Lebesgue Integration Theory: Part I

Part V 17 Introduction: What are measures and why “measurable” sets

Lebesgue Integration Theory Deﬁnition 17.1 (Preliminary). A measure “on” a set [ is a function :2[ $ [0> 4] such that 1. (>)=0 Q 2. If Dl l=1 is a ﬁnite (Q?4) or countable (Q = 4) collection of subsets of {[ which} are pair-wise disjoint (i.e. D _ D = > if l = m) then l m 6 Q Q (^l=1Dl)= (Dl)= l=1 X Example 17.2. Suppose that [ is any set and { 5 [ is a point. For D [> let 1 if { 5 D (D)= { 0 if {@5 D= ½ Then = { is a measure on [ called the Dirac delta measure at {= Example 17.3. Suppose that is a measure on [ and A0> then is also a measure on [= Moreover, if are all measures on [> then· { }5M = 5M > i.e.

P (D)= (D) for all D [ X5M is a measure on [= (See Section 2 for the meaning of this sum.) To prove 4 this we must show that is countably additive. Suppose that Dl l=1 is a collection of pair-wise disjoint subsets of [> then { } 4 4 4 (^l=1Dl)= (Dl)= (Dl) l=1 l=1 5M X 4 X X 4 = (Dl)= (^l=1Dl) 5M l=1 5M X 4X X = (^l=1Dl) 246 17 Introduction: What are measures and why “measurable” sets 17.1 The problem with Lebesgue “measure” 247 wherein the third equality we used Theorem 4.22 and in the fourth we used that is to say hlQ is Q rotated counter clockwise by angle = We now claim that fact that is a measure. that is invariant under these rotations, i.e.

Example 17.4. Suppose that [ is a set : [ $ [0> 4] is a function. Then (}Q)=(Q) (17.2)

1 1 := ({){ for all } 5 V and Q V = To verify this, write Q = !(D) and } = !(w) for some w 5 [0> 1) and D [0> 1)= Then {X5[ is a measure, explicitly !(w)!(D)=!(w + D mod 1) (D)= ({) where for D [0> 1) and 5 [0> 1)> {X5D for all D [= w + D mod 1 := d + w mod 1 5 [0> 1) : d 5 Q { } =(d + D _ d?1 w ) ^ ((w 1) + D _ d 1 w ) = { } { } 17.1 The problem with Lebesgue “measure” Thus

So far all of the examples of measures given above are “counting” type mea- (!(w)!(D)) = (w + D mod 1) sures, i.e. a weighted count of the number of points in a set. We certainly are = ((d + D _ d?1 w ) ^ ((w 1) + D _ d 1 w )) going to want other types of measures too. In particular, it will be of great { } { } = ((d + D _ d?1 w )) + (((w 1) + D _ d 1 w )) interest to have a measure on R (called Lebesgue measure) which measures { } { } the “length” of a subset of = Unfortunately as the next theorem shows, there = (D _ d?1 w )+ (D _ d 1 w ) R { } { } is no such reasonable measure of length if we insist on measuring all subsets = ((D _ d?1 w ) ^ (D _ d 1 w )) of { } { } R= = (D)=(!(D))= Theorem 17.5. There is no measure :2R$[0> 4] such that Therefore it su!ces to prove that no finite non-trivial measure on V1 such 1. ([d> e)) = (e d) for all d?eand that Eq. (17.2) holds. To do this we will “construct” a non-measurable set 2. is translation invariant, i.e. (D + {)=(D) for all { 5 R and D 5 2R> Q = !(D) for some D [0> 1)= Let where U := } = hl2w : w 5 = } = hl2w : w 5 [0> 1) _ D + { := | + { : | 5 D R= Q Q { } { } { } In fact the theorem is still true even if (1) is replaced by the weaker con- — a countable subgroup of V1= As above U acts on V1 by rotations and divides dition that 0 ?((0> 1]) ? 4= V1 up into equivalence classes, where }>z 5 V1 are equivalent if } = uz for some u 5 U= Choose (using the axiom of choice) one representative point q The counting measure (D)=#(D) is translation invariant. However from each of these equivalence classes and let Q V1 be the set of these ((0> 1]) = 4 in this case and so does not satisfy condition 1. representative points. Then every point } 5 V1 may be uniquely written as Proof. First proof. Let us identify [0> 1) with the unit circle V1 := } 5 { } = qu with q 5 Q and u 5 U= That is to say C : } =1 by the map | | } V1 = (uQ) (17.3) !(w)=hl2w =(cos2w + l sin 2w) 5 V1 ua5U for w 5 [0> 1)= Using this identification we may use to define a function on where D is used to denote the union of pair-wise disjoint sets D = By 1 2V by (!(D)) = (D) for all D [0> 1)= This new function is a measure on Eqs. (17.2) and (17.3), { } V1 with the property that 0 ?((0> 1]) ? 4= For } 5 V1 and Q V1 let ` (V1)= (uQ)= (Q)= }Q := }q 5 V1 : q 5 Q > (17.1) { } uX5U uX5U 248 17 Introduction: What are measures and why “measurable” sets 17.1 The problem with Lebesgue “measure” 249

The right member from this equation is either 0 or 4, 0 if (Q)=0and 4 if 1=([0> 1)) = (Q u)= (Q) 1 (Q) A 0= In either case it is not equal (V ) 5 (0> 1)= Thus we have reached u5U u5U X X the desired contradiction. 4 if (Q) A 0 = = Proof. Second proof of Theorem 17.5. For Q [0> 1) and 5 [0> 1)> 0 if (Q)=0 let ½ which is certainly inconsistent. Incidentally we have just produced an example Q = Q + mod 1 of so called “non — measurable” set. = d + mod 1 5 [0> 1) : d 5 Q Because of Theorem 17.5, it is necessary to modify Deﬁnition 17.1. Theo- { } =( + Q _ d?1 ) ^ (( 1) + Q _ d 1 ) = rem 17.5 points out that we will have to give up the idea of trying to measure { } { } all subsets of R but only measure some sub-collections of “measurable” sets. Then This leads us to the notion of — algebra discussed in the next chapter. Our revised notion of a measure will appear in Deﬁnition 19.1 of Chapter 19 below. (Q )= ( + Q _ d?1 )+ (( 1) + Q _ d 1 ) { } { } = (Q _ d?1 )+ (Q _ d 1 ) { } { } = (Q _ d?1 ^ (Q _ d 1 )) { } { } = (Q)= (17.4)

We will now construct a bad set Q which coupled with Eq. (17.4) will lead to a contradiction. Set

T{ := { + u 5 R : u5 Q ={ + Q= { }

Notice that T{ _ T| = > implies that T{ = T|= Let = T{ : { 5 R —the 6 O { } orbit space of the Q action. For all D 5 choose i(D) 5 [0> 1@3) _ D1 and deﬁne Q = i( )= Then observe: O O 1. i(D)=i(E) implies that D _ E = > which implies that D = E so that i is injective. 6 2. = T : q 5 Q = O { q } Let U be the countable set,

U := Q _ [0> 1)= We now claim that

Q u _ Q v = > if u = v and (17.5) 6 u [0> 1) = ^u5UQ = (17.6)

u v Indeed, if { 5 Q _ Q = > then { = u + q mod 1 and { = v + q0 mod 1> then 6 qq0 5 Q> i.e. Tq = Tq0 .Thatistosay,q = i(Tq)=i(Tq0 )=q0 and hence that v = u mod 1> but v> u 5 [0> 1) implies that v = u= Furthermore, if { 5 [0> 1) u mod 1 and q := i(T{)> then { q = u 5 Q and { 5 Q = Now that we have constructed Q> we are ready for the contradiction. By Equations (17.4—17.6) we ﬁnd 1 We have used the Axiom of choice here, i.e. DM (D K [0> 1@3]) = F 6 Q 252 18 Measurability

Proposition 18.4. Let be any collection of subsets of [= Then there exists a unique smallest algebraE ( ) and — algebra ( ) which contains = 18 A E E E Proof. The proof is the same as the analogous Proposition 10.6 for topolo- Measurability gies, i.e. ( ):= : is an algebra such that A E {A A E A} \ and ( ):= : is a — algebra such that = E {M M E M} \

Example 18.5. Suppose [ = 1> 2> 3 and = >>[> 1> 2 > 1> 3 > see Figure 18.1. { } E { { } { }} 18.1 Algebras and —Algebras

Definition 18.1. Acollectionofsubsets of a set [ is an algebra if A 1. >>[ 5 2. D 5 Aimplies that Df 5 A A 3. is closed under finite unions, i.e. if D1>===>Dq 5 then D1 ^ ^Dq 5 A= A ··· InA view of conditions 1. and 2., 3. is equivalent to 30. is closed under finite intersections. A Definition 18.2. Acollectionofsubsets of [ is a — algebra (or sometimes called a — field) if is an algebraM which also closed under countable unions, i.e. if 4 Mthen 4 (Notice that since is also Dl l=1 > ^l=1Dl 5 = Fig. 18.1. A collection of subsets. closed under taking{ } complements,M is also closedM under taking countableM intersections.) A pair ([> )> whereM[ is a set and is a — algebra on [> is called a measurableM space. M Then The reader should compare these definitions with that of a topology in Definition 10.1. Recall that the elements of a topology are called open sets. ( )= >>[> 1 > 1> 2 > 1> 3 Analogously, elements of and algebra or a —algebra will be called E { { } { } { }} A M ( )=( )=2[ = measurable sets. A E E Example 18.3. Here are some examples of algebras. The next proposition is the analogue to Proposition 10.7 for topologies and enables us to give and explicit descriptions of ( )= On the other hand 1. =2[ > then is a topology, an algebra and a —algebra. it should be noted that ( ) typically does not admitA E a simple concrete de- M M 2. Let [ = 1> 2> 3 > then = >>[> 2> 3 is a topology on [ which is not scription. E an algebra.{ } { { }} 3. = = 1 > 2> 3 > >>[ is a topology, an algebra, and a —algebra Proposition 18.6. Let [ be a set and 2[ = Let f := Df : D 5 and A {{ } { } } f E E { E} on [= The sets [> 1 > 2> 3 > > are open and closed. The sets 1> 2 and f := ^ [> > ^ Then E E { } E 1> 3 are neither open{ } nor{ closed} and are not measurable. { } { } ( ):= finite unions of finite intersections of elements from f = (18.1) The reader should compare this example with Example 10.3. A E { E } 18.1 Algebras and —Algebras 253 254 18 Measurability

Proof. Let denote the right member of Eq. (18.1). From the definition of Example 18.9. Let [ = R> then an algebra, it isA clear that ( )= Hence to finish that proof it su!ces to show is an algebra. TheE proofA A ofE these assertions are routine except for := (d> e] _ R : d> e 5 R¯ E possiblyA showing that is closed under complementation. To check is closed = ©(d> e]:d 5 [4> 4)ªand d?e?4 ^ >> R under complementation,A let ] 5 be expressed as A { } { } A is an elementary family. Q N [ \ ] = Dlm Exercise 18.2. Let 2 and 2 be elementary families. Show the l=1 m=1 A B [ \ collection f = = D E : D 5 and E 5 where Dlm 5 f= Therefore, writing Elm = Dlm 5 f> we find that E A×B { × A B} E E is also an elementary family. Q N N f [ ] = Elm = (E1m _ E2m _ _ EQm ) 5 Proposition 18.10. Suppose 2 is an elementary family, then = 1 2 ··· Q A E A l=1 m=1 m1>===>mQ =1 ( ) consists of sets which may be written as finite disjoint unions of sets \ [ [ AfromE = wherein we have used the fact that E1 _ E2 _ _ E is a finite inter- E m1 m2 ··· QmQ section of sets from f= Proof. This could be proved making use of Proposition 18.6. However it E is easier to give a direct proof. Let denote the collection of sets which may Remark 18.7. One might think that in general ( ) may be described as the A E be written as finite disjoint unions of sets from = Clearly ( ) so it countable unions of countable intersections of sets in f= However this is in E E A A E E su!ces to show is an algebra since ( ) is the smallest algebra containing general false, since if A A E 4 4 = By the properties of > we know that >>[ 5 = Now suppose that D = E E A l ] = D I 5 where, for l =1> 2>===>q>l is a finite collection of disjoint sets lm I 5l A l=1 m=1 from = Then [ \ ` E with D 5 > then q q lm Ef Dl = I = (I1 _ I2 _ _ Iq) 4 4 ··· l=1 l=1 ÃI 5l ! (I1>>===>I )51 f = f \ \ a q [ ×···× q ] Dc>mc Ã ! m1=1>m2=1>===mQ =1>=== c=1 and this is a disjoint (you check) union of elements from = Therefore is [ \ E A closed under finite intersections. Similarly, if D = I 5 I with being a which is now an uncountable union. Thus the above description is not cor- f f finite collection of disjoint sets from > then D = I 5 I = Since by assump- rect. In general it is complicated to explicitly describe ( )> see Proposition f E ` 1.23 on page 39 of Folland for details. Also see PropositionE 18.13 below. tion I 5 for I 5 and is closed under finite intersections, it follows thatADf 5 = E A T A Exercise 18.1. Let be a topology on a set [ and = ( ) be the algebra A A Definition 18.11. Let [ be a set. We say that a family of sets 2[ is a generated by = Show is the collection of subsets of [ which may be written F as finite union of setsA of the form I _ Y where I is closed and Y is open. partition of [ if distinct members of are disjoint and if [ is the union of the sets in = F The following notion will be useful in the sequel and plays an analogous F role for algebras as a base (Definition 10.8) does for a topology. Example 18.12. Let [ be a set and = D1>===>Dq where D1>===>Dq is a partition of [= In this case E { } Definition 18.8. Aset 2[ is said to be an elementary family or E elementary class provided that ( )=( )= ( )= ^l5Dl : 1> 2>===>q A E E E { { }}

>5 where ^l5Dl := > when = >= Notice that • isE closed under finite intersections •E 1>2>===>q q if H 5 > then Hf is a finite disjoint union of sets from = (In particular #( ( )) = #(2{ })=2 = A E • [ = >fEis a finite disjoint union of elements from =) E E 18.1 Algebras and —Algebras 255 256 18 Measurability

Proposition 18.13. Suppose that 2[ is a — algebra and is at most (b) All of the standard open subsets of R are in ( )= M M 1 E a countable set. Then there exists a unique finite partition of [ such that (c) { = { q >{ 5 ( ) and every element E 5 is of the form F { } q E F M M (d) [d> e]=Td¡ ^ (d> e]¤5 ( ) { } E E = ^ D 5 : D E = (18.2) (e) Any countable subset of R is in ( ). { F } E q Remark 18.15. In the above example, one may replace by = (d> 4):d 5 In particular is actually a finite set and #( )=2 for some q 5 N= E E { M M Q ^ R> > > in which case ( ) may be described as being those sets which } { } A E Proof. For each { 5 [ let are finite disjoint unions of sets from the following list

D = _ D 5 : { 5 D 5 > (d> 4)> (4>d]> (d> e]:d> e 5 Q ^ >> R = { { M } M { } { } This shows that ( ) is a countable set — a useful fact which will be needed wherein we have used is a countable — algebra to insure D{ 5 = Hence M M later. A E D{ is the smallest set in which contains {= Let F = D{ _ D|= If {@5 F then D F D is an elementM of which contains { and since D is the smallest { \ { M { Notation 18.16 For a general topological space ([> )> the Borel — alge- member of containing {> we must have that F = >= Similarly if |@5 F then q bra is the — algebra [ := ( ) on [= In particular if [ = R > q will F = >= ThereforeM if F = >> then {> | 5 D _ D 5 and D _ D D and B BR 6 { | M { | { be used to denote the Borel —algebraonRq when Rq is equipped with its D{ _D| D| from which it follows that D{ = D{ _D| = D|= This shows that standard Euclidean topology. = D{ : { 5 [ is a (necessarily countable) partition of [ for which F { } M Q Eq. (18.2) holds for all E 5 = Enumerate the elements of as = Sq Exercise 18.3. Verify the — algebra, , is generated by any of the following M F F { }q=1 BR where Q 5 N or Q = 4= If Q = 4> then the correspondence collection of sets:

N d 5 0> 1 $ Dd = ^ Sq : dq =1 5 1. (d> 4):d 5 R > 2. (d> 4):d 5 Q or 3. [d> 4):d 5 Q = { } { } M { } { } { } is bijective and therefore, by Lemma 2.6, is uncountable. Thus any count- Proposition 18.17. If is a second countable topology on [ and is a M E able — algebra is necessarily finite. This finishes the proof modulo the unique- countable collection of subsets of [ such that = ( )> then [ := ( )= E B ness assertion which is left as an exercise to the reader. ( )> i.e. ( ( )) = ( )= E E E Example 18.14. Let [ = and Proof. Let i denote the collection of subsets of [ which are finite inter- R E section of elements from along with [ and >= Notice that i is still countable R E E = (d> 4):d 5 R ^ R> > = (d> 4) _ R : d 5 R¯ 2 = (you prove). A set ] is in ( ) i ] is an arbitrary union of sets from = E { } { } { } E Ei Therefore ] = D for some subset i which is necessarily count- Notice that i = and that is closed under unions, which shows that D5 F E E E E f F ( )= , i.e. is already a topology. Since (d> 4) =(4>d] we find that able. Since i S( ) and ( ) is closed under countable unions it follows E E E E E E f = (d> 4)> (4>d]> 4 d?4 ^ R> > = Noting that that ] 5 ( ) and hence that ( ) ( )= Lastly, since ( ) ( )> E { } { } ( ) ( (E)) ( )= E E E E E (d> 4) _ (4>e]=(d> e] E E E it follows that ( )= ( ˜) where A E A E 18.2 Measurable Functions ˜ := (d> e] _ R : d> e 5 R¯ = E Our notion of a “measurable” function will be analogous to that for a con- ˜ © ª tinuous function. For motivational purposes, suppose ([> >) is a measure Since is an elementary family of subsets of R> Proposition 18.10 implies M ( ) mayE be described as being those sets which are finite disjoint unions of space and i : [ $ R+. Roughly speaking, in the next Chapter we are going setsA E from ˜= The — algebra, ( )> generated by is very complicated. to define ig as a certain limit of sums of the form, E E E [ Here are some sets in ( ) — most of which are not in ( )= R E A E 4 4 1 1 dl(i (dl>dl+1])= (a) (d> e)= (d> e q ] 5 ( )= E 0?d ?d ?d ?=== q=1 1 X2 3 S 18.2 Measurable Functions 257 258 18 Measurability

For this to make sense we will need to require i 1((d> e]) 5 for all d? Lemma 18.22. Suppose that i : [ $ \ is a function and 2\ and D \ e= Because of Lemma 18.22 below, this last condition is equivalentM to the then E condition i 1( ) = BR M i 1( ) = i 1(( )) and (18.3) E E Definition 18.18. Let ([> ) and (\> ) be measurable spaces. A function ¡(( ))D¢ = ( D )= (18.4) i : [ $ \ is measurable Mif i 1( ) F = We will also say that i is @ E E —measurableor( > ) —measurable.F M M F (Similar assertion hold with ( ) being replaced by ( ) =)Moreover,if = M F ( ) and is a —algebraon[>· then i is ( > ) —measurableiA · i 1(F) Example 18.19 (Characteristic Functions). Let ([> ) be a measurable space E= M M F E M M and D [= We define the characteristic function 1D : [ $ R by Proof. By Exercise 18.4, i 1(( )) is a — algebra and since > 1 1 E 1 1 E F 1 if { 5 D i ( ) i (( ))= It now follows that (i ( )) i ( ( ))= For the 1 ({)= E E E E D 0 if {@5 D= reverse inclusion, notice that ½ 1 1 1 R 1 i i ( ) = E \ : i (E) 5 i ( ) If D 5 > then 1D is ( > 2 ) — measurable because 1 (Z ) is either >>[> E E M M D or f for any Conversely, if is any — algebra on containing 1 D D Z R= R is a — algebra which¡ contains¢ © and thus ( ) ¡i i ¢ª( ) = Hence if F f E E E asetZ R such that 1 5 Z and 0 5 Z > then D 5 if 1D is ( > ) — E 5 ( ) we know that i 1(E) 5 i 1( ) > i.e. i 1(( )) i 1( ) measurable. This is because D =11(Z ) 5 = M M F E E ¡ E ¢ E D M and Eq. (18.3) has been proved. Applying Eq. (18.3) with [ = D and i = lD being the inclusion map implies ¡ ¢ ¡ ¢ Exercise 18.4. Suppose i : [ $ \ is a function, 2\ and 2[ = 1 F M 1 1 Show i and i (see Notation 2.7) are algebras ( — algebras) provided (( ))D = lD (( )) = (lD ( )) = ( D)= and Fare algebrasM ( — algebras). E E E E F M Lastly if i 1 > then i 1 ( )= i 1 which shows i is Remark 18.20. Let i : [ $ \ be a function. Given a —algebra 2\ > ( > ) — measurable.E M E E M the —algebra := i 1( ) is the smallest — algebra on [ suchF that i is M F ¡ ¢ M F Corollary 18.23. Suppose that ([> ) is a measurable space. Then the fol- ( > ) - measurable . Similarly, if is a -algebraon[ then = i M isM theF largest — algebra on \ suchM that i is ( > ) - measurableF . M lowing conditions on a function i : [ $ R are equivalent: M F 1. i is ( > ) —measurable, Recall from Definition 2.8 that for 2[ and D [ that M BR E 2. i 1((d> 4)) 5 for all d 5 R> 1 M 1 3. i ((d> 4)) 5 for all d 5 Q> D = lD ( )= D _ H : H 5 M E E { E} 4. i 1((4>d]) 5 for all d 5 R= M where lD : D $ [ is the inclusion map. Because of Exercise 10.3, when Proof. An exercise in using Lemma 18.22 and is the content of Exercise = is an algebra ( — algebra), D is an algebra ( — algebra) on D and 18.8. weE callM the relative or inducedM algebra ( — algebra) on D= MD Here is yet another way to generate — algebras. (Compare with the The next two Lemmas are direct analogues of their topological counter analogous topological Definition 10.20.) parts in Lemmas 10.13 and 10.14. For completeness, the proofs will be given even though they are same as those for Lemmas 10.13 and 10.14. Definition 18.24 ( — Algebras Generated by Functions). Let [ be a set and suppose there is a collection of measurable spaces (\> ): 5 D Lemma 18.21. Suppose that ( ) ( ) and ( ) are measurable { F } [> > \> ]> and functions i : [ $ \ for all 5 D= Let (i : 5 D) denote the spaces. If andM F G are measurable i :([> ) $ (\> ) j :(\> ) $ (]> ) smallest —algebraon[ such that each i is measurable, i.e. functions then j iM:([> ) $F(]> ) is measurableF as well.G M G 1 (i : 5 D)=(î ( ))= Proof. By assumption j1( ) and i 1 ( ) so that F G F F M Proposition 18.25. Assuming the notation in Definition 18.24 and addition- ( )1 ( )= 1 1 ( ) 1 ( ) ally let (]> ) be a measurable space and j : ] $ [ be a function. Then j j i i j i = M G G F M is ( >(i : 5 D)) —measurablei i j is ( > )—measurable for all ¡ ¢ 5MD= M F 18.2 Measurable Functions 259 260 18 Measurability

Proof. This proof is essentially the same as the proof of the topological 2. If E 5 > then F analogue in Proposition 10.21. (,)Ifj is ( >(i : 5 D)) — measurable, 1 4 1 4 1 M i (E)=^ i (E) _ Dq = ^ i (E)= then the composition i j is ( > ) — measurable by Lemma 18.21. (+) q=1 q=1 Dq M F | Let Since each D 5 > ¡ and so the previous¢ displayed equation shows 1 q M MDq M = (i : 5 D)= ^5Di ( ) = i 1(E) 5 = G F M If is ( ) — measurable for all then i j > >¡ ¢ Proposition 18.30. If ([> ) is a measurable space, then M F M 1 1 q j i ( ) ; 5 D =( ): F M i i1>i2>===>iq [ $ R and therefore is ( > q ) —measurablei il : [ $ R is ( > ) — measurable for each M BR M BR l= In particular, a function i : [ $ C is ( > ) —measurablei Re i and 1 1 1 1 M BC j ^5Di ( ) = ^5Dj i ( ) = Im i are ( > ) — measurable. F F M M BR Hence ¡ ¢ Proof. This is formally a consequence of Corollary 18.65 and Proposition 18.60 below. Nevertheless it is instructive to give a direct proof now. Let 1 1 1 1 1 q q ( )= ( ) = ( ( ) q j j ^5Di j ^5Di = R denote the usual topology on R and l : R $ R be projection G F F M th onto the l — factor. Since l is continuous, l is Rq @ R — measurable and which shows that j is¡ (¡ > ) — measurable.¢¢ ¡ ¢ q B B M G therefore if i : [ $ R is measurable then so is il = l i= Now suppose i : [ $ is measurable for all l =1> 2>===>q= Let Deﬁnition 18.26. Afunctioni : [ $ \ between two topological spaces is l R 1 Borel measurable if i ( \ ) [ = := (d> e):d> e 5 Qq d?e > B B E { 3 } q Proposition 18.27. Let [ and \ be two topological spaces and i : [ $ \ where, for d> e 5 R > we write d?ei dl ?el for l =1> 2>===>q and let be a continuous function. Then i is Borel measurable. (d> e)=(d1>e1) (d >e ) = ×···× q q Proof. Using Lemma 18.22 and \ = ( \ )> B Since and every element Y 5 may be written as a (necessarily) 1 1 1 E i ( )=i (( )) = (i ( )) ( )= = countable union of elements from > we have ( ) Rq = ( ) ( ) > i.e. B\ \ \ [ B[ E E B E ( )= Rq = (This part of the proof is essentially a direct proof of Corollary 18.65E below.)B Because

1 1 1 1 Deﬁnition 18.28. Given measurable spaces ([> ) and (\> ) and a subset i ((d> e)) = i1 ((d1>e1)) _ i2 ((d2>e2)) _ _ iq ((dq>eq)) 5 M F ··· M D [= We say a function i : D $ \ is measurable i i is D@ — M F for all d> e 5 Q with d?e>it follows that i 1 and therefore measurable. E M 1 1 1 i q = i ( )= i = Proposition 18.29 (Localizing Measurability). Let ([> ) and (\> ) BR E E M M F be measurable spaces and i : [ $ \ be a function. ¡ ¢

1. If i is measurable and D [ then i D : D $ \ is measurable. Corollary 18.31. Let ([> ) be a measurable space and i>j : [ $ C be | 4 M 2. Suppose there exist Dq 5 such that [ = ^q=1Dq and i Dq is Dq ( > C) —measurablefunctions.Theni j and i j are also ( > C) — measurable for all q> then Mi is —measurable. | M measurable.M B ± · M B M Proof. As the reader will notice, the proof given below is essentially iden- Proof. Deﬁne I : [ $ C C>D : C C $ C and P : C C $ C by × ± × × tical to the proof of Proposition 10.19 which is the topological analogue of I ({)=(i({)>j({))>D (z> })=z } and P(z> })=z}= Then D and P ± ± 1 ± this proposition. 1. If i : [ $ \ is measurable, i (E) 5 for all E 5 are continuous and hence ( C2 > C) — measurable. Also I is ( > C C)= M F B B M B B and therefore ( > C2 ) — measurable since 1 I = i and 2 I = j are ( > C) — measurable.M B Therefore D I = i j and P I = i j> being the compositionM B i 1 (E)=D _ i 1(E) 5 for all E 5 = of measurable functions,± are also measurable.± · |D MD F 18.2 Measurable Functions 261 262 18 Measurability

Lemma 18.32. Let 5 C> ([> ) be a measurable space and i : [ $ C be 1. D 5 ¯ =, D _ R 5 R and M BR¯ B a ( > ) — measurable function. Then 2. if D R is such that D _ R 5 there exists E 5 ¯ such that D _ R = M BC BR BR E _ R= Because DE 4 and 4 > 4 5 ¯ we may conclude 1 {± } { } { } BR if i({) =0 that D 5 ¯ as well. I ({):= i({) 6 R if i({)=0 B ½ This proves Eq. (18.6). is measurable. The proofs of the next two corollaries are left to the reader, see Exercises 18.5 and 18.6. Proof. Define l : C $ C by Corollary 18.34. Let ([> ) be a measurable space and i : [ $ R¯ be a 1 if } =0 function. Then the followingM are equivalent l(})= } 0 if } 6=0= ½ 1. i is ( > ¯ ) -measurable, M BR For any open set Y we have 2. i 1((d> 4]) 5 for all d 5 R> C M 3. i 1((4>d]) 5 for all d 5 R> 1 1 1 M l (Y )=l (Y 0 ) ^ l (Y _ 0 ) 4. i 1( 4 ) 5 >i1( 4 ) 5 and i 0 : [ $ R defined by \{ } { } { } M { } M 1 Because l is continuous except at } =0>l (Y 0 ) is an open set and hence 0 i ({) if i ({) 5 R 1 1\{ } i ({):=1R (i ({)) = in C= Moreover, l (Y _ 0 ) 5 C since l (Y _ 0 ) is either the empty 0 if i ({) 5 4 B { } B 1 { } 1 ½ {± } set or the one point set 0 = Therefore l ( C) C and hence l ( C)= 1 1 { } B B is measurable. l (( C)) = (l ( C)) C which shows that l is Borel measurable. Since I = l i is the compositionB of measurable functions, I is also measurable. Corollary 18.35. Let ([> ) be a measurable space, i>j : [ $ R¯ be func- We will often deal with functions i : [ $ ¯ = ^ 4 = When talking M R R tions and define i j : [ $ R¯ and (i + j):[ $ R¯ using the conventions, about measurability in this context we will refer to the{± }— algebra on ¯ R 0 4 =0and (i +· j)({)=0if i ({)=4 and j ({)=4 or i ({)=4 defined by and· j ({)=4= Then i j and i + j are measurable functions on [ if both i ¯ := ( [d> 4]:d 5 R ) = (18.5) · BR { } and j are measurable. Proposition 18.33 (The Structure of ¯ ). Let and ¯ be as above, BR BR BR Exercise 18.5. Prove Corollary 18.34 noting that the equivalence of items 1. then — 3. is a direct analogue of Corollary 18.23. Use Proposition 18.33 to handle ¯ ¯ = D R : D _ R 5 = (18.6) BR { BR} item 4. In particular 4 > 4 5 ¯ and ¯ = { } { } BR BR BR Exercise 18.6. Prove Corollary 18.35. Proof. Let us first observe that Proposition 18.36 (Closure under sups, infs and limits). Suppose that 4 4 f ([> ) is a measurable space and i :([> ) $ for m 5 is a sequence 4 = _ [4> q)=_ [q> 4] 5 ¯ > m R N { } q=1 q=1 BR of M@ — measurable functions. Then M 4 ¯ R 4 = _ =1[q> 4] 5 ¯ and R = R 4 5 ¯ = M B { } q BR \{± } BR supmim> infmim> lim sup im and lim inf im Letting l : R $ R¯ be the inclusion map, m$4 m$4 1 1 ¯ 1 ¯ are all @ — measurable functions. (Note that this result is in generally l ( R¯ )= l [d> 4]:d 5 R = l ([d> 4]) : d 5 R R B false whenM (B[> ) is a topological space and measurable is replaced by con- = [d> 4] _ R : d 5 R¯ = ( [d> 4):d 5 R )= = M ¡ ¡© ª¢¢ {¡© } BRª¢ tinuous in the statement.) Thus we have shown¡© ª¢ Proof. Define j+({):=supmim({)> then 1 = l ( ¯ )= D _ : D 5 ¯ = R R R R { : j+({) d = { : i ({) d ; m B B { B } { } { m } = _ { : i ({) d 5 This implies: m{ m } M 18.2 Measurable Functions 263 264 18 Measurability so that j+ is measurable. Similarly if j({)=infm im({) then 18.3 — Function Algebras { : j ({) d = _ { : i ({) d 5 = { } m{ m } M In this subsection, we are going to relate — algebras of subsets of a set [ to Since certain algebras of functions on [= We will begin this endeavor after proving the simple but very useful approximation Theorem 18.42 below. lim sup im =infsup im : m q and m$4 q { } Definition 18.41. Let ([> ) be a measurable space. A function ! : [ $ F lim inf im =supinf im : m q M ¯ m$4 q { } (F denotes either R> C or [0> 4] R) is a simple function if ! is — M BF we are done by what we have already proved. measurable and !([) contains only finitely many elements. ¯ Definition 18.37. Given a function i : [ $ R let i+({):=max i({)> 0 Any such simple functions can be written as { } and i ({):=max(i({)> 0) = min (i({)> 0) = Notice that i = i+ i= q Corollary 18.38. Suppose ([> ) is a measurable space and i : [ $ R¯ is ! = l1Dl with Dl 5 and l 5 F= (18.7) M M afunction.Theni is measurable i i are measurable. l=1 ± X Proof. If i is measurable, then Proposition 18.36 implies i are measur- ± Indeed, take 1>2>===>q to be an enumeration of the range of ! and Dl = able. Conversely if i are measurable then so is i = i+ i= 1 ± ! ( l )= Note that this argument shows that any simple function may be written{ } intrinsically as 18.2.1 More general pointwise limits ! = |1!1( | )= (18.8) { } Lemma 18.39. Suppose that ([> ) is a measurable space, (\>g) is a metric |5 XF space and i : [ $ \ is ( > ) M— measurable for all m= Also assume that for m M B\ The next theorem shows that simple functions are “pointwise dense” in each { 5 [> i({) = limq$4 iq({) exists. Then i : [ $ \ is also ( > \ ) — the space of measurable functions. measurable. M B Theorem 18.42 (Approximation Theorem). Let i : [ $ [0> 4] be mea- Proof. Let Y 5 g and Zp := | 5 \ : gY f (|) A 1@p for p =1> 2>==== { } surable and define, see Figure 18.2, Then Zp 5 g>

¯ 2q Zp Zp | 5 \ : gY f (|) 1@p Y 2 1 { } n q !q({):= 1i 1 ( n > n+1 ] ({)+2 1i 1((2q>4])({) for all p and Zp % Y as p $4= The proof will be completed by verifying 2q ( 2q 2q ) =0 the identity, Xn 22q1 1 4 4 1 n q i (Y )=^p=1 ^Q=1 _qQ iq (Zp) 5 = +1 q = 1 n ?i n ({)+2 1 iA2 ({) M 2q 2q 2q { } 1 =0 { } If { 5 i (Y ) then i({) 5 Y and hence i({) 5 Zp for some p= Since iq({) $ nX i({)>i ({) 5 Z for almost all q= That is { 5^4 ^4 _ i 1(Z )= q p p=1 Q=1 qQ q p then ! i for all q> ! ({) % i({) for all { 5 [ and ! % i uniformly on Conversely when { 5^4 ^4 _ i 1(Z ) there exists an p such that q q q p=1 Q=1 qQ q p the sets [ := { 5 [ : i({) P with P?4= Moreover, if i : [ $ i ({) 5 Z Z¯ for almost all q= Since i ({) $ i({) 5 Z¯ Y> it follows P q p p q p is a measurable{ function, then there} exists simple functions ! such that that { 5 i 1(Y )= C q limq$4 !q({)=i({) for all { and !q % i as q $4= Remark 18.40. In the previous Lemma 18.39 it is possible to let (\> ) be any | | | | topological space which has the “regularity” property that if Y 5 there Proof. Since ¯ 4 exists Zp 5 such that Zp Zp Y and Y = ^p=1Zp= Moreover, some n n +1 2n 2n +1 2n +1 2n +2 extra condition is necessary on the topology in order for Lemma 18.39 to ( > ]=( > ] ^ ( > ]> 2q 2q 2q+1 2q+1 2q+1 2q+1 be correct. For example if \ = 1> 2> 3 and = \>>> 1> 2 > 2> 3 > 2 as { } { { } { } { }} 1 2n 2n+1 2n in Example 10.36 and [ = d> e with the trivial — algebra. Let im(d)= if { 5 i ( +1 > +1 ] then ! ({)=! ({)= +1 and if { 5 { } 2q 2q q q+1 2q i (e)=2for all m> then i is constant and hence measurable. Let i(d)=1 1 2n+1 2n+2 2n 2n+1 m m i ( +1 > +1 ] then ! ({)= +1 ? +1 = ! ({)= Similarly 2q 2¡q q¢ 2q 2q q+1 and i(e)=2> then im $ i as m $4with i being non-measurable. Notice that the Borel — algebra on \ is 2\ = ¡ ¢ (2q> 4]=(2q> 2q+1] ^ (2q+1> 4]> 18.3 — Function Algebras 265 266 18 Measurability

Example 18.45. Suppose is a — algebra on [> then M c4 ( > R):= i 5 c4 ([> R): i is @ — measurable (18.9) M { M BR } is a — function algebra. The next theorem will show that these are the only example of — function algebras. (See Exercise 18.7 below for examples of function algebras on [=)

Notation 18.46 If c4 ([> R) be a function algebra, let H ( ):= D [ :1 5 = (18.10) M H { D H} Theorem 18.47. Let be a —functionalgebraonaset[= Then H 1. ( ) is a —algebraon[= M H4 Fig. 18.2. Constructing simple functions approximating a function, i : [ < [0> "]= 2. = c ( ( ) > R) = 3. TheH mapM H

5 —algebrason[ $ c4 ( > ) 5 —functionalgebrason[ 1 q+1 q q+1 R and so for { 5 i ((2 > 4])>!q({)=2 ? 2 = !q+1({) and for { 5 M { } M { (18.11)} 1 q q+1 q i ((2 > 2 ])>!q+1({) 2 = !q({)= Therefore !q !q+1 for all q= It is is bijective with inverse given by $ ( ) = clear by construction that !q({) i({) for all { and that 0 i({)!q({) H M H q 2 if { 5 [2q = Hencewehaveshownthat!q({) % i({) for all { 5 [ and Proof. Let := ( ) = M M H !q % i uniformly on bounded sets. For the second assertion, first assume that 1. Since 0> 1 5 > >>[ 5 = If D 5 then, since is a linear subspace i : [ $ is a measurable function and choose !± to be simple functions R q 4 H M M f H 4 + of c ([> ) > 1Df =1 1D 5 which shows D 5 = If Dq > such that !q± % i as q $4and define !q = !q !q = Then R q=1 ± then since is an algebra, H M { } M + + H !q = !q + !q !q+1 + !q+1 = !q+1 | | | | Q + + and clearly ! = ! +! % i +i = i and ! = ! ! $ i i = i 1_Q D = 1Dq =: iQ 5 q q q + q q q + q=1 q H | | | | q=1 as q $4= Now suppose that i : [ $ C is measurable. We may now choose Y simple function x and y such that x % Re i > y % Im i >x $ Re i q q | q| | | | q| | | q for all Q 5 N= Because is closed under bounded convergence it follows and yq $ Im i as q $4= Let !q = xq + lyq> then that H 2 2 2 2 2 2 1 4 = lim i 5 ! = x + y % Re i + Im i = i _q=1Dq Q | q| q q | | | | | | Q$4 H and this implies _4 D 5 = Hencewehaveshown is a — algebra. and !q = xq + lyq $ Re i + l Im i = i as q $4= q=1 q M M For the rest of this section let [ be a given set. 2. Since is an algebra, s (i) 5 for any i 5 and any polynomial s on R= The WeierstrassH approximationH Theorem 8.34,H asserts that polynomials on Definition 18.43 (Bounded Convergence). We say that a sequence of R are uniformly dense in the space of continuos functions on any compact functions iq from [ to R or C converges boundedly to a function i if subinterval of R= Hence if i 5 and ! 5 F (R) > there exists polynomials H limq$4 iq({)=i({) for all { 5 [ and sq on R such that sq i ({) converges to ! i ({) uniformly (and hence boundedly) in { 5 [ as q $4= Therefore ! i 5 for all i 5 and sup iq({) : { 5 [ and q =1> 2>=== ? 4= iH i H {| | } ! 5 F (R) and in particular i 5 and i := | |±2 5 if i 5 = | | H ± 1@q H H Definition 18.44. A function algebra on [ is a linear subspace of Fix an 5 R and for q 5 N let !q (w):=(w )+ > where (w )+ := 4 H c ([> R) which contains 1 and is closed under pointwise multiplication, i.e. max w > 0 = Then !q 5 F (R) and !q (w) $ 1wA as q $4and the { } is a subalgebra of c4 ([> R) which contains 1= If is further closed under convergence is bounded when w is restricted to any compact subset of R= H H bounded convergence then is said to be a — function algebra. Hence if i 5 it follows that 1iA = limq$4 !q (i) 5 for all 5 R> H H H 18.3 — Function Algebras 267 268 18 Measurability

i.e. iA 5 for all 5 R= Therefore if i 5 then i 5 c4 ( > R) Theorem 18.51 (Dynkin’s Multiplicative System Theorem). Suppose { } M H M andwehaveshown c4 ( > R) = Conversely if i 5 c4 ( > R) > then P c4 ([> R) is a multiplicative system, then H M M for any ?> ?i 5 = ( ) and so by assumption 4 { } M M H (P)= (P)=c ( (P) > ) = (18.12) 1 ?i 5 = Combining this remark with the approximation Theo- R { } H H H rem 18.42 and the fact that is closed under bounded convergence shows In words, the smallest subspace of bounded real valued functions on [ which H 4 that i 5 = Hencewehaveshownc ( > R) which combined with contains P that is closed under bounded convergence is the same as the space 4 H M 4 H c ( > R) already proved shows = c ( ( ) > R) = of bounded real valued (P) — measurable functions on [= 3.H Items 1.M and 2. shows the map in Eq.H (18.11)M is surjective.H To see the map is injective suppose and are two — algebras on [ such that Proof. We begin by proving := (P) is a — function algebra. To do M F H H c4 ( > R)=c4 ( > R) > then this, for any i 5 let M F H 4 := j 5 : ij 5 = D [ :1D 5 c ( > R) i M { M } H { H H} H = :1 4 ( ) = 4 D [ D 5 c > R = and notice that i is a linear subspace of c ([> R) which is closed under { F } F H bounded convergence. Moreover if i 5 P> P i since P is multiplicative. Therefore = andwehaveshownthatijH5 whenever i 5 P and Hi H H 4 j 5 = Given this it now follows that P for any i 5 and by Notation 18.48 Suppose P is a subset of c ([> R) = i the sameH reasoning just used, = = Since iH5 is arbitrary,H we have Hi H H 1. Let (P) denote the smallest subspace of c4 ([> R) which contains P shown ij 5 for all i>j 5 > i.e. is an algebra. Since it is harder to H and the constant functions and is closed under bounded convergence. be an algebraH of functions containingH HP (see Exercise 18.13) than it is to 2. Let (P) denote the smallest — function algebra containing P= be a subspace of functions containing P it follows that (P) (P) = H Butaswehavejustseen (P) is a — function algebraH whichH contains 4 Theorem 18.49. Suppose P is a subset of c ([> ) > then (P)= H R P so we must have (P) (P) because (P) is by deﬁnition the c4 ( (P) > ) or in other words the following diagram commutes:H H H H R smallest such — function algebra. Hence (P)= (P) = The assertion 4 H H that (P)=c ( (P) > R) has already been proved in Theorem 18.49. P $ (P) H P {Multiplicative Subsets} $ { —algebras} Theorem 18.52 (Complex Multiplicative System Theorem). Suppose M && & & is a complex linear subspace of c4([> C) such that: 1 5 > is closed under 4 H H H (P) {—functionalgebras} = {—functionalgebras}c ( > R) = complex conjugation, and is closed under bounded convergence. If P H M is multiplicative system whichH is closed under conjugation, then contains allH Proof. Since 4 ( ( ) ) is — function algebra which contains it 4 H c P > R P bounded complex valued (P)-measurable functions, i.e. c ( (P) > C) = follows that H (P) c4 ( (P) > ) = Proof. Let P0 =span(P ^ 1 ) be the complex span of P= As the R C { } H reader should verify, P0 is an algebra, P0 >P0 is closed under complex For the opposite inclusion, let HR 4 R conjugation and that (P0)= (P) = Let := _ c ([> R) and P0 = 4 R H H R R P _c ([> R)= Then(youverify)P0 is a multiplicative system, P0 and = ( (P)) := D [ :1D 5 (P) = H M M H { H } R is a linear space containing 1 which is closed under bounded convergence. H 4 R R 4 Therefore by Theorem 18.51, c P0 > R = Since and P0 are By Theorem 18.47, P (P)=c ( > R) which implies that every H H i 5 P is — measurable.H This then impliesM (P) and therefore complex linear spaces closed under complex conjugation, for any i 5 or M M 1 ¡ ¡¯ ¢ ¢ 1 ¯ H i 5 P0> the functions Re i = 2 i + i and Im i = 2l i i are in (P0) 4 4 R R R R H R c ( (P) > R) c ( > R)= (P) = or P0 respectively. Therefore = + l >P0 = P0 + lP0 > P0 = M H H¡ H ¢ H ¡ ¢ (P0)= (P) and ¡ ¢ 4 4 R 4 R c ( (P) > C)=c P0 > R + lc P0 > R Deﬁnition 18.50 (Multiplicative System). A collection of bounded real or R + l R = = complex valued functions, P> on a set [ is called a multiplicative system H ¡ ¡H ¢ H¢ ¡ ¡ ¢ ¢ if i j 5 P whenever i and j are in P= · 18.3 — Function Algebras 269 270 18 Measurability

Exercise 18.7 (Algebra analogue of Theorem 18.47). Call a function algebra c4 ([> R) a simple function algebra if the range of each func- Here are some more variants of Corollary 18.55. H tion i 5 is a finite subset of R= Prove there is a one to one correspondence H Proposition 18.56. Let ([> g) be a metric space, [ = ( g) be the Borel between algebras on a set [ and simple function algebras on [= B r A H — algebra and assume there exists compact sets Nn [ such that Nn % [= Definition 18.53. Acollectionofsubsets, > of [ is a multiplicative Suppose that is a subspace of c4([> ) such that F ([> ) (F ([> ) C R f R f R class(or a —class)if is closed under finite intersections. is the space ofH continuous functions with compact support) andH is closed C H 4 under bounded convergence. Then contains all bounded —measurable Corollary 18.54. Suppose is a subspace of c ([> R) which is closed under H B[ bounded convergence and 1H5 = If 2[ is a multiplicative class such real valued functions on [= H C that 1D 5 for all D 5 > then contains all bounded ( ) —measurable Proof. Let n and q be positive integers and set # ({)=min(1>q H C H C q>n · functions. f r g Nr ({))= Then #q>n 5 Ff([> R) and #q>n =0 Nn= Let q>n denote ( n) { 6 } H Proof. Let P = 1 ^ 1D : D 5 = Then P is a multiplicative those bounded [ — measurable functions, i : [ $ R> such that #q>ni 5 = system and the proof is{ completed} { withC} an application ofH Theorem 18.51. It is easily seenB that is closed under bounded convergence and thatH Hq>n q>n contains EF([> R) and therefore by Corollary 18.55, #q>ni 5 for all Corollary 18.55. Suppose that ([> g) is a metric space and [ = ( g) H H 4 B bounded measurable functions i : [ $ R= Since #q>ni $ 1Nr i boundedly is the Borel —algebraon[ and is a subspace of c ([> R) such that n H 1 as q $4> 1Nr i 5 for all n and similarly 1Nr i $ i boundedly as n $4 EF([> R) and is closed under bounded convergence .Then contains n n and therefore H all bounded H —measurablerealvaluedfunctionsonH [= (This mayH be stated i 5 = B[ H as follows: the smallest vector space of bounded functions which is closed under Lemma 18.57. Suppose that ([> ) is a locally compact second countable bounded convergence and contains EF([> R) is the space of bounded [ — Hausdor space.2 Then: measurable real valued functions on [=) B 1. every open subset X [ is —compact. Proof. Let Y 5 g be an open subset of [ and for q 5 let N 2. If I [ isaclosedset,thereexistopensetsYq [ such that Yq & I as iq({):=min(q gY f ({)> 1) for all { 5 [= q $4= · 3. To each open set X [ there exists iq ! X (i.e. iq 5 Ff (X> [0> 1])) such Notice that i = ! g f where ! (w)=min(qw> 1) (see Figure 18.3) which q q Y q that limq$4 iq =1X = is continuous and hence i 5 EF([> ) for all q= Furthermore, i converges q R q 4. [ = (Ff([> R))> i.e. the — algebra generated by Ff([) is the Borel boundedly to 1 =1 as and therefore 1 for all B gY f A0 Y q $4 Y 5 Y 5 = —algebraon[= Since is a — class, the result now follows by an applicationH of Corollary 18.54. Proof. 1. Let X be an open subset of [> be a countable base for and y 1 V X := Z 5 : Z¯ X and Z¯ is compact = 0.75 V { V } For each { 5 X> by Proposition 12.7, there exists an open neighborhood Y of { such that Y¯ X and Y¯ is compact. Since is a base for the 0.5 topology > there exists Z 5 such that { 5 Z Y=VBecause Z¯ Y>¯ it follows that Z¯ is compact andV hence Z 5 X = As { 5 X was arbitrary, 4 V 0.25 X = ^ X = Let Z be an enumeration of X and set N := ^q Z¯ = V { q}q=1 V q n=1 n Then Nq % X as q $4and Nq is compact for each q= 4 f f 0 2. Let Nq q=1 be compact subsets of I such that Nq % I as q $4and 0 0.5 1 1.5 2 { } f set Yq := Nq = [ Nq= Then Yq & I and by Proposition 12.5, Yq is open x for each q= \ Plots of ! >! and ! = 1 2 3 2 For example any separable locally compact metric space and in particular any 1 q Recall that EF([> R) are the bounded continuous functions on [= open subset of R = 18.4 Product —Algebras 271 272 18 Measurability

4 3. Let X [ be an open set and Nq q=1 be compact subsets of X such Moreover, suppose that D is either finite or countably infinite, [ 5 for that N % X= By Lemma 12.8, there{ } exist i ! X such that i =1on each 5 D> and = ( ) for each 5 D= Then the product — algebraE q q q M E Nq= These functions satisfy, 1X = limq$4 iq= satisfies 4. By item 3., 1X is (Ff([> R)) — measurable for all X 5 and hence (Ff([> R))= Therefore [ = ( ) (Ff([> R))= The converse B 5D = H : H 5 for all 5 D = (18.14) inclusion holds because continuous functions are always Borel measurable. M Ã( E )! Y5D

In particular if D = 1> 2>===>q > then [ = [1 [2 [ and Here is a variant of Corollary 18.55. { } × ×···× q

1 2 = ( 1 2 )> Corollary 18.58. Suppose that ([> ) is a second countable locally compact M M ··· Mq M ×M ×···×Mq Hausdor space and [ = ( ) is the Borel —algebraon[= If is a 4 B H where 1 2 is as deﬁned in Notation 10.26. subspace of c ([> R) which is closed under bounded convergence and contains M ×M ×···×Mq Ff([> )> then contains all bounded [ — measurable real valued functions 1 1 R Proof. Since ^ ( ) ^ ( )> it follows that on [= H B E M := ^ 1( ) ^ 1( ) = = Proof. By Item 3. of Lemma 18.57, for every X 5 the characteristic F E M 5DM function, 1X > may be written as a bounded pointwise limit of functions from Conversely, ¡ ¢ ¡ ¢ Ff ([> ) = Therefore 1X 5 for all X 5 = Since is a — class, the proof is 1 1 1 R ( ( )) = (( )) = ( ) ﬁnished with an applicationH of Corollary 18.54 F E E M holds for all implies that

^ 1( ) 18.4 Product —Algebras M F

and hence that 5D = We now prove Eq. (18.14). Since we are Let ( ) be a collection of measurable spaces = = M F [> 5D [ [D [ assuming that [ 5 for each 5 D> we see that { M } 5D E and : [D $ [ be the canonical projection map as in Notation 2.2.Q ^ 1( ) H : H 5 for all 5 D Deﬁnition 18.59 (Product —Algebra).The product —algebra, E E (5D ) > is the smallest —algebraon[ such that each for 5 D is Y 5DM measurable, i.e. and therefore by Eq. (18.13)

1 5D := ( : 5 D)= ^ ( ) = M M 1 5D = ^ ( ) H : H 5 for all 5 D = Applying Proposition 18.25 in this setting¡ implies the following¢ proposi- M E Ã( E )! Y5D tion. ¡ ¢ This last statement is true independent as to whether D is countable or not. Proposition 18.60. Suppose \ is a measurable space and i : \ $ [ = [D For the reverse inclusion it su!ces to notice that since D is countable, is a map. Then i is measurable i i : \ $ [ is measurable for all 1 5 D= In particular if D = 1> 2>===>q so that [ = [1 [2 [ and H = _5D (H) 5 5D { } × ×···× q M i(|)=(i1(|)>i2(|)>===>iq(|)) 5 [1 [2 [q> then i : \ $ [D is Y5D measurable i i : \ $ [ is measurable× for×···× all l= l l and hence Proposition 18.61. Suppose that ([ > ) is a collection of measurable M 5D spaces and generates for each 5 D> then H : H 5 for all 5 D = E M M E 5DM Ã(5D )! = ^ 1( ) (18.13) Y 5DM 5D E ¡ ¢ 18.4 Product —Algebras 273 274 18 Measurability

p q p+q Remark 18.62. One can not relax the assumption that [ 5 in the second Corollary 18.65. After identifying R R with R as above and letting E q × part of Proposition 18.61. For example, if [1 = [2 = 1> 2 and 1 = 2 = q denote the Borel —algebra on R > we have { } E E BR 1 > then ( 1 2)= >>[1 [2> (1> 1) while (( 1) ( 2)) = q—times {{[1}}[2 E ×E { × { }} E × E 2 × = p+q = Rq Rp and Rq = R R= Theorem 18.63. Let [ be a sequence of sets where D is at most BR B B B B ··· B { }5D countable. Suppose for each 5 D we are given a countable set 2[ = Let z }| { E 18.4.1 Factoring of Measurable Maps = ( ) be the topology on [ generated by and [ be the product space E with equipped with the product topologyE := ( ) Then the 5D [ 5D = Lemma 18.66. Suppose that (\> ) is a measurable space and I : [ $ \ is Borel — algebra = ( ) is the same as the product —algebra:E F [ a map. Then to every ( ( ) ¯ ) — measurable function, : ¯ there is Q B I > R K [ $ R> B ¯ = > a ( > R¯ ) — measurable function k : \ $ R such that K = k I= B[ 5DB[ F B Proof. First suppose that K =1 where D 5 (I )=I 1( )= Let where [ = ( ( )) = ( ) for all 5 D= D B E E 1 F In particular if D = 1> 2>===>q and each ([ > ) is a second countable E 5 such that D = I (E) then 1D =1I 1(E) =1E I and hence the { } l l F topological space, then Lemma is valid in this case with k =1E= More generally if K = dl1Dl is a simple function, then there exists El 5 such that 1D =1E I and := ( )= ( )=: F l l P [ 1 2 q [1 [q [1 [q = hence = with := 1 — a simple function on ¯ For general B ··· B ×···×B B ··· B K k I k dl El R= ¯ Proof. By Proposition 10.25, the topology may be described as the ((I )> ) — measurable function, K> from [ $ R> choose simple functions F P ¯ 1 Kq converging to K= Let kq be simple functions on such that Kq = kq I= smallest topology containing = ^5D ( )= Now is the countable union R of countable sets so is still countable.E ThereforeE byE Proposition 18.17 and Then it follows that Proposition 18.61, K = lim Kq = lim sup Kq = lim sup kq I = k I q$4 = ( )=( ( )) = ( )= ( ) q$4 q$4 B[ E E 5D E where := lim sup — a measurable function from to ¯ = 5D( )= 5D [ = k q$4 kq \ R= B The following is an immediate corollary of Proposition 18.25 and Lemma 18.66. Corollary 18.64. If ([ >g ) are separable metric spaces for l =1>===>q>then l l Corollary 18.67. Let [ and D be sets, and suppose for 5 D wearegivea

[1 [q = ([1 [q) measurable space (\> ) and a function i : [ $ \= Let \ := 5D \> B ··· B B ×···× F := 5D be the product —algebraon\ and := (i : 5 D) where [l is the Borel —algebraon[l and ([1 [ ) is the Borel F F M Q B B ×···× q be the smallest —algebraon[ such that each i is measurable. Then the — algebra on [1 [q equipped with the metric topology associ- × ··· × q function I : [ $ \ defined by [I ({)] := i({) for each 5 D is ( > ) ated to the metric g({> |)= l=1 gl({l>|l) where { =({1>{2>===>{q) and ¯ M F —measurableandafunctionK : [ $ R is ( > R¯ ) —measurablei there | =(|1>|2>===>|q)= M B ¯ P exists a ( > R¯ ) — measurable function k from \ to R such that K = k I= Proof. This is a combination of the results in Lemma 10.28, Exercise 10.10 F B and Theorem 18.63. Because all norms on finite dimensional spaces are equivalent, the usual 18.5 Exercises Euclidean norm on Rp Rq is equivalent to the “product” norm defined by × Exercise 18.8. Prove Corollary 18.23. Hint: See Exercise 18.3. ({> |) p q = { p + | q = k kR R k kR k kR × Exercise 18.9. If is the — algebra generated by 2[ > then is the Hence by Lemma 10.28, the Euclidean topology on p+q isthesameasthe R union of the — algebrasM generated by countable subsetsE = M(Folland, product topology on Rp+q = Rp Rq= Here we are identifying Rp Rq with F E × × Problem 1.5 on p.24.) Rp+q by the map p q p+q Exercise 18.10. Let ([> ) be a measure space and iq : [ $ F be a se- ({> |) 5 R R $ ({1>===>{p>|1>===>|q) 5 R = M × quence of measurable functions on [= Show that { : limq$4 iq({) exists in F 5 These comments along with Corollary 18.64 proves the following result. = { } M 18.5 Exercises 275

Exercise 18.11. Show that every monotone function i : R $ R is ( R> R) — measurable. B B 19 Exercise 18.12. Show by example that the supremum of an uncountable family of measurable functions need not be measurable. (Folland problem 2.6 on p. 48.) Measures and Integration

Exercise 18.13. Let [ = 1> 2> 3> 4 >D= 1> 2 >E= 2> 3 and P := 1 > 1 = Show (P) = {(P) in} this case.{ } { } { D E} H 6 H

Definition 19.1. A measure on a measurable space ([> ) is a function : $ [0> 4] such that M M 1. (>)=0and q 2. (Finite Additivity) If Dl l=1 are pairwise disjoint, i.e. Dl _ Dm = > when l = m> then { } M 6 q q ( Dl)= (Dl)= l=1 l=1 [ X 3. (Continuity) If D 5 and D % D> then (D ) % (D)= q M q q We call a triple ([> >)> where ([> ) is a measurable space and : $ [0> 4] is a measure,M a measure space.M M Remark 19.2. Properties 2) and 3) in Definition 19.1 are equivalent to the following condition. If D 4 are pairwise disjoint then { l}l=1 M 4 4 ( Dl)= (Dl)= (19.1) l=1 l=1 [ X To prove this assume that Properties 2) and 3) in Definition 19.1 hold and q 4 4 Dl l=1 are pairwise disjoint. Letting Eq := Dl % E := Dl> we { } M l=1 l=1 have S S 4 q 4 (3) (2) ( Dl)=(E) =lim(Eq) = lim (Dl)= (Dl)= q$4 q$4 l=1 l=1 l=1 [ X X

Conversely, if Eq. (19.1) holds we may take Dm = > for all mAqto see that Property 2) of Deﬁnition 19.1 holds. Also if Dq % D> let Eq := Dq Dq1 with 4 q \ 4 D0 := >= Then Eq q=1 are pairwise disjoint, Dq = ^m=1Em and D = ^m=1Em= So if Eq. (19.1){ holds} we have 278 19 Measures and Integration 19 Measures and Integration 279

4 4 Proposition 19.6. Let ([> >) be a measure space. Set (D)= ^m=1Em = (Em) M m=1 X := Q [ : < I 5 Q I and (I )=0 ¡ q ¢ N { M3 } q = lim (Em)= lim (^m=1Em) = lim (Dq)= and q$4 q$4 q$4 m=1 ¯ = D ^ Q : D 5 >Q 5 > X M { M N} Proposition 19.3 (Basic properties of measures). Suppose that ([> >) see Fig. 19.1. Then ¯ is a —algebra.Define ¯(D ^ Q)=(D),then¯ is 4 M is a measure space and H>I 5 and H > then : M the unique measure on ¯ which extends . M { m}m=1 M M 1. (H) (I ) if H I= Proof. Clearly [> >5 ¯ = Let D 5 and Q 5 and choose I 5 M M N M 2. (^Hm) (Hm)= 3. If (H1) ? 4 and Hm & H, i.e. H1 H2 H3 === and H = _mHm> then P (Hm) & (H) as m $4= Proof. 1. Since I = H ^ (I H)> \ (I)=(H)+(I H) (H)= \ ˜ 2. Let Hm = Hm (H1 ^ ^ Hm1) so that the Hm ’s are pair-wise disjoint \ ···˜ and H = ^Hm= Since Hm Hm it follows from Remark 19.2 and part (1)> that e Fig. 19.1. Completing a —algebra. e (H)= (Hm) (Hm)= 3. Define := then X whichX implies that Gl H1 Hl Gl % H1 eH such that Q I and (I )=0.SinceQ f =(I Q) ^ I f> \ \ \ (H1) (H) = lim (Gl)=(H1) lim (Hl) (D ^ Q)f = Df _ Q f = Df _ (I Q ^ I f) l$4 l$4 \ =[Df _ (I Q)] ^ [Df _ I f] which shows that liml$4 (Hl)=(H). \ where [Df _ (I Q)] 5 and [Df _ I f] 5 = Thus ¯ is closed under \ N M M complements. If Dl 5 and Ql Il 5 such that (Il)=0then M ¯ M Definition 19.4. AsetH [ is a null set if H 5 and (H)=0.IfS is ^(Dl^Ql)=(^Dl)^(^Ql) 5 since ^Dl 5 and ^Ql Îl and (Îl) M M M some “property” which is either true or false for each { 5 [> we will use the (I )=0= Therefore, ¯ is a — algebra. Suppose D ^ Q1 = E ^ Q2 with l M terminology S a.e. (to be read S almost everywhere) to mean D> E 5 and Q1>Q2> 5 .ThenD D ^ Q1 D ^ Q1 ^ I2 = E ^ I2 which showsP thatM N H := { 5 [ : S is false for { (D) (E)+(I )=(E)= { } 2 is a null set. For example if i and j are two measurable functions on Similarly, we show that (E) (D) so that (D)=(E) and hence ¯(D ^ ([> >)>i= j a.e. means that (i = j)=0= Q):=(D) is well defined. It is left as an exercise to show ¯ is a measure, M 6 i.e. that it is countable additive. Definition 19.5. Ameasurespace([> >) is complete if every subset of Many theorems in the sequel will require some control on the size of a M a null set is in ,i.e.forallI [ such that I H 5 with (H)=0 measure = The relevant notion for our purposes (and most purposes) is that M M implies that I 5 . of a — finite measure defined next. M Definition 19.7. Suppose [ is a set, 2[ and : $ [0> 4] is a function. The function is — finiteE on Mif there exists HM5 such that E q E 280 19 Measures and Integration 19.1 Example of Measures 281

(H ) ? 4 and [ = ^4 H = If is a — algebra and is a measure on 4 q q=1 q M ( )=inf ( ( ) ( )) : 4 ( ] (19.4) which is — finite on we will say ([> >) is a — finite measure I D I el I dl D ^l=1 dl>el ( ) Mspace. M M l=1 X4 4 The reader should check that if is a finitely additive measure on an =inf (I(el) I (dl)) : D (dl>el] = (19.5) (l=1 l=1 ) algebra, > then is — finite on i there exists [q 5 such that X a [ % [ andM ([ ) ? 4= M M q q In fact the map I $ I is a one to one correspondence between right continuous functions I with I(0) = 0 on one hand and measures on R such that (M) ? 4 on any bounded set M 5 on the other. B 19.1 Example of Measures BR Proof. This follows directly from Proposition 26.18 and Theorem 26.2 Most — algebras and -additive measures are somewhat di!cult to describe below. It can also be easily derived from Theorem 26.26 below. and define. However, one special case is fairly easy to understand. Namely Example 19.9. The most important special case of Theorem 19.8 is when suppose that 2[ is a countable or finite partition of [ and 2[ is the — algebraF which consists of the collection of sets D [ suchM that I ({)={> in which case we write p for I = The measure p is called Lebesgue measure. D = ^ 5 : D = (19.2) { F } Theorem 19.10. Lebesgue measure p is invariant under translations, i.e. for E 5 and { 5 R> It is easily seen that is a — algebra. BR Any measure : M $ [0> 4] is determined uniquely by its values on = p({ + E)=p(E)= (19.6) M F Conversely, if we are given any function : $ [0> 4] we may define, for Moreover, p istheuniquemeasureon such that p((0> 1]) = 1 and Eq. F BR D 5 > (19.6) holds for E 5 and { 5 R= Moreover, p has the scaling property M BR (D)= ()= ()1D 5 D 5 p(E)= p(E) (19.7) F3X XF | | where 1D is one if D and zero otherwise. We may check that is a 4 where 5 R>E5 R and E := { : { 5 E = measure on = Indeed, if D = l=1 Dl and 5 > then D i Dl for B { } M F 4 one and hence exactly one Dl= Therefore 1D = l=1 1Dl and hence Proof. Let p (E):=p({ + E)> then one easily shows that p is a ` { { measure on such that p ((d> e]) = e d for all d?e=Therefore, p = p P4 R { { by the uniquenessB assertion in Theorem 19.8. For the converse, suppose that (D)= ()1D = () 1D l is translation invariant and ((0 1]) = 1 Given we have 5 5 l=1 p p > = q 5 N> F F X4 X 4 X 1 1 1 = ()1 = (D ) q n n q n Dl l (0> 1] = ^n=1( > ]=^n=1 +(0> ] = l=1 5 l=1 q q q q X XF X µ ¶ as desired. Thus we have shown that there is a one to one correspondence Therefore, between measures on and functions : $ [0> 4]= q M F n 1 1 The construction of measures will be covered in Chapters 27 — 29 below. 1=p((0> 1]) = p +(0> ] q q However, let us record here the existence of an interesting class of measures. =1 nX µ ¶ q 1 1 Theorem 19.8. To every right continuous non-decreasing function I : = p((0> ]) = q p((0> ])= there exists a unique measure on such that q · q R $ R I R =1 B nX I ((d> e]) = I (e) I (d) ;4?d e?4 (19.3) That is to say 1 ((0 ]) = 1 Moreover, if D 5 then p > @q= BR q 282 19 Measures and Integration 19.2 Integrals of Simple functions 283

o Similarly, p((0> q ]) = o@q for all o> q 5 N and therefore by the translation Proposition 19.12. Let 5 F and ! and # be two simple functions, then L invariance of p> satisﬁes:

p((d> e]) = e d for all d> e 5 Q with d?e= 1. L(!)=L(!)= (19.8) Finally for d> e 5 R such that d?e>choose dq>eq 5 Q such that eq & e and 2. dq % d> then (dq>eq] & (d> e] and thus L(! + #)=L(#)+L(!)= p((d> e]) = lim p((dq>eq]) = lim (eq dq)=e d> q$4 q$4 3. If ! and # are non-negative simple functions such that ! # then i.e. p is Lebesgue measure. To prove Eq. (19.7) we may assume that =0 L(!) L(#)= 1 6 since this case is trivial to prove. Now let p(E):= p(E)= It is easily | | Proof. Let us write = for the set 1( ) and ( = ) for checked that p is again a measure on R which satisfies ! | ! | [ ! | B ( ! = | )=(!1 ( | {)) so that} { } 1 1 { } { } p((d> e]) = p ((d> e]) = (e d)=e d L(!)= |(! = |)= if A0 and X|5F (( ]) = 1 ([ )) = 1 ( )= p d> e p e> d e d e d We will also write ! = d> # = e for !1( d ) _ #1( e )= This notation is | | | | { } { } { } more intuitive for the purposes of this proof. Suppose that 5 F then if ?0= Hence p = p= We are now going to develop integration theory relative to a measure. The L (!)= |(! = |)= |(! = |@) integral defined in the case for Lebesgue measure, p> will be an extension of |5 |5 the standard Riemann integral on R= XF XF = } (! = })=L(!) 19.1.1 ADD: Examples of Measures X}5F provided that =0= The case =0is clear, so we have proved 1. Suppose BRUCE: ADD details. that ! and # are6 two simple functions, then 1. Product measure for the flipping of a coin. 2. Haar Measure L(! + #)= }(! + # = }) 3. Measure on embedded submanifolds, i.e. Hausdor measure. X}5F 4. Wiener measure. = }(^ ! = z> # = } z ) z5F { } 5. Gibbs states. }5F 6. Measure associated to self-adjoint operators and classifying them. X = } (! = z> # = } z) X}5F zX5F 19.2 Integrals of Simple functions = (} + z)(! = z> # = }) }>zX5F Let ([> >) be a fixed measure space in this section. = }(# = })+ z(! = z) M }5F z5F Definition 19.11. Let F = C or [0> 4) and suppose that ! : [ $ F is X X = L (#)+L (!)= a simple function as in Definition 18.41. If F = C assume further that 1 (! ( | )) ? 4 for all | =0in C= For such functions !> define L(!) by { } 6 which proves 2. For 3. if ! and # are non-negative simple functions such that ! # L (!)= |(!1( | ))= { } X|5F 284 19 Measures and Integration 19.3 Integrals of positive functions 285

L(!)= d(! = d)= d(! = d> # = e) Proof. Suppose that ! : [ $ [0> 4) is a simple function, then ! = d0 d>e0 }5[0>4) }1 !=} and X X { } e(! = d> # = e)= e(# = e)=L(#)> P 0 0 ! = ({) }1 !=} ({)= } ({)1 !=} ({) d>eX Xe { } { } X[ {X5[ }5X[0>4) }5X[0>4) {X5[ wherein the third inequality we have used ! = d> # = e = > if dAe= { } = }( ! = } )= !g= { } [ }5X[0>4) Z 19.3 Integrals of positive functions So if ! : [ $ [0> 4) is a simple function such that ! i> then

Definition 19.13. Let O+ = O+ ( )= i : [ $ [0> 4]:i is measurable = = Define M { } !g ! i= [ Z X[ X[ i ({) g ({)= ig := sup L (!):! is simple and ! i = Taking the sup over ! in this last equation then shows that { } Z[ Z[ + ig i= We say the i 5 O is integrable if [ ig? 4= If D 5 > let M Z[ [ R X i ({) g ({)= ig := 1Dig= For the reverse inequality, let [ be a finite set and Q 5 (0> 4)= D D [ Q Z Z Z Set i ({)=min Q>i({) and let !Q> be the simple function given by Q{ } Remark 19.14. Because of item 3. of Proposition 19.12, if ! is a non-negative !Q>({):=1({)i ({)= Because !Q>({) i({)> simple function, [ !g = L(!) so that [ is an extension of L= This exten- Q sion still has the monotonicity property if L : namely if 0 i j then i = !Q> = !Q>g ig= R R [ [ X X[ Z Z ig =sup L (!):! is simple and ! i Q { } Since i % i as Q $4> we may let Q $4in this last equation to concluded Z[ sup L(!):! is simple and ! j jg= i ig= { } [ Z [ X Z Similarly if fA0> Since is arbitrary, this implies fig = f ig= [ [ Z Z i ig= Also notice that if i is integrable, then ( i = 4 )=0= [ { } X[ Z Lemma 19.15 (Sums as Integrals). Let [ be a set and : [ $ [0> 4] be [ afunction,let = ({){ on =2 > i.e. {5[ M + Theorem 19.16 (Monotone Convergence Theorem). Suppose iq 5 O P + (D)= ({)= is a sequence of functions such that iq % i (i is necessarily in O ) then {X5D iq % i as q $4= If i : [ $ [0> 4] is a function (which is necessarily measurable), then Z Z Proof. Since iq ip i> for all q p?4> ig = i= Z[ [ X iq ip i Z Z Z 286 19 Measures and Integration 19.3 Integrals of positive functions 287 from which if follows iq is increasing in q and If P := [ ig ? 4> then R R P lim iq i= (19.9) (i = 4) (i q) $ 0 as q $4 q$4 q Z Z For the opposite inequality, let ! : [ $ [0> 4) be a simple function such and i 1@q % iA0 with (i 1@q) qP ? 4 for all q= { } { } that 0 ! i> 5 (0> 1) and [q := iq ! = Notice that [q % [ and + { } Corollary 19.18. If iq 5 O is a sequence of functions then iq 1[q ! and so by definition of iq> R 4 4 iq 1[q ! = 1[q != (19.10) iq = iq= q=1 q=1 Z Z Z Z X X Z Then using the continuity property of > 4 4 In particular, if q=1 iq ? 4 then q=1 iq ? 4 a.e.

lim 1[q ! = lim 1[q |1 !=| Proof. FirstP o weR show that P q$4 q$4 { } |A0 Z Z X ( + )= + = lim |([q _ ! = | )= | lim ([q _ ! = | ) i1 i2 i1 i2 q$4 { } q$4 { } |A0 |A0 Z Z Z X X by choosing non-negative simple function !q and #q such that !q % i1 and = | lim ( ! = | )= != q$4 { } #q % i2= Then (!q + #q) is simple as well and (!q + #q) % (i1 + i2) so by the |A0 X Z monotone convergence theorem, Thisidentityallowsustoletq $4in Eq. (19.10) to conclude (i1 + i2) = lim (!q + #q) = lim !q + #q 1 q$4 q$4 ! lim iq= Z Z µZ Z ¶ [ q$4 Z Z = lim !q + lim #q = i1 + i2= q$4 q$4 Since this is true for all non-negative simple functions ! with ! i; Z Z Z Z Q 4 1 Now to the general case. Let jQ := iq and j = iq> then jQ % j and so i =sup ! : ! is simple and ! i lim iq= q$4 q=1 1 Z ½Z[ ¾ Z again by monotone convergence theoremP and the additivityP just proved, Because 5 (0> 1) was arbitrary, it follows that i lim iq which com- q$4 4 Q Q bined with Eq. (19.9) proves the theorem. iq := lim iq = lim iq R R Q$4 Q$4 The following simple lemma will be use often in the sequel. q=1 Z q=1 Z Z q=1 X X 4X Lemma 19.17 (Chebyshev’s Inequality). Suppose that i 0 is a mea- = lim jQ = j =: iq= Q$4 surable function, then for any %A0> q=1 Z Z Z X 1 (i %) ig= (19.11) % Z[ Remark 19.19. It is in the proof of this corollary (i.e. the linearity of the integral) that we really make use of the assumption that all of our functions are In particular if [ ig ? 4 then (i = 4)=0(i.e. i?4 a.e.) and the set iA0 is — ﬁnite. measurable. In fact the deﬁnition ig makes sense for all functions i : [ $ { } R [0> 4] not just measurable functions. Moreover the monotone convergence 1 1 R Proof. Since 1 i% 1 i% i i> theorem holds in this generality with no change in the proof. However, in { } { } % % the proof of Corollary 19.18, we use the approximation Theorem 18.42 which 1 1 (i %)= 1 i% g 1 i% ig ig= relies heavily on the measurability of the functions to be approximated. { } { } % % Z[ Z[ Z[ 288 19 Measures and Integration 19.3 Integrals of positive functions 289

The following Lemma and the next Corollary are simple applications of it su!ces to show 4 Corollary 19.18. 1 =1 4 — a.e. (19.12) Dq ^q=1Dq q=1 Lemma 19.20 (The First Borell — Carntelli Lemma). Let ([> >) be X M 4 4 a measure space, Dq 5 > and set Now 1D 1^4 D and 1D ({) =1^4 D ({) i { 5 Dl _ Dm for M q=1 q q=1 q q=1 q q=1 q some l = m> that is 6 4 P6 P D i.o. = { 5 [ : { 5 D for infinitely many q’s = D = { q } { q } q 4 Q=1 qQ { : 1D ({) =1^4 D ({) = ^l?m Dl _ Dm \ [ q 6 q=1 q 4 ( q=1 ) If q=1 (Dq) ? 4 then ( Dq i.o. )=0. X { } and the latter set has measure 0 being the countable union of sets of measure PProof. (First Proof.) Let us first observe that zero. This proves Eq. (19.12) and hence the corollary. 4 Notation 19.22 If p is Lebesgue measure on R>iis a non-negative Borel Dq i.o. = { 5 [ : 1D ({)=4 = B e { } q measurable function and d?ewith d> e 5 ¯> we will often write i ({) g{ ( q=1 ) R d X or e igp for igp= 4 d (d>e]_R R Hence if q=1 (Dq) ? 4 then ExampleR 19.23.RSuppose 4 ?d?e?4>i5 F([d> e]> [0> 4)) and p be P 4 4 4 Lebesgue measure on = Also let = d = dn ?dn ? ?dn = e be a ( )= 1 = 1 R n 0 1 qn 4 A Dq Dq g Dq g { ··· } q=1 q=1 [ [ q=1 sequence of refining partitions (i.e. n n+1 for all n) such that X X Z Z X 4 n n+1 mesh(n):=max dm dm1 : m =1>===>qn $ 0 as n $4= implies that 1Dq ({) ? 4 for -a.e.{= That is to say ( Dq i.o. )=0. { } =1 { } q For each let ¯ ¯ (Second Proof.)P Of course we may give a strictly measure theoretic proof of n> ¯ ¯ this fact: qn1 3 4 n n in({)=i(d)1 d + min i({):do { do+1 1(dn>dn ]({) { } o o+1 o=0 (Dq i.o.)= lim C DqD Q$4 X © ª qQ [ then in % i as n $4and so by the monotone convergence theorem, lim (Dq) e e Q$4 qQ := = lim X igp igp in gp d [d>e] n$4 d 4 Z Z Z and the last limit is zero since q=1 (Dq) ? 4= qn1 n n n n 4 = lim min i({):do { do+1 p (do >do+1] Corollary 19.21. Suppose thatP([> >) is a measure space and Dq n$4 M { }q=1 o=0 is a collection of sets such that (Dl _ Dm)=0for all l = m> then X © ª ¡ ¢ M 6 e 4 = i({)g{= 4 d (^q=1Dq)= (Dq)= Z q=1 The latter integral being the Riemann integral. X Proof. Since We can use the above result to integrate some non-Riemann integrable functions: 4 ( )= 1 4 and ^q=1Dq ^q=1Dq g [ Example 19.24. For all A0> 4 Z 4 (D )= 1 g 4 q Dq { 1 1 [ h gp({)= and gp({)== q=1 Z q=1 1+{2 X X Z0 ZR 290 19 Measures and Integration 19.3 Integrals of positive functions 291

The proof of these identities are similar. By the monotone convergence the- we find orem, Example 19.23 and the fundamental theorem of calculus for Riemann 4 1 4 integrals (or see Theorem 8.13 above or Theorem 19.40 below), i({)gp({)= 2q g{ 2q4=4? 4= [0>1] [0>1] { uq 4 Q Q Z q=1 Z | | q=1 { { { X X h gp({) = lim h gp({)= lim h g{ p 0 Q$4 0 Q$4 0 In particular, p(i = 4)=0> i.e. that i?4 for almost every { 5 [0> 1] and Z Z Z this implies that 1 { Q 1 = lim h 0 = Q$4 | 4 q 1 and 2 ? 4 for a.e. { 5 [0> 1]= { u q=1 | q| 1 Q 1 Q 1 X gp({) = lim gp({) = lim g{ This result is somewhat surprisingp since the singularities of the summands 1+{2 Q$4 1+{2 Q$4 1+{2 ZR ZQ ZQ form a dense subset of [0> 1]= = lim tan1(Q) tan1(Q) = = Q$4 Proposition 19.26. Suppose that i 0 is a measurable function. Then £ s ¤ Let us also consider the functions { > [ ig =0i i =0a.e. Also if i>j 0 are measurable functions such that i j a.e. then ig jg= In particular if i = j a.e. then ig = jg= 1 1 1 R gp({)= lim 1 1 ({) gp({) s ( q >1] s Proof. If iR=0a.e.R and ! i is a simple function thenR ! =0R a.e. (0>1] { q$4 0 { Z Z This implies that (!1( | )) = 0 for all |A0 and hence !g =0 1 +1 1 [ 1 {s and therefore =0{ }Conversely, if =0 then by Chebyshev’s =lim =lim [ ig = ig > s g{ R q$4 1 { q$4 1 s Inequality (Lemma 19.17), Z q ¯1@q ¯ R R 1 if s?1 ¯ = 1s ¯ 4 if sA1 (i 1@q) q ig =0for all q= ½ Z If s =1we find 4 Therefore, (iA0) q=1 (i 1@q)=0> i.e. i =0a.e. For the second 1 assertion let H be the exceptional set where iAj>i.e. H := { 5 [ : i({) A 1 1 1 P { ( ) = lim = lim ln( ) = f f s gp { g{ { 1@q 4= j({) = By assumption H is a null set and 1H i 1H j everywhere. Because (0>1] { q$4 1 { q$4 | } Z Z q j =1Hf j +1Hj and 1Hj =0a.e., 4 Example 19.25. Let uq =1 be an enumeration of the points in Q _ [0> 1] and { }q define jg = 1Hf jg + 1Hjg = 1Hf jg 4 1 Z Z Z Z i({)= 2q f f f q=1 { uq and similarly ig = 1H ig= Since 1H i 1H j everywhere, X | | with the convention that p R R ig = 1 f ig 1 f jg = jg= 1 H H =5if { = uq= Z Z Z Z { u | q| Since, By Theorem 19.40, p Corollary 19.27. Suppose that i is a sequence of non-negative measurable { q} 1 1 functions and i is a measurable function such that i % i o anullset,then 1 1 uq 1 q g{ = s g{ + s g{ 0 { uq uq { uq 0 uq { Z | | Zs sZ s s iq % i as q $4= =2 { u 1 2 u { uq =2 1 u u p q|uq q |0 q q Z Z 4> ¡ ¢ 292 19 Measures and Integration 19.4 Integrals of Complex Valued Functions 293

Proof. Let H [ be a null set such that iq1Hf % i1Hf as q $4= Then Remark 19.31. Since by the monotone convergence theorem and Proposition 19.26, i i i+ + i> ± | | a measurable function i is integrable i i g ? 4= Hence iq = iq1Hf % i1Hf = i as q $4= | | R Z Z Z Z 1 L (; R):= i : [ $ R¯ : i is measurable and i g ? 4 = | | ½ Z[ ¾ Lemma 19.28 (Fatou’s Lemma). If i : [ $ [0> 4] is a sequence of q If i>j 5 L1 (; R) and i = j a.e. then i = j a.e. and so it follows from measurable functions then ± ± Proposition 19.26 that ig = jg= In particular if i>j 5 L1 (; R) we may define lim inf iq lim inf iq R R q$4 q$4 (i + j) g = kg Z Z Z[ Z[ Proof. Define jn := inf iq so that jn % lim infq$4 iq as n $4= Since qn where k is any element of i + j= j i for all n q> n q Proposition 19.32. The map for all jn iq q n 1 i 5 L (; R) $ ig 5 R Z Z [ and therefore Z is linear and has the monotonicity property: ig jg for all i>j 5 jn lim inf iq for all n= q$4 L1 (; R) such that i j a.e. Z Z R R We may now use the monotone convergence theorem to let n $4to find Proof. Let i>j 5 L1 (; R) and d> e 5 R= By modifying i and j on a null MCT set, we may assume that i>j are real valued functions. We have di + ej 5 lim inf iq = lim jn = lim jn lim inf iq= 1 q$4 n$4 n$4 q$4 L (; R) because Z Z Z Z 1 di + ej d i + e j 5 L (; R) = | | | || | | || | If d?0> then 19.4 Integrals of Complex Valued Functions (di)+ = di and (di) = di+ ¯ so that Definition 19.29. A measurable function i : [ $ R is integrable if i+ := 1 i1 i0 and i = i 1 i0 are integrable.WewriteL (; R) for the space { } { } 1 di = d i + d i = d( i i )=d i= of real valued integrable functions. For i 5 L (; R) > let + + Z Z Z Z Z Z A similar calculation works for dA0 and the case d =0is trivial so we have ig = i+g ig shown that Z Z Z di = d i= Convention: If i>j : [ $ R¯ are two measurable functions, let i + j denote the collection of measurable functions k : [ $ ¯ such that k({)= Z Z R Now set k = i + j= Since k = k k > i({)+j({) whenever i({)+j({) is well defined, i.e. is not of the form 44 or + 4 + 4= We use a similar convention for i j= Notice that if i>j 5 L1 (; ) R k+ k = i+ i + j+ j and k1>k2 5 i + j> then k1 = k2 a.e. because i ? 4 and j ? 4 a.e. | | | | or Notation 19.30 (Abuse of notation) We will sometimes denote the in- k+ + i + j = k + i+ + j+= tegral [ ig by (i) = With this notation we have (D)= (1D) for all D 5 = Therefore, MR 294 19 Measures and Integration 19.4 Integrals of Complex Valued Functions 295

Proof. Start by writing ig= Uhl with U 0= We may assume that k+ + i + j = k + i+ + j+ [ U = ig A 0 since otherwise there is nothing to prove. Since Z Z Z Z Z Z [ R and hence ¯R ¯ U =¯ hl ¯ ig= hlig= Re hli g + l Im hli g> k = k+ k = i+ + j+ i j = i + j= Z[ Z[ Z[ Z[ ¡ ¢ ¡ ¢ Z Z Z Z Z Z Z Z Z l it must be that [ Im h i g =0= Using the monotonicity in Proposition Finally if i+ i = i j = j+ j then i+ + j j+ + i which implies 19.26, that R £ ¤ i+ + j j+ + i ig = Re hli g Re hli g i g= | | Z Z Z Z ¯Z[ ¯ Z[ Z[ Z[ or equivalently that ¯ ¯ ¡ ¢ ¯ ¡ ¢¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 i = i+ i j+ j = j= Proposition 19.35. Let i>j 5 L () > then Z Z Z Z Z Z 1 1. The set i =0 is — finite, in fact i q % i =0 and ( i The monotonicity property is also a consequence of the linearity of the inte- 1 { 6 } {| | } { 6 } | | q ) ? 4 for all q. gral, the fact that i j a.e. implies 0 j i a.e. and Proposition 19.26. 2. The following are equivalent a) i = j for all H 5 H H M Definition 19.33. A measurable function i : [ $ is integrable if b) i j =0 C [R | R| i g ? 4= Analogously to the real case, let [ | | c) iR = j a.e= R 1 Proof. 1. By Chebyshev’s inequality, Lemma 19.17, L (; C):= i : [ $ C : i is measurable and i g ? 4 = | | ½ Z[ ¾ 1 ( i ) q i g ? 4 denote the complex valued integrable functions. Because, max ( Re i > Im i ) | | q | | s Z[ i 2max(Re i > Im i ) > i g ? 4 i | | | | | | | | | | | | for all q= 2. (a) =, (c) Notice that R Re i g + Im i g ? 4= | | | | i = j / (i j)=0 Z Z ZH ZH ZH For i 5 L1 (; ) define C for all H 5 .TakingH = Re(i j) A 0 and using 1H Re(i j) 0> we learn that M { } ig= Re ig+ l Im ig= Z Z Z 0=Re (i j)g = 1H Re(i j)=, 1H Re(i j)=0a.e. H It is routine to show the integral is still linear on L1 (; C) (prove!). In the Z Z 1 1 1 remainder of this section, let L () be either L (; C) or L (; R) = If D 5 This implies that 1H =0a.e. which happens i M and i 5 L1 (; C) or i : [ $ [0> 4] is a measurable function, let ( Re(i j) A 0 )=(H)=0= { } ig := 1Dig= Similar (Re(i j) ? 0) = 0 so that Re(i j)=0a.e. Similarly, Im(i j)=0 ZD Z[ a.e and hence i j =0a.e., i.e. i = j a.e. (c) =, (b) is clear and so is (b) Proposition 19.34. Suppose that i 5 L1 (; C) > then =, (a) since i j i j =0= | | ig i g= (19.13) ¯ZH ZH ¯ Z | | ¯ ¯ ¯Z[ ¯ Z[ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 296 19 Measures and Integration 19.4 Integrals of Complex Valued Functions 297

1 1 Deﬁnition 19.36. Let ([> >) be a measure space and O ()=O ([> >) lim infq$4 [ iqg 1 M M jg ig jg + denote the set of L () functions modulo the equivalence relation; i j i ± lim sup iqg Z[ Z[ Z[ ½ q$4R [ i = j a.e. We make this into a normed space using the norm and therefore R

i j 1 = i j g lim sup iqg ig lim inf iqg= k kO | | q$4 Z q$4 Z[ Z[ Z[ andintoametricspaceusing1(i>j)= i j 1 = This shows that lim iqg exists and is equal to ig= k kO q$4 [ [ Warning: in the future we will often not make much of a distinction Exercise 19.1. Give anotherR proof of Proposition 19.34R by ﬁrst proving Eq. between O1() and L1 () = On occasion this can be dangerous and this danger (19.13) with i being a cylinder function in which case the triangle inequality will be pointed out when necessary. for complex numbers will do the trick. Then use the approximation Theorem 18.42 along with the dominated convergence Theorem 19.38 to handle the Remark 19.37. More generally we may deﬁne Os()=Os([> >) for s 5 general case. M [1> 4) as the set of measurable functions i such that 4 1 4 Corollary 19.39. Let i L () be a sequence such that i 1 ? { q}q=1 q=1 k qkL () 4> then 4 i is convergent a.e. and i s g ? 4 q=1 q P | | Z[ P 4 4 modulo the equivalence relation; i j i i = j a.e. iq g = iqg= [ Ãq=1 ! q=1 [ Z X X Z We will see in Chapter 21 that 4 4 Proof. The condition q=1 iq L1() ? 4 is equivalent to q=1 iq 5 1 4 k k | | 1@s L () = Hence q=1 iq is almost everywhere convergent and if VQ := s s Q P P i s = i g for i 5 O () i > then k kO | | q=1 q P µZ ¶ Q 4 1 s P VQ iq iq 5 L () = is a norm and (O ()> Os ) is a Banach space in this norm. | | | | | | k·k q=1 q=1 X X Theorem 19.38 (Dominated Convergence Theorem). Suppose iq>jq>j 5 So by the dominated convergence theorem, 1 1 L () >iq $ i a.e., iq jq 5 L () >jq $ j a.e. and [ jqg $ [ jg= 4 Then L1 ( ) and | | i 5 iq g = lim VQ g =lim VQ g R R Q$4 Q$4 [ Ãq=1 ! [ [ ig = lim iqg= Z Z Z k$4 X Z[ Z[ Q 4 1 = lim iqg = iqg= (In most typical applications of this theorem jq = j 5 L () for all q=) Q$4 q=1 [ q=1 [ X Z X Z Proof. Notice that i = limq$4 iq limq$4 jq j a.e. so that i 5 L1 () = By considering| | the real and| imaginary| parts| | of i separately, it Theorem 19.40 (The Fundamental Theorem of Calculus). Suppose su!ces to prove the theorem in the case where i is real. By Fatou’s Lemma, 1 { 4 ?d?e?4>i5 F((d> e)> R)_O ((d> e)>p) and I({):= d i(|)gp(|)= Then (j i)g = lim inf (jq iq) g lim inf (jq iq) g R ± q$4 ± q$4 ± 1 Z[ Z[ Z[ 1. I 5 F([d> e]> R) _ F ((d> e)> R)= 2. I 0({)=i({) for all { 5 (d> e)= = lim jqg +liminf iqg q$4 q$4 ± 3. If J 5 F([d> e]> ) _ F1((d> e)> ) is an anti-derivative of i on (d> e) (i.e. Z[ µ Z[ ¶ R R i = J0 (d>e)) then = jg + lim inf iqg | q$4 ± e Z[ µ Z[ ¶ i({)gp({)=J(e) J(d)= Since lim inf ( )= lim sup we have shown, d q$4 dq q$4 dq> Z 298 19 Measures and Integration 19.4 Integrals of Complex Valued Functions 299

q { 4 Proof. Since I({):= 1(d>{)(|)i(|)gp(|)> lim{$} 1(d>{)(|)=1(d>})(|) q R lim (1 ) gp({)= lim iq({)gp({) 1 q$4 q$4 for p —a.e.| and 1(d>{)(|)i(|) 1(d>e)(|) i(|) is an O — function, it 0 q 0 R | | Z 4 Z 4 follows from the dominated convergence Theorem 19.38 that I is continuous { ¯ ¯ = lim iq({)gp({)= h gp({)=1= on [d> e]= Simple manipulations¯ show,¯ 0 q$4 0 ; Z Z {+k 4 ? [i(|) i({)] gp(|) if kA0 Example 19.42 (Integration of Power Series). Suppose UA0 and dq q=0 is I ({ + k) I({) 1 { 4 q { } i({) = a sequence of complex numbers such that =0 dq u ? 4 for all u 5 (0>U)= k k = ¯ { [i(|) i({)] gp(|)¯ if k?0 q | | ¯ ¯ | | ¯R{+k ¯ Then ¯ ¯ ¯ ¯ P ¯ ¯ ¯ {+k ¯ ¯ ¯ 1 ¯R i(|) i({) gp(|) if¯ kA0 4 4 4 q+1 q+1 ¯{ ¯ q q { | ( ) ( )| ( ) if 0 dq{ gp({)= dq { gp({)= dq k ( {+k i | i { gp | k? q +1 | | R | | Ãq=0 ! q=0 q=0 Z X X Z X sup iR(|) i({) : | 5 [{ k >{+ k ] {| | | | | | } for all U???U=Indeed this follows from Corollary 19.39 since and the latter expression, by the continuity of i> goes to zero as k $ 0 = This 4 4 shows I 0 = i on (d> e)= For the converse direction, we have by assumption q | | q | | q dq { gp({) dq { gp({)+ dq { gp({) that J ({)=I ({) for { 5 (d> e)= Therefore by the mean value theorem, | || | | || | | || | 0 0 q=0 Z q=0 ÃZ0 Z0 ! I J = F for some constant F= Hence X X4 4 q+1 + q+1 d | | | | 2u d uq ? 4 e | q| q +1 | q| ( ) ( )= ( )= ( ) ( ) q=0 q=0 i { gp { I e I e I d X X d Z where u =max( > )= =(J(e)+F) (J(d)+F)=J(e) J(d)= | | | | Corollary 19.43 (Dierentiation Under the Integral). Suppose that M R is an open interval and i : M [ $ C is a function such that × Example 19.41. The following limit holds, 1. { $ i(w> {) is measurable for each w 5 M= q 1 { q 2. i(w0> ) 5 O () for some w0 5 M= lim (1 ) gp({)=1= 3. Ci (w>· {) exists for all (w> {)= q$4 0 q Cw Z 1 Ci 1 { q { 4. There is a function j 5 L () such that Cw (w> ) j 5 L () for each Let iq({)=(1 q ) 1[0>q]({) and notice that limq$4 iq({)=h = We will · now show w 5 M= ¯ ¯ 1 ¯ ¯ { Then i(w> ) 5 L () for all w 5 M (i.e.¯ i(w>¯ {) g({) ? 4)>w$ 0 iq({) h for all { 0= [ i(w> {)g· ({) is a dierentiable function on |M and | { [ It su!ces to consider { 5 [0>q]= Let j({)=h iq({)> then for { 5 (0>q)> R R g Ci g 1 1 1 i(w> {)g({)= (w> {)g({)= ln j({)=1+q ( )=1 0 gw Cw { { Z[ Z[ g{ (1 q ) q (1 q ) Proof. (The proof is essentially the same as for sums.) By considering the which shows that ln j({) and hence j({) is decreasing on [0>q]= Therefore real and imaginary parts of i separately, we may assume that i is real. Also j({) j(0) = 1> i.e. notice that 0 i ({) h{= Ci q (w> {) = lim q(i(w + q1>{) i(w> {)) From Example 19.24, we know Cw q$4 Ci 4 and therefore, for { $ Cw (w> {) is a sequential limit of measurable functions h{gp({)=1? 4> and hence is measurable for all w 5 M= By the mean value theorem, Z0 i(w> {) i(w0>{) j({) w w0 for all w 5 M (19.14) so that h{ is an integrable function on [0> 4)= Hence by the dominated con- | | | | vergence theorem, and hence 300 19 Measures and Integration 19.5 Measurability on Complete Measure Spaces 301

q q { i(w> {) i(w> {) i(w0>{) + i(w0>{) j({) w w0 + i(w0>{) = That is q!= { h gp({)= Recall that | | | | | | | | | | [0>4) 1 R This shows i(w> ) 5 L () for all w 5 M= Let J(w):= [ i(w> {)g({)> then · (w):= {w1h{g{ for wA0= J(w) J(w ) i(w> {) i(w >{R) [0>4) 0 = 0 g({)= Z w w w w 0 Z[ 0 (The reader should check that (w) ? 4 for all wA0=) We have just shown By assumption, that (q +1)=q! for all q 5 N=

i(w> {) i(w0>{) Ci Remark 19.45. Corollary 19.43 may be generalized by allowing the hypothesis lim = (w> {) for all { 5 [ w$w0 w w0 Cw to hold for { 5 [ H where H 5 is a ﬁxed null set, i.e. H must be independent of w= Consider\ what happensM if we formally apply Corollary 19.43 and by Eq. (19.14), 4 to j(w):= 0 1{wgp({)> i(w> {) i(w >{) 0 j({) for all w 5 M and { 5 [= R 4 4 w w g ? C ¯ 0 ¯ j˙(w)= 1{wgp({) = 1{wgp({)= ¯ ¯ gw 0 0 Cw Therefore, we¯ may apply the dominated¯ convergence theorem to conclude Z Z ¯ ¯ The last integral is zero since C 1 =0unless w = { in which case it is J(w ) J(w ) i(w >{) i(w >{) Cw {w lim q 0 = lim q 0 g({) not deﬁned. On the other hand j(w)=w so that j˙(w)=1= (The reader should q$4 w w q$4 w w q 0 Z[ q 0 decide which hypothesis of Corollary 19.43 has been violated in this example.) i(w >{) i(w >{) = lim q 0 g({) q$4 w w Z[ q 0 Ci 19.5 Measurability on Complete Measure Spaces = (w0>{)g({) [ Cw Z In this subsection we will discuss a couple of measurability results concerning for all sequences w 5 M w0 such that w $ w0= Therefore, J˙ (w0)= q \{ } q completions of measure spaces. lim J(w)J(w0) exists and w$w0 ww0 Proposition 19.46. Suppose that ([> >) is a complete measure space1 Ci and i : [ $ is measurable. M J˙ (w0)= (w0>{)g({)= R [ Cw Z 1. If j : [ $ R is a function such that i({)=j({) for —a.e.{> then j is measurable. Example 19.44. Recall from Example 19.24 that 2. If iq : [ $ R aremeasurableandi : [ $ R is a function such that limq$4 iq = i> - a.e., then i is measurable as well. 1 = h{gp({) for all A0= [0>4) Proof. 1. Let H = { : i({) = j({) which is assumed to be in and Z { 6 } f M (H)=0.Thenj =1Hf i +1Hj since i = j on H .Now1Hf i is measurable Let 0 For 2 0 and there exists ( ) such that %A = %A q 5 N Fq % ? 4 so j will be measurable if we show 1Hj is measurable. For this consider, q g { q { %{ Hf ^ (1 j)1(D 0 )if05 D 0 h = { h F(%)h = (1 j)1(D)= H (19.15) g H (1 j)1(D)if0\{ } 5@ D µ ¶ ½ H Using this fact, Corollary 19.43 and induction gives 1 Since (1Hj) (E) H if 0 5@ E and (H)=0, it follow by completeness of q q 1 g g that (1Hj) (E) 5 if 0 5@ E= Therefore Eq. (19.15) shows that 1Hj is q!q1 = 1 = h{gp({) M M g g measurable. 2. Let H = { : lim iq({) = i({) by assumption H 5 and µ ¶ Z[0>4) µ ¶ { q$4 6 } M = {qh{gp({)= 1 Recall this means that if Q a [ is a set such that Q a D M and (D)=0> M [0>4) then Q M as well. Z M 302 19 Measures and Integration 19.6 Comparison of the Lebesgue and the Riemann Integral 303

(H)=0.Sincej := 1Hi =limq$4 1Hf iq, j is measurable. Because i = j 19.6 Comparison of the Lebesgue and the Riemann on Hf and (H)=0>i= j a.e. so by part 1. i is also measurable. Integral The above results are in general false if ([> >) is not complete. For M example, let [ = 0> 1> 2 > = 0 > 1> 2 >[>! and = 0= Take j(0) = For the rest of this chapter, let and :[ ] be a { } M {{ } { } } 4 ?d?e?4 i d> e $ R 0>j(1) = 1>j(2) = 2> then j =0a.e. yet j is not measurable. bounded function. A partition of [d> e] is a ﬁnite subset [d> e] containing d> e = To each partition Lemma 19.47. Suppose that ([> >) is a measure space and ¯ is the { } completion of relative to andM¯ is the extension of to ¯M= Then a M M = d = w0 ?w1 ? ?wq = e (19.17) function i : [ $ R is ( ¯ > = ) —measurablei there exists a function { ··· } M B BR j : [ $ R that is ( > ) —measurablesuchH = { : i({) = j({) 5 ¯ and of [d> e] let M B { 6 } M ¯ (H)=0> i.e. i({)=j({) for ¯ —a.e.{= Moreover for such a pair i and j> mesh():=max wm wm1 : m =1>===>q > i 5 O1(¯) i j 5 O1() and in which case {| | }

P =sup i({):w { w 1 >p=inf i({):w { w 1 m { m m } m { m m } ig¯ = jg= q q [ [ Z Z J = i(d)1 d + Pm1(w 1>w ]>j = i(d)1 d + pm1(w 1>w ] and { } m m { } m m 1 1 Proof. Suppose first that such a function j exists so that ¯(H)=0= Since X X j is also ( ¯ > ) — measurable, we see from Proposition 19.46 that i is ( ¯ > ) M B M B — measurable. Conversely if i is ( ¯ > ) — measurable, by considering i we Vi = Pm(wm wm1) and vi = pm(wm wm1)= M¯ B ± may assume that i 0= Choose ( > ) — measurable simple function !q 0 Notice that X X M B e e such that !q % i as q $4= Writing Vi = Jgp and vi = jgp= d d !q = dn1D Z Z n The upper and lower Riemann integrals are defined respectively by ¯ X with Dn 5 > we may choose En 5 such that En Dn and ¯(Dn En)=0= e d Letting M M \ i({)g{ =infVi and i({)g{ =sup vi= ˜ := 1 d !q dn En Z Z e X ˜ e e we have produced a ( > ) — measurable simple function !q 0 such that Definition 19.48. The function i is Riemann integrable i i = i 5 ˜ M B d d Hq := !q = !q has zero ¯ —measure.Since¯ (^qHq) q ¯ (Hq) > there e { 6 } R and which case the Riemann integral i is definedtobethecommonvalue: exists I 5 such that ^qHq I and (I )=0= It now follows that d R R M P R 1 !˜ =1 ! % j := 1 i as q $4= e e e I q I q I i({)g{ = i({)g{ = i({)g{= d d d This shows that j =1 i is ( > ) — measurable and that i = j I has Z Z Z I M B { 6 } ¯ — measure zero. Since i = j, ¯ — a.e., [ ig¯ = [ jg¯ so to prove Eq. The proof of the following Lemma is left to the reader as Exercise 19.20. (19.16) it su!ces to prove R R Lemma 19.49. If 0 and are two partitions of [d> e] and 0 then jg¯ = jg= (19.16) J J i j j and Z[ Z[ 0 0 Vi V i v i vi= Because ¯ = on > Eq. (19.16) is easily verified for non-negative — 0 0 measurable simple functions.M Then by the monotone convergence theoremM and 4 There exists an increasing sequence of partitions n n=1 such that mesh(n) & the approximation Theorem 18.42 it holds for all — measurable functions 0 and { } j : [ $ [0> 4]= The rest of the assertions followM in the standard way by e e and as considering (Re j) and (Im j) = Vn i & i vn i % i n $4= ± ± Zd Z d 304 19 Measures and Integration 19.6 Comparison of the Lebesgue and the Riemann Integral 305

If we let K({)=limsupi(|) J({) |${ J := lim J and j := lim j (19.18) n$4 n n$4 n for all {@5 = Combining this equation with Eq. (19.22) then implies then by the dominated convergence theorem, K({)=J({) if {@5 = A similar argument shows that k({)=j({) if e {@5 and hence Eq. (19.21) is proved.

jgp = lim jn = lim vn i = i({)g{ (19.19) 3. The functions J and j are limits of measurable functions and hence mea- n$4 n$4 Z[d>e] Z[d>e] Zd surable. Since K = J and k = j except possibly on the countable set > and both K and k are also Borel measurable. (You justify this statement.) e

Jgp = lim Jn = lim Vn i = i({)g{= (19.20) [d>e] n$4 [d>e] n$4 d Z Z Z Theorem 19.52. Let i :[d> e] $ R be a bounded function. Then Notation 19.50 For { 5 [d> e]> let e e = and = (19.23) and i Kgp i kgp K({) = lim sup i(|) := lim sup i(|): | { %> | 5 [d> e] d [d>e] d [d>e] |${ %&0 { | | } Z Z Z Z k({) = lim inf i(|) := lim inf i(|): | { %> | 5 [d> e] = and the following statements are equivalent: |${ %&0 { | | } 1. K({)=k({) for p -a.e. {> Lemma 19.51. The functions K> k :[d> e] $ R satisfy: 2. the set H := { 5 [d> e]:i is discontinuous at { 1. k({) i({) K({) for all { 5 [d> e] and k({)=K({) i i is continuous { } at {= is an p¯ —nullset. 4 2. If n n=1 is any increasing sequence of partitions such that mesh(n) & 0 3. i is Riemann integrable. and{ J}and j are deﬁned as in Eq. (19.18), then If i is Riemann integrable then i is Lebesgue measurable2 , i.e. i is @ — 4 measurable where is the Lebesgue —algebraand is the Borel — algebraL B J({)=K({) i({) k({)=j({) ; {@5 := ^n=1n= (19.21) on [d> e].MoreoverifweletL p¯ denote the completionB of p> then (Note is a countable set.) 3. and are Borel measurable. e K k Kgp = i({)g{ = igp¯ = kgp= (19.24) Z[d>e] Zd Z[d>e] Z[d>e] Proof. Let Jn := Jn & J and jn := jn % j= Proof. Let 4 be an increasing sequence of partitions of [d> e] as 1. It is clear that k({) i({) K({) for all { and K({)=k({) i lim i(|) { n}n=1 |${ described in Lemma 19.49 and let J and j be deﬁned as in Lemma 19.51. exists and is equal to i({)= That is K({)=k({) i i is continuous at {= Since p()=0>K= J a.e., Eq. (19.23) is a consequence of Eqs. (19.19) and 2. For {@5 > (19.20). From Eq. (19.23), i is Riemann integrable i Jn({) K({) i({) k({) jn({) ; n and letting n $4in this equation implies Kgp = kgp Z[d>e] Z[d>e] J({) K({) i({) k({) j({) ; {@5 = (19.22) and because k i K this happens i k({)=K({) for p -a.e.{= Since H = { : K({) = k({) > this last condition is equivalent to H being a p Moreover, given 0 and %A {@5 > — null{ set. In light6 of these} results and Eq. (19.21), the remaining assertions sup i(|): | { %> | 5 [d> e] J ({) including Eq. (19.24) are now consequences of Lemma 19.47. { | | } n Notation 19.53 In view of this theorem we will often write e i({)g{ for for all n large enough, since eventually Jn({) is the supremum of i(|) d e over some interval contained in [{%> { +%]= Again letting n $4implies d igp= R sup i(|) J({) and therefore, that 2 |{ % R i need not be Borel measurable. | | 306 19 Measures and Integration 19.7 Determining Classes of Measures 307

19.7 Determining Classes of Measures Proposition 19.58. Suppose that ([> g) is a metric space, and are two measures on [ := ( g) which are finite on bounded measurable subsets of Definition 19.54 ( — finite). Let [ be a set and 2[ = We say [ and B that a function : $ [0> 4] is — finite on if thereE existF [ 5 such F E q E ig = ig (19.25) that [q % [ and ([q) ? 4 for all q= Z[ Z[ for all i 5 EFe([> R) where Theorem 19.55 (Uniqueness). Suppose that 2[ is a — class (see Definition 18.53), = ( ) and and aretwomeasureonC If and = EFe([> R)= i 5 EF([> R):supp(i) is bounded = (19.26) are — finite on Mand =C on > then = on = M { } C C M Then = Proof. We begin first with the special case where ([) ? 4 and therefore also Proof. To prove this fixar 5 [ and let ([) = lim ([q) = lim ([q)=([) ? 4= q$4 q$4 #U({)=([U +1 g({> r)] a 1) b 0 (19.27) Let 4 so that ( [0 1]) supp( ) ( +2)and 1 as := i 5 c ( > R): (i)= (i) = #U 5 EFe [> > > #U E r> U #U % U $4= H { M } Let U denote the space of bounded real valued [ — measurable functions Then is a linear subspace which is closed under bounded convergence (by the i suchH that B dominatedH convergence theorem), contains 1 and contains the multiplicative system, P := 1 : F 5 = Therefore, by Theorem 18.51 or Corollary 18.54, #Uig = #Uig= (19.28) F [ [ 4 { C} 1 2 Z Z = c ( > R) and hence = = For the general case, let [q>[q 5 be H M 1 2 1 2 C Then U is closed under bounded convergence and because of Eq. (19.25) chosen so that [q % [ and [q % [ as q $4and [q + [q ? 4 for H 1 2 contains EF([> R)= Therefore by Corollary 18.55, U contains all bounded all q= Then [q := [q _ [q 5 increases to [ and ([q)= ([q) ? 4 for H C ¡ ¢ ¡ ¢ measurable functions on [= Take i =1D in Eq. (19.28) with D 5 [ > and all q= For each q 5 N> define two measures q and q on by B M then use the monotone convergence theorem to let U $4= The result is (D)=(D) for all D 5 = q(D):=(D _ [q) and q(D)=(D _ [q)= [ Here is another versionB of Proposition 19.58. Then, as the reader should verify, and are finite measure on such q q M Proposition 19.59. Suppose that ([> g) is a metric space, and are two that q = q on = Therefore, by the special case just proved, q = q on = C M measures on [ = ( g) which are both finite on compact sets. Further assume Finally, using the continuity properties of measures, B r there exists compact sets Nn [ such that Nn % [= If (D) = lim (D _ [q) = lim (D _ [q)=(D) q$4 q$4 ig = ig (19.29) for all D 5 = Z[ Z[ As an immediateM consequence we have the following corollaries. for all i 5 Ff([> R) then =

Corollary 19.56. Suppose that ([> ) is a topological space, [ = ( ) is Proof. Let # be defined as in the proof of Proposition 18.56 and let B q>n the Borel —algebraon[ and and are two measures on [ which are q>n denote those bounded [ — measurable functions, i : [ $ R such that — finite on = If = on then = on > i.e. = B H B B[ i#q>ng = i#q>ng= Corollary 19.57. Suppose that and are two measures on q which are R [ [ finite on bounded sets and such that (D)=(D) for all sets DBof the form Z Z By assumption EF([> R) q>n and one easily checks that q>n is closed H H D =(d> e]=(d1>e1] (dq>eq] under bounded convergence. Therefore, by Corollary 18.55, contains all ×···× Hq>n bounded measurable function. In particular for D 5 [ > q B with d> e 5 R and d?e>i.e. dl ?el for all l= Then = on q = BR 1D#q>ng = 1D#q>ng= Z[ Z[ 308 19 Measures and Integration 19.8 Exercises 309

Letting q $4in this equation, using the dominated convergence theorem, 2. Let i : [ $ [0> 4] be a measurable function, show one shows ig = ig= (19.30) 1D1Nr g = 1D1Nr g n n [ [ Z[ Z[ Z Z holds for n= Finally using the monotone convergence theorem we may let Hint: first prove the relationship for characteristic functions, then for n $4to conclude simple functions, and then for general positive measurable functions. 3. Show that a measurable function i : [ $ C is in O1() i i 5 O1() and if i 5 O1() then Eq. (19.30) still holds. | | (D)= 1Dg = 1Dg = (D) Z[ Z[ Notation 19.60 It is customary to informally describe defined in Exercise for all D 5 = 19.6 by writing g = g= B[ Exercise 19.7. Let ([> >) be a measure space, (\> ) be a measurable space and i : [ $ \ beM a measurable map. Define a functionF : $ [0> 4] 19.8 Exercises by (D):=(i 1(D)) for all D 5 = F F 1 Exercise 19.2. Let be a measure on an algebra 2[ > then (D)+ 1. Show is a measure. (We will write = i or = i =) (E)=(D ^ E)+(D _ E) for all D> E 5 = A 2. Show A jg = (j i) g (19.31) Exercise 19.3 (From problem 12 on p. 27 of Folland.). Let ([> >) Z\ Z[ be a finite measure space and for D> E 5 let (D> E)=(DE) Mwhere for all measurable functions j : \ $ [0> 4]= Hint: see the hint from DE =(D E) ^ (E D) = It is clear that M(D> E)= (E>D) = Show: Exercise 19.6. \ \ 1 1 3. Show a measurable function j : \ $ C is in O () i j i 5 O () and 1. satisfies the triangle inequality: that Eq. (19.31) holds for all j 5 L1()= 1 (D> F) (D> E)+ (E>F) for all D> E> F 5 = Exercise 19.8. Let I : R $ R be a F -function such that I 0({) A 0 for all M { 5 R and lim{$ 4 I({)= 4= (Notice that I is strictly increasing so that ± ± 2. Define D E i (DE)=0and notice that (D> E)=0i D E= I 1 : R $ R exists and moreover, by the inverse function theorem that I 1 1 Show “ ” is an equivalence relation. is a F — function.) Let p be Lebesgue measure on R and 3. Let @ denote modulo the equivalence relation, > and let B M M 1 1 1 [D]:= E 5 : E D = Show that ¯([D] > [E]) := (D> E) is gives a (D)=p(I (D)) = p( I (D)) = I p (D) well defi{ned metricM on }@ = for all D 5 R= Show g = I 0gp=¡ Use¢ this result¡ to prove¢ the change of 4. Similarly show ˜ ([D])M = (D) is a well defined function on @ and B M variable formula, show ˜ :( @ ) $ R+ is ¯ — continuous. M k I I 0gp = kgp (19.32) R · R Exercise 19.4. Suppose that q : $ [0> 4] are measures on for q 5 Z Z M M which is valid for all Borel measurable functions k : R $ [0> 4]= N= Also suppose that q(D) is increasing in q for all D 5 = Prove that M Hint: Start by showing g = I 0gp on sets of the form D =(d> e] with : $ [0> 4] defined by (D) := limq$4 q(D) is also a measure. M d> e 5 R and d?e=Then use the uniqueness assertions in Theorem 19.8 (or Exercise 19.5. Now suppose that is some index set and for each 5 > see Corollary 19.57) to conclude g = I 0gp on all of = To prove Eq. (19.32) BR : $ [0> 4] is a measure on = Define : $ [0> 4] by (D)= apply Exercise 19.7 with j = k I and i = I 1= M M M ( ) for each Show that is also a measure. 4 5 D D 5 = Exercise 19.9. Let ([> >) be a measure space and D > show M M { q}q=1 M P Exercise 19.6. Let ([> >) be a measure space and : [ $ [0> 4] be a ( Dq a.a. ) lim inf (Dq) measurable function. ForMD 5 > set (D):= g= { } q$4 M D and if (^pqDp) ? 4 for some q> then 1. Show : $ [0> 4] is a measure. R M ( Dq i.o. ) lim sup (Dq) = { } q$4 310 19 Measures and Integration

Exercise 19.10. BRUCE: Delete this exercise which is contained in Lemma 19.17. Suppose ([> >) be a measure space and i : [ $ [0 4] be a mea- M surable function such that [ ig ? 4= Show ( i = 4 )=0and the set 20 iA0 is — ﬁnite. { } { } R Exercise 19.11. Folland 2.13 on p. 52. Hint: “Fatou times two.” Multiple Integrals

Exercise 19.12. Folland 2.14 on p. 52. BRUCE: delete this exercise

Exercise 19.13. Give examples of measurable functions iq on R such that { } iq decreases to 0 uniformly yet iqgp = 4 for all q= Also give an example of a sequence of measurable functions j on [0> 1] such that j $ 0 while R { q} q jqgp =1for all q= ExerciseR 19.14. Folland 2.19 on p. 59. (This problem is essentially covered in the previous exercise.) In this chapter we will introduce iterated integrals and product measures. We 4 are particularly interested in when it is permissible to interchange the order Exercise 19.15. Suppose dq C is a summable sequence (i.e. { }q=4 of integration in multiple integrals. 4 d ? 4)> then i():= 4 d hlq is a continuous function q=4 | q| q=4 q for 5 R and Example 20.1. As an example let [ =[1> 4) and \ =[0> 1] equipped with P P 1 lq their Borel - algebras and let = = p> where p is Lebesgue measure. dq = i()h g= 2 The iterated integrals of the function i ({> |):=h{| 2h2{| satisfy, Z Exercise 19.16. For any function i 5 O1 (p) > show { 5 R $ i (w) gp (w) 1 4 1 1 h| (4>{] (h{| 2h2{|)g{ g| = h| g| 5 (0> 4) is continuous in {= Also find a finite measure, > on R such that { $ | B R Z0 ·Z1 ¸ Z0 µ ¶ ( ] i (w) g (w) is not continuous. 4>{ and ExerciseR 19.17. Folland 2.28 on p. 60. 4 1 4 1 h{ (h{| 2h2{|)g| g{ = h{ g{ 5 (4> 0) Exercise 19.18. Folland 2.31b and 2.31e on p. 60. (The answer in 2.13b is { Z1 ·Z0 ¸ Z1 · ¸ wrong by a factor of 1 and the sum is on n =1to 4= In part e, v should be and therefore are not equal. Hence it is not always true that order of integra- taken to be d. You may also freely use the Taylor series expansion tion is irrelevant. 4 4 (2q 1)!! (2q)! (1 })1@2 = }q = }q for } ? 1= Lemma 20.2. Let F be either [0> 4)> R or C= Suppose ([> ) and (\> ) 2qq! q 2 | | M N q=0 q=0 4 (q!) are two measurable spaces and i : [ \ $ F is a ( > F) —measurable X X function, then for each | 5 \> × M N B Exercise 19.19. There exists a meager (see Definition 13.4 and Proposition 13.3) subsets of R which have full Lebesgue measure, i.e. whose complement { $ i({> |) is ( > F) measurable, (20.1) is a Lebesgue null set. (This is Folland 5.27. Hint: Consider the generalized M B Cantor sets discussed on p. 39 of Folland.) for each { 5 [> | $ i({> |) is ( > F) measurable. (20.2) Exercise 19.20. Prove Lemma 19.49. N B Proof. Suppose that H = D E 5 := and i =1 = Then × E M×N H

i({> |)=1D E({> |)=1D({)1E(|) × from which it follows that Eqs. (20.1) and (20.2) for this function. Let be the collection of all bounded ( > ) — measurable functions on [ H\ such M N BF × 312 20 Multiple Integrals 20.1 Fubini-Tonelli’s Theorem and Product Measure 313 that Eqs. (20.1) and (20.2) hold, here we assume F = R or C= Because mea- and surable functions are closed under taking linear combinations and pointwise g({) g(|)i({> |)= g(|) g({)i({> |)= (20.7) 4 limits, is linear subspace of c ( > F) which is closed under bounded Z[ Z\ Z\ Z[ convergenceH and contain 1 5 forM all HNin the —class, = Therefore by by H H E Proof. Suppose that H = D E 5 := and i =1H= Then Corollary 18.54, that = c4 ( > F) = × E M×N H M N For the general ( > R) — measurable functions i : [ \ $ F and i({> |)=1D E({> |)=1D({)1E(|) M N B 4 × × P 5 N> let iP := 1 i P i 5 c ( > F) = Then Eqs. (20.1) and (20.2) | | M N and one sees that Eqs. (20.3) and (20.4) hold. Moreover hold with i replaced by iP and hence for i as well by letting P $4= Notation 20.3 (Iterated Integrals) If ([> >) and (\> >) are two M N i({> |)g(|)= 1D({)1E(|)g(|)=1D({)(E)> measure spaces and : is a —measurablefunction, \ \ i [ \ $ C Z Z the iterated integrals of ×(when they makeM sense)N are: i so that Eq. (20.5) holds and we have

( ) ( ) ( ):= ( ) ( ) ( ) g { g | i {> | i {> | g | g { g({) g(|)i({> |)=(E)(D)= (20.8) [ \ [ · \ ¸ Z Z Z Z Z[ Z\ and Similarly, g(|) g({)i({> |):= i({> |)g({) g(|)= \ [ \ [ Z Z Z ·Z ¸ i({> |)g({)=(D)1E(|) and Notation 20.4 Suppose that i : [ $ C and j : \ $ C are functions, let Z[ i j denote the function on [ \ given by × g(|) g({)i({> |)=(E)(D) i j({> |)=i({)j(|)= Z\ Z[ from which it follows that Eqs. (20.6) and (20.7) hold in this case as well. For Notice that if i>j are measurable, then i j is ( > C) — measurable. the moment let us further assume that ([) ? 4 and (\ ) ? 4 and let To prove this let I ({> |)=i({) and J({> |)=j(|)Mso thatN Bi j = I J will · be the collection of all bounded ( > R) — measurable functions on be measurable provided that I and J are measurable. Now I = i 1 where [H \ such that Eqs. (20.3) — (20.7)M hold.N UsingB the fact that measurable 1 : [ \ $ [ is the projection map. This shows that I is the composition × × functions are closed under pointwise limits and the dominated convergence of measurable functions and hence measurable. Similarly one shows that J is theorem (the dominating function always being a constant), one easily shows measurable. that closed under bounded convergence. Since we have just verified that H 1H 5 for all H in the —class, > it follows by Corollary 18.54 that is the spaceH of all bounded ( > E) — measurable functions on [ \= FinallyH if 20.1 Fubini-Tonelli’s Theorem and Product Measure M N BR × i : [ \ $ [0> 4] is a ( > R¯ ) — measurable function, let iP = P a i so that× i % i as P $4M andN Eqs.B (20.3) — (20.7) hold with i replaced by Theorem 20.5. Suppose ([> >) and (\> >) are -finite measure spaces P iP for all P 5 N= Repeated use of the monotone convergence theorem allows and i is a nonnegative ( M > ) — measurableN function, then for each R us to pass to the limit P $4in these equations to deduce the theorem in | 5 \> M N B the case and are finite measures. For the — finite case, choose [q 5 > { $ i({> |) is — [0>4] measurable, (20.3) M M B \q 5 such that [q % [> \q % \> ([q) ? 4 and (\q) ? 4 for all N for each { 5 [> p> q 5 N= Then define p(D)=([p _ D) and q(E)=(\q _ E) for all

D 5 and E 5 or equivalently gp =1[p g and gq =1\q g= By what | $ i({> |) is — [0>4] measurable, (20.4) M N N B we have just proved Eqs. (20.3) — (20.7) with replaced by p and by q for all ( > R¯ ) — measurable functions, i : [ \ $ [0> 4]= The validity of Eqs.M (20.3)N — (20.7)B then follows by passing to the× limits p $4and then { $ i({> |)g(|) is — measurable, (20.5) M B[0>4] q $4making use of the monotone convergence theorem in the form, Z\ | $ i({> |)g({) is — [0>4] measurable, (20.6) = 1 as [ N B xgp x [p g % xg p $4 Z Z[ Z[ Z[ 314 20 Multiple Integrals 20.1 Fubini-Tonelli’s Theorem and Product Measure 315 and The following convention will be in force for the rest of this chapter. yg = y1 g % yg as q $4 Convention: If ([> >) is a measure space and i : [ $ C is a measur- q \q M \ \ \ able but non-integrable function, i.e. = by convention we will de- Z Z Z [ i g 4> + + | | for all x 5 O ([> ) and y 5 O (\> )= fine [ ig := 0= However if i is a non-negative function (i.e. i : [ $ [0> 4]) M N R is a non-integrable function we will still write [ ig = 4= Corollary 20.6. Suppose ([> >) and (\> >) are — finite measure R spaces. Then there exists a uniqueM measure onN such that (D E)= Theorem 20.9 (Fubini’s Theorem). SupposeR ([> >) and (\> >) are M N × — finite measure spaces, = is the product measureM on N and (D)(E) for all D 5 and E 5 = Moreover is given by M N M N i : [ \ $ C is a — measurable function. Then the following three conditions× are equivalent:M N (H)= g({) g(|)1H({> |)= g(|) g({)1H({> |) (20.9) Z[ Z\ Z\ Z[ i g ? 4, i.e. i 5 O1()> (20.12) for all H 5 and is — finite. [ \ | | M N Z × Proof. Notice that any measure such that (D E)=(D)(E) for i({> |) g(|) g({) ? 4 and (20.13) × [ \ | | all D 5 and E 5 is necessarily — finite. Indeed, let [q 5 and Z µZ ¶ M N M \q 5 be chosen so that ([q) ? 4>(\q) ? 4>[q % [ and \q % \> then i({> |) g({) g(|) ? 4= (20.14) [ N\ 5 >[ \ % [ \ and ([ \ ) ? 4 for all q= The | | q q q q q q Z\ µZ[ ¶ uniqueness× assertionM N is a× consequence× of Theorem 19.55× or see Theorem 25.6 1 below with = = For the existence, it su!ces to observe, using the If any one (and hence all) of these condition hold, then i({> ) 5 O () for E M×N a.e. ( ) 1( ) for a.e. ( ) ( ) 1( ) · ( ) ( ) monotone convergence theorem, that defined in Eq. (20.9) is a measure on {> i >| 5 O |> \ i >| gy | 5 O > [ i {> g { 5 O1() and· Eqs. (20.10) and (20.11) are still· valid. · = Moreover this measure satisfies (D E)=(D)(E) for all D 5 R R andM EN5 from Eq. (20.8 × M N Proof. The equivalence of Eqs. (20.12) — (20.14) is a direct consequence 1 Notation 20.7 The measure is called the product measure of and and of Tonelli’s Theorem 20.8. Now suppose i 5 O () is a real valued function will be denoted by = and let H := { 5 [ : i ({> |) g (|)=4 = (20.15) Theorem 20.8 (Tonelli’s Theorem). Suppose ([> >) and (\> >) are \ | | M N ½ Z ¾ — finite measure spaces and = istheproductmeasureon = Then by Tonelli’s theorem, { $ i ({> |) g (|) is measurable and hence + + M N \ If i 5 O ([ \> )> then i( >|) 5 O ([> ) for all | 5 \> i({> ) 5 H 5 = Moreover Tonelli’s theorem| implies | O+(\> ) for× all {M5 [>N · M · M R N i ({> |) g (|) g ({)= i g ? 4 + + i( >|)g(|) 5 O ([> )> i({> )g({) 5 O (\> ) [ \ | | [ \ | | · M · N Z ·Z ¸ Z × \Z [Z which implies that (H)=0= Let i be the positive and negative parts of i> ± and then using the above convention we have

ig= g({) g(|)i({> |) (20.10) i ({> |) g (|)= 1H ({) i ({> |) g (|) [ \ [ \ \ \ Z × Z Z Z Z = g(|) g({)i({> |)= (20.11) = 1H ({)[i+ ({> |) i ({> |)] g (|) \ Z\ Z[ Z Proof. By Theorem 20.5 and Corollary 20.6, the theorem holds when = 1H ({) i+ ({> |) g (|) 1H ({) i ({> |) g (|) = Z\ Z\ i =1H with H 5 = Using the linearity of all of the statements, the the- (20.16) orem is also true forM non-negativeN simple functions. Then using the monotone convergence theorem repeatedly along with the approximation Theorem 18.42, Noting that 1H ({) i ({> |)=(1H 1\ i )({> |) is a positive — one deduces the theorem for general i 5 O+([ \> )= measurable function,± it follows from another· ± application of Tonelli’sM theoremN × M N 316 20 Multiple Integrals 20.1 Fubini-Tonelli’s Theorem and Product Measure 317 that { $ \ i ({> |) g (|) is — measurable, being the dierence of two Theorem 20.11. Suppose ([> >) and (\> >) are complete — ﬁnite measurable functions. MoreoverM measure spaces. Let ([ \> >M) be the completionN of ([ \> > ). R If i is — measurable× andL (a) i 0 or (b) i 5 O1(×) thenM iN is — L { N i ({> |) g (|) g ({) i ({> |) g (|) g ({) ? 4> measurable for a.e. { and i | is —measurablefor a.e. | and in case (b) | | 1 | 1 M Z[ ¯Z\ ¯ Z[ ·Z\ ¸ i{ 5 O () and i 5 O () for a.e. { and a.e. | respectively. Moreover, ¯ ¯ ¯ ¯ 1 which shows¯ i( >|)gy(|¯ ) 5 O ()= Integrating Eq. (20.16) on { and using \ · 1 | 1 Tonelli’s theorem repeatedly implies, { $ i{g 5 O () and | $ i g 5 O () R µ Z\ ¶ µ Z[ ¶ i ({> |) g (|) g ({) and Z[ ·Z\ ¸ ig = g g i = g g i= [ \ \ [ [ \ = g ({) g (|)1H ({) i+ ({> |) g ({) g (|)1H ({) i ({> |) Z × Z Z Z Z Z[ Z\ Z[ Z\ Proof. If H 5 is a null set (i.e. ( )(H)=0), then M N = g (|) g ({)1H ({) i+ ({> |) g (|) g ({)1H ({) i ({> |) \ [ \ [ Z Z Z Z 0=( )(H)= ({H)g({)= (H|)g(|)= = g (|) g ({) i+ ({> |) g (|) g ({) i ({> |) [Z [Z Z\ Z[ Z\ Z[ This shows that = i+g ig = (i+ i) g = ig (20.17) [ \ [ \ [ \ [ \ Z × Z × Z × Z × ( { : ( H) =0 )=0and ( | : (H ) =0 )=0> { { 6 } { | 6 } which proves Eq. (20.10) holds. i.e. ( H)=0for a.e. { and (H )=0for a.e. |= If k is measurable and Now suppose that i = x + ly is complex valued and again let H be as { | L in Eq. (20.15). Just as above we still have H 5 and (H) ? 4= By our k =0for — a.e., then there exists H 5 such that ({> |):k({> |) = M M N { 6 convention, 0 H and ( )(H)=0= Therefore k({> |) 1H({> |) and ( )(H)=0= Since} | |

i ({> |) g (|)= 1H ({) i ({> |) g (|)= 1H ({)[x ({> |)+ly ({> |)] g (|) k{ =0 = | 5 \ : k({> |) =0 {H and Z\ Z\ Z\ { 6 } { 6 } k| =0 = { 5 [ : k({> |) =0 H| = 1H ({) x ({> |) g (|)+l 1H ({) y ({> |) g (|) { 6 } { 6 } Z\ Z\ we learn that for a.e. { and a.e. | that k =0 5 > k =0 5 > { { 6 } M { | 6 } N which is measurable in { by what we have just proved. Similarly one shows ( k{ =0 )=0and a.e. and ( k| =0 )=0= This implies \ k({> |)g(|) 1 exists{ and6 } equals 0 for a.e. and{ similarly6 } that ( ) ( ) exists and \ i ( >|) g (|) 5 O () and Eq. (20.10) still holds by a computation similar { [ k {> | g { to that· done in Eq. (20.17). The assertions pertaining to Eq. (20.11) may be equals 0 for a.e. |= Therefore R Rproved in the same way. R Notation 20.10 Given and let 0= kg = kg g = kg g= H [ \ { 5 [> [ \ \ [ [ \ × Z × Z µZ ¶ Z µZ ¶ {H := | 5 \ :({> |) 5 H = For general 1( ) we may choose 1( ) such that { } i 5 O > j 5 O > i({> |)=j({> |) for a.e. ({> |)= Deﬁne k := i M j= ThenN k =0> a.e. Similarly if | 5 \ is given let Hence by what we have just proved and Theorem 20.8 i = j + k has the following properties: H := { 5 [ :({> |) 5 H = | { } 1. For a.e. ( )= ( )+ ( ) is in 1( ) and | {> | $ i {> | j {> | k {> | O If i : [ \ $ C is a function let i{ = i({> ) and i := i( >|) so that × | · · i{ : \ $ C and i : [ $ C= i({> |)g(|)= j({> |)g(|)= Z\ Z\ 318 20 Multiple Integrals 20.1 Fubini-Tonelli’s Theorem and Product Measure 319

2. For a.e. |> { $ i({> |)=j({> |)+k({> |) is in O1() and Example 20.13. In this example we will show

P sin { i({> |)g({)= j({> |)g({)= lim g{ = @2= (20.19) [ [ P$4 { Z Z Z0 From these assertions and Theorem 20.8, it follows that 1 4 w{ To see this write { = 0 h gw and use Fubini-Tonelli to conclude that

g({) g(|)i({> |)= g({) g(|)j({> |) P sin { R P 4 g{ = hw{ sin {gw g{ Z[ Z\ Z[ Z\ { Z0 Z0 ·Z0 ¸ = g(|) g({)j({> |) 4 P w{ Z\ Z\ = h sin {g{ gw 0 " 0 # = j({> |)g( )({> |) Z Z 4 [ \ 1 Z × = 1 whPw sin P hPw cos P gw 1+ 2 = ( ) ( ) 0 w i {> | g {> | = Z 4 [ \ 1 ¡ ¢ Z × $ gw = as P $4> 1+w2 2 Similarly it is shown that Z0 wherein we have used the dominated convergence theorem to pass to the limit. g(|) g({)i({> |)= i({> |)g({> |)= \ [ [ \ The next example is a refinement of this result. Z Z Z × Example 20.14. We have The previous theorems have obvious generalizations to products of any finite number of — finite measure spaces. For example the following theorem 4 sin { 1 h{g{ = arctan for all A0 (20.20) holds. { 2 Z0 q Theorem 20.12. Suppose ([l> l>l) are — finite measure spaces and for> P 5 [0> 4)> { M }l=1 and [ := [1 [q= Then there exists a unique measure, > on × ··· × P ([> 1 q) such that (D1 Dq)=1(D1) ===q(Dq) for all sin { 1 hP M ··· M ×···× h{g{ + arctan F (20.21) Dl 5 l= (This measure and its completion will be denote by 1 q=) M ··· ¯ 0 { 2 ¯ P If i : [ $ [0> 4] is a 1 — measurable function then ¯Z ¯ M ··· Mq ¯ ¯ ¯ 1+{ 1 ¯ where F =max 2 = s = 1=2= In particular Eq. (20.19) is valid. ¯{0 1+{ 2 22 ¯ ig = g(1)({(1)) === g(q)({(q)) i({1>===>{q) (20.18) Z[ Z[(1) Z[(q) To verify these assertions, first notice that by the fundamental theorem of calculus, where is any permutation of 1> 2>===>q = This equation also holds for any { } i 5 O1() and moreover, i 5 O1() i { { { sin { = cos |g| cos | g| 1g| = { | | | | | | ¯Z0 ¯ ¯Z0 ¯ ¯Z0 ¯ g(1)({(1)) === g(q)({(q)) i({1>===>{q) ? 4 ¯ ¯ ¯ ¯ ¯ ¯ | | sin { ¯ ¯ ¯ ¯ ¯ ¯ [(1) [(q) so 1 for all¯{ =0= Making¯ ¯ use of the identity¯ ¯ ¯ Z Z { 6 for some (and hence all) permutations, = ¯ ¯ 4 ¯ ¯ hw{gw =1@{ This theorem can be proved by the same methods as in the two factor case, Z0 see Exercise 20.4. Alternatively, one can use the theorems already proved and and Fubini’s theorem, induction on q> see Exercise 20.5 in this regard. 320 20 Multiple Integrals 20.2 Lebesgue Measure on Rg and the Change of Variables Theorem 321

P P 4 sin { g h{g{ = g{ sin {h{ hw{gw i ({) g{ = igp = igp = { g g g Z0 Z0 Z0 ZR ZR ZR 4 P = gw g{ sin {h(+w){ Hopefully the reader will understand the meaning from the context. 0 0 Z Z g g 4 P(+w) Theorem 20.18. Lebesgue measure p is translation invariant. Moreover p 1 (cos P +( + w)sinP) h g g = 2 gw is the unique translation invariant measure on Rg such that p ((0> 1] )=1= 0 ( + w) +1 B Z 4 4 Proof. Let = with and g Then 1 cos P +( + w)sinP P(+w) D M1 Mg Ml 5 R { 5 R = = 2 gw 2 h gw ×···× B 0 ( + w) +1 0 ( + w) +1 Z Z { + D =({1 + M1) ({2 + M2) ({ + M ) 1 × ×···× g g = arctan %(P>) (20.22) 2 and therefore by translation invariance of p on we ﬁnd that BR where g g 4 p ({ + D)=p({1 + M1) ===p({g + Mg)=p(M1) ===p(Mg)=p (D) cos P +( + w)sinP P(+w) %(P>)= 2 h gw= 0 ( + ) +1 g g w and hence p ({ + D)=p (D) for all D 5 g by Corollary 19.57. From this Z BR Since fact we see that the measure pg({ + ) and pg( ) have the same null sets. · · cos P +( + w)sinP 1+( + w) Using this it is easily seen that p({ + D)=p(D) for all D 5 g= The proof L 2 2 F> of the second assertion is Exercise 20.6. ¯ ( + w) +1 ¯ ( + w) +1 ¯ ¯ ¯ ¯ g ¯ 4 ¯ hP Theorem 20.19 (Change of Variables Theorem). Let r R be an ¯ %(P>) hP¯(+w)gw = F = g 1 1 open set and W : $ W ( ) r R be a F —dieomorphism, see Figure | | 0 P Z 20.1. Then for any Borel measurable function, i : W ( ) $ [0> 4]> This estimate along with Eq. (20.22) proves Eq. (20.21) from which Eq. (20.19) follows by taking $4and Eq. (20.20) follows (using the dominated con- i (W ({)) det W 0 ({) g{ = i (|) g|> (20.23) vergence theorem again) by letting P $4= | | Z W (Z )

g g d where W 0({) is the linear transformation on R deﬁned by W 0({)y := gw 0W ({+ 20.2 Lebesgue Measure on R and the Change of g | wy)= More explicitly, viewing vectors in R as columns, W 0 ({) may be repre- Variables Theorem sented by the matrix 5 6 Notation 20.15 Let C1W1 ({) === CgW1 ({) 9 . . . : g times g times W 0 ({)=7 . .. . 8 > (20.24) g p := p p on g = C1Wg ({) === CgWg ({) ··· BR BR ··· BR z }| { z }| { g be the g — fold product of Lebesgue measure p on R= We will also use p i.e. the l - m — matrix entry of W 0({) is given by W 0({)lm = ClWm({) where B g tr to denote its completion and let g be the completion of Rg relative to p = W ({)=(W1({)>===>Wg({)) and Cl = C@C{l= L B g AsubsetD 5 g is called a Lebesgue measurable set and p is called g — dimensional LebesgueL measure, or just Lebesgue measure for short. Remark 20.20. Theorem 20.19 is best remembered as the statement: if we g make the change of variables | = W ({) > then g| = det W 0 ({) g{= As usual, Deﬁnition 20.16. Afunctioni : R $ R is Lebesgue measurable if | | 1 you must also change the limits of integration appropriately, i.e. if { ranges i ( R) g= B L through then | must range through W ( ) = Notation 20.17 I will often be sloppy in the sequel and write p for pg and 1 That is : ( ) g is a continuously di erentiable bijection and the g{ for gp({)=gpg({)> i.e. W < W ar R inverse map W 31 : W ( ) < is also continuously dierentiable. 322 20 Multiple Integrals 20.2 Lebesgue Measure on Rg and the Change of Variables Theorem 323

whenever i is of the form i =1W ((>]) with d???e=An application of Dynkin’s multiplicative system Theorem 18.51 then implies that Eq. (20.25) holds for every bounded measurable function i : W ([d> e]) $ R= (Observe that W 0 ({) is continuous and hence bounded for { in the compact interval, | | Q [d> e] = From Exercise 10.12, = (dq>eq) where dq>eq 5 R^ 4 for q=1 {± } q =1> 2> ?Q with Q = 4 possible. Hence if i : W ( ) $ R + is a Borel ··· ` measurable function and dq ?n ?n ?eq with n & dq and n % eq> then by what we have already proved and the monotone convergence theorem

1 (i W ) W 0 gp = 1 i W W 0 gp (dq>eq) · ·| | W ((dq>eq)) · ·| | Z Z ¡ ¢

=lim 1W ([ > ]) i W W 0 gp n$4 n n · ·| | Z ¡ ¢

=lim 1W ([ > ]) igp n$4 n n · Fig. 20.1. The geometric setup of Theorem 20.19. W (Z )

= 1 igp= W ((dq>eq)) · Z Proof. The proof will be by induction on g= The case g =1was essentially W ( ) done in Exercise 19.8. Nevertheless, for the sake of completeness let us give Summing this equality on q> then shows Eq. (20.23) holds. a proof here. Suppose g =1>d???esuch that [d> e] is a compact To carry out the induction step, we now suppose gA1 and suppose the subinterval of = Then det W 0 = W 0 and | | | | theorem is valid with g being replaced by g1= For notational compactness, let g us write vectors in R as row vectors rather than column vectors. Nevertheless, 1 (W ({)) W 0 ({) g{ = 1 ({) W 0 ({) g{ = W 0 ({) g{= the matrix associated to the dierential, W ({) > will always be taken to be W ((>]) | | (>] | | | | 0 Z[d>e] Z[d>e] Z given as in Eq. (20.24). Case 1. Suppose W ({) has the form If W 0 ({) A 0 on [d> e] > then

W ({)=({l>W2 ({) >===>Wg ({)) (20.26) W 0 ({) g{ = W 0 ({) g{ = W () W () | | Z Z or W ({)=(W1 ({) >===>Wg1 ({) >{l) (20.27) = p (W ((> ])) = 1W ((>]) (|) g| ZW ([d>e]) for some l 5 1>===>g = For definiteness we will assume W is as in Eq. (20.26), { } the case of W in Eq. (20.27) may be handled similarly. For w 5 > let lw : while if W 0 ({) ? 0 on [d> e] > then R Rg1 $ Rg be the inclusion map defined by W 0 ({) g{ = W 0 ({) g{ = W () W () l (z):=z := (z >===>z >w>z >===>z ) > | | w w 1 l1 l+1 g1 Z Z g1 w be the (possibly empty) open subset of R defined by = p (W ((> ])) = 1W ((>]) (|) g|= W ([d>e]) Z g1 w := z 5 R :(z1>===>zl1>w>zl+1>===>zg1) 5 Combining the previous three equations shows ©g1 ª and Ww : w $ R be defined by

i (W ({)) W 0 ({) g{ = i (|) g| (20.25) | | Ww (z)=(W2 (zw) >===>Wg (zw)) > Z[d>e] ZW ([d>e]) 324 20 Multiple Integrals 20.2 Lebesgue Measure on Rg and the Change of Variables Theorem 325

W ( )= W (l ( )) = (w> }):} 5 W ( ) = w { w w } : wa5R wa5R Case 2. (Eq. (20.23) is true locally.) Suppose that W : $ Rg is a general : map as in the statement of the theorem and {0 5 is an arbitrary point. We will now show there exists an open neighborhood Z of {0 such that :

i W det W 0 gp = igp | | W (Z ) ZZ Z

: holds for all Borel measurable function, i : W (Z ) $ [0> 4]= Let P be the 1-l : l : th minor of W 0 ({0) > i.e. the determinant of W 0 ({0) with the ﬁrst row and l — column removed. Since

Fig. 20.2. In this picture g = l =3and is an egg-shaped region with an egg- g l+1 shaped hole. The picture indicates the geometry associated with the map W and 0 =detW 0 ({0)= (1) ClWm ({0) Pl> 6 · slicing the set along planes where {3 = w= l=1 X theremustbesomel such that P =0= Fix an l such that P =0and let, l 6 l 6 see Figure 20.2. Expanding det W 0 (zw) along the ﬁrst row of the matrix W 0 (zw) V ({):=({l>W2 ({) >===>Wg ({)) = (20.28) shows det W 0 (z ) = det W 0 (z) = Observe that det V0 ({0) = P =0= Hence by the inverse function Theorem | w | | w | | | | l| 6 16.25, there exist an open neighborhood Z of {0 such that Z r and Now by the Fubini-Tonelli Theorem and the induction hypothesis, g 1 V (Z ) r R and V : Z $ V (Z ) is a F —dieomorphism. Let U : V (Z) $ g 1 W (Z ) r R to be the F —dieomorphism deﬁned by i W det W 0 gp = 1 i W det W 0 gp | | · | | 1 Z RZg U (}):=W V (}) for all } 5 V (Z ) =

= 1 (z )(i W )(z ) det W 0 (z ) gzgw w w | w | Because Zg R 5 6 (W1 ({) >===>Wg ({)) = W ({)=U (V ({)) = U (({l>W2 ({) >===>Wg ({)))

= 7 (i W )(z ) det W 0 (z ) gz8 gw w w for all { 5 Z> if R | | Z Z 5 w 6 (}1>}2>===>}g)=V ({)=({l>W2 ({) >===>Wg ({)) 7 8 = i (w> Ww (z)) det Ww0 (z) gz gw R | | then Z Zw 1 5 6 5 6 U (})= W1 V (}) >}2>===>}g = (20.29) 9 : Observe that V is a map of the¡ form¡ in Eq.¢ (20.26), U¢ is a map of the form = i (w> }) g} gw = 7 1 (w> }) i (w> }) g}8 gw 7 8 W ( ) in Eq. (20.27), W ({)=U (V ({)) V ({) (by the chain rule) and (by the mul- R R 0 0 0 Z Z Z gZ1 Ww( w) R tiplicative property of the determinant) = i (|) g| det W 0 ({) = det U0 (V ({)) det V0 ({) ; { 5 Z= | | | || | W (Z ) So if i : W (Z ) $ [0> 4] is a Borel measurable function, two applications of wherein the last two equalities we have used Fubini-Tonelli along with the the results in Case 1. shows, identity; 326 20 Multiple Integrals 20.2 Lebesgue Measure on Rg and the Change of Variables Theorem 327

W (d) i W det W 0 gp = (i U det U0 ) V det V0 gp ·| | ·| | ·| | i (W ({)) ( W 0 ({) ) g{ = i (|) g| = i (|) g| ( ) | | ( ) (( )) ZZ ZZ Z d>e ZW e ZW d>e

= i U det U0 gp = igp which is again implies Eq. (20.23). On the other hand Eq. Eq. (20.30) is ·| | more general than Eq. (20.23) since it does not require W to be injective. The (Z ) ( Z( )) V Z U V Z standard proof of Eq. (20.30) is as follows. For } 5 W ([d> e]) > let = igp } ZW (Z ) I (}):= i (|) g|= W (d) and Case 2. is proved. Z Case 3. (General Case.) Let i : $ [0> 4] be a general non-negative Then by the chain rule and the fundamental theorem of calculus, Borel measurable function and let e e e g ( ( )) ( ) = ( ( )) ( ) = [ ( ( ))] f i W { W 0 { g{ I 0 W { W 0 { g{ I W { g{ N := { 5 : dist({> ) 1@q and { q = d d d g{ q Z Z Z { | | } W (e) e Then each Nq is a compact subset of and Nq % as q $4= Using the = I (W ({)) d = i (|) g|= | W (d) compactness of Nq and case 2, for each q 5 N> there is a ﬁnite open cover Z q of Nq such that Z and Eq. (20.23) holds with replaced by Z for An application of Dynkin’s multiplicative systems theorem (in the form of W 4 4 ˜ each Z 5 q= Let Zl l=1 be an enumeration of ^q=1 q and set Z1 = Z1 Corollary 18.55) now shows that Eq. (20.30) holds for all bounded measurable ˜ W { } W 4 ˜ and Zl := Zl (Z1 ^ ^ Zl1) for all l 2= Then = =1 Zl and by functions i on (d> e) = Then by the usual truncation argument, it also holds \ ··· l repeated use of case 2., for all positive measurable functions on (d> e) = ` 4 Example 20.22. Continuing the setup in Theorem 20.19, if D 5 > then i W det W 0 gp = 1 ˜ (i W ) det W 0 gp B | | Zl · ·| | =1 Z l Z X p (W (D)) = 1W (D) (|) g| = 1W (D) (W{) det W 0 ({) g{ 4 g g | | ZR ZR = 1 ˜ i W det W 0 gp W (Zl) ·| | = 1 ( ) det ( ) =1 D { W 0 { g{ l Z g XZl h³ ´ i R | | 4 q Z wherein the second equality we have made the change of variables, | = W ({) = = 1 ˜ igp= 1 ˜ igp W (Zl) · W (Zl) · Hencewehaveshown l=1 Z l=1 Z XW (Zl) XW ( ) g (p W )= det W 0 ( ) gp= = igp= | · | g W (Z ) In particular if W 5 JO(g> R)=JO(R ) —thespaceofg g invertible matrices, then p W = det W p> i.e. × | |

p (W (D)) = det W p (D) for allD 5 Rg = (20.31) Remark 20.21. When g =1> one often learns the change of variables formula | | B as This equation also shows that p W and p have the same null sets and hence e W (e) the equality in Eq. (20.31) is valid for any D 5 g= i (W ({)) W 0 ({) g{ = i (|) g| (20.30) L Zd ZW (d) Exercise 20.1. Show that i 5 O1 W ( ) >pg i where i :[d> e] $ R is a continuous function and W is F1 — function deﬁned ¡ ¢ in a neighborhood of [d> e] = If W 0 A 0 on (d> e) then W ((d> e)) = (W (d) >W (e)) i W det W 0 gp ? 4 and Eq. (20.30) is implies Eq. (20.23) with =(d> e) = On the other hand if | || | Z W 0 ? 0 on (d> e) then W ((d> e)) = (W (e) >W (d)) and Eq. (20.30) is equivalent to and if i 5 O1 W ( ) >pg > then Eq. (20.23) holds. ¡ ¢ 328 20 Multiple Integrals 20.2 Lebesgue Measure on Rg and the Change of Variables Theorem 329

Example 20.23 (Polar Coordinates). Suppose W :(0> 4) (0> 2) $ R2 is × deﬁned by y 2 { = W (u> )=(u cos > u sin ) > i.e. we are making the change of variable, 1

= cos and = sin for 0 and 0 2 0 {1 u {2 u ?u?4 ?? = -2 -1 0 1 2

x In this case -1 cos u sin W 0(u> )= sin ucos µ ¶ -2 and therefore g{ = det W 0(u> ) gug = ugug= | | Observing that Fig. 20.3. The region consists of the two curved rectangular regions shown.

2 R W ((0> 4) (0> 2)) = c := ({> 0) : { 0 \ × { } Example 20.25. In this example we will evaluate the integral has p2 — measure zero, it follows from the change of variables Theorem 20.19 that 2 4 L := {4 |4 g{g| ( ) = ( (cos sin )) (20.32) i { g{ g gu u i u > ZZ R2 0 0 · ¡ ¢ Z Z Z where for any Borel measurable function i : 2 $ [0> 4]= R = ({> |):1?{2 |2 ? 2> 0 ?{|?1 > Example 20.24 (Holomorphic Change of Variables). Suppose that i : r see Figure 20.3 We are© going to do this by making the changeª of variables, 2 C = R $ C is an injective holomorphic function such that i 0 (}) =0for all 6 } 5 = We may express i as (x> y):=W ({> |)= {2 |2>{| >

i ({ + l|)=X ({> |)+lY ({> |) in which case ¡ ¢ for all = + Hence if we make the change of variables, 2{ 2| 2 2 } { l| 5 = gxgy = det g{g| =2 { + | g{g| |{ ¯ · ¸¯ z = x + ly = i ({ + l|)=X ({> |)+lY ({> |) ¯ ¯ ¡ ¢ ¯ ¯ Notice that ¯ ¯ then 1 X{ X| {4 |4 = {2 |2 {2 + |2 = x {2 + |2 = xgxgy= gxgy = det g{g| = X{Y| X|Y{ g{g|= 2 Y{ Y| | | ¯ · ¸¯ ¯ ¯ ¡ ¢ ¡ ¢¡ ¢ ¡ ¢ Recalling that X and¯Y satisfy the¯ Cauchy Riemann equations, X{ = Y| and The function W is not injective on but it is injective on each of its connected ¯ ¯ X| = Y{ with i 0 = X{ + lY{> we learn components. Let G be the connected component in the ﬁrst quadrant so that = G ^ G and W ( G)=(1> 2) (0> 1) The change of variables theorem 2 2 2 ± × X Y X Y = X + Y = i 0 = then implies { | | { { { | | 2 Therefore 4 4 1 1 x 2 3 2 L := { | g{g| = xgxgy = 1 1= gxgy = i 0 ({ + l|) g{g|= ± G 2 (1>2) (0>1) 2 2 | · 4 ZZ± ZZ × | | ¡ ¢ and therefore L = L+ + L =2 (3@4) = 3@2= · 330 20 Multiple Integrals 20.3 The Polar Decomposition of Lebesgue Measure 331

Exercise 20.2 (Spherical Coordinates). Let W :(0> 4) (0>) (0> 2) $ Using polar coordinates, see Eq. (20.32), we ﬁnd, 3 × × R be deﬁned by 4 2 4 du2 du2 L2(d)= gu u g h =2 uh gu W (u> !> )=(u sin ! cos > u sin ! sin > u cos !) Z0 Z0 Z0 P du2 P = u (sin ! cos > sin ! sin > cos !) > 2 h 2 =2 lim uhdu gu =2 lim = = @d= P$4 0 P$4 2d 0 2d see Figure 20.4. By making the change of variables { = W (u> !> ) > show Z Z This shows that L2(d)=@d and the result now follows from Eq. (20.33).

20.3 The Polar Decomposition of Lebesgue Measure

Let g M g1 g 2 2 V = { 5 R : { := { =1 { | | l } l=1 X g T be the unit sphere in R equipped with its Borel — algebra, Vg1 and 1 B : Rg 0 $ (0> 4) Vg1 be deﬁned by ({):=({ > { {)= The inverse \{ } × | | | | map, 1 :(0> 4) Vg1 $ Rg 0 > is given by 1(u> $)=u$= Since 1 × \{ } and are continuous, they are both Borel measurable. For H 5 g1 and Fig. 20.4. The relation of { to (u> !> ) in spherical coordinates. V dA0> let B 1 H := u$ : u 5 (0>d] and $ 5 H = ((0>d] H) 5 g = d { } × BR

2 4 Definition 20.27. For H 5 Vg1 ,let(H):=g p(H1)= We call the g1 B · i({)g{ = g! g gu u2 sin ! i(W (u> !> )) surface measure on V = R3 0 0 0 · Z Z Z Z It is easy to check that is a measure. Indeed if H 5 Vg1 > then H1 = 3 1 B 4 for any Borel measurable function, i : R $ [0> 4]= ((0> 1] H) 5 Rg so that p(H1) is well defined. Moreover if H = l=1 Hl> × 4 B then H1 = l=1 (Hl)1 and Lemma 20.26. Let dA0 and ` ` 4 4 d { 2 (H)=g p(H1)= p ((Hl)1)= (Hl)= L (d):= h | | gp({)= · g l=1 l=1 Zg X X R The intuition behind this definition is as follows. If H Vg1 is a set and g@2 Then Lg(d)=(@d) = %A0 is a small number, then the volume of (1> 1+%] H = u$ : u 5 (1> 1+%] and $ 5 H Proof. By Tonelli’s theorem and induction, · { } should be approximately given by p ((1> 1+%] H) = (H)%> seeFigure20.5 d | 2 dw2 · Lg(d)= h | | h pg1(g|) gw below. On the other hand Rg1 R Z × g g p ((1> 1+%]H)=p (H1+% H1)= (1 + %) 1 p(H1)= = Lg1(d)L1(d)=L1 (d)= (20.33) \ Therefore we expect the area of H should be© given by ª So it su!ces to compute: g (1 + %) 1 p(H1) 2 2 2 (H) = lim = g p(H1)= d { d({1+{2) L2(d)= h | | gp({)= h g{1g{2= %&0 © % ª · RZ2 R2 Z 0 The following theorem is motivated by Example 20.23 and Exercise 20.2. \{ } 332 20 Multiple Integrals 20.3 The Polar Decomposition of Lebesgue Measure 333

( p)((d> e] H)=((d> e]) (H)=( )((d> e] H) = (20.39) × · × Since = (d> e] H :0?d?eand H 5 g1 > E { × BV } is a class (in fact it is an elementary class) such that ( )= g1 > E B(0>4) BV it follows from Theorem 19.55 and Eq. (20.39) that p = = Using this result in Eq. (20.35) gives

igp = i 1 g ( ) Fig. 20.5. Motivating the definition of surface measure for a sphere. RZg (0>4)Z Vg1 × ¡ ¢ which combined with Tonelli’s Theorem 20.8 proves Eq. (20.35). g Theorem 20.28 (Polar Coordinates). If i : $ [0> 4] is a ( g > )— R U Corollary 20.29. The surface area (Vg1) of the unit sphere Vg1 g is measurable function then B B R 2g@2 g1 (Vg1)= (20.40) i({)gp({)= i(u$)u gug($)= (20.34) (g@2) RZg (0>4)Z Vg1 × where is the gamma function given by Proof. By Exercise 19.7, 4 ({):= x{1hxgu (20.41) 1 1 igp = i gp= i g ( p) (20.35) Z0 s RZg Rg Z 0 (0>4)Z Vg1 \{ } ¡ ¢ × ¡ ¢ Moreover, (1@2) = > (1) = 1 and ({ +1)={ ({) for {A0= and therefore to prove Eq. (20.34) we must work out the measure p on Proof. Using Theorem 20.28 we find (0>4) Vg1 defined by B B 4 4 g1 u2 g1 g1 u2 1 Lg(1) = gu u h g = (V ) u h gu= p(D):=p (D) ; D 5 (0>4) Vg1 = (20.36) 0 0 B B Z VgZ1 Z ¡ ¢ If D =(d> e] H with 0 ?d?eand H 5 Vg1 > then × B We simplify this last integral by making the change of variables x = u2 so 1 1@2 1 1@2 (D)= u$ : u 5 (d> e] and $ 5 H = eH1 dH1 that u = x and gu = x gx= The result is { } \ 2 4 4 wherein we have used Hd = dH1 in the last equality. Therefore by the basic g1 u2 g1 x 1 1@2 u h gu = x 2 h x gx scaling properties of p and the fundamental theorem of calculus, 0 0 2 Z Z 4 1 g 1 x 1 ( p)((d> e] H)=p (eH1 dH1)=p(eH1) p(dH1) = x 2 h gx = (g@2)= (20.42) × \ 2 0 2 e Z g g g1 = e p(H1) d p(H1)=g p(H1) u gu= (20.37) · Combing the the last two equations with Lemma 20.26 which states that d g@2 Z Lg(1) = > we conclude that Letting g(u)=ug1gu> i.e. 1 g@2 = L (1) = (Vg1) (g@2) g 2 (M)= ug1gu ; M 5 > (20.38) B(0>4) ZM which proves Eq. (20.40). Example 19.24 implies (1) = 1 and from Eq. Eq. (20.37) may be written as (20.42), 334 20 Multiple Integrals 20.4 More proofs of the classical Weierstrass approximation Theorem 8.34 335

4 4 u2 u2 q 1 q g q (1@2) = 2 h gu = h gu Vqg{ := Vqg{>===> Vqg{ =({1>===>{g)={ (20.46) 0 q q q Z s Z4 Z µZ Z ¶ = L1(1) = = and 1 g The relation, ( +1) = ( ) is the consequence of the following integration 2 q g { { { Vq { g{ = {l(1 {l) = (20.47) q | | q q by parts argument: l=1 Z X 4 gx 4 g From these equations it follows that Vq is concentrating near { as q $4> a ({ +1)= hx x{+1 = x{ hx gx manifestation of the law of large numbers. Therefore it is reasonable to expect 0 x 0 gx Z 4 Z µ ¶ {1 x q = { x h gx = {({)= sq({):= i(Vq)g{ (20.48) q Z0 Z should approach i({) as q $4= Let %A0 be given, P =sup i({) : { 5 [0> 1] {| | } BRUCE: add Morrey’s Inequality 72.1 here. and =sup i(|) i({) : {> | 5 [0> 1] and | { % = % {| | | | } By uniform continuity of i on [0> 1]> lim%&0 % =0= Using these definitions and 20.4 More proofs of the classical Weierstrass q q the fact that {( )=1> approximation Theorem 8.34 q q i({) sq({) = (i({) i(Vq)) g{ i({) i(Vq) g{ In each of these proofs we will use the reduction explained the previous proof | | q q | | ¯Z ¯ Z ofTheorem8.34toreducetothecasewherei 5 F([0> 1]g)= The first proof we ¯ ¯ ¯ ( ) ( ¯) q + ( ) ( ) q will give here is based on the “weak law” of large numbers. The second will ¯ i { i Vq¯ g{ i { i Vq g{ Vq{ A% | | Vq{ % | | be another approximate — function argument. Z{| | } Z{| | } 2Pq ( V { A%)+ = (20.49) Proof. of Theorem 8.34. Let 0 :=(0> 0>===>0)> 1 :=(1> 1>===>1) and { | q | % [0> 1]:=[0> 1]g= By considering the real and imaginary parts of i separately, By Chebyshev’s inequality, it su!ces to assume i 5 F([0> 1]> R)= For { 5 [0> 1]> let { be the measure on q 1 2 q g 0> 1 such that { ( 0 )=1 { and { ( 1 )={= Then { ( Vq { A%) 2 (Vq {) g{ = 2 > { } { } { } | | % q q% Z |g{(|)=0 (1 {)+1 { = { and (20.43) and therefore, Eq. (20.49) yields the estimate 0>1 · · Z{ } 2 2 2 2 gP (| {) g{(|)={ (1 {)+(1 {) { = {(1 {)= (20.44) i sq 4 2 + % 0>1 · k k q% Z{ } and hence For { 5 [0> 1] let = be the product of >===> on { {1 {g {1 {g lim sup 0 as 0 g ··· i sq 4 % $ % & = := 0> 1 = Alternatively the measure may be described by q$4 k k { } { This completes the proof since, using Eq. (20.45), g 1%l %l q { ( % )= (1 {l) { (20.45) { } l q l=1 sq({)= i(Vq($)) ( $ )= i(Vq($)) {( $l )> Y { { } { } $5 q $5 q l=1 for % 5 = Notice that { ( % ) is a degree g polynomial in { for each % 5 = X X Y { } q g For q 5 N and { 5 [0> 1]> let { denote the q — fold product of { with itself is an qg — degree polynomial in { 5 R )= on q>[($)=$ 5 g for $ 5 q and let l l R Exercise 20.3. Verify Eqs. (20.46) and (20.47). This is most easily done using 1 g Eqs. (20.43) and (20.44) and Fubini’s theorem repeatedly. (Of course Fubini’s Vq =(Vq>===>Vq):=([1 + [2 + + [q)@q> ··· theorem here is over kill since these are only finite sums after all. Nevertheless q g so Vq : $ R = The reader is asked to verify (Exercise 20.3) that it is convenient to use this formulation.) 336 20 Multiple Integrals 20.5 More Spherical Coordinates 337

The second proof requires the next two lemmas. Since the product in the above integrand is a polynomial if ({> |) 5 Rg Rg> × 4 it follows easily that s ({) is polynomial in {= Lemma 20.30 (Approximate — sequences). Suppose that T is a q { q}q=1 sequence of positive functions on Rg such that 20.5 More Spherical Coordinates Tq({) g{ =1and (20.50) g ZR In this section we will define spherical coordinates in all dimensions. Along lim Tq({)g{ =0for all %A0= (20.51) q$4 the way we will develop an explicit method for computing surface integrals on {Z% spheres. As usual when q =2define spherical coordinates (u> ) 5 (0> 4) | | × g g [0> 2) so that For i 5 EF(R )>Tq i converges to i uniformly on compact subsets of R = {1 u cos = = W2(> u)= Proof. The proof is exactly the same as the proof of Lemma 8.28, it is {2 u sin only necessary to replace R by Rg everywhere in the proof. µ ¶ µ ¶ Define For q =3we let {3 = u cos !1 and then q Tq : R $ [0> 4) by Tq({)=tq({1) ===tq({g)= (20.52) {1 = W2(> u sin !1)> where tq is defined in Eq. (8.23). {2 µ ¶ Lemma 20.31. The sequence 4 is an approximate — sequence, i.e. Tq q=1 as can be seen from Figure 20.6, so that they satisfy Eqs. (20.50) and (20.51).{ }

Proof. The fact that Tq integrates to one is an easy consequence of Tonelli’s theorem and the fact that tq integrates to one. Since all norms on g R are equivalent, we may assume that { =max {l : l =1> 2>===>g when proving Eq. (20.51). With this norm | | {| | }

g g g { 5 R : { % = ^ =1 { 5 R : {l % | | l | | and therefore by© Tonelli’s theorem,ª © ª

g Tq({)g{ Tq({)g{ = g tq(w)gw l=1 Fig. 20.6. Setting up polar coordinates in two and three dimensions. {Z% {lZ% {5RZ{ % {| | } X{| | } { | | } which tends to zero as q $4by Lemma 8.29. Proof. Proof of Theorem 8.34. Again we assume i 5 F Rg> C and i 0 3 4 3 4 on f where := (0 1)g Let ( ) be definedasinEq.(20.52).Thenby Tg Tg > = Tq { ¡ ¢ {1 u sin !1 cos W2(> u sin !1) Lemma 20.31 and 20.30, sq({):=(Tq I )({) $ I ({) uniformly for { 5 [0> 1] C {2 D = = C u sin !1 sin D =: W3(> !1>u>)= u cos !1 as q $4= So to finish the proof it only remains to show sq({) is a polynomial {3 µ ¶ u cos !1 when { 5 [0> 1]= For { 5 [0> 1]> We continue to work inductively this way to define sq({)= Tq({ |)i(|)g| 3 4 g ZR {1 g E . F 1 1 2 q E . F Wq(> !1>===>!q2>usin !q1> ) = i(|) f (1 ({ | ) ) 1 g| E F = = Wq+1(> !1>===>!q2>!q1>u)= q l l {l|l 1 cos fq [0>1] | | C { D u !q1 Z l=1 q µ ¶ Y £ ¤ {q+1 1 g = i(|) f1(1 ({ | )2)q g|= f q l l So for example, q Z[0>1] l=1 Y £ ¤ 338 20 Multiple Integrals 20.5 More Spherical Coordinates 339

{1 = u sin !2 sin !1 cos q+1(>!1>===>!q2>!q1>u)

CWq CWq CWq CWq CWq {2 = u sin !2 sin !1 sin === u cos !q1 sin !q1 = C C!1 C!q2 C C { = u sin ! cos ! 00=== 0 u sin ! cos ! 3 2 1 ¯· q1 q1 ¸¯ ¯ 2 2 ¯ {4 = u cos !2 = u¯ cos !q1 +sin !q1 q(>>!1>===>!q2>) ¯ ¯ ¯ = ( sin ) and more generally, u¡ q > !1>===>!q2>u ¢ !q1 > i.e. {1 = u sin !q2 ===sin !2 sin !1 cos

{2 = u sin !q2 ===sin !2 sin !1 sin q+1(> !1>===>!q2>!q1>u)=uq(> !1>===>!q2>usin !q1)= (20.57) {3 = u sin !q2 ===sin !2 cos !1 To arrive at this result we have expanded the determinant along the bottom . . row. Staring with 2(> u)=u already derived in Example 20.23, Eq. (20.57) implies, {q2 = u sin !q2 sin !q3 cos !q4 2 {q1 = u sin !q2 cos !q3 3(> !1>u)=u2(> u sin !1)=u sin !1 = cos (20.53) 3 2 {q u !q2= 4(> !1>!2>u)=u3(> !1>usin !2)=u sin !2 sin !1 . By the change of variables formula, . (> ! >===>! >u)=uq1 sinq2 ! ===sin2 ! sin ! i({)gp({) q 1 q2 q2 2 1 Rq Z 4 which proves Eq. (20.55). Eq. (20.56) now follows from Eqs. (50.3), (20.54) = gu g!1 ===g!q2gq(> !1>===>!q2>u)i(Wq(> !1>===>!q2>u)) and (20.55). Z0 Z0!l>02 As a simple application, Eq. (20.56) implies (20.54)

q1 q2 2 where (V )= sin !q2 ===sin !2 sin !1g!1 ===g!q2g 0!l>02 q(> !1>===>!q2>u):= det Wq0 (> !1>===>!q2>u) = Z | | q2 =2 = ( q2) (20.58) Proposition 20.32. The Jacobian, q is given by n V q2 =1 nY (> ! >===>! >u)=uq1 sinq2 ! ===sin2 ! sin ! = (20.55) q 1 q2 q2 2 1 n where n := 0 sin !g!= If n 1> we have by integration by parts that, If i is a function on uVq1 — the sphere of radius u centered at 0 inside of R q> then n n1 n2 2 R n = sin !g! = sin !gcos ! =2n>1 +(n 1) sin ! cos !g! Z0 Z0 Z0 q1 i({)g({)=u i(u$)g($) n2 2 =2n>1 +(n 1) sin ! 1 sin ! g! =2n>1 +(n 1) [n2 n] uVq1 Vq1 Z Z Z0 ¡ ¢ = i(Wq(> !1>===>!q2>u))q(> !1>===>!q2>u)g!1 ===g!q2g and hence n satisﬁes 0 = > 1 =2and the recursion relation Z0!l>02 (20.56) n 1 = 2 for n 2= n n n Proof. We are going to compute q inductively. Letting := u sin !q1 CWq CWq and writing C for C (> !1>===>!q2>) we have Hencewemayconclude 1 2 3 1 4 2 5 3 1 = > =2> = > = 2> = > = 2> = 0 1 2 2 3 3 4 4 2 5 5 3 6 6 4 2 340 20 Multiple Integrals 20.6 Sard’s Theorem 341

2q+1 2q+1 and more generally by induction that we may write (V )= q! which shows that Eqs. (50.9) and (20.60 in agreement. We may also write the formula in Eq. (20.60) as (2n 1)!! (2n)!! 2n = and 2n+1 =2 = ; q@2 (2n)!! (2n +1)!! ? 2(2) for q even q (q1)!! (V )= q+1 Indeed, = (2) 2 (q1)!! for q odd. 2n +2 2n +2 (2n)!! [2(n + 1)]!! = 2 +1 = 2 =2 2(n+1)+1 2n +3 n 2n +3 (2n + 1)!! (2(n +1)+1)!! 20.6 Sard’s Theorem and 2n +1 2n +1 (2n 1)!! (2n +1)!! See p. 538 of Taylor and references. Also see Milnor’s topology book. Add 2(n+1) = 2n = = = 2n +1 2n +2 (2n)!! (2n +2)!! in the Brower’s Fixed point theorem here as well. Also Spivak’s calculus on The recursion relation in Eq. (20.58) may be written as manifolds. p 4 g q q1 Theorem 20.33. Let X r >i 5 F (X> ) and F := { 5 X : rank(i 0({)) ?q (V )= V q1 (20.59) R R be the set of critical points of i= Then the critical values, i({F)> is a Borel mea- } g which combined with V1 =2 implies¡ ¢ surable subset of R of Lebesgue measure 0=

V1 =2¡ >¢ Remark 20.34. This result clearly extends to manifolds. 2 ¡(V ¢)=2 1 =2 2> For simplicity in the proof given below it will be convenient to use the · · 1 g 2 2 norm, { := maxl {l = Recall that if i 5 F (X> R ) and s 5 X> then 3 1 2 | | | | (V )=2 2 2 =2 2 = > · · · · 2 2!! 1 1 2 2 2 2 3 2 4 2 2 2 2 i(s+{)=i(s)+ i 0(s+w{){gw = i(s)+i 0(s){+ [i 0(s + w{) i 0(s)] {gw (V )= 3 = 2 = 2!! · 2!! · 3 3!! Z0 Z0 1 2 3 1 233 so that if (V5)=2 2 2 = > · · 2 · 3 · 4 2 4!! 1 4 3 1 2 3 1 4 2 2 U(s> {):=i(s + {) i(s) i 0(s){ = [i 0(s + w{) i 0(s)] {gw (V6)=2 2 2 2= · · 2 · 3 · 4 2 · 5 3 5!! Z0 and more generally that we have 1 q q+1 2 2(2) 2 +1 (2) U(s> {) { i 0(s + w{) i 0(s) gw = { %(s> {)= (V q)= and (V q )= (20.60) | | | | | | | | (2q 1)!! (2q)!! Z0 By uniform continuity, it follows for any compact subset N X that which is veriﬁed inductively using Eq. (20.59). Indeed, sup %(s> {) : s 5 N and { $ 0 as & 0= 2(2)q (2q 1)!! (2)q+1 {| | | | } (V2q+1)=(V2q) = = 2q p q (2q 1)!! (2q)!! (2q)!! Proof. Notice that if { 5 X F> then i 0({):R $ R is surjective, whichisanopencondition,sothat\ X F is an open subset of X= This shows and \ ˜ p ˜ F is relatively closed in X> i.e. there exists F @ R such that F = F _ X= q+1 q+1 Let Nq X be compact subsets of X such that Nq % X> then Nq _ F % F (q+1) 2q+2 2q+1 (2) (2q)!! 2(2) (V )=(V )=(V )2 +1 = 2 = = ˜ q (2q)!! (2q + 1)!! (2q +1)!! and Nq _ F = Nq _ F is compact for each q= Therefore, i(Nq _ F) % i(F) i.e. i(F)=^qi(Nq _ F) is a countable union of compact sets and therefore Using is Borel measurable. Moreover, since p(i(F)) = limq$4 p(i(Nq _ F))> it (2q)!! = 2q (2(q 1)) ===(2 1) = 2qq! su!ces to show p(i(N)) = 0 for all compact subsets N F= Case 1. (q p) · 342 20 Multiple Integrals 20.6 Sard’s Theorem 343

Let N =[d> d + ] be a cube contained in X and by scaling the domain we 1. p(i(F F1)) = 0= q \ may assume =(1> 1> 1>===>1)= For Q 5 N and m 5 VQ := 0> 1>===>Q 1 2. p(i(Fl Fl+1)) = 0 for all l 1= let N = m@Q +[d> d + @Q] so that N = ^ N with{Nr _ Nr = >}if 3. p(i(F ))\ = 0 for all l su!ciently large. m m5VQ m m m0 l m = m0= Let Tm : m =1===>P be the collection of those Nm : m 5 VQ which 6 { } { } Step 1. If p =1> there is nothing to prove since F = F1 so we may assume intersect F= For each m> let sm 5 Tm _ F and for { 5 Tm sm we have p 2= Suppose that { 5 F F1> then i 0(s) =0and so by reordering the \ 6 i(sm + {)=i(sm)+i 0(sm){ + Um({) components of { and i(s) if necessary we may assume that C1i1 (s) =0where 6 we are writing Ci(s)@C{l as Cli (s) = The map k({):=(i1({)>{2>===>{q) has where Um({) %m(Q)@Q and %(Q):=maxm %m(Q) $ 0 as Q $4= Now | | dierential 5 6 p (i(Tm)) = p (i(sm)+(i 0(sm)+Um)(Tm sm)) C1i1 (s) C2i1 (s) === Cqi1 (s) 9 0100: = p ((i 0(sm)+Um)(Tm sm)) 9 : k0(s)=9 . . . . : 7 . . . . 8 = p (Rm (i 0(sm)+Um)(Tm sm)) (20.61) . . . . q p1 0001 where Rm 5 VR(q) is chosen so that Rmi 0(sm)R R 0 = Now p1 ×{ } which is not singular. So by the implicit function theorem, there exists there Rmi 0(sm)(Tm sm) is contained in 0 where R is a cube cen- p1 ×{ } exists Y 5 such that k : Y $ k(Y ) 5 is a dieomorphism and in tered at 0 5 R with side length at most 2 i 0(sm) @Q 2P@Q where s k(s) | | particular ( ) =0for and hence Consider the P =maxs5N i 0(s) = It now follows that Rm (i 0(sm)+Um)(Tm sm) is con- Ci1 { @C{1 { 5 Y Y X F1= | | 1 6 p \ tained the set of all points within %(Q)@Q of 0 and in particular map j := i k : Y 0 := k(Y ) $ R > which satisﬁes ×{ } R (i 0(s )+U )(T s ) (1 + %(Q)@Q ) [%(Q)@Q> %(Q)@Q ]= (i1({)>i2({)>===>ip({)) = i({)=j(k({)) = j((i1({)>{2>===>{q)) m m m m m × From this inclusion and Eq. (20.61) it follows that which implies j(w> |)=(w> x(w> |)) for (w> |) 5 Y 0 := k(Y ) 5 k(s)> see Figure p1 20.7 below where s =¯{ and p = s= Since P p (i(T )) 2 (1 + %(Q)@Q ) 2%(Q)@Q m Q · ¸ 1 =2pP p1 [(1 + %(Q)@Q )]p1 %(Q) Q p and therefore, 1 p (i(F _ N)) p (i(T )) Q q2pP p1 [(1 + %(Q)@Q )]p1 %(Q) m Q p m X 1 =2qP q1 [(1 + %(Q)@Q )]q1 %(Q) $ 0 as Q $4 Q pq since p q= This proves the easy case since we may write X as a countable union of cubes N as above. Remark.Thecase(p?q) also follows from the case p = q as follows. When p?q>F= X and we must show p(i(X)) = 0= qp q Letting I : X R $ R be the map I ({> |)=i({)= Then I 0({> |)(y> z)= × qp Fig. 20.7. Making a change of variable so as to apply induction. i 0({)y> and hence FI := X R = So if the assertion holds for p = q we have × qp p(i(X)) = p(I(X R )) = 0= × Case 2. (pAq) This is the hard case and the case we will need in the co-area 10 formula to be proved later. Here I will follow the proof in Milnor. Let j0(w> |)= Cwx(w> |) C|x(w> |) · ¸ Fl := { 5 X : C i({)=0when l { | | } it follows that (w> |) is a critical point of j i | 5 Fw0 —thesetofcriticalpoints so that F F1 F2 F3 ==== The proof is by induction on q and goes by of | $ x(w> |)= Since k is a dieomorphism we have F0 := k(F _ Y ) are the the following steps: critical points of j in Y 0 and 344 20 Multiple Integrals 20.7 Exercises 345

i(F _ Y )=j(F0)=^ [ w x (F0)] = 20.7 Exercises w { }× w w

By the induction hypothesis, pp1(xw(Fw0)) = 0 for all w> and therefore by Exercise 20.4. Prove Theorem 20.12. Suggestion, to get started deﬁne Fubini’s theorem,

(D):= g ({1) === g ({q)1D ({1>===>{q) p(i(F _ Y )) = pp1(xw(F0))1 gw =0= [1 [q w Yw0 => Z Z R 6 Z and then show Eq. (20.18) holds. Use the case of two factors as the model of Since F F1 may be covered by a countable collection of open sets Y as above, \ your proof. it follows that p(i(F F1)) = 0= Step 2. Suppose that s 5 Fn Fn+1> then \ \ there is an such that = n +1such that C i(s)=0while C i(s)=0 Exercise 20.5. Let ([m> m>m) for m =1> 2> 3 be — finite measure spaces. | | M for all n= Again by permuting coordinates we may assume that 1 =0 Let I :([1 [2) [3 $ [1 [2 [3 be defined by | | h 6 × × × × and C i1(s) =0= Let z({):=C 1 i1({)> then z(s)=0while C1z(s) =0= 6 6 I (({1>{2)>{3)=({1>{2>{3)= So again the implicit function theorem there exists Y 5 s such that k({):= (z ({) >{2>===>{q) maps Y $ Y 0 := k(Y ) 5 k(s) in a dieomorphic way and 1. Show is (( ) ) — measurable and 1 1 I 1 2 3> 1 2 3 I in particular C1z({) =0on Y so that Y X Fn+1= As before, let j := i k is ( M M( M M) M) — measurable.M That is 6 q\1 1 2 3> 1 2 3 and notice that F0 := k(Fn _ Y ) 0 R and M M M M M M n { }× I :(([1 [2) [3> ( 1 2) 3) $ ([1 [2 [3> 1 2 3) i(Fn _ Y )=j(Fn0 )=¯j (Fn0 ) × × M M M × × M M M is a “measure theoretic isomorphism.” where j¯ := j ( 0 q1) = Clearly F0 is contained in the critical points of j>¯ 1 R _Y 0 n 2. Let := I [(1 2) 3] > i.e. (D)=[(1 2) 3](I (D)) for all and therefore,| { by}× induction D 5 1 2 3= Then is the unique measure on 1 2 3 suchM that M M M M M 0=p(¯j(Fn0 )) = p(i(Fn _ Y ))= (D1 D2 D3)=1(D1)2(D2)3(D3) × × Since Fn Fn+1 is covered by a countable collection of such open sets, it follows for all Dl 5 l= We will write := 1 2 3= that \ M 3. Let i : [1 [2 [3 $ [0> 4] be a ( 1 2 3> R¯ ) — measurable p(i(F F +1)) = 0 for all n 1= × × M M M B n \ n function. Verify the identity, Step 3. Suppose that T is a closed cube with edge length contained in X and nAq@p 1= We will show p(i(T _ Fn)) = 0 and since T is arbitrary it ig = g3({3) g2({2) g1({1)i({1>{2>{3)> [1 [2 [3 [3 [2 [1 will follows that p(i(Fn)) = 0 as desired. By Taylor’s theorem with (integral) Z × × Z Z Z remainder, it follows for { 5 T _ Fn and k such that { + k 5 T that makes sense and is correct. 4. (Optional.) Also show the above identity holds for any one of the six i({ + k)=i({)+U({> k) possible orderings of the iterated integrals. where n+1 Exercise 20.6. Prove the second assertion of Theorem 20.18. That is show U({> k) f k g g g | | k k p is the unique translation invariant measure on Rg such that p ((0> 1] )= where f = f(T> n)= Now subdivide T into uq cubes of edge size @u and let 1= Hint: Look at the proof of Theorem 19.10. B T0 be one of the cubes in this subdivision such that T0 _ Fn = > and let 6 Exercise 20.7. (Part of Folland Problem 2.46 on p. 69.) Let [ =[0> 1]> { 5 T0 _ Fn= It then follows that i(T0) is contained in a cube centered at n+1 = be the Borel — field on [> p be Lebesgue measure on [0> 1] and i({) 5 p with side length at most 2f (@u) and hence volume at most [0>1] R Mbe countingB measure, (D)=#(D)= Finally let G = ({> {) 5 [2 : { 5 [ (2 )p ( )p(n+1) Therefore, ( ) is contained in the union of at most f @u = i T _ Fn be the diagonal in [2= Show { } uq cubes of volume (2f)p (@u)p(n+1) and hence peach

p p(n+1) q p p(n+1) qp(n+1) 1G({> |)g(|) gp({) = 1G({> |)gp({) g(|) p (i(T _ Fn)) (2f) (@u) u =(2f) u $ 0 as u %4 6 Z[ ·Z[ ¸ Z[ ·Z[ ¸ provided that q p(n +1)? 0> i.e. provided nAq@p 1= by explicitly computing both sides of this equation. 346 20 Multiple Integrals

Exercise 20.8. Folland Problem 2.48 on p. 69. (Counter example related to Fubini Theorem involving counting measures.) 21 Exercise 20.9. Folland Problem 2.50 on p. 69 pertaining to area under a curve. (Note the should be ¯ in this problem.) p M×BR M BR L -spaces Exercise 20.10. Folland Problem 2.55 on p. 77. (Explicit integrations.)

Exercise 20.11. Folland Problem 2.56 on p. 77. Let i 5 O1((0>d)>gp)> d i(w) 1 j({)= { w gw for { 5 (0>d)> show j 5 O ((0>d)>gp) and R d d j({)g{ = i(w)gw= Z0 Z0 4 sin { sin { 1 Exercise 20.12. Show 0 { gp({)=4= So { 5@ O ([0> 4)>p) and 4 sin { gp({) is not deﬁned as a Lebesgue integral. Let ([> >) be a measure space and for 0 ?s?4 and a measurable 0 { R ¯ ¯ M ¯ ¯ function i : [ $ C let ExerciseR 20.13. Folland Problem 2.57 on p. 77. 1@s i := i s g = (21.1) Exercise 20.14. Folland Problem 2.58 on p. 77. k ks | | µZ[ ¶ Exercise 20.15. Folland Problem 2.60 on p. 77. Properties of the — func- When s = 4> let tion. i =inf d 0:( i Ad)=0 (21.2) k k4 { | | } Exercise 20.16. Folland Problem 2.61 on p. 77. Fractional integration. For 0 ?s4> let s Exercise 20.17. Folland Problem 2.62 on p. 80. Rotation invariance of sur- O ([> >)= i : [ $ C : i is measurable and i s ? 4 @ q M { k k } face measure on Vq1= where i q j i i = j a.e. Notice that i j s =0i i q j and if i q j then i = j = In general we will (byk abusek of notation) use i to denote Exercise 20.18. Folland Problem 2.64 on p. 80. On the integrability of k ks k ks d e both the function i and the equivalence class containing i= { log { for { near 0 and { near 4 in Rq= | | | | || Remark 21.1. Suppose that i P> then for all dAP>( i Ad)=0 k k4 | | and therefore ( i AP)=limq$4 ( i AP+1@q)=0> i.e. i({) P for -a.e.{= Conversely,| | if i P a.e.| and| dAP then ( i Ad| )=0| and hence i P= This leads| to| the identity: | | k k4 i =inf d 0: i({) d for —a.e.{ = k k4 { | | } The next theorem is a generalization Theorem 5.6 to general integrals and the proof is essentially identical to the proof of Theorem 5.6. Theorem 21.2 (Hölder’s inequality). Suppose that 1 s 4and t := s 1 1 s1 > or equivalently s + t =1= If i and j aremeasurablefunctionsthen

ij 1 i j = (21.3) k k k ks ·k kt s Assuming s 5 (1> 4) and i s j t ? 4> equality holds in Eq. (21.3) i i and j t are linearly dependentk k ·k ask elements of O1 which happens i | | | | j t i s = j t i s a.e. (21.4) | | k ks k kt | | 348 21 Os-spaces 21 Os-spaces 349

Proof. The cases where i =0or 4 or j =0or 4 are easy to deal Theorem 21.4 (Minkowski’s Inequality). If 1 s 4and i>j 5 Os k kt k ks with and are left to the reader. So we will now assume that 0 ? i t> j s ? then 4= Let v = i @ i and w = j @ j then Lemma 5.5 implies k k k k i + j i + j = (21.6) | | k ks | | k kt k ks k ks k ks ij 1 i s 1 j t Moreover, assuming i and j are not identically zero, equality holds in Eq. | | | | + | | (21.5) (21.6) i sgn(i) sgn(j) a.e. (see the notation in Definition 5.7) when s =1 i j s i t j t $ k ksk kt k ks k k and i = fj a.e. for some fA0 for s 5 (1> 4)= s1 (s1) s@t s@t t s with equality i j@ j t = i @ i s = i @ i s , i.e. j i s = Proof. When s = 4> i i a.e. and j j a.e. so that i + j t s | k k | | | k k | | k k | | k k | | k k4 | | k k4 | | j t i = Integrating Eq. (21.5) implies i + j i + j a.e. and therefore k k | | | | | | k k4 k k4 ij 1 1 1 i + j i + j = k k + =1 k k4 k k4 k k4 i s j t s t k k k k When s?4> with equality i Eq. (21.4) holds. The proof is finished since it is easily checked that equality holds in Eq. (21.3) when i s = f j t of j t = f i s for some i + j s (2 max ( i > j ))s =2s max ( i s > j s) 2s ( i s + j s) > constant f= | | | | | | | | | | | | | | | | | | | | | | s s s s The following corollary is an easy extension of Hölder’s inequality. i + j s 2 i s + j s ? 4= k k k k k k In case s =1> ¡ ¢ Corollary 21.3. Suppose that il : [ $ C aremeasurablefunctionsforl = q 1 1 1>===>q and s1>===>sq and u arepositivenumberssuchthat l=1 sl = u > then i + j 1 = i + j g i g + j g q q q P k k [ | | [ | | [ | | 1 1 Z Z Z il il where s = u = k ksl l with equality i i + j = i + j a.e. which happens i sgn(i) sgn(j) a.e. ° =1 ° =1 =1 $ °l °u l l In case (1 |)| we| may| | assume| + and are all positive °Y ° Y X s 5 > 4 > i j s> i s j s ° ° k k k k k k Proof. To prove° this° inequality, start with q =2> then for any s 5 [1> 4]> since otherwise the theorem is easily verified. Now s s1 s1 u u u u u i + j = i + j i + j ( i + j ) i + j ij u = i j g i s j s | | | || | | | | | | | k k [ | | | | k k k k Z with equality i sgn(i) $ sgn(j)= Integrating this equation and applying s where s = s1 is the conjugate exponent. Let s1 = su and s2 = s u so that Holder’s inequality with t = s@(s 1) gives 1 1 1 s1 + s2 = u as desired. Then the previous equation states that i + j sg i i + j s1g + j i + j s1g ij i j [ | | [ | || | [ | || | k ku k ks1 k ks2 Z Z Z ( i + j ) i + j s1 (21.7) as desired. The general case is now proved by induction. Indeed, k ks k ks k| | kt

q+1 q q with equality i il = il iq+1 il iq+1 · k ksq+1 sgn(i) $ sgn(j) and ° l=1 ° °l=1 ° °l=1 ° ° °u ° °u ° °t s s °Y ° °Y ° °Y ° i i + j s j 1 °1 °1 ° q °1 °1 ° | | = | | s = | | a.e. (21.8) where t +s°q+1 = u° = Since° l=1 sl ° = t° > we° may now use the induction i s i + j s j s hypothesis to conclude µk k ¶ k k µk k ¶ qP q Therefore i i > l l sl k k s1 t s1 t s °l=1 °t l=1 + = ( + ) = + (21.9) °Y ° Y i j t i j g i j g= ° ° k| | k [ | | [ | | which combined with the previous° ° displayed equation proves the generalized Z Z ° ° form of Holder’s inequality. Combining Eqs. (21.7) and (21.9) implies 350 21 Os-spaces 21.1 Jensen’s Inequality 351

i + j s i i + j s@t + j i + j s@t (21.10) wherein we used Tonelli’s theorem in third line. From this it follows that k ks k ksk ks k ksk ks \ n({> |)i(|) g(|) ? 4 for -a.e.{> with equality i Eq. (21.8) holds which happens i i = fj a.e. with fA0= | | Solving for i + j in Eq. (21.10) gives Eq. (21.6). R s { $ Ni({):= n({> |)i(|)g(|) 5 Os() The nextk theoremk gives another example of using Hölder’s inequality Z\ Theorem 21.5. Suppose that ([> >) and (\> >) be — ﬁnite measure and that Eq. (21.11) holds. M N spaces, s 5 [1> 4]>t= s@(s1) and n : [ \ $ C be a —measurable Similarly if s = 4> × M N function. Assume there exist ﬁnite constants F1 and F2 such that

n({> |)i(|) g(|) i 4 n({> |) g(|) F2 i 4 for —a.e.{= | | k kO ()· | | k kO () n({> |) g({) F1 for a.e. | and \ \ | | Z Z Z[ so that Ni 4 F2 i 4 = If s =1> then k kO () k kO () n({> |) g(|) F2 for a.e. {. | | Z\ g({) g(|) n({> |)i(|) = g(|) i(|) g({) n({> |) If s( ) then | | | | | | i 5 O > Z[ Z\ Z\ Z[ F1 g(|) i(|) n({> |)i(|) g(|) ? 4 for —a.e.{> \ | | | | Z Z\ which shows Ni 1 F1 i 1 = s k kO () k kO () { $ Ni({):= \ n({> |)i(|)g(|) 5 O () and

R 1@s 1@t Ni s F F i s (21.11) k kO () 1 2 k kO () 21.1 Jensen’s Inequality Proof. Suppose s 5 (1> 4) to begin with and let t = s@(s 1)> then by Deﬁnition 21.6. Afunction! :(d> e) $ R is convex if for all d?{0 ?{1 ? Hölder’s inequality, e and w 5 [0> 1] !({w) w!({1)+(1 w)!({0) where {w = w{1 +(1 w){0=

n({> |)i(|) g(|)= n({> |) 1@t n({> |) 1@s i(|) g(|) Example 21.7. The functions exp({) and log({) are convex and {s is | | | | | | | | Z\ Z\ convex i s 1 as follows from Corollary 21.9 below which in part states 1@t 1@s 2 that any ! 5 F ((d> e) > R) such that !00 0 is convex. n({> |) g(|) n({> |) i(|) s g(|) | | | || | ·Z\ ¸ ·Z\ ¸ The following Proposition is clearly motivated by Figure 21.1. 1@s F1@t n({> |) i(|) s g(|) = 2 | || | Proposition 21.8. Suppose ! :(d> e) $ R is a convex function, then ·Z\ ¸ Therefore, 1. For all x> y> z> } 5 (d> e) such that x?}>z5 [x> }) and y 5 (x> }]> s s !(y) !(x) !(}) !(z) = (21.12) n( >|)i(|) g(|) = g({) n({> |)i(|) g(|) y x } z | · | s | | °Z\ °O () Z[ ·Z\ ¸ ° ° ° ° s@t s 2. For each f 5 (d> e)> the right and left sided derivatives !0 (f) exists in R ( ) ( ) ( ) ( ) ± ° ° F2 g { g | n {> | i | and if d?x?y?e>then ! (x) ! (y) ! (y)= [ \ | || | +0 0 +0 Z Z 3. The function ! is continuous. s@t s = F2 g(|) i(|) g({) n({> |) 4. For all w 5 (d> e) and 5 [! (w)>! (w)]>!({) !(w)+({ w) for all | | | | 0 +0 Z\ Z[ { 5 (d> e)= In particular, s@t s s@t s F2 F1 g(|) i(|) = F2 F1 i Os() > \ | | k k ( ) ( )+ ( )( ) for all ( ) Z ! { ! w !0 w { w {> w 5 d> e = 352 21 Os-spaces 21.1 Jensen’s Inequality 353 z x } z !(z) !(}) + !(x) 20 y } x } x which again is true by the convexity of != 1) General case. If x?z?y?}> 15 then by 1a-1c)

10 !(}) !(z) !(y) !(z) !(y) !(x) } z y z y x 5 and if x?y?z?}

0 !(}) !(z) !(z) !(y) !(z) !(x) -6 -4 -2 0 2 = } z z y z x x We have now taken care of all possible cases. 2) On the set d?z?}?e> Fig. 21.1. A convex function along with two cords corresponding to {0 = 32 and Eq. (21.12) shows that (!(}) !(z)) @ (} z) is a decreasing function in z {1 =4and {0 = 35 and {1 = 32= and an increasing function in } and therefore !0 ({) exists for all { 5 (d> e)= Also from Eq. (21.12) we learn that ±

Proof. 1a) Suppose first that x?y= z?}>in which case Eq. (21.12) is !(}) !(z) !0 (x) for all d?x?z?}?e> (21.13) equivalent to + } z !(y) !(x) (!(y) !(x)) (} y) (!(}) !(y)) (y x) !0 (}) for all d?x?y?}?e> (21.14) y x which after solving for !(y) is equivalent to the following equations holding: and letting z % } in the first equation also implies that y x } y !(y) !(}) + !(x) = !0 (x) !0 (}) for all d?x?}?e= } x } x + The inequality, ! (}) ! (})> is also an easy consequence of Eq. (21.12). 3) Butthislastequationstatesthat!(y) !(})w + !(x)(1 w) where w = yx 0 +0 }x Since !({) has both left and right finite derivatives, it follows that ! is contin- and y = w} +(1 w)x and hence is valid by the definition of ! being convex. uous. (For an alternative proof, see Rudin.) 4) Given w> let 5 [! (w)>! (w)]> 1b) Now assume x = z?y?}>in which case Eq. (21.12) is equivalent to 0 +0 then by Eqs. (21.13) and (21.14), ( ( ) ( )) ( ) ( ( ) ( )) ( ) ! y ! x } x ! } ! x y x !(w) !(x) !(}) !(w) !0 (w) !+0 (w) which after solving for !(y) is equivalent to w x } w for all d?x?w?}?e=Item 4. now follows. !(y)(} x) !(})(y x)+!(x)(} y) Corollary 21.9. Suppose ! :(d> e) $ R is dierential then ! is convex i !0 which is equivalent to 2 is non decreasing. In particular if ! 5 F (d> e) then ! is convex i !00 0= y x } y !(y) !(}) + !(x) = Proof. By Proposition 21.8, if ! is convex then !0 is non-decreasing. Con- } x } x versely if !0 is increasing then by the mean value theorem, Again this equation is valid by the convexity of != 1c) x?z?y= }> in !({1) !(f) which case Eq. (21.12) is equivalent to = !0(1) for some 1 5 (f> {1) {1 f (!(}) !(x)) (} z) (!(}) !(z)) (} x) and !(f) !({0) and this is equivalent to the inequality, = !0(2) for some 2 5 ({0>f)= f {0 354 21 Os-spaces 21.2 Modes of Convergence 355

q q s sl Hence q q 1 l 1 sl v ln vl l=1 s ln vl ln v l ( ) ( ) ( ) ( ) v1 ===vq = h l=1 = h l h l = (21.16) ! {1 ! f ! f ! {0 s s P P l=1 l l=1 l {1 f f {0 X X for all {0 ?f?{1= Solving this inequality for !(f) gives where the inequality follows from Eq. (21.15) with [ = 1> 2>===>q >= q 1 and i (l):=lnvsl = Of course Eq. (21.16) may be{ proved directly} l=1 sl l l f {0 {1 f !(f) !({1)+ !({0) using the convexity of the exponential function. {1 {0 {1 {0 P showing ! is convex. 21.2 Modes of Convergence Theorem 21.10 (Jensen’s Inequality). Suppose that ([> >) is a prob- M ability space, i.e. is a positive measure and ([)=1= Also suppose that As usual let ([> >) be a fixed measure space, assume 1 s 4and let 1 4 M i 5 O ()>i: [ $ (d> e)> and ! :(d> e) $ R is a convex function. Then iq q=1 ^ i be a collection of complex valued measurable functions on [= {We} have the{ following} notions of convergence and Cauchy sequences. ! ig !(i)g Definition 21.12. 1. iq $ i a.e. if there is a set H 5 such that (H)= µZ[ ¶ Z[ M 0 and limq$4 1Hf iq =1Hf i= 1 where if ! i@5 O ()> then ! i is integrable in the extended sense and 2. iq $ i in —measureiflimq$4 ( iq i A%)=0for all %A0= We | 0 | [ !(i)g = 4= will abbreviate this by saying iq $ i in O or by iq $ i= 3. in s i s and s for all and lim =0 R iq $ i O i 5 O iq 5 O q> q$4 iq i s = Proof. Let w = [ ig 5 (d> e) and let 5 R be such that !(v) !(w) k k (vw) for all v 5 (d> e)= Then integrating the inequality, !(i)!(w) (i w)> Definition 21.13. 1. i is a.e. Cauchy if there is a set H 5 such that R { q} M implies that (H)=0and 1Hf iq is a pointwise Cauchy sequences. 2. i is Cauchy{ in }— measure (or O0 —Cauchy)iflim ( i { q} p>q$4 | q 0 !(i)g !(w)= !(i)g !( ig)= ip A%)=0for all %A0= [ [ [ | s Z Z Z 3. iq is Cauchy in O if limp>q$4 iq ip =0= { } k ks Moreover, if !(i) is not integrable, then !(i) !(w)+(i w) which shows Lemma 21.14 (Chebyshev’s inequality again). Let s 5 [1> 4) and i 5 that negative part of !(i) is integrable. Therefore, !(i)g = 4 in this [ Os> then case. 1 R ( ) s for all 0 i % s i s %A = Example 21.11. The convex functions in Example 21.7 lead to the following | | % k k s s inequalities, In particular if iq O is O — convergent (Cauchy) then iq is also convergent (Cauchy){ } in measure. { } i exp ig h g> (21.15) Proof. By Chebyshev’s inequality (19.11), µZ[ ¶ Z[ 1 1 log( i )g log i g s s s s ( i %)= ( i % ) i g = i s [ | | [ | | | | | | %s | | %s k k Z µZ ¶ Z[ and for s 1> and therefore if i is Os —Cauchy,then { q} s s s 1 ig i g i g= ( i i %) i i s $ 0 as p> q $4 | | | | | q p| %s k q pks ¯Z[ ¯ µZ[ ¶ Z[ ¯ ¯ ¯ ¯ 0 s The last equation¯ may also¯ easily be derived using Hölder’s inequality. As a showing iq is O — Cauchy. A similar argument holds for the O — convergent special case of the first equation, we get another proof of Lemma 5.5. Indeed, case. { } more generally, suppose s >v A 0 for l =1> 2>===>q and q 1 =1> then l l l=1 sl P 356 21 Os-spaces 21.2 Modes of Convergence 357

Proof. (This is a special case of Exercise 6.9.) Let pAqthen y 1.5

1.25

1 p1 p1 4

0.75 dp dq = (dn+1 dn) dn+1 dn %n := q = (21.17) 0.5 | | | | ¯ n=q ¯ n=q n=q 0.25 ¯ P ¯ P P 0 ¯ ¯ 0 1.25 2.5 3.75 5 So dp dq ¯min(p>q) $ 0 as¯>p>q$4> i.e. dq is Cauchy. Let p $4 x in (21.17)| to| ﬁnd d d = { } 1 p q q Here is a sequence of functions where iq $ 0 a.e., iq 9 0 in O >iq $ 0= | | 0 Theorem 21.16. Suppose iq is O -Cauchy. Then there exists a subse- { } quence jm = iqm of iq such that lim jm := i exists a.e. and iq $ i y 1.5 y 1.5 y 1.5 y 1.5 { } 1.25 1.25 1.25 1.25 as q $4= Moreover if j is a measurable function such that iq $ j as q $4> 1 1 1 1

0.75 0.75 0.75 0.75 then i = j a.e. 0.5 0.5 0.5 0.5

0.25 0.25 0.25 0.25 4 0 0 0 0 0 1.25 2.5 3.75 5 0 1.25 2.5 3.75 5 0 1.25 2.5 3.75 5 0 1.25 2.5 3.75 5 q x x x x Proof. Let %q A 0 such that %q ? 4 (%q =2 would do) and set q=1 4 P 1 = % = Choose j = i such that q is a subsequence of and Above is a sequence of functions where i $ 0 a.e., yet i 0 in O = or in q n m qm m N q q 9 n=q { } measure. P ( jm+1 jm A%m ) %m= y 5 {| | }

3.75 Let Hm = jm+1 jm A%m > 2.5 {| | } 1.25 4 4

0 0 0.25 0.5 0.75 1 1.25 1.5 x I = H = j +1 j A% Q m {| m m| m} p 1 m=Q m=Q Here is a sequence of functions where iq $ 0 a.e., iq $ 0 but iq 9 0 in O = [ [ and 4 4 4 y 1.5 y 1.5 := = = i.o. 1.25 H I H j j A% = 1.25 Q m m+1 m m

1 1 {| | } Q=1 Q=1 m=Q 0.75 0.75

0.5 \ \ [ 0.5 0.25 0.25 Then (H)=0by Lemma 19.20 or the computation 0 0 0 0.5 1 1.5 2 0 0.5 1 1.5 2

x x 4 4 (H) (Hm) %m = Q $ 0 as Q $4= m=Q m=Q

y 1.5 X X y 1.5 y 1.5 y 1.5 1.25 1.25 1.25 1.25

1 1 1 If {@5 I > i.e. j ({) j ({) % for all m Q> then by Lemma 21.15, 1 Q m+1 m m

0.75 0.75 0.75 0.75 | | 0.5 0.5 ( ) = lim ( ) exists and ( ) ( ) for all Therefore, 0.5 0.5 i { jm { i { jm { m m Q= 0.25 0.25 m$4 0.25 0.25 | |

0 0 0 0 0 0.5 1 1.5 2 4 0 0.5 1 1.5 2 0 0.5 1 1.5 2 0 0.5 1 1.5 2 x x x x f f since H = IQ > lim jm({)=i({) exists for all {@5 H. Moreover, { : Q=1 m$4 { i({) j ({) A I for all m Q and hence 1 m S m m Above is a sequence of functions where iq $ 0 in O >iq 9 0 a.e., and | | } p iq $ 0= ( i j A ) (I ) $ 0 as m $4= | m| m m m 4 Lemma 21.15. Suppose dq 5 C and dq+1 dq %q and %q ? 4.Then Therefore jm $ i as m $4= Since | | q=1 4 P iq i A% = i jm + jm iq A% lim dq = d 5 C exists and d dq q := %n. {| | } {| | } q$4 | | n=q i jm A%@2 ^ jm iq A%@2 > P {| | } {| | } 358 21 Os-spaces 21.3 Completeness of Os —spaces 359

( iq i A% ) ( i jm A%@2 )+( jm iq A%@2) Theorem 21.18 (Egoro’s Theorem). Suppose ([) ? 4 and iq $ i {| | } {| | } | | a.e. Then for all %A0 there exists H 5 such that (H) ?%and i $ i and q f M uniformly on H = In particular iq $ i as q $4=

( iq i A% ) lim sup ( jm iq A%@2) $ 0 as q $4= 1 m$4 {| | } | | Proof. Let iq $ i a.e. Then ( iq i A n i.o. q )=0for all nA0> i.e. {| | } If there is another function j such that iq $ j as q $4> then arguing as 3 4 3 4 above 1 4 1 lim C iq i A D = C iq i A D =0= Q$4 {| | n } {| | n } ( i j A%) ( i i A%@2 )+( j i A%@2) $ 0 as q $4= qQ Q=1 qQ | | {| q| } | q| [ \ [ 1 4 Hence Let Hn := iq i A n and choose an increasing sequence Qn n=1 qQn{| | } { } 4 S n n 4 1 1 such that (Hn) ?%2 for all n= Setting H := ^Hn>(H) ? n %2 = % ( i j A 0) = (^q=1 i j A ) ( i j A )=0> 1 | | {| | q} | | q and if {@5 H> then iq i n for all q Qn and all n= That is iq $ i q=1 f | | P X uniformly on H = i.e. i = j a.e. Exercise 21.2. Show that Egoro’s Theorem remains valid when the as- 1 Corollary 21.17 (Dominated Convergence Theorem). Suppose iq > sumption ([) ? 4 is replaced by the assumption that iq j 5 O for all q= 1 0 { } | | 1 j > and j are in O and i 5 O are functions such that Hint: make use of Theorem 21.18 applied to iq [n where [n := j n = { q} | | | © ª i j a.e.>i $ i> j $ j> and j $ j as q $4= | q| q q q q p Z Z 21.3 Completeness of L —spaces 1 1 Then i 5 O and limq$4 i iq 1 =0> i.e. iq $ i in O = In particular 4 k k Theorem 21.19. Let 4 be as defined in Eq. (21.2), then (O ([> >)> 4) is limq$4 iq = i= k·k 4 4 4 M k·k a Banach space. A sequence iq q=1 O converges to i 5 O i there ex- { } f Proof.R FirstR notice that i j a.e. and hence i 5 O1 since j 5 O1= To ists H 5 such that (H)=0and iq $ i uniformly on H = Moreover, | | bounded simpleM functions are dense in O4= see that i j> use Theorem 21.16 to find subsequences iqn and jqn of i and| |j respectively which are almost everywhere convergent.{ } { Then} { q} { q} Proof. By Minkowski’s Theorem 21.4, satisfies the triangle inequal- k·k4 ity. The reader may easily check the remaining conditions that ensure 4 i = lim iqn lim jqn = j a.e. 4 k·k | | n$4 | | n$4 is a norm. Suppose that i O4 is a sequence such i $ i 5 O4> i.e. { q}q=1 q i iq $ 0 as q $4= Then for all n 5 N> there exists Qn ? 4 such that If (for sake of contradiction) lim i i =0there exists %A0 and a k k4 q$4 k qk1 6 subsequence of such that 1 iqn iq i i An =0for all q Q = { } { } | q| n Let ¡ ¢ i iqn % for all n= (21.18) | | H = ^4 ^ i i An1 = Z n=1 qQn | q| f 1 Using Theorem 21.16 again, we may assume (by passing to a further subse- Then (H)=0and for { 5 H > i({)© iq({) n ª for all q Qn= This | f | quences if necessary) that iqn $ i and jqn $ j almost everywhere. Noting, shows that iq $ i uniformly on H = Conversely, if there exists H 5 such f M i iq j + jq $ 2j and (j + jq ) $ 2j> an application of the that (H)=0and i $ i uniformly on H > then for any %A0> | n | n n q dominated convergence Theorem 19.38 implies limn$4 i iqn =0which R R | | f contradicts Eq. (21.18). ( i iq %)= ( i iq % _ H )=0 R | | {| | } Exercise 21.1 (Fatou’s Lemma). If i 0 and i $ i in measure, then for all q su!ciently large. That is to say lim sup i i % for q q q$4 k qk4 i lim infq$4 iq= all %A0= The density of simple functions follows from the approximation R R 360 21 Os-spaces 21.3 Completeness of Os —spaces 361

Theorem 18.42. So the last item to prove is the completeness of O4 for which there are two notable exceptions. (1) If ([) ? 4> then there is no behavior we will use Theorem 7.13. at infinity to worry about and Ot() Os() for all t s as is shown in 4 4 4 Suppose that iq q=1 O is a sequence such that q=1 iq 4 ? 4= Corollary 21.21 below. (2) If is counting measure, i.e. (D)=#(D)> then { } 4 k k s Let Pq := iq 4 >Hq := iq APq > and H := ^q=1Hq so that (H)=0= all functions in O () for any s can not blow up on a set of positive measure, Then k k {| | } P so there are no local singularities. In this case Os() Ot() for all t s> 4 4 see Corollary 21.25 below. sup iq({) Pq ? 4 f | | q=1 {5H q=1 t s X X Corollary 21.21. If ([) ? 4 and 0 ?s?t4> then O () O ()> Q the inclusion map is bounded and in fact which shows that VQ ({)= q=1 iq({) converges uniformly to V({):= 4 ( ) on f i.e. lim =0 q=1 iq { H > q$4 V Vq 4 = 1 1 Pk k ( s t ) Alternatively, suppose %p>q := ip iq $ 0 as p> q $4= Let i [([)] i = P k k4 k ks k kt Hp>q = iq ip A%p>q and H := ^Hp>q> then (H)=0and {| | } Proof. Take d 5 [1> 4] such that sup ip ({) iq ({) %p>q $ 0 as p> q $4= {5Hf | | 1 1 1 st = + > i.e. d = = f f s d t t s Therefore, i := limq$4 iq exists on H and the limit is uniform on H = Letting i = limq$4 1Hf iq> it then follows that limq$4 iq i =0= Then by Corollary 21.3, k k4 s s Theorem 21.20 (Completeness of O ()). For 1 s 4>O () equipped 1 ( 1 1 ) i = i 1 i 1 = ([) @d i = ([) s t i = with the Os —norm, (see Eq. (21.1)), is a Banach space. k ks k · ks k kt ·k kd k kt k kt k·ks The reader may easily check this final formula is correct even when t = 4 Proof. By Minkowski’s Theorem 21.4, s satisfies the triangle inequality. As above the reader may easily check thek·k remaining conditions that ensure provided we interpret 1@s 1@4 to be 1@s= is a norm. So we are left to prove the completeness of Os() for 1 s?4> k·ks Proposition 21.22. Suppose that 0 ?s0 ?s1 4>5 (0> 1) and s 5 the case s = 4 being done in Theorem 21.19. ( ) be defined by 4 s s0>s1 Let iq q=1 O () be a Cauchy sequence. By Chebyshev’s inequality 1 1 { } 0 = + (21.19) (Lemma 21.14), iq is O -Cauchy (i.e. Cauchy in measure) and by Theorem { } s s0 s1 21.16 there exists a subsequence jm of iq such that jm $ i a.e. By Fatou’s { } { } 1 s s0 s1 Lemma, with the interpretation that @s1 =0if s1 = 4= Then O O + O > i.e. every function i 5 Os may be written as i = j +k with j 5 Os0 and k 5 Os1 = s0 s1 s s s For 1 s0 ?s1 4and i 5 O + O let jm i s = lim inf jm jn g lim inf jm jn g k k n$4 | | n$4 | | Z Z s := inf + : = + = lim inf jm jn s $ 0 as m $4= i j s0 k s1 i j k = n$4 k k k k k k k k n o s s0 s1 s s O Then (O + O > ) is a Banach space and the inclusion map from O to In particular, i s jm i s + jm s ? 4 so the i 5 O and jm $ i.The k·k k k k k k k Os0 + Os1 is bounded; in fact i 2 i for all i 5 Os = proof is finished because, k k k ks Proof. Let PA0> then the local singularities of i are contained in the i i i j + j i $ 0 as m> q $4= k q ks k q mks k m ks set H := i AP and the behavior of i at “infinity” is solely determined {|f | } by i on H = Hence let j = i1H and k = i1Hf so that i = j +k= By our earlier s s The Os() — norm controls two types of behaviors of i> namely the “be- discussion we expect that j 5 O 0 and k 5 O 1 and this is the case since, havior at infinity” and the behavior of “local singularities.” So in particular, if 1 A little algebra shows that maybecomputedintermsofs0>s and s1 by i is blows up at a point {0 5 [> then locally near {0 it is harder for i to be in s s s0 s1 3 s O () as s increases. On the other hand a function i 5 O () is allowed to de- = = cay at “infinity”slowerandslowerass increases. With these insights in mind, s · s1 3 s0 we should not in general expect Os() Ot() or Ot() Os()= However, 362 21 Os-spaces 21.4 Converse of Hölder’s Inequality 363

s0 s0 s1 s0 s1 s0 s0 s0 i iq is both O and O — Cauchy. Hence there exist i 5 O and j 5 O such j s = i 1 i AP = P 1 i AP { } k k 0 | | | | P | | that lim i i =0and lim j i =0= By Chebyshev’s q$4 q s0 q$4 q s Z s Z ¯ ¯ k k k k i ¯ ¯ inequality (Lemma 21.14) iq $ i and iq $ j in measure and therefore by s0 ¯ s0¯ s s P 1 i AP P¯ ¯ i s ? 4 Theorem 21.16, i = j a.e. It now is clear that lim i i =0= The P | | k k q$4 q Z ¯ ¯ k k ¯ ¯ estimate in Eq. (21.21) is left as Exercise 21.6 to the reader. and ¯ ¯ ¯ ¯ Remark 21.24. Combining Proposition 21.22 and Corollary 21.23 gives s1 s1 s1 s1 s1 i k s = i1 i P = i 1 i P = P 1 i P s0 s1 s s0 s1 k k 1 | | s1 | | | | P | | O _ O O O + O Z Z ¯ ¯ ° i° s ¯ ¯ ° s1 ° s1s s ¯ ¯ for 0 (0 1) and ( ) as in Eq. (21.19). P 1 i P P i s ?¯4=¯ ?s0 ?s1 4>5 > s 5 s0>s1 P | | k k Z ¯ ¯ ¯ ¯ Corollary 21.25. Suppose now that is counting measure on [= Then Moreover this shows ¯ ¯ ¯ ¯ Os() Ot() for all 0 ?s?t4and i i = k kt k ks i P 1s@s0 i s@s0 + P 1s@s1 i s@s1 = k k k ks k ks Proof. Suppose that 0 ?s?t= 4> then Taking P = i then gives k ks i s =sup i({) s : { 5 [ i({) s = i s > k k4 {| | } | | k ks {5[ i 1s@s0 + 1s@s1 i X k k k ks ³ ´ i.e. i 4 i s for all 0 ?s?4= For 0 ?s t 4> apply Corollary andthentaking =1shows i 2 i = The the proof that 21.23k withk s k=ks and s = 4 to find k k k ks 0 1 (Os0 + Os1 > ) is a Banach space is left as Exercise 21.7 to the reader. k·k s@t 1s@t s@t 1s@t = s i t i s i 4 i s i s i s = Corollary 21.23 (Interpolation of O —norms).Suppose that 0 ?s0 ? k k k k k k k k k k k k s1 4>5 (0> 1) and s 5 (s0>s1) be defined as in Eq. (21.19), then Os0 _ Os1 Os and i i i 1 = (21.20) k ks k ks0 k ks1 21.3.1 Summary: s0 s1 Further assume 1 s0 ?s ?s1 4> and for i 5 O _ O let s s s 0 1. Since ( i A%) % i s >O — convergence implies O — convergence. i := i + i = 0 | | k k k k k ks0 k ks1 2. O — convergence implies almost everywhere convergence for some subsequence. Then (Os0 _ Os1 > ) is a Banach space and the inclusion map of Os0 _ Os1 k·k 3. If ([) ? 4 then almost everywhere convergence implies uniform con- into Os is bounded, in fact vergence o certain sets of small measure and in particular we have O0 — 1 1 convergence. i max > (1 ) i + i = (21.21) k ks k ks0 k ks1 4. If ([) ? 4> then Ot Os for all s t and Ot — convergence implies Os ³ ´ — convergence. The heuristic explanation¡ of this corollary¢ is that if i 5 Os0 _ Os1 > then i s0 s1 t s0 s1 has local singularities no worse than an s1 function and behavior at infinity 5. O _ O O O + O for any t 5 (s0>s1)= O s t s0 s 6. If s t> then c c and i i = no worse than an O function. Hence i 5 O for any s between s0 and s1= k kt k ks Proof. Let be determined as above, d = s0@ and e = s1@(1 ),then by Corollary 21.3, 21.4 Converse of Hölder’s Inequality i = i i 1 i i 1 = i i 1 = s s0 s1 k k | | | | s | | d | | e k k k k Throughout this section we assume ([> >) is a — finite measure space, ° ° ° ° ° ° M 1 1 ° ° ° ° ° s0 ° s1 t 5 [1> 4] and s 5 [1> 4] are conjugate exponents, i.e. s + t =1= For It is easily checked° that °is a norm° ° on°O _ O° = To show this space is k·k t s complete, suppose that i Os0 _ Os1 is a — Cauchy sequence. Then j 5 O > let !j 5 (O ) be given by { q} k·k 364 21 Os-spaces 21.4 Converse of Hölder’s Inequality 365

The last case to consider is s =1and t = 4= Let P := j 4 and choose !j(i)= ji g= (21.22) [ 5 such that [ % [ as q $4and ([ ) ? 4 fork k all q= For any Z q M q q By Hölder’s inequality %A0>( j P %) A 0 and [q _ j P % % j P % = Therefore, ([ _ |j| P % ) A 0 for q su{|!ciently| large.} Let{| | } q {| | } !j(i) ji g j t i s (21.23) = sgn( )1 | | | | k k k k i j [q_ j P% > Z {| | } which implies that then i 1 = ([q _ j P % ) 5 (0> 4) s k k {| | } !j (O ) := sup !j(i) : i s =1 j t= (21.24) k k {| | k k } k k and Proposition 21.26 (Converse of Hölder’s Inequality). Let ([> >) be t M a — ﬁnite measure space and 1 s 4as above. For all j 5 O > !j(i) = sgn(j)jg = j g | | [ _ j P% [ _ j P% | | Z q {| | } Z q {| | } j t = !j (Os) := sup !j(i) : i =1 (21.25) (P %)([ _ j P % )=(P %) i 1= k k k k | | k ks q {| | } k k n o and for any measurable function j : [ $ C> Since %A0 is arbitrary, it follows from this equation that !j (O1) P = j = k k k k4 j t =sup j ig : i =1and i 0 = (21.26) Now for the proof of Eq. (21.26). The key new point is that we no longer k k | | k ks t ½Z[ ¾ are assuming that j 5 O = Let P(j) denote the right member in Eq. (21.26) and set j := 1 j= Then j % j as q $4and it is clear that Proof. We begin by proving Eq. (21.25). Assume ﬁrst that t?4 so q [q_ j q q ( ) is increasing{| in| } Therefore| using| | | Lemma 4.10 and the monotone sA1= Then P jq q= convergence theorem,

!j(i) = ji g ji g j t i s | | | | k k k k lim P(j )=supP(j )=supsup j ig : i =1and i 0 ¯Z ¯ Z q q q s ¯ ¯ q$4 q q [ | | k k ¯ ¯ ½Z ¾ and equality occurs in the¯ﬁrst inequality¯ when sgn(ji) is constant a.e. while s t equality in the second occurs, by Theorem 21.2, when i = f j for some =sup sup jq ig : i s =1and i 0 q | | k k constant fA0= So let i := sgn(j) j t@s which for s = 4| is| to be| | interpreted ½ Z[ ¾ t@4 | | as i = sgn(j)> i.e. j 1= When s = 4> =sup lim jq ig : i =1and i 0 | | q$4 | | k ks ½ Z[ ¾

!j(i) = j sgn(j)g = j 1 = j i O () 1 4 =sup j ig : i s =1and i 0 = P(j)= | | [ k k k k k k | | k k Z ½Z[ ¾ which shows that !j ( 4) j 1= If s?4> then t O Since j 5 O for all q and P(j )= ! (as you should verify), it k k k k q q k jq k(Os) follows from Eq. (21.25) that P(jq)= jq = When t?4 (by the monotone s s t t k kt i s = i = j = j t convergence theorem) and when = (directlyfromthedeﬁnitions) one k k | | | | k k t 4 Z Z learns that limq$4 jq t = j t = Combining this fact with limq$4 P(jq)= while P(j) just proved showsk k P(jk)=k j = k kt ! (i)= jig = j j t@sg = j tg = j t= As an application we can derive a sweeping generalization of Minkowski’s j | || | | | k kt Z Z Z inequality. (See Reed and Simon, Vol II. Appendix IX.4 for a more thorough Hence discussion of complex interpolation theory.) t ! (i) j t(1 1 ) | j | = k kt = j s = j = t@s t t Theorem 21.27 (Minkowski’s Inequality for Integrals). Let ([> >) i s j t k k k k k k k k and (\> >) be — ﬁnite measure spaces and 1 s 4= If i is a M This shows that !j j t which combined with Eq. (21.24) implies Eq. N M N measurable function, then | $ i( >|) Os() is measurable and (21.25). || k k k k · k 366 21 Os-spaces 21.4 Converse of Hölder’s Inequality 367

1. if i is a positive measurable function, then i({> |) g(|) ? 4 for —a.e.{> M N | | Z\ i( >|)g(|) Os() i( >|) Os()g(|)= (21.27) 1 k \ · k \ k · k i.e. i({> ) 5 O () for the —a.e. {= Since \ i({> |)g(|) \ i({> |) g(|) Z Z it follows· by item 1) that | | ¯R ¯ R 2. If i : [ \ $ C is a measurable function and \ i( >|) Os()g(|) ? ¯ ¯ 4 then× M N k · k i( >|)g(|) i( >|) g(|) i( >|) s g(|)= 1 R O () a) for —a.e.{> i({> ) 5 O ()> \ · s \ | · | s \ k · k °Z °O () °Z °O () Z b) the —a.e. defined function,· { $ i({> |)g(|)> is in Os() and ° ° ° ° \ ° ° ° ° s c) the bound in Eq. (21.27) holds. Hence° the function,° { 5 [° $ \ i({> |)g(°|)> is in O () and the bound in R Eq. (21.27) holds. R Proof. For s 5 [1> 4]> let Is(|):= i( >|) s( )= If s 5 [1> 4) Here is an application of Minkowski’s inequality for integrals. In this the- k · kO orem we will be using the convention that {1@4 := 1= 1@s s Is(|)= i( >|) Os() = i({> |) g({) Theorem 21.28 (Theorem 6.20 in Folland). Suppose that n :(0> 4) k · k [ | | × µZ ¶ (0> 4) $ C is a measurable function such that n is homogenous of degree 1> 1 is a measurable function on \ by Fubini’s theorem. To see that I4 is mea- i.e. n({> |)= n({> |) for all A0= If, for some s 5 [1> 4]> surable, let [ 5 such that [ % [ and ([ ) ? 4 for all q= Then by q M q q 4 Exercise 21.5, 1@s Fs := n({> 1) { g{ ? 4 I4(|)= lim lim i( >|)1[ Os() 0 | | q$4 s$4 k · q k Z then for i 5 Os((0> 4)>p)>n({> )i( ) 5 O1((0> 4)>p) for p —a.e.{= More- which shows that I4 is (\> ) — measurable as well. This shows that integral · · on the right side of Eq. (21.27)N is well defined. over, the p —a.e.defined function t Now suppose that i 0>t= s@(s 1)and j 5 O () such that j 0 and 4

j t =1= Then by Tonelli’s theorem and Hölder’s inequality, (Ni)({)= n({> |)i(|)g| (21.28) k kO () Z0 s i({> |)g(|) j({)g({)= g(|) g({)i({> |)j({) is in O ((0> 4)>p) and Z[ ·Z\ ¸ Z\ Z[ Ni Os((0>4)>p) Fs i Os((0>4)>p)= j t i( >|) s g(|) k k k k k kO () k · kO () Z\ 1 | Proof. By the homogeneity of n> n({> |)={ n(1> { )= Using this relation = i( >|) s g(|)= and making the change of variables, | = }{> gives k · kO () Z\ 4 4 | Therefore by the converse to Hölder’s inequality (Proposition 21.26), n({> |)i(|) g| = {1 n(1> )i(|) g| 0 | | 0 { Z Z 4 ¯ ¯ 4 1 ¯ ¯ i( >|)g(|) s( ) = { ¯n(1>})i({})¯ {g} = n(1>})i({}) g}= k · kO | | | | Z\ Z0 Z0

=sup i({> |)g(|) j({)g({): j t =1and j 0 Since k kO () 4 4 g{ ½Z[ ·Z\ ¸ ¾ s s s i( }) Os((0>4)>p) = i(|}) g| = i({) > k · k 0 | | 0 | | } i( >|) Os()g(|) Z Z \ k · k 1@s Z i( }) s = } i s = k · kO ((0>4)>p) k kO ((0>4)>p) proving Eq. (21.27) in this case. Using Minkowski’s inequality for integrals then shows Now let i : [ \ $ C be as in item 2) of the theorem. Applying the ﬁrst part of the theorem× to i shows | | 368 21 Os-spaces 21.5 Uniform Integrability 369

4 4 1 showing j 5 O () and j 1 Pt(j)= Since Holder’s inequality implies that n( >|)i(|) g| n(1>}) i( }) Os((0>4)>p) g} k k 0 | · | Os((0>4)>p) 0 | |k · k Pt(j) j 1 > we have proved the theorem in case t =1= For tA1> we will °Z ° Z k k t ° ° 4 begin by assuming that j 5 O ()= Since s 5 [1> 4) we know that Si is a ° ° 1@s s s ° ° = i Os((0>4)>p) n(1>}) } g} dense subspace of O () and therefore, using ! is continuous on O ()> k k | | j Z0 = Fs i Os((0>4)>p) ? 4= k k P (j)=sup !jg : ! 5 Os() with ! =1 = j t k ks k kt This shows that Ni in Eq. (21.28) is well defined from p —a.e.{= The proof ½¯Z[ ¯ ¾ ¯ ¯ is finished by observing ¯ ¯ where the last equality¯ follows¯ by Proposition 21.26. So it remains to show 1 t 4 that if !j 5 O for all ! 5 Si and Pt(j) ? 4 then j 5 O ()= For q 5 N> let t 1@t Ni s n( >|)i(|) g| Fs i s((0 ) ) jq := 1[q 1 j qj= Then jq 5 O ()> in fact jq t q([q) ? 4= So by O ((0>4)>p) O >4 >p | | k k 0 | · | s((0 ) ) k k k k °Z °O >4 >p the previous paragraph, jq t = Pt(jq) and hence ° ° k k ° ° for all i 5 Os((0> 4)>p°)= ° s The following theorem is a strengthening of Proposition 21.26. It may be jq t =sup !1[q 1 j qjg : ! 5 O () with ! s =1 k k | | k k skipped on the first reading. ½¯Z[ ¯ ¾ ¯ ¯ Pt(j)¯ !1[q 1 j q P¯ t(j) 1=Pt(j) | | s · Theorem 21.29 (Converse of Hölder’s Inequality II). Assume that ¯ ¯ ° ° ([> >) is a — finite measure space, t>s 5 [1> 4] are conjugate exponents wherein the second to° last inequality° we have made use of the definition of M ( ) and the fact that 1 1 If (1 ) an application of the and let Si denote the set of simple functions ! on [ such that (! =0)? 4= Pt j ! [q j q 5 Si = t 5 > 4 > 1 6 2 | | Let j : [ $ C be a measurable function such that !j 5 O () for all ! 5 Si > monotone convergence theorem (or Fatou’s Lemma) along with the continuity and define of the norm, > implies k·ks

j t = lim jq t Pt(j) ? 4= Pt(j):=sup !jg : ! 5 Si with ! =1 = (21.29) k k q$4 k k k ks ½¯Z[ ¯ ¾ ¯ ¯ If t = 4> then jq 4 Pt(j) ? 4 for all q implies jq Pt(j) a.e. which ¯t ¯ k k | | If Pt(j) ? 4 then j 5 O () and Pt(j)= j = then implies that ( ) a.e. since = lim That is 4( ) ¯ ¯ k kt j Pt j j q$4 jq = j 5 O and j P (|j)|= | | | | Proof. Let [ 5 be sets such that ([ ) ? 4 and [ % [ as q %4= k k4 4 q M q q Suppose that t =1and hence s = 4= Choose simple functions !q on [ such that ! 1 and sgn(j) = lim ! in the pointwise sense. Then | q| q$4 q 21.5 Uniform Integrability 1[p !q 5 Si and therefore This section will address the question as to what extra conditions are needed 1 ( ) 0 s [p !qjg Pt j in order that an O — convergent sequence is O — convergent. ¯Z[ ¯ ¯ ¯ 1 ¯ 1 ¯ Notation 21.30 For i 5 O () and H 5 > let for all p> q= By assumption¯ 1[p j 5 O (¯) and therefore by the dominated M convergence theorem we may let q $4in this equation to ﬁnd (i : H):= ig= ZH 1[p j g Pt(j) [ | | and more generally if D> E 5 let Z M for all p= The monotone convergence theorem then implies that (i : D> E):= ig= ZD_E j g = lim 1[p j g Pt(j) | | p$4 | | 1 Z[ Z[ Lemma 21.31. Suppose j 5 O ()> then for any %A0 there exist a A0 2 1 such that ( : ) whenever ( ) This is equivalent to requiring 1Dj M O () for all D M such that (D) ? "= j H ?% H ?= M | | 370 21 Os-spaces 21.5 Uniform Integrability 371

Proof. If the Lemma is false, there would exist %A0 and sets Hq such ( iq )=( iq : iq P)+( iq : iq ?P) that (H ) $ 0 while ( j : H ) % for all q= Since 1 j j 5 O1 and | | | | | | | | | | q q Hq sup ( iq : iq P)+P([) ? 4= | | | | | | q | | | | for any 5 (0> 1)>(1Hq j A) (Hq) $ 0 as q $4> the dominated convergence theorem of Corollary| | 21.17 implies lim ( j : H )=0= This q$4 | | q Moreover, contradicts ( j : Hq) % for all q and the proof is complete. | | 4 Suppose that iq is a sequence of measurable functions which con- { }q=1 ( iq : H)=( iq : iq P>H)+( iq : iq ?P>H) verge in O1() to a function i= Then for H 5 and q 5 N> | | | | | | | | | | M sup ( i : i P)+P(H)= | q| | q| (i : H) (i i : H) + (i : H) i i + (i : H) = q | q | | q | | | k qk1 | | So given %A0 choose P so large that supq ( iq : iq P) ?%@2 and then Let %Q := sup i iq > then %Q & 0 as Q %4and | | | | qAQ k k1 take = %@ (2P) =

sup (iq : H) sup (iq : H) b (%Q + (i : H) ) %Q + (jQ : H) > 1 q | | qQ | | | | Remark 21.35. It is not in general true that if i O () is uniformly { q} (21.30) integrable then supq ( iq ) ? 4= For example take [ = and ( )=1= Q 1 Let ( )= Since for| | 1 aset such that { (} ) {is} in fact where jQ = i + q=1 iq 5 O = From Lemma 21.31 and Eq. (21.30) one iq q= ? H [ H ? easily concludes,| | | | the empty set, we see that Eq. (21.32) holds in this example. However, for P finite measure spaces with out “atoms”, for every A0 we may find a finite ; %A0 < A0 sup (iq : H) ?%when (H) ?= (21.31) partition of [ by sets H n with (H ) ?=Then if Eq. (21.32) holds with 3 q | | c c=1 c 2% =1,then { } Definition 21.32. Functions i 4 O1() satisfying Eq. (21.31) are n { q}q=1 saidtobeuniformly integrable. ( iq )= ( iq : Hc) n | | | | c=1 Remark 21.33. Let iq be real functions satisfying Eq. (21.31), H be a set X { } + showing that ( iq ) n for all q= where (H) ?and Hq = H _ iq 0 = Then (Hq) ?so that (iq : | | )= ( : ) and similarly{ ( } : ) Therefore if Eq. (21.31) H iq Hq ?% iq H ?%= The following Lemmas gives a concrete necessary and su!cient conditions holds then for verifying a sequence of functions is uniformly bounded and uniformly in- sup ( iq : H) ? 2% when (H) ?= (21.32) q | | tegrable. Similar arguments work for the complex case by looking at the real and imag- Lemma 21.36. Suppose that ([) ? 4> and O0([) is a collection of inary parts of i = Therefore i 4 O1() is uniformly integrable i q { q}q=1 functions. ; %A0 < A0 sup ( iq : H) ?%when (H) ?= (21.33) 1. If there exists a non decreasing function : such that 3 q | | ! R+ $ R+ lim{$4 !({)@{ = 4 and Lemma 21.34. Assume that ([) ? 4> then iq is uniformly bounded in 1 { } O () (i.e. N =supq iq 1 ? 4) and iq is uniformly integrable i N := sup (!( i )) ? 4 (21.35) k k { } i5 | | lim sup ( iq : iq P)=0= (21.34) P$4 q | | | | then Proof. Since i is uniformly bounded in O1()>( i P) N@P= lim sup i 1 i P =0= (21.36) { q} | q| P$4 i5 | | | | So if (21.33) holds and %A0 is given, we may choose P su!ciently large so ¡ ¢ 2. Conversely if Eq. (21.36) holds, there exists a non-decreasing continuous that ( iq P) ?(%) for all q and therefore, | | function ! : R+ $ R+ such that !(0) = 0> lim{$4 !({)@{ = 4 and Eq. sup ( iq : iq P) %= (21.35) is valid. q | | | | { Since % is arbitrary, we concluded that Eq. (21.34) must hold. Conversely, Proof. 1. Let ! be as in item 1. above and set %P := sup{P !({) $ 0 as by assumption. Then for suppose that Eq. (21.34) holds, then automatically N =supq ( iq ) ? 4 P $4 i 5 because | | 372 21 Os-spaces 21.5 Uniform Integrability 373

i Theorem 21.37 (Vitali Convergence Theorem). (Folland 6.15) Suppose ( i : i P)=( | | ! ( i ): i P) %P (! ( i ): i P) | | | | ! ( i ) | | | | | | | | that 1 s?4= Asequence i Os is Cauchy i | | { q} %P (! ( i )) N%P 0 | | 1. iq is O —Cauchy, { }s and hence 2. iq — is uniformly integrable. {| | } lim sup i 1 lim N% =0= 3. For all %A0> there exists a set H 5 such that (H) ? 4 and i P P s P$4 i5 | | | | P$4 M Hf iq g ? % for all q= (This condition is vacuous when ([) ? 4=) ¡ ¢ | | 2. By assumption, %P := supi5 i 1 i P $ 0 as P $4= Therefore s 0 | | | | Proof.R (=,) Suppose iq O is Cauchy. Then (1) iq is O — we may choose P %4such that { } { } q ¡ ¢ Cauchy by Lemma 21.14. (2) By completeness of Os,thereexistsi 5 Os such 4 that i i $ 0 as q $4= By the mean value theorem, k q ks (q +1)%Pq ? 4 q=0 i s i s s(max( i > i ))s1 i i s( i + i )s1 i i X || | | q| | | | | q| || | | q|| | | | q| || | | q|| where by convention P := 0= Now deﬁne ! so that !(0) = 0 and 0 and therefore by Hölder’s inequality, 4 !0({)= (q +1)1( ]({)> s s s1 s1 Pq>Pq+1 i iq g s ( i + iq ) i iq g s ( i + iq ) i iq g q=0 || | | | | | | | | || | | || | | | | | | X Z Z s1 Z s@t i.e. s i iq s ( i + iq ) t = s i + iq s i iq s { 4 k k k | | | | k k| | | |k k k s( i + i )s@t i i !({)= !0(|)g| = (q +1)({ a Pq+1 { a Pq) = k ks k qks k qks 0 =0 Z q s s 3 X where t := s@(s 1)= This shows that i iq g $ 0 as q $4= By By construction ! is continuous, !(0) = 0>!0({) is increasing (so ! is convex) ||s | | | | the remarks prior to Deﬁnition 21.32, iq is uniformly integrable. To verify and !0({) (q +1)for { Pq= In particular {|R | } (3), for PA0 and q 5 N let HP = i P and HP (q)= iq P = 1 s {| | } {| | } !({) !(P )+(q +1){ Then (HP ) P s i s ? 4 and by the dominated convergence theorem, q q +1for { P k || { { q s s i g = i 1 i ?P g $ 0 as P $ 0= from which we conclude lim{$4 !({)@{ = 4= We also have !0({) (q +1) | | | | | | HZf Z on [0>Pq+1] and therefore P

!({) (q +1){ for { Pq+1= Moreover,

So for i 5 > iq1Hf i1Hf + (iq i)1Hf i1Hf + iq i = (21.37) P s P s P s P s k ks 4 ° ° ° ° ° ° ° ° So given %A0,chooseQ su!ciently large such that for all q Q> i (!( i )) = !( i )1( ]( i ) ° ° ° ° ° ° ° ° | | | | Pq>Pq+1 | | s s k q=0 iq s ?%. Then choose P su!ciently small such that Hf i g ? % and k P | | X4 ¡ ¢ s Hf (q) i g ? % for all q =1> 2>===>Q 1.LettingH := HP ^ HP (1) ^ P | | R (q +1) i 1(P >P +1]( i ) | | q q | | ^ HP (Q 1)> we have q=0 R··· X4 ¡ 4 ¢ (H) ? 4> i s g ? % for q Q 1 (q +1) i 1 i Pq (q +1)%Pq q | | | | Hf | | q=0 q=0 Z X ¡ ¢ X and hence and by Eq. (21.37) 4 3 Here is an alternative proof. Let k i s 3 i s $ i s + i s =: j M O1 and sup (!( i )) (q +1)%Pq ? 4= q q q q i5 | | s || | | | | | | | | q=0 j 2 i .Thenjq < j> kq < 0 and jq < j= Therefore by the dominated X | | convergence theorem in Corollary 21.17, lim kq g =0= R q<" R R 374 21 Os-spaces 21.5 Uniform Integrability 375

i s g ? (%1@s + %1@s)s 2s% for q Q= Proof. Let %A0 be given and choose A0 as in the Eq. (21.32). Now use q f Hf | | Egoro’s Theorem 21.18 to choose a set H where i converges uniformly on Z q Hf and (H) ?=By uniform convergence on Hf{> there} is an integer Q?4 ThereforewehavefoundH 5 such that (H) ? 4 and f M such that iq ip 1 on H for all p> q Q= Letting p $4> we learn | | s s that sup iq g 2 % i i 1 on Hf= q Hf | | | Q | Z f s Therefore i iQ +1on H and hence which veriﬁes (3) since %A0 was arbitrary. (+=) Now suppose iq O | | | | { } satisﬁes conditions (1) - (3). Let %A0>Hbe as in (3) and ( i )=( i : Hf)+( i : H) | | | | | | ( )+ ( )+ ( : ) D := { 5 H i ({) i ({) % = iQ [ i H = pq { | p q | } | | | | Now by Fatou’s lemma, Then 1@s (iq ip)1Hf s iq1Hf s + ip 1Hf s ? 2% ( i : H) lim inf ( iq : H) 2%?4 k k k k k k | | q$4 | | and by Eq. (21.32). This shows that i 5 O1= Finally

i i = (i i )1 f + (i i )1 f q p s q p H s q p H Dpq s ( i i )=( i i : H )+( i i : H) k k k k k \ k | q| | q| | q| + (iq ip)1D s f k pq k ( i iq : H )+( i + iq : H) 1 | | | | | | ( )1 + ( )1 +2 @s (21.38) f iq ip H Dpq s iq ip Dpq s % = ( i i : H )+4% k \ k k k | q| s s 1 Using properties (1) and (3) and 1H_ ipiq ?% ip iq % 1H 5 O > the and so by the Dominated convergence theorem we learn that dominated convergence theorem in Corollary{| | 21.17}| implies| lim sup ( i iq ) 4%= q$4 | | s s (iq ip)1H Dpq s = 1H_ ipiq ?% ip iq $ 0= k \ k {| | } | | p>q$4 Since %A0 was arbitrary this completes the proof. Z which combined with Eq. (21.38) implies Theorem 21.39 (Vitali again). Suppose that iq $ i in measure and Eq. 1 (21.34) holds, then iq $ i in O = 1@s lim sup iq ip s lim sup (iq ip)1Dpq s +2% = p>q$4 k k p>q$4 k k Proof. This could of course be proved using 21.38 after passing to subsequences to get i to converge a.s. However I wish to give another proof. Finally q First o, by Fatou’s{ } lemma, i 5 O1()= Now let

(iq ip)1D s iq1D s + ip 1D s 2(%) pq pq pq !N ({)={1 { N + N1 { AN = k k k k k k | | | | where then !N (iq) $ !N (i) because !N (i) !N (iq) i iq and since (%):=sup sup iq 1H s : H 5 (H) % | | | | q {k k M3 } i iq i !N (i) + !N (i) !N (iq) + !N (iq) iq By property (2), (%) $ 0 as % $ 0. Therefore | | | | | | | | we have that 1@s lim sup iq ip s 2% +0+2(%) $ 0 as % & 0 k k i iq i !N (i) + !N (i) !N (iq) + !N (iq) iq p>q$4 | | | | | | | | s = ( i : i N)+ !N (i) !N (iq) + ( iq : iq N)= and therefore iq is O -Cauchy. | | | | | | | | | | { } Here is another version of Vitali’s Convergence Theorem. Therefore by the dominated convergence theorem

Theorem 21.38 (Vitali Convergence Theorem). (Thisisproblem9on lim sup i iq ( i : i N) + lim sup ( iq : iq N)= q$4 | | | | | | q$4 | | | | p. 133 in Rudin.) Assume that ([) ? 4> iq is uniformly integrable, iq $ i a.e. and i ? 4 a.e., then i 5 O1() and{ i} $ i in O1()= This last expression goes to zero as N $4by uniform integrability. | | q 376 21 Os-spaces 21.6 Exercises 377

21.6 Exercises Exercise 21.9. By making the change of variables, x =ln{> prove the following facts: Deﬁnition 21.40. The essential range of i> essran(i)> consists of those 1@2 5 C such that ( i ?%) A 0 for all %A0= d e | | { ln { g{ ? 4+,d?1 or d =1and e?1 0 | | Deﬁnition 21.41. Let ([> ) be a topological space and be a measure on Z 4 = ( )= The support of > supp(), consists of those { 5 [ such that {d ln { e g{ ? 4+,dA1 or d =1and e?1 [ | | B(Y ) A 0 for all open neighborhoods, Y> of {= 2 Z 1 {d ln { e g{ ? 4+,d?1 and eA1 Exercise 21.3. Let ([> ) be a second countable topological space and be | | Z0 a measure on [ — the Borel — algebra on [= Show 4 B {d ln { e g{ ? 4+,dA1 and eA1= | | 1. supp() is a closed set. (This is actually true on all topological spaces.) Z1 2. ([ supp()) = 0 and use this to conclude that Z := [ supp() Suppose 0 ?s ?s 4and p is Lebesgue measure on (0> 4) = Use is the\ largest open set in [ such that (Z )=0= Hint: \ be a 0 1 the above results to manufacture a function i on (0> 4) such that i 5 countable base for the topology = Show that Z may be writtenU as a Os ((0> 4) >p) i (a) s 5 (s >s ) > (b) s 5 [s >s ] and (c) s = s = union of elements from Y 5 with the property that (Y )=0= 0 1 0 1 0 V Exercise 21.10. Folland 6.9 on p. 186. Exercise 21.4. Prove the following facts about essran(i)=

1 Exercise 21.11. Folland 6.10 on p. 186. Use the strong form of Theorem 1. Let = i := i — a Borel measure on C= Show essran(i) = supp()= 19.38. 2. essran(i) is a closed set and i({) 5 essran(i) for almost every {> i.e. (i@5 essran(i)) = 0= Exercise 21.12. Let ([> >) and (\> >) be — ﬁnite measure spaces, 3. If I C is a closed set such that i({) 5 I for almost every { then i 5 O2() and n 5 O2( M)= Show N essran(i) I= So essran(i) is the smallest closed set I such that i({) 5 I for almost every {= n({> |)i(|) g(|) ? 4 for —a.e.{= 4. i 4 =sup : 5 essran(i) = | | k k {| | } Z Exercise 21.5. Let i 5 Os _ O4 for some s?4= Show i = 4 Let Ni({):= \ n({> |)i(|)g(|) when the integral is deﬁned. Show Ni 5 k k 2 2 2 limt$4 i t = If we further assume ([) ? 4> show i 4 = limt$4 i t for O () and N : O () $ O () is a bounded operator with N k k k k 4 k k R k krs all measurable functions i : [ $ C= In particular, i 5 O i limt$4 i t ? n 2 = k k k kO ( ) 4= Hints: Use Corollary 21.23 to show lim supt$4 i t i 4 and to k k k k Exercise 21.13. Folland 6.27 on p. 196. Hint: Theorem 21.28. show lim inft$4 i t i 4 > let P? i 4 and make use of Chebyshev’s inequality. k k k k k k Exercise 21.14. Folland 2.32 on p. 63. Exercise 21.6. Prove Eq. (21.21) in Corollary 21.23. (Part of Folland 6.3 on p. 186.) Hint: Use the inequality, with d> e 1 with d1 + e1 =1chosen Exercise 21.15. Folland 2.38 on p. 63. appropriately, vd we vw + > d e (see Lemma 5.5 for Eq. (21.16)) applied to the right side of Eq. (21.20).

Exercise 21.7. Complete the proof of Proposition 21.22 by showing (Os + Ou> ) is a Banach space. (Part of Folland 6.4 on p. 186.) k·k Exercise 21.8. Folland 6.5 on p. 186.