Construction of a general measure structure

Transcription

1 Chapter 4 Construction of a general measure structure We turn to the development of general measure theory. The ingredients are a set describing the universe of points, a class of measurable subsets along with permissible operations, and the measure itself. In the next chapter, we apply the general theory to Euclidean space. In later chapters, we apply the general theory to probability spaces. In this chapter, we assume that the set representing the master set or universe is given. It is just a set, but the choice of is important. For example, as with and I, it may be related to modeling a physical situation. Its properties have a strong impact on its possible measure structures. For example, whether or not is a metric space and whether or not is bounded are important in terms of properties of measures. In the next chapter, we investigate the choice = n in detail. In this chapter, we choose various to illuminate various theoretical issues. 4.1 Sigma algebras We begin by constructing the class of measurable sets and the permissible set operations. For convenience, we repeat Definition Definition Let be a non-empty set. The family of all subsets of is called the power set of and is denoted by Algebras and σ- algebras Recall that in Chapter 3, we found that natural questions about sequences of coin flips lead to considering countable unions and intersections as well as complements of sets. Definition Let be a non-empty set. An algebra on is a non-empty collection of subsets with the following properties, 1. If A then A c. (Closed under complements) 2. If A 1, A 2,...,A n then n A i. (Closed under finite unions) Definition Let be a non-empty set. A sigma algebra (σ- algebra) on is a nonempty collection of subsets with the properties, 49

2 50 Chapter 4. Construction of a general measure structure 1. If A then A c. (Closed under complements) 2. If {A i } then A i. (Closed under countable unions) Theorem Let be a non-empty set. 1. A σ- algebra is an algebra. 2. If is an algebra or σ- algebra, then,. n 3. If is an algebra and {A i } n then A i. (Closed under finite intersections) 4. If is a σ- algebra and {A n } then A i. (Closed under countable intersections) 5. If is an algebra or σ- algebra and if A, B, then A B. Proof. We prove these in order. Result 1 Follows from the definitions. Result 2 Since is nonempty there is a A such that A. Thus A c and = A A c. Since, c =. Result 3 Result 4 Since {A c i }n and n n Ac i, Since {A c i } and Ac i, Result 5 From above, A B = A B c. A i = n Ac i c. A i = Ac i c. Definition is a σ- algebra called the maximal σ- algebra. Definition {,} is a σ- algebra on called the trivial or minimal σ- algebra. Example If B, the collection {, B, B c,} is a σ- algebra. Example If is uncountable, then = {A : Ais countable or A c is countable} is a σ- algebra. First note that (A c ) c = A. Thus, if A, it is either countable or A c is countable. So, A c. Let {A n }. If all the sets are countable then the countable union is countable and so is in. If not all the sets are countable, there exist an index j + such that A c is j c countable. Thus, A i = A c i Ac is countable. j The following theorem provides a convenient way to generate new σ- algebras from an existing σ- algebra.

3 4.1. Sigma algebras 51 Theorem Let be a σ- algebra on and B. Then, the collection B = {B A : A } is a σ- algebra on. Proof. Let B 1 B, then B 1 = B A 1 for some A 1. The complement of B 1 in B is B B1 c. Now, B Bc 1 = B (Bc A c 1 ) and B Bc 1 = (B Bc ) (B A c 1 ). So B Bc 1 = B Ac 1 B. Let {B i } be a sequence in B. Each B i = B A i for some A i. Therefore, B i = B A i B. The next result gives conditions establishing an algebra is a σ- algebra. Theorem An algebra of sets on is a σ- algebra if and only if 1. If {A i } is disjoint then. (Closed under countable disjoint unions) A i 2. If A then A c. (Closed under complements) Proof. follows by definition. For, consider a sequence of sets {A i }. Following Theorem 1.1.5, we can construct a disjoint sequence {B i } such that A i = B i Generating σ- algebras There is little chance that an arbitrary set is a σ- algebra. It is natural to wonder if it is possible to construct a σ- algebra starting with a given set. The following result gives a partial answer. Theorem Let be a non-empty set. 1. The intersection of any collection of σ- algebras on is a σ- algebra. 2. If is a collection of subsets, then is a unique smallest σ- algebra containing in the sense that any σ- algebra containing also contains. This result says that the unique smallest σ- algebra exists, but the proof does not give a practical procedure to construct it. This is a major topic in the application of measure theory to Euclidean space. Proof. Result 1 Let be a collection of σ- algebras on. Define, = = {E : E for all }. Suppose A. Then, A c for all and therefore A c. In a similar way, if {A i }, then A i.

4 52 Chapter 4. Construction of a general measure structure Result 2 Define to be the intersection of all σ- algebras containing. The intersection is nonempty and is itself a σ- algebra by 1. By definition, is contained in any σ- algebra that contains. Definition Let be any non-empty set and let be a collection of subsets. The unique smallest σ- algebra containing is denoted by σ () and is said to be the σ- algebra generated by. Proving the following is a good exercise. Theorem Let be a non-empty set. 1. If is a σ- algebra, then = σ ( ). 2. For any A, we have σ ({A}) = {, A, A c,}. 3. If are collections of subsets, then σ () σ ( ). The following useful result gives conditions under which the σ- algebra generated by a family of sets inherits some desirable property of those sets. Theorem (Principle of Good Sets). Let be a nonempty set and a a collection of subsets. Define to be the class of good sets, = {B σ ( ) : B has a desired property}. If, so all sets in the generating set are good, and is a σ- algebra, then σ ( ). In words, all the sets in the σ- algebra generated by the original family are good. Proof. σ ( ) σ ( ) = σ- algebras and maps Another important method to generate a σ-algebra from another σ-algebra is using an inverse to a map. The reason this possible is due to the following set result. Theorem Let A β : β be a family of subsets of a set and let f be a map from a set to a set. Then, \ A β = (Y \ A β ), \ A β = ( \ A β ). β β β β f 1 A β = f 1 A β, β β f 1 A β = f 1 A β β β Proving this theorem is a good exercise.

5 4.1. Sigma algebras 53 Theorem Let f 1. If is a σ- algebra on, then is a σ- algebra on. : be a map, where and are nonempty sets. f 1 ( ) := f 1 (B) : B 2. If is a σ- algebra on, then B : f 1 (B) is a σ- algebra on. 3. If is a collection of sets in, then σ f 1 () = f 1 (σ ()) Proof. Result 1 Let A f 1 ( ), i.e. A = f 1 (B) for some B. A c = ( f 1 (B)) c = f 1 (B c ), so A c f 1 ( ). Let {A i } be a sequence of sets in f 1 ( ). Then A i = f 1 (B i ) for B i and A i = f 1 (B i ) = f 1 B i f 1 ( ). If B is in the collection, f 1 (B), hence ( f 1 (B)) c = f 1 (B c ). Exer- Result 2 cise: show that a countable union is treated in the same fashion. Result 3 We first show σ f 1 () f 1 (σ ()). If B then B σ (), hence f 1 () f 1 (σ ()). By 1., f 1 (σ ()) is a σ- algebra, so σ f 1 () f 1 (σ ()). Next, we show f 1 (σ ()) σ f 1 () using Theorem Define the collection of good sets to be, = B σ () : f 1 (B) σ f 1 (). First, since f 1 () σ f 1 (). Next, we show that is a σ- algebra. For example, assume B. This means that f 1 (B) σ f 1 (), which in turn implies that B c since f 1 (B c ) = ( f 1 (B)) c σ f 1 (). Countable unions are treated with the same approach. Thus, σ (), which proves the result Rings Monotone classes There are different approaches to constructing the family of sets used for measure theory, as the reader can easily see by looking in two or more textbooks. These all yield ultimately equivalent results, but various formulations are more or less convenient for different applications and for proving different results. In this section and the next, we discuss two of the most important alternative approaches. These sections can be skipped at the first reading. Definition Let be a non-empty set. A monotone class is a non-empty collection of subsets of satisfying,

6 54 Chapter 4. Construction of a general measure structure 1. If the sequence {A i } is an increasing sequence, i.e., A i A i+1 for each i, then A i. 2. If the sequence {A i } is a decreasing sequence, i.e., A i A i+1 for each i, then A i. We say the monotone class is closed under monotone sequences of sets. Theorem An algebra is a σ- algebra if and only if it is a monotone class. Proof. A σ- algebra is closed under countable unions so it is a monotone class. Let be an algebra that is a monotone class and consider a sequence {A i }. From Theorem 1.1.5, we construct an increasing sequence {B i } such that B i = A i, thus A i. Hence is a σ- algebra. Following the analogous concept for σ- algebras, Definition Let be a non-empty set and also be nonempty. The minimal monotone class containing or monotone class generated by is the intersection of all monotone classes containing. We denote it by ς(). The proof of the next theorem illustrates a useful technique. Theorem Let be an algebra in non-empty set. Then, ς() = σ (). Proof. First, σ () is a monotone class containing so ς() σ (). Thus, the proof is complete if we show that ς() is a σ- algebra containing. To do this, we have to show that ς() is an algebra. We use the Principle of Good Sets The good sets property is: 1 = {A ς() : A c ς() and A B ς() for all B ς()}. 1 is an algebra, so if we show that 1 ς(), the proof is complete. We can do this by showing that: 1. 1 is a monotone class and To show 1., let {A i } be an increasing sequence of sets in 1, so A i ς(). Let B be any set in ς(), so A i B is in 1 for all i. Hence, {A i B} is an increasing sequence in ς() and so A i B = (A i B) ς().

7 4.1. Sigma algebras 55 Thus, A i 1. Also, {A c } is a decreasing sequence of sets and i c A i = A c i ς(). Since {A i } is an arbitrary increasing sequence, this proves the result. To prove 2., we show that 1 2 for another good sets monotone class 2, which implies that 1 2 ς(). The second good sets property is: 2 = {A ς() : A c ς() and A B ς() for all B }. Since is an algebra, 2 1. Exercise: argue as above to show that 2 is a monotone class The π λ Theorem Next we discuss another useful approach for construction of sets relevant for measure theory. As with Monotone Classes, which is a closely related idea, this formulation is very useful for certain proofs. For example, it can be used to show that two measures which agree on a certain class of sets in a σ- algebra actually agree on all sets in the σ- algebra. Definition Let be a non-empty set. A collection of subsets is a π class if A B for any A, B. This is not an algebra since we do not mention complement. Definition Let be a non-empty set. A collection of subsets is a λ class if, 1., 2. A implies A c, 3. If {A i } is a disjoint sequence, then A i. It is a good exercise to show the relation to σ- algebras. Theorem A λ class that is also a π class is a σ- algebra. The following equivalent formulation of the λ class is sometimes useful. Definition Let be a non-empty set. A collection of subsets is a D class if, 1., 2. A, B and A B implies A B, 3. If {A i } is a disjoint sequence, then A i.

8 56 Chapter 4. Construction of a general measure structure It is another good exercise to show the equivalence. Theorem A collection of subsets of a nonempty set is a D class if and only if it is a λ class. Definition Let be a nonempty set and a nonempty collection of subsets. We use ζ () to denote the λ class given by the intersection of all λ classes containing. Theorem Let be a nonempty set and a nonempty collection of subsets. Then, ζ () σ (). Proof. σ () is a λ class containing and so by minimality contains (). Theorem Let be a non-empty set and let be a non-empty collection of subsets of. If is a π class, then ζ () = σ (). Proof. First, ζ () is a σ- algebra. Since, a σ- algebra is a λ class, ζ () σ (). We show σ () ζ (). This follows if we show that ζ () is a π class. We define the good sets to be 1 = {A ζ () : A B ζ () for all B ζ ()}. By construction 1 ζ () and is a π class. If we show that ζ () is a λ class and 1, we are done. It is clear that is in 1. Let A be in 1, and let B be any set in ζ (). Then, A c B = (A B c ) c. Since, A, A ζ () and since B ζ (), B c ζ (). Thus, (A B c ) c ζ (). Hence, for any B ζ (), A c B 1 whenever A 1. Let {A i } be a disjoint sequence of sets in 1 and let B be any set in ζ (). Then, A i B = (A i B). Since 1 ζ (), A i B ζ for each i and so the countable union is in ζ (). To show that 1, we construct a family 2 such that 1 2 and 2 is a λ class. Define the good sets 2 = {A ζ () : A B ζ () for all B }. Since is a π class, 2. If we show 2 is a λ class, then 2 ζ (). But since by construction, 2 1 we get 1 ζ (). Proving that 2 is a λ class follows arguments analogous to those used for 1. Theorem (π λ Theorem). implies σ (). If is a π class and is a λ class, then Proof. By definition, ζ (). We showed that σ () = ζ () and so σ ().

9 4.2. Measure 57 References Exercises 4.2 Measure In this section, we develop the basic properties of measure quantifying the size of the sets in a given σ- algebra on a set. Size could be the analog of length, area or volume as discussed in Chapter 3. But, it can be something else entirely. Much of the later focus in this book is on using measure to evaluate probability. In Chapter 3, the probability measure coincided with the measure of length, which worked out because of the length of the unit interval is 1 and we require the probability of the entire space to be 1 as well. But, that is a special case. In this section, we develop general properties and consequences. We leave a discussion of how measure is actually computed to later. We begin by recognizing the measure structure we have spent so much time creating. Definition Let be a set on which there is a σ- algebra. We call (, ) a measurable space. The sets in are called measurable sets. A measure is defined on a measurable space. Definition Let (, ) be a measurable space. An additive measure on (, ) is a function µ : [0, ] satisfying, µ() = 0, If {A i } n is a set of disjoint sets in, then n µ A i = n µ(a i ). A (countably additive) measure on (, ) is a function µ : [0, ] satisfying, µ() = 0, If {A i } is a sequence of disjoint sets in, then µ A i = µ(a i ). We use measure to refer to a countably additive measure. Finite additivity fits intuition of how a function that measures the size of sets should behave, and countable additivity is the extension to countable collection of sets. We are almost entirely interested in measures, but use additive measures as a stepping stone. Definition If (, ) is a measurable space on which there is a measure µ, then the triple (,,µ) is called a measure space. All of these terms are abused regularly. We might say µ is a measure on, where is taken to be the natural domain for µ. For example, there is a natural choice in a metric

10 58 Chapter 4. Construction of a general measure structure space. If we know, then we also know since is a maximal element of. So may not be mentioned explicitly. There is a significant difference between the cases when the space has finite measure and when it does not. In the latter situation, we distinguish two particular cases. Definition Let (,,µ) be a measure space. We say that µ is a finite measure if µ() < and that (,,µ) is a finite measure space. We say that µ is a probability measure if µ() = 1 and (,,µ) is a probability space. We say that µ is a σ- finite measure if contains an increasing sequence A 1 A 2 A 3... such that A i = and µ(a i ) < for every i +. We say that (,,µ) is a σ- finite measure space. The idea behind σ- finite follows naturally from considering that = i= (i 1, i]. In this book, we deal with finite and σ- finite measures. Dealing with non-σ- finite measures require additional assumptions and work. We note above that defining a measure is not the same as giving a practical recipe for computing the measure of sets. Even direct verification of the properties of a measure is difficult outside simple examples. A natural approach is to start with a measure-like function that satisfies the key properties on an algebra of relatively simple sets and then undertake some kind of limiting process to expand the domain and range of the protomeasure. We discuss that below. In the meantime, we consider some simple examples that are nonetheless illustrative. Definition Let be a non-empty set, =, and f Then, f determines a measure µ on by the formula : [0, ] any function. µ(e) = f (x). x E In the case of uncountable sums, one has to interpret the meaning of this definition. We avoid this and consider only the countable case. It is a good exercise to show that µ is σ-finite if and only if f (x) < for all x, and the set {x : f (x) > 0} is countable. If f (x) = 1 for all x, we call µ the counting measure. If there is a point x 0 with f (x 0 ) = 1 and f (x) = 0 for x x 0, then µ is called the point mass at x 0. Example Let be an uncountable set and let be the σ- algebra, {A : Ais countable or A c is countable}. Then, it is a good exercise to show that the set function, 1, A is countable, µ(a) = 0, A c is countable, is a finitely additive measure but is not countably additive.

11 4.2. Measure 59 Example Let be an infinite set, =, and define 0, A finite, µ(a) =, A infinite. It is a good exercise to show that µ is a finitely additive measure but it is not countably additive. As with the definition of σ- algebras, the assumptions for a measure have significant consequences. Theorem (Properties of measure). Let (,,µ) be a measure space. Then, 1. µ is finitely additive. 2. If A, B and A B then µ(a) µ(b). 3. If A, B and A B where µ(b) <, then µ(b A) = µ(b) µ(a). 4. If {A i } is a sequence of sets in, then µ A i µ(a i ). 5. If {A i } is a sequence of sets in, and A 1 A 2 A 3..., then µ A i = lim µ(a i ). i 6. If {A i } is a sequence of sets in such that A 1 A 2 A 3... and µ(a n ) < for some n +, then µ A i = lim µ(a i ). i Definition Property 2 is called monotonicity and Property 4 is called subadditivity. Property 5 is called continuity from below and Property 6 is called continuity from above. Proof. We prove in order. Result 1 Let A 1 = A and A i = for all i 2. Result 2 If A B, then µ(b) = µ(a) + µ(b A c ) µ(a). We use the fact that A and B A c are disjoint. Result 3 If µ(b) <, then µ(a) < and so we can subtract the µ(a) to obtain µ(b) µ(a) = µ(b A c ) = µ(b A). Result 4 Following Theorem 1.1.5, we construct a sequence of disjoint sets: B 1 = A 1 and j 1 B j = A j A i for j 2. Note that each µ B j µ Aj and A i = B i. Hence, µ A i = µ B i = µ(b i ) µ(a i ).

12 60 Chapter 4. Construction of a general measure structure Result 5 Set A 0 =. Then µ A i = µ A i \A i 1 = lim n n µ A i \A i 1 = lim n µ(a n ). Result 6 Let B j = A n A j for j > n, and B j = for j n. Then, B n+1 B n+2..., µ(a n ) = µ(b j ) + µ(a j ) for j > n, and B j = A n \ A j. By Result 5, µ(a n ) = µ j =n j =n+1 A j + lim i µ(b i ) = µ j =n j =n Since µ(a n ) <, we can subtract it from both sides. A j + lim i µ(an ) µ(a i ). The extra assumption in proving continuity from above is necessary. It is easy to make examples of decreasing sequences {A i } with µ(a i ) = for all i, but A i =. Remark 4.1. The proof of Result 2 shows that additive measures are monotone as well. The assumption of σ- finite implies that the space cannot have too much content. Theorem Let (,,µ) be a σ- finite measure space. Then, cannot contain an uncountable, disjoint collection of sets of positive measure. Proof. Let = {A α } α be the collection of subsets of such that for each α, µ(a α ) > 0. We show that is countable. There is a sequence of sets {B i } such that B i and µ(b i ) < for each i. For any A, A = (A B i ). For j, define i, j = A : µ(a B i ) > 1. j For all i, j, i, j and for any A there is a i, j such that A i, j. Hence, = i, j. i, j =1 The result follows if we show that i, j is finite for all indices. Consider a finite sequence of sets {C k } m in a k=1 i, j. Since, these sets are disjoint, m m m µ(c j k B i ) = µ (C k B i ) µ(b i ). k=1 k=1 Thus, m j µ(b i ), i.e. m is finite, and i, j is finite.

13 4.3. Sets of measure zero and completion of measure 61 References Exercises 4.3 Sets of measure zero and completion of measure The previous discussion hints at the importance of dealing with sets of measure zero. There is a technical issue about such sets that we settle in this section. Definition Let (,,µ) be a measure space. A set E with µ(e) = 0 is called a set of measure zero. If a statement about points x is true except for x in a set of measure zero, we say that the statement is true almost everywhere (a.e.). It is necessary to reconcile this definition with Definition introduced in Chapter 3 for Lebesgue measure. In that chapter, we built up the idea of measure based on measuring the size of intervals. Now, we are dealing with an abstract construction of measure. Note that if we prove there is a Lebesgue measure (!) and if A is a Lebesgue measurable set that is covered by a countable set of intervals {I i }, then Theorem implies µ (A) µ I i µ (I i ). Now, if we can make the sum on the right smaller than any given δ > 0 by a suitable choice of {I i }, then µ (A) = 0. So, Definition implies that Definition holds. The reverse implication requires some proof, and we discuss that later. First, recall this claim from Chapter 3. Its proof is a good exercise. Theorem Let (,,µ) be a measure space. A finite or countable union of sets of measure zero has measure zero. If A is a measurable set of measure zero then any measurable subset of A has measure zero. However, a subset of a measurable set of measure zero is not necessarily measurable! Example Consider a set, the σ- algebra {,} and the measure µ that is zero on this σ- algebra. Then µ is not defined on any proper, non-empty subset of. We resolve this annoying point by adding all those subsets of sets in a σ- algebra that have measure 0 to the σ- algebra. Definition If (,,µ) is a measure space such that contains all subsets of sets in with measure 0, then (,,µ) is complete. Being complete eliminates some annoying issues and it can always be obtained by enlarging the domain of a given measure to obtain an equivalent measure in the following sense:

14 62 Chapter 4. Construction of a general measure structure Theorem (Completion of a measure). = {N : µ(n) = 0}. Define, Let (,,µ) be a measure space and let = {A B : A and B N for some N }. Then, is a σ- algebra on that contains. Moreover, the unique measure µ on defined by µ(e F ) = µ(e) for all E makes,,µ a complete measure space. We are literally adding all the subsets of measurable sets of measure zero to the σ- algebra. Definition µ is an example of an extension of µ from to. This particular extension is called the completion of µ and is the completion of with respect to µ. Proof. Clearly so we show that is a σ- algebra. Let C, so C = A B with A and B N for some N. Since B N, N c B c and B c is equal to the disjoint union B c = N c N B. This implies, C c = (A c N c ) (A c N B c ). Now, (A c N c ) since both E, N. Also, (A c N B c ) N. Hence, C c. Let {C i } be a sequence of sets in. For each i, C i = A i B i with A i and B i N i for some N i. Thus, where A i and C i = B i A i. Thus is a σ- algebra. A i B i, N i. From Theorem 4.3.1, N i and hence We next verify that µ(a) is a well defined function. If C can be written as C = A 1 B 1 = A 2 B 2 with A i and B i N i for some N i, then we want to show that µ(a 1 ) = µ(a 2 ). To see this note that A 1 A 1 B 1 = A 2 B 2 A 2 N 2. Thus µ(a 1 ) µ(a 2 ) + µ(b 2 ) = µ(a 2 ). Similarly, µ(a 2 ) µ(a 1 ). To show that,,µ is complete, we show that if C satisfies µ(c ) = 0, then D for any D C. With C = A B with A and B N for some N, we have µ(a) = µ(c ) = 0. This implies A. But, D = D where and D A B. Thus D. Example Returning to Example 4.3.1, to complete we add all subsets of to the σ- algebra. Thus, =.

15 4.4. Construction of measures 63 References Exercises 4.4 Construction of measures We have developed the basic properties of a measure on a σ- algebra, but we have presented only illustrative examples. In this section, we develop a systematic method for constructing a measure based on specifying the values of the measure on a class of simple sets. This is the same idea that drove the intuitive development of measure in Chapter 3, which was based on defining the measure of an interval I = (a, b] to be µ(i ) = b a n and for a finite union of disjoint intervals {I i }, µ I i = n µ(i i ). That is actually sufficient for the probability computations in Chapter 3. But, the discussion in Chapter 3 also hints that even in one dimension we are likely to want to consider sets that are more complicated that just finite collections of disjoint intervals, e.g. recall the Cantor set and the set of non-normal numbers. Working in higher dimensions introduces the possibility of further complications in boundaries and interior structure of sets. The technical difficulties involved in construction of measure, as well as computing its values, arise mainly from these complications. There are various ways to approach construction of measures that appear to be quite different, though they all end up with the same result. The technical challenges involved with measure means that all of the approaches have aspects that are not very intuitive. We describe the outer measure approach developed by Carathéodory based on Lebesgue s development. But, we first briefly describe an earlier approach due to Peano and Jordan that developed the ancient idea of approximating complex geometric shapes with simple shapes. Definition An Jordan elementary set in n is a finite disjoint union of n-dimensional cubes. We define the measure µ(r) of n-dimensional cube R to be its Euclidean volume and m the measure of a finite disjoint collection of n-dimensional cubes {R i } m to be µ R i = m µ(r i ). A set A n is Jordan measurable if for any δ > 0, there are elementary sets A δ and A δ with A δ A A δ and µ A δ \ A δ < δ. This uses the fact that the set difference of two Jordan elementary sets is another Jordan elementary set. The idea of this definition is that if we take a sequence of δ 0, we obtain a sequence of elementary sets {A δ, A δ } such that the Jordan measures of A δ and A δ converge to the same number. We can think of µ(a δ ) and µ A δ as the inner and outer content of A. We define the common limit to be the Jordan measure of A. This is a good idea and it can be used to improve the Riemann integral in particular. But it has the problem that the collection of Jordan measurable sets is too restrictive. For example, it is a good exercise to show that the set of rational numbers in the unit interval I is not Jordan measurable. Lebesgue s development aimed at preserving the properties of the Jordan measure, though extended to countable disjoint collections of Jordan measurable sets. But, it altered the Jordan approach to set approximation. It defined an outer measure µ (A) of any set A n to be the infimum of the measures of countable covers of A consisting of Jordan elementary sets. Lebesgue s definition of measurability can be expressed as saying

16 64 Chapter 4. Construction of a general measure structure that for any δ > 0, there is a Jordan elementary set A δ such that µ (A A δ ) < δ. A notion of inner measure can also be defined using the outer measure of the complement of A. Since no approximation of a set by included elementary sets is involved, this approach greatly increases the size of the collection of measurable sets. Indeed, the construction of non-measurable sets becomes quite difficult and depends on some set axioms. We chose to use Carathéodory s generalization of Lebesgue s approach because it provides a way to construct measures on a wide variety of spaces Outer measure We use an approach similar to that used for the original definition of a set of measure zero Def , which employs countable covers. Definition If is a non-empty set, an outer measure on is a function µ [0, ] such that: 1. µ () = 0; 2. For any A B, µ (A) µ (B); Monotonicity 3. For any sequence {A} i in, µ A i µ (A i ). Subadditivity : Note that an outer measure is defined on all of, which is the largest σ- algebra on. So, provided we prove that outer measures exist, they place no restrictions on the subsets. But, this is also a disadvantage in the sense that computationally, we prefer to start with a function defined on some smaller collection of sets and build up the values on more complicated sets as opposed to having to specify values for every subset of. Theorem Assume that be a non-empty set, a non-empty family of subsets of such that,, and there is a set function f : [0, ] such that f () = 0. For any subset A, define µ (A) = µ f (A) = inf Then, µ f is an outer measure. See Fig f (A i ) : {A i } and A A i. (4.1) Definition We say that the outer measure µ in Theorem is induced by f. We use f the subscript f only when it is important to indicate the function f. Proof. We begin by showing the infimum exists. The collection of countable covers in (4.1) is not empty since and covers A. We are computing an infimum over positive real numbers that are bounded below by 0, so the infimum exists. Next, µ () = 0 because is contained in a countable collection of empty sets and f () = 0. Next, we verify monotonicity. If A B, then µ (A) µ (B) since the collection of countable covers over which the infimum is computed for A includes the collection of countable covers of B.

17 4.4. Construction of measures 65 A A A Figure 4.1. Illustration of computing the infimum for the outer measure for a set in 2 in the case that the set function f gives the area of a square. Let {A i } be a sequence of sets in. Given ε > 0, for each i there is a sequence of sets {B i, j } j =1 such that A i j =1 B i, j and f (B i, j ) µ (A i ) + ε2 i. j =1 If A = A i, then A i, j =1 B i, j, and f (B i, j ) = f (B i, j ) µ (A i ) + 2 i ε µ (A i ) + ε. i, j =1 j =1 Since ε is arbitrary, the result follows. Proof Comment 4.1. Analogs of the trick of specifying 2 i ε for the accuracy of each successive member of a countable cover of A i is employed frequently in measure theory proofs. Theorem implies Theorem Any measure whose associated σ- algebra is is an outer measure. Theorem suggests there are many ways to construct an outer measure corresponding to many possible choices of f. Moreover, as noted, outer measures have the advantage that they are defined on all of. Example It is a good exercise to show that if = and f is defined as f ([a, b]) = b a for a, b and f () = 0, then the corresponding outer measure satisfies µ ([a, b]) = b a f for any a, b. This example makes a subtle point. The outer measure µ f induced by a set function f is not the same as f in general. After all, the outer measure is defined on all subsets of, not just the family on which f is defined. This example indicates that induced outer measure reduces to f on in the special case of lengths of intervals in. One disadvantage of outer measures is that there is no assumption of countable additivity. The next step is to identify a subset of on which an outer measure satisfies additional properties that make it into a measure. There are several ways to do this. We use Carathédory s approach.

18 66 Chapter 4. Construction of a general measure structure Definition (Carathédory s Condition). set A is µ - measurable if Given an outer measure µ on a set, a µ (E) = µ (E A) + µ (E A c ), (4.2) for all E. Note that the sets E used to test (4.2) are not restricted to µ - measurable sets. Below, we use E to denote the testing sets in order to make things a little more readable. There is no other significance to the choice of label. Any set A together with its complement A c forms a decomposition of in the sense that if E then E is equal to the disjoint union (E A) (E A c ). Therefore, A is µ - measurable if µ is additive with respect to all disjoint unions formed using A. On one hand, µ (E A) is an outer measure of the part of E that lies inside A. By varying E, we can measure different parts of the interior of A. On the other hand, µ (E) µ (E A c ) can be interpreted as another way to compute the outer measure of the interior of A. It is the outer measure of that part of E that does not lie outside A. So, (4.2) is the requirement that these ways to compute the outer measure of part of the interior of A should match. Example Let be an infinite set and define, A µ (A) = if A is finite, if A is infinite, Then, we claim that µ is an outer measure and all finite sets are µ -measurable. Clearly µ : [0, ] and µ () = 0. For monotonicity: consider that if A B and both are finite we have A B and so µ (A) µ (B). On the other hand, if either set is infinite then monotonicity also holds. Consider a sequence {A i } of sets in. If any of the A i is infinite then µ A i = = µ (A i ). Here we treat terms like + to be equal to. If all A i are finite then there are two cases. Either A i is finite or it is infinite. (For example, A i = i, then A i is infinite if A i are pairwise disjoint). If A i is finite then we know that A i A i, so the result n 1 follows. If A i is infinite, then we choose a sequence of disjoint finite sets B n = A n A i, so A i = B i. Thus, µ B i = µ (B i ) =, since we have finite disjoint sets, but µ (B i ) µ (A i ). Hence, µ A i µ (A i ). The rest is an exercise. We want to build a measure out of an outer measure. First a useful result: Theorem Let be a nonempty set and µ an outer measure. A set Ais µ - measurable if and only if µ (E) µ (E A) + µ (E A c ), (4.3) for any subset E.

19 4.4. Construction of measures 67 The point is that we only have to check (4.3) not (4.2). Proof. Since µ is sub-additive, we always have µ (E) µ (E A) + µ (E A c ). We usually say measurable instead of µ - measurable if µ is clear from the context. Theorem (Carathéodory Theorem). Let be a non-empty set with an outer measure µ defined on. The collection = {A : Ais µ - measurable} of µ - measurable sets is a σ- algebra and the restriction µ of µ to is a measure. Thus,,,µ is a measure space. Proof. is nonempty. We show that any set Aof outer measure 0 is outer measurable so since µ () = 0. Choose E. Since E E A c, µ (E) E A c. Similarly, A E A, so µ (A) E A. Addition gives µ (E) + µ (A) = µ (E) µ (E A) + µ (E A c ), since µ (A) = 0. If we show that is a σ- algebra, this argument also shows that it is complete. is an algebra. The definition of outer measurability is symmetric with respect to a set and its complement, so is closed under complements. Let A 1, A 2. Then for any E 1, E 2, µ (E 1 ) = µ (E 1 A 1 ) + µ (E 1 A c 1 ), µ (E 2 ) = µ (E 2 A 2 ) + µ (E 2 A c 2 ). For E, set E 1 = E and E 2 = E A c 1 to get or, µ (E) = µ (E A 1 ) + µ (E A c 1 ) µ (E A c 1 ) = µ (E A c 1 A 2 ) + µ (E A c 1 Ac 2 ). µ (E) = µ (E A 1 ) + µ (E A c 1 A 2 ) + µ (E (A 1 A 2 ) c ). Now, E (A 1 A 2 ) = (E A 1 ) (E A c 1 A 2 ) is a disjoint union, so Hence, µ (E A 1 ) + µ (E A 1 c A 2 ) µ (E (A 1 A 2 )). µ (E) µ (E (A 1 A 2 )) + µ (E (A 1 A 2 ) c ). Since E is an arbitrary set in, induction shows that µ is finitely additive on. is a σ- algebra. We show that is closed under countable disjoint unions, so the result follows from Theorem Let {A i } be a disjoint sequence in and set B n = n A i and B = A i. For E, 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, n µ (E B n ) = µ (E A i ).

20 68 Chapter 4. Construction of a general measure structure Thus, µ (E) = µ (E B n ) + µ (E B c n ) = n µ (E A i ) + µ (E B c n ). (4.4) Now {B n } is an increasing sequence of sets, so {Bn c } is a decreasing sequence of sets. Taking the limit as n in (4.4) (noting that all the terms are nonnegative!), we obtain µ (E) µ (E A i ) + µ (E B c ) µ A i ) + µ (E (E B c ) = µ (E B) + µ (E B c ) µ (E). All the inequalities are therefore equalities, so B. Taking E = B yields, µ A i = µ (A i ). is complete. Choose A with µ(a) = 0 and let B A. Since, B, µ (B) µ (A) = µ(a) = 0. As proved above, this implies that B and µ(b) = Metric outer measures Before continuing with the construction of measures, we present an important example of outer measure. In this section, we assume that is a nonempty metric space with metric d. Recall that the class of open sets (the topology) is centrally important to metric spaces. Definition The σ- algebra generated by the open sets of is called the Borel σ- algebra and is denoted =. The members of are called Borel sets. We could just as well use the closed sets of to generate. The Borel sets include countable unions and intersections of open and closed sets, and countable unions and intersections of those sets, and so on. Definition In reference to the Borel σ- algebra of a metric space, a countable intersection of open sets is called a G δ set, a countable union of closed sets is called a F σ set, a countable intersection of G δ sets is called a G δδ set, a countable union of G δ sets is called a G δσ set, a countable intersection of F σ sets is called a F σδ set, and so on. Note that taking countable unions, intersections and complements starting with open (or closed) sets does not generate all the members of. The point is that these set operations can be repeated in an unlimited fashion, not simply a countable number of times. We see below that the Borel σ- algebra is centrally important to the construction of a measure on n. The existence of a metric makes it possible to separate sets.

21 4.4. Construction of measures 69 G A A 1 A 2 1 1/2 A 3 1/3 G c X Figure 4.2. Illustration of computing a metric outer measure from the inside. Definition Let A, B. The distance between A and B is defined, d(a, B) = inf{d(a, b) : a A, b B}. This is well defined since d is nonnegative. In the next definition, we add a condition to the definition of an outer measure. Definition Let µ be an outer measure on. We say that µ is a metric outer measure if µ (A B) = µ (A) + µ (B), (4.5) for all A, B with d(a, B) > 0. Note that if d(a, B) > 0, then A B =. But more than that, the two sets are well separated. This separation should mean that we can compute the outer measure of the union by summing the outer measures of each set. Before stating the main result, we prove a theorem by Carathédory that discusses the approximation of outer measure from within, see Fig Theorem Let µ be a metric outer measure on, let G be open, and assume A G. For each i 1, set A i = {x A : d(x, G c ) 1/i}. Then, lim i µ (A i ) = µ (A). Proof. Since A i is an increasing sequence and A i A for all i, we just have to show that limµ (A i ) µ (A). Each point of A is an interior point of G therefore since A A i, so each point of A must belong to A i for i sufficiently large. Thus, A A i or A = A i. Set B i = A i+1 \ A i for i 1. For each n, A = A 2n B i = A 2n B 2i B 2i+1. i=2n i=n i=n

22 70 Chapter 4. Construction of a general measure structure Therefore, µ (A) µ (A 2n ) + µ (B 2i ) + µ (B 2i+1 ). (4.6) i=n If both of the series in (4.6) converge, then letting n and noting that limµ (A 2n ) = limµ (A n ), shows µ (A) limµ (A n ). Otherwise, at least one of the series diverges. Without loss of generality, assume the first series in (4.6) diverges. Since d(b 2i, B 2i+2 ) 1 2i+1 1 2i+2 > 0, So, limµ (A n ) = µ (A). i=n n 1 n 1 µ (A 2n ) µ B 2i = µ (B 2i ) as n. Theorem Let be a metric space with metric d. µ is a metric outer measure if and only if every Borel set is µ -measurable. Proof. First, we assume µ is a metric outer measure and show that every closed set F is µ -measurable. Let E. E \ F is contained in the open set F c, so there is a sequence {A i } of subsets of E \ F such that d(a i, F ) 1/i and limµ (A i ) = µ (E \ F ). Therefore, µ (E) µ ((E F ) A i ) = µ (E F ) + µ (A i ) µ (E F ) + µ (E \ F ). Now we assume that every Borel set is µ -measurable. Let A, B satisfy d(a, B) > 0. Choose an open set G A such that G B =. By assumption, G is µ -measurable, so µ (A B) = µ ((A B) G) + µ ((A B) \ G) = µ (A) + µ (B) Premeasures Now we use Carathédory s theorem to devise a process for measuring the size of sets in a large class by using the sizes of a simple collection of sets. For example, later we build a measure for general sets of n by starting with the volume of the n-dimensional cube. Specifically, we use Carathédory s Theorem to extend measures defined on an algebra to a specific σ- algebra. Definition Let be a non-empty set and let be an algebra. A set function µ 0 : [0, ] is a pre-measure if 1. µ 0 () = If {A i } is a sequence of disjoint sets with A i, then µ 0 A i = µ 0 (A i )

23 4.4. Construction of measures 71 Note that we have to assume that A i. It is a good idea to go back and study the definition of outer measure before proceeding. Remark 4.2. It follows that pre-measures are additive and monotone. As with measures, we distinguish two kinds of domains: Definition If is a non-empty set, an algebra, and µ 0 a pre-measure, then µ 0 is a finite pre-measure if µ 0 () < and a σ-finite pre-measure if there is an increasing sequence {A i } with = A i and µ 0 (A i ) < for all i. By Theorem 4.4.1, if µ 0 is a pre-measure on an algebra, then it induces an outer measure via µ (A) = inf µ 0 (A i ) : Aj, A A j, A. (4.7) j =1 Theorem Let be a nonempty set, be an algebra, and µ 0 be a pre-measure on. If µ is the induced outer measure (4.7), then 1. µ (A) = µ 0 (A), A. 2. Every set in is µ - measurable. j =1 Proof. We prove in order. Result 1 We show µ µ 0 and µ 0 µ. If A and A A i with {A i }, define B n = A A n \ n 1 A i, n 1. {B i } is a disjoint sequence whose union is A, and B i A i, so µ 0 (A) = µ 0 (B i ) µ 0 (A i ). Therefore, µ 0 (A) µ (A). However, A A i, where A 1 = A and A i = for i 2. So µ (A) µ 0 (A). Result 2 Let E. For ε > 0, there is a collection {A i } such that E A i and, µ 0 (A i ) < µ (E) + ε. On the other hand, for any E, A i = (A i E c ) (A i E) is a disjoint union of sets in, and µ 0 (A i ) = µ 0 (A i E c ) + µ 0 (A i E). Thus, µ (E) + ε µ 0 (A i E) + µ 0 (A i E c ) µ (A E) + µ (A E c ).

24 72 Chapter 4. Construction of a general measure structure Since ε was arbitrary, we get the result. Example Let = {1,2,3}, = {,{1},{1,2},{1,2,3}} and let µ 0 with, : [0, ] µ 0 () = 0, µ 0 ({1}) = 2, µ 0 ({1,2}) = 1. µ 0 ({1,2,3}) = 3. Then, µ 0 is not a pre-measure because is not an algebra. The outer measure generated by µ 0 has values µ () = 0, since and ρ() = 0. µ ({1}) = 1, since {1} {1,2} and that has the lowest value. µ ({2}) = 1. µ ({3}) = 3, since {3} {1,2,3}. µ ({1,2}) = 1. µ ({2,3}) = µ ({1,3}) = 3. µ ({1,2,3}) = 3. Checking case by case shows that the only µ - measurable set is the. For example, if we try A = {1}, choose E = 1,2,3, then µ (E) = 3 while, µ (A E) + µ (A c E) = µ ({1}) + µ ({2,3}) = = 4. Example Let = {1,2,3}, = {,{1},{2,3},{1,2,3}} and let µ 0 with, : [0, ] µ 0 () = 0, µ 0 ({1}) = 2, µ 0 ({2,3}) = 3. µ 0 ({1,2,3}) = 5. Then, µ 0 is a pre-measure. The outer measure generated by µ 0 has values µ () = 0. µ ({1}) = 2. µ ({2}) = µ ({3}) = 3. µ ({1,2}) = µ ({1,3}) = 5.

25 4.4. Construction of measures 73 µ ({2,3}) = 3. µ ({1,2,3}) = 5. All the sets in are µ - measurable. The set {2} is not measurable since µ ({2,3}) = 3 while µ ({2} {2,3}) + µ ({1,3} {2,3}) = 6. Now we are ready to construct a measure by starting with a pre-measure on an algebra, inducing an outer measure, and then restricting the outer measure to get a measure. Theorem (Hahn-Kolmogorov Extension Theorem). Let be a non-empty set, be an algebra, and µ 0 be a σ finite pre-measure defined on. Then, there exists a unique measure µ on σ ( ) whose restriction to is µ 0. Proof. Existence of the extension We have done all the hard work here. We begin by extending the pre-measure µ 0 on to the outer measure µ on all of using Theorem and Theorem Then, we restrict µ to obtain a measure µ on the µ -measurable sets σ- algebra using Theorem Since contains, it contains σ ( ). Thus, we have obtained a measure µ on σ ( ) such that, µ = µ σ( ) and µ = µ = µ 0. Uniqueness Assume that ν is another measure on σ ( ) that extends µ 0. We begin by showing that ν(a) µ(a) for any A σ ( ). First, note that ν = µ 0 = µ, so for A, ν(a) = µ 0 (A) = µ(a). However, if {A i } is a collection of sets in, then we know that A = A i is in σ( ), but we do not know if A is in. So, it is not immediately clear that ν(a) = µ(a). For this, we use the fact that n A i is an increasing sequence of sets in. n=1 Since ν and µ are measures, we use monotonicity (Theorem 4.2.1) to conclude, n n ν(a) = lim ν A n i = lim µ A n i = µ(a). Now choose A σ ( ) and {A i } with A A i. Then, ν(a) ν A i ν(a i ) = µ 0 (A i ). But this means that, ν(a) inf µ 0 (A i ) : A A i = µ(a), since A σ (). Next, we show that if in addition µ(a) <, then ν(a) = µ(a) for any A σ ( ). Choose A σ ( ) and, for ε > 0, choose {A i } with A A i so that µ 0 (A i ) <

26 74 Chapter 4. Construction of a general measure structure µ(a) + ε. Thus, µ A i \ A < ε since µ(a) <. This implies µ(a) µ A i = ν A i = ν(a) + ν A i \ A Since ε is arbitrary, ν(a) = µ(a). ν(a) + µ A i \ A ν(a) + ε. Finally, we assume that µ 0 is σ- finite, so that = A i with µ 0 (A i ) <. We can assume that {A i } is disjoint (by the argument that should now be familiar). For any A σ ( ), A is given by the disjoint union, A = A A i. Hence, µ(a) = µ(a A i ) = ν(a A i ) = ν(a). Proof Comment 4.2. Note how we approached the uniqueness by assuming the existence of another object, then showing it has to be the same as the original object A zoology of measure creatures We tabulate the properties of all the measures we have seen in the Table 4.1. Table 4.1. Properties of measure, pre-measure and outer-measure Properties Pre-Measure µ 0 Outer-Measure µ Measure µ domain algebra σ- algebra monotonicity finite-additivity - finite-sub-additivity countable-additivity - - countable-sub-additivity -