let A be an algebra of subsets of X, generating the σ- algebra M = σ(a). Let µ 0 be a premeasure on A, and let µ be the measure on M given by Theorem 5.4, extending µ 0. If ν is another measure on M which agrees with µ 0 on A, then for all S M, ν(s) µ(s), and (5.25) µ(s) < = µ(s) = ν(s). Furthermore, if there is a countable family A j such that (5.26) X = j 1 A j, A j M, µ(a j ) <, then µ(s) = ν(s) for all S M. Proof. The implication (5.25) was established above. If (5.26) holds, we can assume the A j are disjoint. If S M, we have the disjoint union S = S j, S j = S A j. By (5.25), µ(s j ) = ν(s j ), and then µ(s) = ν(s) follows by countable additivity. If (5.26) holds, we say (X, M, µ) is a σ-finite measure space. See Exercise 15 for another proof of Proposition 5.5. If Theorem 5.4 is applied to the example mentioned after (5.16), it yields the fact that finite unions of intervals in I = [a, b] are Lebesgue measurable and hence so are sets in the σ-algebra σ(a) generated by this algebra of subsets of I. In particular, all open sets in I are measurable, since they are countable unions of open intervals. Hence all closed sets in I are measurable. Of course, this recaptures results obtained in Chapter 2. Recall that, in our treatment of Lebesgue measure on an interval, we gave a definition of measurability different from (5.6) and then showed in Proposition 2.12 that the definition of measurability given there implied (5.6). We now give a result in counterpoint to Proposition 2.12. Proposition 5.6. Suppose we are given an algebra A of subsets of X and a premeasure µ 0 on A, with associated outer measure µ, defined by (5.17). Assume Z X is µ -measurable and µ (Z) <. Then a set S Z is µ -measurable if and only if (5.27) µ (S) + µ (Z \ S) = µ (Z). Proof. We need to show that, if (5.27) holds, then, for any Y X, (5.28) µ (Y ) = µ (Y S) + µ (Y S c ).

5. The Caratheodory Construction of Measures 63 Only the part needs to be established, so we can assume µ (Y ) <. Write Y = Y 0 Y 1, a disjoint union, where Y 0 = Y Z, Y 1 = Y Z c. Using the measurability of Z, we have (5.29) µ (Y ) = µ (Y Z) + µ (Y Z c ) = µ (Y 0 ) + µ (Y 1 ) and (5.30) µ (Y S c ) = µ (Y S c Z)+µ (Y S c Z c ) = µ (Y 0 S c )+µ (Y 1 ), provided S Z, while (5.31) µ (Y S) = µ (Y 0 S). Consequently it suffices to prove (5.28) for Y = Y 0, i.e., for Y Z. For that, we bring in a lemma Lemma 5.7. Let µ arise from a premeasure on an algebra A. consist of countable unions of sets in A. Then Let A σ (5.32) S X = µ (S) = inf{µ (E) : E A σ, S E}. Proof. We leave this to the reader; see Exercise 3 at the end of this chapter. We continue the proof of Proposition 5.6. Given ε > 0, there exists à A σ such that à Y and µ (Ã) µ (Y ) + ε. Set A = à Z. Then A M (the σ-algebra of µ -measurable sets), A Y (if Y Z), and µ (A) µ (Y ) + ε. We claim that (5.33) µ (A) = µ (A S) + µ (A S c ). Suppose this is known. Then we have (5.34) µ (Y ) + ε µ (Y S) + µ (Y S c ), for all ε > 0. In the limit ε 0, we obtain the part of (5.28), and the problem is solved. Thus it remains to establish (5.32), given A M, A Z, µ (Z) <, and given the hypothesis (5.27). Using the measurability of A, we have (5.35) µ (Z) = µ (A) + µ (Z A c ).

64 5. The Caratheodory Construction of Measures Now the left side of (5.35) is equal to µ (S) + µ (Z \ S), by hypothesis. By subadditivity of outer measure, the right side of (5.35) is (5.36) µ (A S) + µ (A S c ) + µ ( (Z \ A) S ) + µ ( (Z \ A) S c) = µ (S A) + µ ( (Z \ S) A ) + µ (S \ A) + µ ( (Z \ S) \ A ) = µ (S) + µ (Z \ S), the last identity following by grouping together the odd terms and the even terms on the second line of (5.36) and using measurability of A. Since the bottom line of (5.36) is equal to the left side of (5.35), the in (5.36) must be equality. That inequality arose from the sum of two inequalities, and so both of them must be equalities. One of them is the desired result (5.33). This finishes the proof of Proposition 5.6. We next describe an important class of outer measures on metric spaces, for which all open sets and all closed sets can be shown to be measurable. This result will play an important role in Chapter 12, on Hausdorff measures, and in Chapter 13, on Radon measures. Let X be a metric space, with distance function d(x, y). An outer measure µ on X is called a metric outer measure provided (5.37) ρ(s 1, S 2 ) = inf {d(x 1, x 2 ) : x j S j } > 0 = µ (S 1 S 2 ) = µ (S 1 ) + µ (S 2 ). Note that the part (2.15) (2.16) of Lemma 2.4 is the statement that Lebesgue outer measure is a metric outer measure on I = [a, b]. The following is another result of Caratheodory: Proposition 5.8. If µ is a metric outer measure on a metric space X, then every closed subset of X is µ -measurable. Proof. We must show that if F X is closed and Y X satisfies µ (Y ) <, then (5.38) µ (Y ) µ (Y F ) + µ (Y \ F ). Consider B n = { x Y \ F : ρ(x, F ) 1 }. n Note that B n Y \ F. Also ρ(b n, F ) 1/n, so, by (5.37), (5.39) µ (Y F ) + µ (B n ) = µ ( (Y F ) B n ) µ (Y ).

5. The Caratheodory Construction of Measures 65 Thus it will suffice to show that (5.40) µ (B n ) µ (Y \ F ). Let C n = B n+1 \ B n. Note that j k 2 = ρ(c j, C k ) > 0. Hence, for any N, one obtains inductively (using (5.37)) that (5.41) N µ (C 2j ) = µ ( N ) C 2j µ (Y ), N µ (C 2j+1 ) = µ ( N ) C 2j+1 µ (Y ), and consequently j 1 µ (C j ) <. Now countable subadditivity implies (5.42) µ (Y \ F ) µ (B n ) + j n µ (C j ), so as n, the last sum tends to zero, and we obtain (5.43) µ (Y \ F ) lim inf n µ (B n ) lim sup n µ (B n ) µ (Y \ F ), the last inequality by monotonicity. proof. This implies (5.40) and finishes the As one may have noticed, at several points in this chapter and previous chapters, we have considered a class of sets with certain desirable properties and have wanted to prove it was a σ-algebra. The following result, called the Monotone Class Lemma, sometimes furnishes a convenient tool for doing this. We define a monotone class on a set X to be a collection C P(X) having the properties (5.44) E j C, E j E = E C, E j C, E j E = E C. The smallest monotone class containing a collection E P(X) is called the monotone class generated by E. Proposition 5.9. If A is an algebra of subsets of X and C is the monotone class generated by A, then C = σ(a).

66 5. The Caratheodory Construction of Measures Proof. Clearly C σ(a). We will show that C is a σ-algebra. If E C, let (5.45) C(E) = {F C : E \ F, F \ E, E F C}. Clearly, E C(E), and C(E) is easily verified to be a monotone class. Also, E C(F ) F C(E). Since A is an algebra, we have (5.46) E, F A = F C(E). In other words, if E A, then A C(E); hence C C(E). Thus, if F C, then F C(E) for all E A. This implies (5.47) E C(F ), E A, F C. Thus A C(F ) and hence C C(F ), for all F C. In other words, (5.48) E, F C = E \ F, E F C. Since X A C, it follows that C is an algebra of sets. We finally note that C must be a σ-algebra. Indeed, if E j C, then F j = n j E j C, and (5.44) implies F j F = j 1 E j C. This finishes the proof. The proof of Proposition 6.2 in the next chapter will provide a nice application of the Monotone Class Lemma. Exercises 1. Let I = [a, b] and let A be the algebra of subsets of I consisting of finite unions of intervals, as in the example after (5.16). Let ϕ : I R be a monotonically increasing function, i.e., assume that x < y ϕ(x) ϕ(y). Set (5.49) ϕ (x) = lim u x ϕ(u), (5.50) ϕ + (x) = lim u x ϕ(u), with the convention that ϕ (0) = ϕ(0) and ϕ + (1) = ϕ(1). Define µ 0 : A R + as follows. If S = N k=1 J k is a union of mutually disjoint intervals, set µ 0 (S) = N k=1 µ 0(J k ), where µ 0 is defined on a single interval J I as follows: µ 0 ( [x, y] ) = ϕ+ (y) ϕ (x), µ 0 ( [x, y) ) = ϕ (y) ϕ (x), µ 0 ( (x, y] ) = ϕ+ (y) ϕ + (x), µ 0 ( (x, y) ) = ϕ (y) ϕ + (x).

5. The Caratheodory Construction of Measures 67 Also, set µ 0 ( ) = 0. Show that µ 0 is a premeasure. The measure then produced via Theorem 5.4 is called Lebesgue-Stieltjes measure. Hint. Say J is a closed interval and J = k 1 J k, with disjoint intervals J k. First show that m k=1 µ 0(J k ) µ 0 (J) for each m, so µ 0 (J) k 1 µ 0(J k ). For the reverse inequality, one can argue as follows. Pick ε > 0. Say the endpoints of J k are a k and b j, so (a k, b k ) J k [a k, b k ]. If a k J k, pick ã k < a k such that 0 ϕ (a k ) ϕ + (ã k ) < ε2 k. If a k / J k, let ã k = a k. Make an appropriate analogous choice of b k, and let J k = (ã k, b k ). Note that µ 0 ( J k ) < µ 0 (J k ) + ε2 k+1. Now J J k 1 k, and this is an open cover. Take a finite subcover. For Exercises 2 6, suppose we are given an algebra A of subsets of X and a premeasure µ 0 on A, with associated outer measure µ, defined by (5.17). Let A σ consist of countable unions of sets in A. 2. Show that (5.51) E A σ = µ (E) = sup {µ 0 (A) : A A, A E}. 3. Show that (5.52) S X = µ (S) = inf {µ (E) : E A σ, S E}. 4. Let A δ consist of countable intersections of sets in A, and set (5.53) µ (S) = sup {µ (E) : E A δ, E S}, S X. Show that µ (S) µ (S). If µ (S) <, show that S is µ -measurable if and only if µ (S) = µ (S). Hint. Show that, if S Z A σ, µ (Z) <, then µ (Z) = µ (S) + µ (Z \ S). Then use Proposition 5.6. 5. Given S X, show that there exist S 0 A σδ and S 1 A δσ such that (5.54) S 0 S S 1, µ(s 1 ) = µ (S), µ(s 0 ) = µ (S). Show that, if µ (S) <, then S is µ -measurable if and only if µ(s 1 \ S 0 ) = 0. Here, µ is the measure on σ(a) given by Theorem 5.4. 6. Let µ # be the outer measure on X obtained via (5.17), with (A, µ 0 ) replaced by (σ(a), µ). Equivalently, (5.55) µ # (S) = inf {µ(a) : S A σ(a)}.

68 5. The Caratheodory Construction of Measures Clearly µ # (S) µ (S) for all S X. Show that µ # = µ. Hint. If S A σ(a), show that, for any ε > 0, there exists B A σ such that A B and µ(b) µ(a) + ε. 7. If µ is an outer measure on X, then clearly (5.56) S 1 S 2 S 3 S = µ (S j ) M µ (S). If µ arises from a premeasure, via (5.17), show that M = µ (S). Hint. First note that A n σ(a), A n A µ (A n ) µ (A). Then show that, for any ε > 0, there exist Z n σ(a) such that Z n S n, Z n, and µ (Z n ) = µ (S n ). Make use of (5.54). 8. As in Exercise 5 of Chapter 3, a measure µ on (X, M) is said to be complete provided (5.57) A M, µ(a) = 0, S A = S M, and µ(s) = 0. Show that the measures provided by Theorem 5.2, from outer measures, are all complete. In particular, Lebesgue measure is complete. 9. If µ is a measure on (X, F) that is not complete, form the outer measure µ by (5.17), with A replaced by F. Let M be the σ-algebra of µ - measurable sets. Show that Theorem 5.2 produces a measure µ on (X, M) which is complete and such that F M and µ = µ on F. Show that M = F, given by Exercise 5 of Chapter 3, and that µ coincides with the measure produced there, the completion of µ. In Exercises 10 14, suppose X is a compact metric space, B the σ- algebra of Borel subsets of X, and µ a finite measure on (X, B). Construct the outer measure µ via (5.17), with A replaced by B. (Note that Proposition 5.3 applies in this case.) 10. Show that µ must be a metric outer measure. Hint. If S 1, S 2 satisfy (5.37), construct compact K X such that S 1 K X \ S 2. Apply (5.6) with A = K, Y = S 1 S 2. 11. We know that B = σ(a) where A is the algebra generated by the compact sets in X. Show that A consists of sets of the form (5.58) N l M l α l =1 β l =1 N 1 M 1 α 1 =1 β 1 =1 E αβ,

5. The Caratheodory Construction of Measures 69 where each E αβ X is either open or closed. 12. Show that (5.59) S X = µ (S) = inf {µ (O) : S O, O open}. Hint. By Exercise 6, µ (S) is given by (5.52) or, equivalently, by (5.17) (with some changes in notation). Show that, given A j A, ε > 0, there exist open sets O j such that A j O j and µ(o j ) µ(a j ) + 2 j ε. It suffices to make such a construction for each E αβ in (5.58), and we need worry only about the case when E αβ is compact; i.e., we need to verify (5.59) when S is compact. For this, keep in mind that X is a metric space. 13. Show that (5.60) S B = µ(s) = sup {µ(k) : K S, K compact}. 14. If f L 1 (X, µ) and ε > 0, show that there exists a compact K X such that µ(x \ K) < ε and f K is continuous. Hint. Using Proposition 4.5 (or otherwise), produce f ν C(X) such that f ν f, µ-a.e. Then use Egoroff s Theorem. Then use (5.60). This result is Lusin s Theorem. A special case was stated in Exercise 16 of Chapter 3. 15. Use the Monotone Class Lemma to give another proof of Proposition 5.5. Furthermore, establish the following variant of Proposition 5.5. Proposition 5.5A. Let A be an algebra of subsets of X, generating the σ-algebra M = σ(a). Let µ and ν be measures on M. Assume there exist A j such that X = j 1 A j, A j A, µ(a j ) <. If µ = ν on A, then µ = ν on M. How do these two propositions differ? 16. Suppose E is a collection of subsets of X having the following properties: (1) The intersection of any two elements of E belongs to E. (2) The collection of finite disjoint unions of elements of E is an algebra A.

70 5. The Caratheodory Construction of Measures Let µ b : E [0, ] satisfy (a) µ b ( ) = 0, (b) E j E countable, disjoint, j E j = E E µ b (E) = j µ b(e j ). Show that µ b extends to a premeasure µ 0 on A, satisfying E 1,..., E K E disjoint µ 0 (E 1 E K ) = K µ b (E j ). Show that if µ is defined by (5.17), then also, for E X, { µ (E) = inf j 0 µ b (A j ) : A j E, E j 0 A j }. Hint. Start with uniqueness; if also F 1,..., F L E are disjoint and L l=1 F l = K k=1 E k, write this set as k,l (E k F l ) to show µ 0 is well defined. Note. An example is X = [0, 1], E = the collection of intervals in [0, 1], µ b (J) = l(j), the length. Another family of examples arises at the beginning of Chapter 6. 17. Suppose E is a collection of subsets of X satisfying property (1) in Exercise 16 and also satisfying (2 ) E E X \ E is a finite disjoint union of elements of E. Show that E satisfies property (2) in Exercise 16.