A := A i : {A i } S. is an algebra. The same object is obtained when the union in required to be disjoint.

Transcription

1 59 6. ABSTRACT MEASURE THEORY Having developed the Lebesgue integral with respect to the general easures, we now have a general concept with few specific exaples to actually test it on. Indeed, so far we have constructed only one easure; naely, the Lebesgue easure on R d.) In this chapter we will end this by providing tools how to construct easures on abstract easurable spaces. As for the Lebesgue easure, the key tool will be outer easures. 6.1 Fro seialgebras to outer easures. Consider an arbitrary set X. Our goal is to develop structures needed for putting a easure on X. For this, let us recall how the construction of the Lebesgue easure went: We began by defining the easure of half-open boxes. Then we extended this to eleentary sets, which were finite unions of half-open boxes. Next, we constructed the Lebesgue outer easure by coverings by countable collections of half-open boxes. Finally, we used topology to define easurable sets and checked that the outer easure restricted to these is actually a easure. We will proceed along this plan even for a general set X except, in the absence of topology, we will define easurability differently. As is easy to check, the class of half-open boxes in R d fors the following structure: Definition 6.1 A class S 2 X is a seialgebra on X if it satisfies the following requireents: 1) /0,X 2 S, 2) A,B 2 S iplies A \ B 2 S, 3) if A 2 S, then there are disjoint B 1,...,B n 2 S such that n[ A c = B i. 6.1) Recall that an algebra is a subset of 2 X that contains /0 and X and is closed under finite unions and copleents. Moving up to our generalization of eleentary sets, we observe: Lea 6.2 If S is a seialgebra, then n [ n o A := A i : {A i } S is an algebra. The sae object is obtained when the union in required to be disjoint. Proof. Let A 1,...,A n S. By the property 3) of the seialgebra, for each i = 1,...,n there are {B i, j : j = 1,...,} S such that [ A c i = B i, j. 6.3) j=1 We can always add epty sets to ake the union the sae size for all i.) Fro de Morgan s laws and the distributive property of union around intersection we then get [ n c n\ n\ [ [ [ A i = A c \ n i = B i, j = B i, ji. 6.4) j=1 j 1 =1 j n =1 6.2)

2 60 But T n B i, ji 2 S by property 2) of the seialgebra and so S n A i 2 A. It follows that A is closed under copleents. As A is obviously closed under finite unions and contains /0 and X because it includes S which contains these) it is an algebra. To see that we can define the algebra using disjoint unions, let A = S n A i with {A i } S disjoint and let B 2 S. By property 3) of the seialgebra, there are disjoint sets {B j } S such that B c = S j=1 B j. Then B [ A = B [ A \ B c )=B [ n[ j=1 [ A i \ B j ). 6.5) Now {A i \B j } are disjoint and they are being subsets of B c ) disjoint fro B. Hence, if any union of n sets fro S can be disjointified, also any union of n + 1 sets fro S can be disjointified. By induction, A can be defined by disjoint unions as well. Now consider a set function µ : S! R [ {± }, which we perit, for the tie being, to take both positive and negative and even infinite) values. Our goal is to extend it to a finitely additive set function on A in accord with the following definition: Definition 6.3 A function µ : A! R [ {± } on an algebra A is finitely additive if for any disjoint {A i } A, [ n n µ A i = µa i ) 6.6) and the su on the right-hand side akes sense. The only way the su would not ake sense is that both + and appears aong soe of the ters. If this happens for disjoint sets then, indeed, we have a proble indeed. That disjointness does not provide enough protection for both infinities to coexists is the content of: Lea 6.4 If µ is a finitely additive set function on an algebra, then µ cannot take both + and there. Proof. Let A,B be sets in the algebra such that µa)= and µb)=. Then µa)=µa \ B)+µA r B) and µb)=µa \ B)+µB r A) 6.7) with both sus on the right aking sense. This iplies that µa \ B) ust be finite while µa r B)= and µb r A)=. But then µ is certainly not additive on the disjoint union of A r B and B r A as that would lead to a eaningless expression. Thus we ay assue fro the start that µ takes at ost one of the infinities. There is, however, still another proble for the extension fro S to A : Finite additivity ay be violated already on S. That this is the only obstruction is the content of: Lea 6.5 Extension to algebra) Let S 2 X be a seialgebra and let µ : S! R [{ } be a set function that is finitely additive on S in the sense that 8{A i } S disjoint: n[ [ n n A i 2 S ) µ A i = µa i ). 6.8)

3 61 Then for any two disjoint failies {A i } S and {B j } S, [ A i = n[ B j ) j=1 µa i )= n µb j ). 6.9) j=1 Under such conditions µ extends uniquely to a finitely additive set function on A. Proof. Let {A i } S and {B j } S be two disjoint failies with S A i = S n j=1 B j. Note that {A i \B j } are disjoint. Then A i = S n j=1 A i \B j ) and B j = S A i \B j ) and so, by the assuption, Then µa i )= µa i )= n j=1 µa i ) and µb j )= µa i \ B j )= n j=1 n µb j ). 6.10) j=1 µa i \ B j )= n µb j ) 6.11) j=1 thus proving the first part of the clai. By Lea 5.2, every set A 2 A can be written as a disjoint union S n A i for {A i } S.We then define µa) := µa i ) 6.12) and note that, by the first part of the clai, this does not depend on the representation of A. If now B 1,...,B 2 A are disjoint sets and B i = S n j=1 B ij is their representation as a union of a disjoint faily {B ij : j = 1,...,n} S again, we use epty sets to ake all these unions have the sae nuber of ters), we have [ [ n[ n µ B i = µ B ij = µb ij )= j=1 j=1 [ n µ j=1 B ij = µb i ). 6.13) Here in the first equality we used that by disjointness of {B i }) the whole collection {B ij : i = 1,...,, j = 1,...,n} consists of pairwise disjoint sets, then in the second equality) we applied the definition 6.12) and then in the third equality) we used 6.12) again to contract the second su into a union. As 6.13) is a stateent of finite additivity of µ on A, we are done. Keeping in ind our plan fro the construction of the Lebesgue easure, we now go back to non-negative set functions only and axioatize the properties fro Proposition 3.2: Definition 6.6 Outer easure) Given a set X, an outer easure on X is a ap µ? :2 X! [0, ] satisfying the following properties: 1) µ? /0 )=0, 2) onotonicity) if A B, then µ? A) apple µ? B), 3) countable subadditivity) for any sets {A i : i 2 N}, µ? [ A i apple µ? A i ). 6.14) An iitation of the construction of the Lebesgue outer easure yields:

4 62 Proposition 6.7 is an outer easure on X. Let C 2 X obey /0 2 C and let µ 0 : C! [0, ] satisfy µ 0 C )=0. Then µ? A) := inf µ 0 A i ): {A i } C, [ A i A 6.15) Proof. The proof is alost identical to that for the Lebesgue easure. First off, µ? 0 because µ 0 0. Hence, µ? /0 ) apple µ 0 /0 )=0. Next if {B i } C covers B then it also covers any A with A B and so µ? A) apple µ? B). Finally, if {A i } is a collection of sets with µ? A i ) < otherwise there is nothing to prove), then for each e > 0 we ay find {B ij } with A i j2nb [ ij and µ? A i )+e2 i µ 0 B ij ) 6.16) j2n valid for all i. But then S i, j2n B ij S A i and thus µ? [ A i apple µ 0 B ij )= i, j2n apple j2n µb ij ) µ? A i )+e2 i = e + µ? A i ) Since this holds for all e > 0, µ? is countably subadditive and thus an outer easure. 6.17) The class C class of sets in the above construction will soeties be referred to as the covering class. Thus, for the Lebesgue easure, the covering class where half-open boxes.) There are three natural questions that surround this construction: 1) Is µ? A)=µ 0 A) for sets in the covering class? 2) Is there a restriction of µ? to a s-algebra where it becoes a easure? 3) If 2) is answered affiratively, are the sets in the covering class easurable? In the forthcoing sections we will answers all these at various level of detail. 6.2 Carathéodory easurability and extension theore. When µ? X) <, a natural idea how to define easurable sets goes via the inner easure. Indeed, we could set µ? A) := µ? X) µ? X r A) 6.18) and try to define A to be easurable when µ? A)=µ? A). However, this is quickly found to run into a dead end. Indeed, consider the exaple of X := {1,2,3} with µ? given by µ? /0 )=0, µ? X)=3, µa)=1 for A 6= /0,X. 6.19) Then µ? A)=µ? A) for all sets A X and yet µ? is not a easure on 2 X e.g., for for B := {1,2} and A := {1} we have µ? B)=1 and µ? B \ A)+µ? B r A)=2. The oral of this exaple is that, for A to be easurable, it has to satisfy ore than just the condition µ? X)=µ? X \ A)+µ? X r A). As it turns out, we have to require that {A,A c } partitions not just X additively, but in fact all subsets thereof:

5 Definition 6.8 Carathéodory easurability) A X Carathéodory) easurable if Let µ? be an outer easure on a set X. We call 8E X : µ? E)=µ? E \ A)+µ? E r A). 6.20) We will write Sµ? ) to denote the class of all Carathéodory) easurable sets. Note that, thanks to the subadditivity of µ?, the nontrivial condition to check is that The principal result of this section is then: Theore 6.9 Carathéodory) 1) Sµ? ) is a s-algebra, and 2) µ? is a easure on Sµ? ). Proof. The proof coes in four steps. STEP 1: Sµ? ) is an algebra µ? E) µ? E \ A)+µ? E r A). 6.21) Let µ? be an outer easure on a set X. Then: Obviously, /0,X 2 Sµ? ) and if A 2 Sµ? ) then also A c 2 Sµ? ). Now let A,B 2 Sµ? ) and pick an arbitrary set E X. Then and so by subadditivity E r A [ B)=E \ A c \ B c and E \ A [ B)=E \ A) [ E \ A c \ B) 6.22) µ? E \ A [ B) + µ? E r A [ B) apple µ? E \ A)+µ? E \ A c \ B c )+µ? E \ A c \ B) = µ? E \ A)+µ? E \ A c )=µ? E) ) where the first equality is because A 2 Sµ? ) and the second is due to B 2 Sµ? ). This shows 6.21) and thus 6.20) for A [ B. Hence, Sµ? ) is an algebra. STEP 2: For any A 1,...,A n 2 Sµ? ) disjoint and any E X, µ? n[ n E \ A i = µ? E \ A i ). 6.24) In particular, µ? is finitely additive on Sµ? ). We observe that, since A n+1 2 Sµ? ) and A n+1 is disjoint fro S n A i, µ? n+1 [ E \ n+1 A i = µ? [ E \ = µ? E \ A n+1 )+µ? E \ n+1 A i \ A n+1 + µ? [ E \ n[ A i A i r A n ) Since the clai holds trivially for n = 1, the identity 6.24) follows by induction on n. Setting E := X then shows that µ? is finitely additive. STEP 3: Sµ? ) is a s-algebra

6 64 Let {A n : n 2 N} Sµ? ). Then S A n = S A 0 n where A 0 n := A n r S n 1 A i are now disjoint fro each other. But Sµ? ) is an algebra and so {A 0 n}2sµ? ) as well. Hence, we ay suppose that {A n } are disjoint. Then 6.24) gives µ? E) = STEP 1 µ? E \ n STEP 2 n[ A i + µ? n[ E r A i µ? E \ A i )+µ? E r [ 6.26) A i where we also used onotonicity of µ? to replace the finite union by infinite union in the second ter on the right. Taking n! and using subadditivity of µ? now shows µ? E) µ? E \ A i )+µ? E r [ A i µ? E \ [ A i + µ? E r [ A i 6.27) and so 6.21) holds for the countable union S A i. Hence Sµ? ) is closed under countable unions and, being an algebra, it is a s-algebra. STEP 4: µ? is a easure on Sµ? ) Let {A n } Sµ? ) be disjoint. Taking E := S A i in the previous arguent shows µ? [ A i µ? A i ) 6.28) and, by countable subadditivity of µ?, equality ust in fact hold here. Hence µ? is countably additive on Sµ? ) and, since µ? /0 )=0, it is a easure. Note that any A X with µ? A)=0 is autoatically Carathéodory easurable. As a consequence, the easure space X,Sµ? ), µ? ) is autoatically coplete see Definition 4.16). The Carathéodory theore copletely answers question 2) fro the end of the class section. To address the reaining two questions, and to ake practical use of Carathéodory extension easier, we state and prove: Theore 6.10 Hahn-Kologorov) Let A 2 X be an algebra and assue that µ 0 : A! [0, ] obeys µ 0 /0 )=0 and that the following holds: 1) finite additivity) A,B 2 A,A\ B = /0 iplies µ 0 A [ B)=µ 0 A)+µ 0 B), 2) countable subadditivity) {A i } A and S A i 2 A iply µ 0 S A i ) apple µ 0 A i ). Define the outer easure generated by the covering class A and set function µ 0 by µ? A) := inf µ 0 A i ): {A i } A, [ A i A. 6.29) Then In particular, µ? extends µ 0 to a easure on Sµ? ). Proof. The proof coes again in two steps. STEP 1: A Sµ? ) A Sµ? ) and µ? = µ 0 on A. 6.30)

7 Let A 2 A and pick any E X. We need to show 6.21). For this we ay assue µ? E) <. By 6.29), there are sets {B i } A such that E [ B i and µ? E)+e µ 0 B i ). 6.31) But µ 0 is finitely additive and so µb i )=µb i \ A)+µ 0 B i r A) and so µ? E)+e µ 0 B i )=µ 0 B i \ A)+ µ 0 B i r A) µ? [ B i \ A + µ? [ B i r A µ? E \ A)+µ? E r A), ) where we used {B i \ A} and {B i r A} cover their own unions and then applied 6.29) and the onotonicity of µ?. As e was arbitrary, we have A 2 Sµ? ). STEP 2: µ? = µ 0 on A Pick A 2 A. We have µ? A) apple µ 0 A) because /0 2 A and µ 0 /0 )=0. For the opposite inequality let {B i } A be any collection of sets such that A S B i. Then S B i \ A)=A and so, by the onotonicity and countable subadditivity of µ 0 on A, µ 0 B i ) µ 0 B i \ A) µ 0 A). 6.33) Taking infiu over all such {B i } gives µ? A) µ 0 A) as desired. To deonstrate the power of this theore, let us consider a couple of exaples. The first one will be quite failiar: A := A R d : Peano-Jordan easurable and µ 0 A) := ca). 6.34) Then 6.29) yields µ? = l? and so we have a definition of the Lebesgue easure on the class Sl? ) of Carathéodory easurable sets. However, the relation between Sl? ) and L R d ), and the values of l? thereupon, is unclear at this point we will return to this later). As another exaple, pick any set X and consider the collection of sets A := A X : finite or co-finite. 6.35) This is easily verified to be an algebra and so we set 0, if A finite, µ 0 A) := 1, if A co-finite. Then µ 0 is additive and countably additive on A. The definition 6.29) yields µ? 0, if A countably infinite, A) := 1, else 6.36) 6.37) and, as is again checked easily, Sµ? )={A X : countable or co-countable}. Further exaples will be produced in the sections on Lebesgue-Stieltjes easures and product spaces. In soe cases verifying the conditions of the Hahn-Kologorov theore in an algebra can be quite a nuisance. Invariably as we did in the context of the Lebesgue easure in Lea 3.4) this is usually easier at the level of seialgebras. We then need the following general fact:

8 66 Lea 6.11 Let S be a seialgebra and µ 0 : S! [0, ] obey µ 0 /0 )=0. If µ 0 is finitely additive and countably subadditive on S, then so is also its unique extension guaranteed by Lea 6.5) to the algebra A generated by taking all finite unions of sets fro S. Proof. That additivity of µ 0 on S iplies additivity on A was the subject of Lea 6.5. To show that countable additivity on S iplies countable additivity on S, we note that, since A is an algebra, it suffices to check this for disjoint unions. Let {A n : n 2 N} A be disjoint sets with A := S A n 2 A. By Lea 6.2, each A n is a finite disjoint union of sets {B nj : j = 1,...,kn)} and A is a finite disjoint union of sets {C i : i = 1,...,}. By the disjointness assuptions, and setting C i \ B nj 6= /0 ) B nj C i 6.38) S i := n, j): n 2 N, j = 1,...,kn), B nj C i 6.39) we thus have C i = [ B nj and µ 0 C i ) apple µ 0 B nj ) 6.40) n, j)2s i n, j)2s i where the second conclusion follows by countable subadditivity of µ 0 on S. On the other hand, A n can be written as the finite disjoint union [ [ A n = B nj 6.41) j : n, j)2s i and so finite additivity of µ 0 iplies Putting this together, we conclude as desired. µ 0 A)= 6.3 Uniqueness of extension. µa n )= µ 0 C i ) apple = j : n, j)2s i µ 0 B nj ). 6.42) n, j)2s i µ 0 B nj ) µ 0 B nj )= µ 0 A n ) j : n, j)2s i 6.43) Having constructed extensions of non-negative) additive set functions on algebras to easures on s-algebras, we need to address the question of uniqueness. As already noted, at this point we have two possibly equivalent) definitions of the Lebesgue easure: one on the Lebesgue s-algebra L R d ) of sets that can be approxiated, in the sense of outer easure, by open sets fro outside, and the other by way of Carathéodory easurability. A question is then how to relate these to each other. That the extension ay not be unique is in fact fairly easy to check by the following exaple. Let A be the algebra consisting of finite unions of all half-open intervals in R and consider the set

9 function µ 0 defined by µ 0 A)= if A 6= /0 and µ 0 /0 )=0. Now consider three easures on 2 R : first, the counting easure carda), if A finite, µ 1 A) := 6.44), else, second, the countability-detection easure 0, if A finite or countably infinite, µ 2 A) :=, else, and, third, the identically-infinite easure µ 3 A) := 0, if A = /0,, else. Since every non-epty set in A is autoatically infinite and uncountable, we have ) 6.46) µ 1 = µ 2 = µ 3 = µ 0 on A. 6.47) Yet, clearly, all three easures are different. Note also that all sets are of course Carathéodory easurable with respect to these easures, regarded as outer easures.) It is iportant to note the highlight the issues that underpin the conclusion of this exaple. First, either the class A is too sall or the class 2 X is too large to ake all its sets reachable by approxiations using sets fro A. So the relative size of the extension s-algebra copared to that of the generating algebra ust soehow atter. Also, the fact that the easures are infinite ust play a role because the exaples rely on the fact that reoval of a finite-easure set has no effect on the easure of an infinite-easure set. We will first address the fact that soe restrictions on the size of the target s-algebra ust be ade should the extension be unique. Noting the easy fact that an intersection of any nuber of s-algebras on X is a s-algebra on X, we introduce: Definition 6.12 For M 2 X, let be the least s-algebra containing M. As an exaple, define sm ) := \ F : s-algebra F M F 6.48) BR d ) := s {O R d : open} 6.49) to be the class of Borel sets in R d. Here the open sets are defined using the Euclidean etric.) As is not hard to check, the sae object will be obtained if we replace open sets by half-open boxes. Thus, BR d ) L R d ) \ Sl? ) 6.50) and, in fact, BR d ) is the least s-algebra on which the Lebesgue easure can be defined. It is obvious that, if we have a easure on a s-algebra, then it naturally restricts to a easure on any sub-s-algebra thereof. So if there is a hope to prove uniqueness at all, it should be priarily attepted on sa ). This is the conclusion of:

10 68 Theore 6.13 Uniqueness criterion) easures on sa ). Then Let A 2 X be an algebra and let µ and n be finite µ = n on A ) µ = n on sa ). 6.51) The sae is true even when µ and n are infinite provided that there are sets {X n } A such that [ X n = X and µx n )=nx n ) <, 8n 2 N. 6.52) The proof is abstract but also quite instructive as its ideas show up in any other reincarnations elsewhere. First we define: Definition 6.14 Monotone class) 1) 8{A n } M : A n " A iplies A 2 M, 2) 8{A n } M : A n # A iplies A 2 M. A class of sets M 2 X is a onotone class if In words, a onotone class is class that is closed under increasing and decreasing liits. It is easy to check that any s-algebra is a onotone class. The class of all intervals in R open, half-open and closed including degenerate ones is a onotone class but not a s- algebra. However, once a onotone class contains an algebra, it contains also the least s-algebra generated thereby: Lea 6.15 Monotone Class Lea) Let X be any set. If M 2 X is a onotone class and A 2 X an algebra, then A M ) sa ) M. 6.53) Proof. As is easy to check, the intersection of any nuber of onotone classes is a onotone class, so it akes sense to define \ A ) := M 6.54) M : onotone class M A Now sa ) is a onotone class containing A and so A ) sa ). Our goal is to show that sa ) A ) because that, by the iniality of A ), iplies the clai. First we note that /0,X 2 A ) by their containent in A. Also, since M 0 := {A c : A 2 M } 6.55) is a onotone class if M is a onotone class, we have A 2 A ) if and only if A c 2 A ). We will now show that A ) is closed under finite intersection. For this we let M A := C 2 A ): C \ A 2 A ). 6.56) Now we observe that, since A is itself closed under intersections and A ) is a inial onotone class containing A, But the conclusion can be rewritten as 8A 2 A : A M A and so A ) M A. 6.57) 8B 2 A ): A M B and so A ) M B. 6.58)

11 This eans that if A,B 2 A ), then also A \ B 2 A ), i.e., A ) is closed under finite intersections and, in light of the above, it is an algebra. But A ) is also a onotone class and so it is closed under countable intersections as well. It follows that A ) is a s-algebra containing A and so A ) sa ) by the iniality of sa ). Proof of Theore Let µ and n be finite easures on a s-algebra F. Then, by the Monotone Convergence Theore for sets, M := A 2 F : µa)=na) 6.59) is a onotone class. So if A is an algebra and µ = n on A, we have A M and, by the Monotonce Class Lea, sa ) M. Hence, µ = n on sa ). To address the infinite-easure case, consider the sets {X n }2A such that µx n )=nx n ) <. Without loss of generality take finite unions), we ay assue that X n " X. Define easures µ n A) := µa \ X n ) and n n A) := na \ X n ). 6.60) These are finite easures and µ n = n n on A. Hence, µ n = n n on sa ) by the first part of the clai. But µ n A) " µa) and n n A) " na) by the Monotone Convergence Theore for sets, and so µ = n on sa ) as well. 6.4 Lebesgue-Stieltjes easures. The Hahn-Kologorov Theore gives us a way to construct a large class of easures on the real line. Recall that BR) denotes the class of Borel sets in R see 6.49)). Then we have: Theore 6.16 Let F : R! R be a non-decreasing, right-continuous function. Then there exits a unique easure µ F on R,BR)) such that The key point is to show: 8a < b: µ F a,b] = Fb) Fa). 6.61) Lea 6.17 Let S := {a,b]: apple a apple b apple }. Then S is a seialgebra on R and is finitely additive and countably subadditive on S. µ 0 a,b] := Fb) Fa) 6.62) Proof. That S is a seialgebra is verified easily. It is also iediate that µ 0 is finitely additive see the first part of the proof of Lea 2.4). To prove countable additivity, thanks to Lea 6.11, we need to verify that for any disjoint collection {a n,b n ]: n 2 N} S and any a,b] 2 S, a,b]= [ a n,b n ] ) Fb) Fa) apple Fb n ) Fa n ). 6.63) By intersection with a finite interval we ay assue that a,b] is finite. Then Fb) Fa) < and so, by the right-continuity of F, given e > 0 there is a 0 > a so that Fa 0 ) apple Fa)+e. Siilarly, for each n 2 N, we can find b 0 n > b n so that Fb 0 n) apple Fb n )+e2 n. Since [a 0,b] a,b]= [ a n,b n ] [ a n,b 0 n) 6.64) 69

12 70 the open intervals {a n,b 0 n)} constitute a cover of a copact interval [a 0,b]. By the Heine-Borel Theore, there is a finite subcover. Let a 1,b1 0),...,a N,bn) 0 denote the intervals in a finite subcover so that each interval contains a point that is not contain by the other intervals. We ay assue that the intervals are labeled so that a 1 < a 2 < < a N, where the strict inequality ust hold by disjointness of {a n,b n ]}. Since each a i,bi 0 ) contains a point which does not lie in any other interval, we ust have a 1 < a 0, b N > b and b 0 i > a i+1, i = 1,...,N ) By the onotonicity of F, our choices of a 0 and b 0 n and 6.65) we thus have Fb) Fa) e apple Fb) Fa 0 ) apple Fb 0 N) Fa 1 ) = FbN) 0 N 1 Fa N )+ apple N apple Since e was arbitrary, we get 6.63). Fb 0 i ) Fa i ) Fb 0 n) Fa n ) apple e + Fa i+1 ) Fa i ) Fb n ) Fa n ). 6.66) Proof of Theore By Lea 6.11 and the Hahn-Kologorov Theore, µ 0 fro the previous lea extends to a easure on ss )=BR) which agrees with µ 0 on S, i.e., 6.61) holds. Since µ 0 n,n+1]) = Fn+1) Fn) < for all n 2 N, the second part of Theore 6.13 applies and so µ F is unique. We reark that the construction actually gives an extension to the s-algebra of Carathéodory easurable sets Sµ F? ). This s-algebra ay vary quite draatically depending on F. Here are soe natural exaples of easures that can be identified via the above construction: 1) if Fx) := x then µ F = l, the Lebesgue easure on BR), 2) if Fx) := 1 {x a} then µ F is the Dirac point ass at a, often denoted by d a, such that 1, if a 2 A, d a A) := 0, else. Note that Sµ? F )=2R in this case. 3) if Fx) := p i 1 {x ai } for soe {a i } R and p i 2 0, ) with p i <, then 6.67) µ F A)= p i d ai A), 6.68) i.e., µ F is a collection of point asses with ass at a i equal to p 1. Note that, in general, µ F will charge a singleton {a} if and only if F is discontinuous at a.) Again Sµ F?)=2R, i.e., all subsets of R are easurable. 4) if F is continuously differentiable, then Fb) Fa) = R b a F0 x)dx = R a,b] F0 dl and µ F is then the easure given by Z µ F A)= F 0 dl. 6.69) A

13 71 1 3/4 1/2 1/4 1/3 2/3 1 FIG 1. The Cantor-Lebesgue function associated with the iddle-third Cantor set arked on the x-axis). This function is soeties referred to as the devil s staircase in the physics literature. Note that, reversing this process gives us a way to define a eaning of f dl, i.e., ultiplying the Lebesgue easure by an integrable function. 5) F is the Cantor-Lebesgue function that is defined as follows: Let C n be the n-th iteration of the process of reoval of iddle-third intervals defining to the iddle-third Cantor set in C 0 :=[0,1]. Then C n 1 rc n is a collection of 2 n open intervals of length 3 n. Exaining these intervals left to right, F is constant on each of these and takes values in increasing non-negative odd ultiples of 2 n. See Fig. 6.4 Let C := T n 1C n be the iddle-third Cantor set. Then µ F [0,1] rc)=0 6.70) because this set is a union of open intervals on which F is piecewise constant. Yet µ F [0,1]) = F1) F0)=1 6.71) Since F is also continuous, µ F {a})=0 for all a. Note that lc)=0. The defining property of the easure µ F suggests a natural connection to the Rieann-Stieltjes integration. And, indeed, the Lebesgue integral Z f dµ g 6.72) [a,b] bears the sae connection to the Rieann-Stieltjes integral Z b f dg 6.73) a

14 72 as the Lebesgue integral with respect to the Lebesgue easure) bears to the Rieann integral. Thus, for instance, it can be checked that, for a non-decreasing, right-continuous function, the function f is Rieann-Stieltjes integrable with respect to dg on [a,b] if and only if f is µ g -a.e. continuous. Thanks to this connection, we have: Definition 6.18 A easure µ on R,BR)) is a Lebesgue-Stieltjes easure if there is a nondecreasing, right-continuous function F such that 6.61) holds for µ. Then, of course, µ = µ F.) The class of Lebesgue-Stieltjes easures is quite rich. Indeed, every easure µ on R, BR)) that assigns finite values to copact sets such Borel easures are called Radon easures is a Lebesgue-Stieltjes easure. The function F such that that µ = µ F is then given by µ[0,x]), if x 0, Fx) := 6.74) µx,0)), if x < 0. An exaple of a easure that is not of this kind is the counting easure of the rationals, carda \ Q), if A \ Q finite, µa) :=, else, Obviously, there is no F : R! R that could ake a unit upward jup at every rational. 6.75)