4.1 Bounded integral.

Transcription

1 33 4. LEBESGUE INTEGRAL: BOUNDED CASE Hvig developed the cocept of the Lebesgue mesure, we will ow move to defiig the Lebesgue itegrl. It would be wste to restrict ttetio to itegrls with respect to the Lebesgue mesure oly; ideed, workig with geerl mesure dds oly little extr work. (We will still cll the itegrl the Lebesgue itegrl, though.) The theory is best pprecited if strted from itegrtig bouded fuctios o sets of fiite mesure. We refer to this s the bouded itegrl lthough tht me is ot stdrd i the literture. 4.1 Bouded itegrl. Recll the defiitios of the s-lgebr d geerl mesure from Defiitios 3.13 d We strt with some stdrd otios: Defiitio 4.1 A pir (X,F ), where F is s-lgebr of subsets of X, is clled mesurble spce. If µ is mesure o (X,F ), we cll (X,F, µ) mesure spce. The mesure spce (X, F, µ) is fiite if µ(x) <. It is probbility spce if µ(x)=1. Let (X,F ) be mesurble spce d cosider bouded fuctio f : X! R. The defiitio of the Lebesgue itegrl we will give is similr to Drboux versio of Riem s itegrl: We will pproximte f from below d bove by piece-wise costt fuctios to be clled simple fuctios for which the itegrl is coiclly defied. If the resultig upper d lower itegrls coicide, we proclim the commo vlue the itegrl of f. The simple fuctios re defied s follows: Defiitio 4.2 Let (X,F ) be mesurble spce. A mp f : X! R is simple fuctio if there re disjoit sets {A 1,...,A } F d vlues 1,..., 2 R such tht f = i 1 Ai. (4.1) Obviously, we my eve require tht the vlues { i } re distict d o-zero. Uder this restrictio, the represettio (4.1) of f is uique. The itegrl of simple fuctios with respect to mesure µ o (X,F ) is the defied by the sme expressio s i Riem s cse: f dµ := i µ(a i ). (4.2) The distictess restrictio o { i } re sometimes icoveiet to impose. However, without this restrictio the represettio of the simple fuctio is o loger uique d oe the eeds to check tht the itegrl does ot deped o the represettio. It suffices to prove: Lemm 4.3 (Idepedece of represettio) Suppose µ(x) < d let 1,...,,b 1,...,b m 2 R d A 1,...,A,B 1,...,B m 2 F be such tht both fmilies {A i } d {B j } re disjoit d {b j } distict. If i 1 Ai (x)= m b j 1 B j (x), x 2 X, (4.3) j=1

2 34 the lso i µ(a i )= m b j µ(b j ). (4.4) j=1 I prticulr, (4.2) pplies wheever (4.1) holds, regrdless of distictess restrictio o { i }. Proof. A term with i = 0 or b j = 0 c be left out of the expressio so let us lso ssume tht 1,...,,b 1,...,b m 6= 0. The (4.3) shows S A i = S m j=1 B j. Hece {A i } d {B j } re both disjoit prtitios of this uio d so Sice lso we thus hve A i = m[ (A i \ B j ) d B j = j=1 i µ(a i )= [ (A i \ B j ). (4.5) A i \ B j 6= /0 ) i = b j (4.6) = m j=1 m j=1 where we lso used fiite dditivity of µ. i µ(a i \ B j ) b j µ(a i \ B j ),= The itegrl i (4.2) hs the followig immedite properties: Lemm 4.4 Suppose µ(x) <. The we hve: (1) If f 1, f 2 re simple the so is f 1 + f 2 d (f 1 + f 2 )dµ = (2) If f is simple d c 2 R, the cf is simple d (cf)dµ = c (3) If f 1, f 2 re simple with f 1 pple f 2, the f 1 dµ pple (4) If f is simple, the f 1 dµ + m b j µ(b j ) j=1 (4.7) f 2 dµ. (4.8) f dµ. (4.9) f 2 dµ. (4.10) f dµ pple sup f(x) µ(x). (4.11) x2x Proof. Property (2) is immedite from the defiitio of the itegrl. A importt techicl poit for the proof of (1) d (3) is tht if f 1 d f 2 re simple fuctios the there re {A i } disjoit d 1,..., d b 1,...,b such tht f 1 = i 1 Ai d f 2 = b i 1 Ai (4.12)

3 For (1) we the ote tht f 1 + f 2 = ( i + b i )1 Ai d so by Lemm 4.3, (f 1 + f 2 )dµ = ( i + b i ) µ(a i ) = i µ(a i )+ b i µ(a i )= f 1 dµ + f 2 dµ. 35 (4.13) For (3) we i tur ote tht f 1 pple f 2 the implies i pple b i wheever A i 6= /0 d so f 1 dµ = i µ(a i ) pple b i µ(a i )= f 2 dµ (4.14) Filly, for property (4) we use the fct tht f 1 := sup x2x f(x) is simple d f 1 pple f pple f 1. So, by (2) d (3) we hve sup f(x) µ(x)= ( f 1 )dµ pple f dµ pple f 1 dµ = sup f(x) µ(x). (4.15) x2x x2x The clim i (4) ow follows. Now let us ssume tht f : X! R is bouded fuctio d tht µ(x) <. Similrly to the Drboux pproch to Riem itegrtio, we defie the lower Lebesgue itegrl of f with respect to µ by f dµ := sup o f dµ : f simple, f pple f (4.16) d the upper Lebesgue itegrl of f with respect to µ by f dµ := if f dµ : f simple, f o f. (4.17) By Lemm 4.6 f 1 pple f pple f 2 implies R f 1 dµ pple R f 2 dµ d so f dµ pple f dµ. (4.18) Lemm 4.6(4) lso shows tht both itegrls re fiite. We the defie: Defiitio 4.5 (Lebesgue itegrbility) Assumig µ(x) <, bouded fuctio f : X! R is sid to be Lebesgue itegrble with respect to µ, or just itegrble, if f dµ = f dµ. (4.19) For itegrble f we the defie its Lebesgue itegrl with respect to µ by f dµ := f dµ = f dµ. (4.20) The itegrl thus defied turlly exteds the itegrl of simple fuctios d lso iherits my properties we lredy estblished for itegrls of simple fuctios: Lemm 4.6 Suppose µ(x) < d let f, g: X! R below be bouded fuctios. The: (0) If f is simple the f is itegrble d its Lebesgue itegrl coicides with tht i (4.2).

4 36 (1) If f,g re itegrble the so is f + g d ( f + g)dµ = f dµ + gdµ. (4.21) (2) If f is itegrble d c 2 R, the c f is simple d (cf)dµ = c f dµ. (4.22) (3) If f, g re itegrble with f pple g, the f dµ pple gdµ. (4.23) (4) If f is itegrble, the f dµ pple sup f (x) µ(x). (4.24) x2x Proof. (0) If f is simple d f pple f pple f 0, the R fdµ pple R f dµ pple R f 0 dµ. This shows tht the upper itegrl of f is t lest R f dµ while the lower itegrl is t most R f dµ. But f is upper/lower boud o itself, so the two itegrls ctully coicide d equl R f dµ s climed. (1) If f 1,f 2 re simple fuctios such tht f 1 pple f f 2 d f1 0,f0 2 re simple fuctios such tht f1 0 pple g pple f 2 0, the f 1 + f 2 pple f + g pple f1 0 + f 2 0. Hece we get f dµ + gdµ pple ( f + g)dµ pple ( f + g)dµ pple f dµ + gdµ, (4.25) where ll itegrls re fiite. So if both f d g re itegrble, the extreme left d right hd sides coicide d the clim follows. (2) If c 0 d f 1 pple f pple f 2 the cf 1 pple cf pple cf 2. This shows c f dµ pple (cf)dµ pple (cf)dµ pple c f dµ. (4.26) If f is itegrble, we right d left hd sides re equl. For c < 0 the proof is completely logous except tht f 1 pple f pple f 2 ow implies cf 2 pple cf pple cf 1. The sme rgumet follows with the roles of upper/lower itegrls iterchged i pproprite plces. (3) Let f pple g. If f is simple with f pple f the lso f pple g. It follows tht f dµ pple gdµ. (4.27) For itegrble f d g, this iduces similr compriso of the Lebesgue itegrls. (4) If f := sup x2x f (x), the f is simple with f pple f pple f. It follows tht sup f (x) µ(x)= ( f)dµ pple f dµ pple f dµ pple f dµ = sup f (x) µ(x). (4.28) x2x x2x For f itegrble, this wrps up ito the climed iequlity. The properties i the bove lemm c be summrized by syig tht the mp f 7! R f dµ is positive bouded lier fuctiol o the lier spce of bouded itegrble fuctios. I itegrtio theory oe sometimes wts to itegrte f oly o subset of the uderlyig spce. For this we ote:

5 Lemm 4.7 Let µ(x) < d let f : X! R be bouded itegrble fuctio. The f 1 A is itegrble for ll A 2 F s well. Proof. Let A 2 F. Sice f is itegrble, for ech e > 0 there re simple fuctios f,f 0 such tht f pple f pple f 0 d R f 0 dµ R f dµ < e. But the both f1 A d f 0 1 A re simple fuctios with f1 A pple f 1 A pple f 0 1 A d (f 0 1 A )dµ (f1 A )dµ = ((f 0 f)1 A )dµ pple (f 0 f)dµ < e. (4.29) So f 1 A is itegrble s well. I light of this observtio, we put forwrd: Defiitio 4.8 Give y A 2 F, we will sy tht f : X! R is itegrble o A wheever f 1 A is itegrble. I this cse we defie f dµ := ( f 1 A )dµ. (4.30) A It is strightforwrd exercise, lbeit somewht legthy, to show tht we my ltertively thik of this s itegrl of ( restrictio to A of) f o the mesure spce (A,F A, µ A ), where F A := {B \ A: B 2 F } d µ A is restrictio of µ to F A. 4.2 Itegrbility vs mesurbility. We ote tht, i ll of the bove sttemets, it would suffice to hve µ oly fiitely dditive o F with µ(x) <. (Reformultig the fiiteess coditio ppropritely, oe c eve use fiitely dditive set fuctios tkig both positive d egtive vlues; cf the forthcomig chpters.) However, eve if we use tht µ is ctully coutbly dditive, it is ot cler tht the itegrl behves turlly uder poit-wise limits. Explicitly, if uiformly bouded itegrble fuctios f coverge to bouded fuctio f, it is ot ppret whether or ot f is itegrble d whether or ot the limit of itegrls is the itegrl of the limit. Aother turl questio is whether mgeble regulrity coditios c t ll be produced tht gurtee itegrbility. (Remember tht this ws serious issue surroudig the otherwise very elegt defiitio of the Riem itegrl.) As it turs out, the proper coditio is tht of mesurbility. Oe c rrive t this cocept quite turlly by recllig oe of the mi iovtio tht Lebesgue brought to itegrtio theory: Isted of defiig the itegrl by prtitioig the domi of f (tht is, the set X) we prtitio the rge of vlues of f d use the pre-imge mp, f 1 (A) := {x 2 X : f (x) 2 A, (4.31) to iduce prtitio of X. Nturlly, if oe is the to ssig meig to the resultig piecewise costt pproximtio of f, ll sets i the iduced prtitio eed to be mesurble. This i tur forces f to belog to the the clss of fuctios described by: Defiitio 4.9 Let (X,F ) be mesurble spce d f : X! R fuctio. We sy tht f is mesurble (or F -mesurble, if the uderlyig s-lgebr eeds to be emphsized) if 37 8 < b: f 1 (,b]) 2 F. (4.32)

6 38 If f tkes vlues i the geerlized rel umbers, R [ {± }, the we require this for < b icludig = d b =. The restrictio to (preimges of) hlf-ope itervls is quite rbitrry; ideed, we just eed clss of sets tht geertes eough sets by coutble uios d complemets. Ideed, we hve: Lemm 4.10 is s-lgebr. Let (X,F ) be mesurble spce d f : X! R. The B R: f 1 (B) 2 F (4.33) Proof. We hve f 1 (A c )= f 1 (A) c d f 1 ( S i2n A i )= S i2n f 1 (A i ). Hece the stted collectio is closed uder complemets d coutble uios. Sice lso f 1 (/0 )=/0 d f 1 (R)=X, this collectio is s-lgebr. or As cosequece, the sme cocept would be defied if we just required f f 1 (,] 2 F, 2 R. (4.34) 1 (,) 2 F, 2 R. (4.35) (I fct, requirig this just for 2 Q would suffice.) Notice tht, i prticulr, every level set f 1 ({}) of mesurble fuctio f is mesurble. Obviously, ll simple fuctios re utomticlly mesurble. My turl opertios o fuctios preserve mesurbility. Ideed, we hve: Lemm 4.11 We hve: (1) If f,g re mesurble the so is f + g. (2) If f is mesurble d c 2 R, the c f is lso mesurble. (3) If f,g re mesurble d A 2 F, the lso f 1 A + g1 A c is mesurble. Proof. (1) Let us write f 1 (A)={ f 2 A}. The { f + g < } = [ p,q2q p+q< f < p}\{g < q}. (4.36) The sets { f < p},{g < q} 2F whe f,g re mesurble, d sice the uio is coutble, { f + g < }2F for y 2 R s well. It follows tht f + g is mesurble. (2) Here we just ote tht, for c > 0, we hve {cf < } = { f < /c}, while for c < 0 we hve {cf < } = { f > /c}. Sice f := 0 is lso mesurble, we re doe. (3) Let h := f 1 A + g1 A c. The d ll sets o the right re clerly mesurble. {h < } = A \{f < } [ A c \{g < } (4.37) The clss of mesurble fuctios is vector spce. Perhps more importtly, mesurbility behves quite well uder poitwise limits. This icludes the cses whe the limit is plus/mius ifiity, but for this we eed to use the proviso i the the bove defiitio tht dels with fuctios tkig vlues i geerlized rels.

7 39 Lemm 4.12 If { f } re mesurble, the so re the fuctios sup f, 2N if f, 2N limsup! I prticulr, if f := lim! f exists, the f is lso mesurble. Proof. For y 2 R [ {± } we hve sup 2N f, limif! f. (4.38) f > = [ { f > } (4.39) d so the supremum of f is mesurble (s fuctio tkig vlues i R [ {± }) if { f } re mesurble. Similrly, we get lso tht the ifimum is mesurble s well. As lim sup! f = if sup 1 m 2N f m d limsup! f = sup if f m (4.40) m these re mesurble s well. If limit f := lim! f exists, the it equls the limes superior (s well s the limes iferior) d so it is mesurble s well. We re ow redy to tie mesurbility with the cocept of itegrbility itroduced erlier. Oe directio is quite immedite: Propositio 4.13 (Mesurbility implies itegrbility) bouded fuctio. The 1 Assume µ(x) < d let f : X! R be f mesurble ) f itegrble. (4.41) Proof. Let f : X! R be bouded mesurble fuctio. Pick 2 N d cosider the fuctios i f := f 1 i i2, i+1 d f 0 i + 1 := f i2 1 i, i+1 (4.42) We write these s ifiite sums but, sice f is bouded, the preimge set is o-empty oly for fiite umber if i s. Sice f is mesurble, the preimges lie i F d so both f d f 0 re simple. Now f pple f pple f 0 d so, by the defiitio of the upper/lower Lebesgue itegrls, f dµ pple f dµ pple f dµ pple f 0 dµ. (4.43) Moreover, 0 pple f 0 f pple 1 d Lemm 4.6(1,2) thus shows 0 pple f 0 dµ f dµ pple 1 µ(x) (4.44) Puttig these together, we coclude 0 pple f dµ f dµ pple 1 µ(x). (4.45) Tkig! shows tht f is itegrble. Propositio 4.13 ow permits us to stte iequlity tht will be useful i the followig:

8 40 Lemm 4.14 (Mrkov s iequlity) mesurble. The for ll l > 0, Suppose µ(x) < d let f : X! [0, ) be bouded d µ {x: f (x) l} pple 1 l f dµ. (4.46) Proof. Let l > 0. The fuctio f := l1 { f l} is simple becuse f is mesurble d obeys f pple f. Hece lµ {x: f (x) l} = f dµ pple f dµ. (4.47) The clim follows by dividig both sides by l. As it turs out, coverse to Propositio 4.13 holds s well ssumig oe dditiol property of the uderlyig mesure spce. This coditio is give i: Defiitio 4.15 (Completeess) A mesure spce (X,F, µ) or just the mesure µ or the s-lgebr F is sid to be complete if, wheever N 2 F obeys µ(n)=0, we hve A 2 F for every A N. A set N 2 F with µ(n)=0 is clled ull set. We remrk tht the Lebesgue mesure we costructed erlier is utomticlly complete. As we will see, this is commo for ll costructios ivolvig outer mesures. Propositio 4.16 (Itegrbility d completeess imply mesurbility) let f : X! R be bouded fuctio. If (X,F, µ) is complete, the Assume µ(x) < d f itegrble ) f mesurble. (4.48) Proof. Suppose tht f is itegrble. The, for ech 2 N, there re simple fuctios f d f 0 (ot ecessrily those bove, of course) with f pple f pple f 0 d 0 pple f 0 dµ f dµ pple 1. (4.49) The mximum of y fiite umber of simple fuctios is simple d so we my replce f by mx{f 1,...,f } d get tht f is icresig. Similrly, we my ssume tht f 0 is decresig d the boud (4.49) still holds. The limits g := lim! f d h := lim! f 0 (4.50) the exist d obey g pple f pple h. Sice ll simple fuctios re mesurble, Lemm 4.12 implies tht both g d h re mesurble. As they squeeze f i betwee, d were obtied by mootoe limits of simple fuctios, both g d h re lso bouded. Propositio 4.13 esures tht they re lso itegrble. But f pple g pple h pple f 0 imply f dµ pple gdµ pple hdµ pple f 0 dµ (4.51) d so, by (4.49), gdµ = hdµ. (4.52)

9 But h g is o-egtive d mesurble by Lemm 4.11, so Mrkov s iequlity (Lemm 4.14), Lemm 4.6 d (4.52) yield µ(h g e) pple 1 (h g)dµ = 1 1 hdµ gdµ = 0, e > 0. (4.53) e e e It follows tht µ(h g e)=0 for y e > 0 d, by tkig e # 0 log sequece d pplyig the Upwrd Mootoe Covergece Theorem for sets, µ(h g > 0)=0. (4.54) But f is squeezed betwee g d h d so f equls g (d h) o set of full mesure. To prove tht f is mesurble, we ow observe tht, for y 2 R, { f < } = {g < }\{h g = 0} [ { f < }\{h g > 0}. (4.55) The first set i the uio is mesurble becuse g d h g re mesurble. The secod set i the uio is subset of {h g > 0}, which is set of zero µ-mesure. Sice (X,F, µ) is complete, y subset of this set is mesurble s well. It follows tht f is mesurble. I complete mesure spces, itegrbility is thus equivlet to mesurbility. I o-complete spces this is ot true i geerl, lthough the proof does imply tht every itegrble fuctio c be modified o ull set to produce mesurble fuctio. As we will lter (prticulrly, whe fuctio spces re beig cosidered) regrd fuctios tht differ oly o subsets of ull sets s equivlet, the otios of itegrbility d mesurbility bled together s well. 4.3 Bouded Covergece Theorem. With the bove chrcteriztio i hd, we c ow exmie the behvior of the bouded itegrl uder poitwise limits. We i fct do ot eed sequeces of fuctios to coverge everywhere, it suffices to hve weker versios of this covergece. We first itroduce geerl cocept of property beig true lmost everywhere: Defiitio 4.17 Let P(x) be property i.e., boole vrible ssiged to poit x d let µ be mesure o mesurble spce (X,F ). We sy tht property P holds µ-lmost everywhere, or just µ-.e., if the set {x 2 X : P(x) fils} is subset of µ-ull set. Note tht i complete spces we my simply require tht {x 2 X : P(x) fils} is ull set. We c ow pply this geerl otio to the cocept of covergece: Defiitio 4.18 is subset of ull set. We sy tht f coverges to f lmost everywhere, or f! f.e., if the set o x 2 X : limsup! f (x) f (x) > 0 (4.56) A yet weker versio of covergece is: Defiitio 4.19 Let { f } d f be mesurble fuctios. We sy tht f! f i mesure if 8e > 0: µ f f > e!! 0. (4.57) 41

10 42 Here we ote tht { f f > e} is mesurble set becuse (by Lemm 4.11) if f is mesurble the so is f = f 1 { f 0} f 1 { f <0}. We ow ote: Lemm 4.20 Let { f } d f be mesurble fuctios o mesure spce (X,F, µ). The: (1) If f! f poitwise, the f! f lmost everywhere. (2) If f! f lmost everywhere, the f! f i mesure. Proof. (1) is true trivilly. For (2) we ote tht µ f f e pple µ sup f m f e! µ limsup f f e, (4.58) m!! where we used the Mootoce Covergece Theorem for sets to get the lst step. If f! f.e., the the right-hd side is zero for every e > 0. It is ot hrd to check tht ll three types of covergece discussed bove re distict from ech other. We leve the costructio of requisite couterexmples to the reder. A key poit is tht our ext covergece theorem pplies to the wekest of these three: Theorem 4.21 (Bouded Covergece Theorem) Suppose µ(x) < d let { f } d f be bouded mesurble fuctios with sup 2N sup x2x f (x) <. The f! f i mesure ) f dµ = lim f dµ. (4.59)! Proof. Let K be costt such tht f pplek d f pplek. The for y e > 0, f dµ f dµ = ( f f )dµ pple ( f f )1 { f f >e} dµ + ( f f )1 { f f pplee} dµ (4.60) pple 2Kµ f f > e + eµ(x), where we first used dditivity of the itegrl, the wrote 1 = 1 { f f >e} +1 { f f pplee} d pplied dditivity of itegrl d subdditivity of bsolute vlue. The, i the first bsolute vlue, we bouded ( f f )1 { f f >e} betwee f := 2K1 { f f >e} d f, while i the secod prt we pplied Lemm 4.6(4) log with the fct tht the itegrted fuctio is t most e. The clim ow follows by tkig! followed by e # 0. Eve if the limit fuctio f is bouded, the uiform boudedess of the sequece { f } cot be omitted. Ideed, i the mesure spce ([0,1],L ([0,1]),l), the fuctios f (x) := 1 [0,1/] (x) coverge to zero i Lebesgue mesure d yet R f dl = Chrcteriztio of Riem itegrbility. A turl questio to sk t this poit is to wht exted does the Lebesgue itegrl geerlize the Riem itegrl. This will be swered quite elegtly by rgumet similr to tht used i the proof of Propositio We will heceforth work with fuctios o fiite itervl R which we turlly edow with the s-lgebr L () of Lebesgue-mesurble subsets thereof d the Lebesgue mesure l. To reduce clutter of ottio, we write the resultig Lebesgue itegrl s R f dl; see (4.30).

11 We begi by sttig the mi theorem. Recll tht Defiitio 4.17 ssigs uique meig to the phrse f is cotiuous.e. Usig this cocept, we hve: Theorem 4.22 (Lebesgue s chrcteriztio of Riem itegrbility) bouded fuctio. The 43 Let f : [, b]! R be f is Riem itegrble, f is cotiuous Lebesgue.e. (4.61) The proof will cosist of severl observtios tht we stte s lemms. Cosider bouded fuctio f :! R. We will hve to develop good hdle of the set {x: f is cotiuous t x}. For this we defie m d (x) : = if f (y): y 2, y x < d (4.62) M d (x) : = sup f (y): y 2, y x < d d set m(x) := lim d#0 m d (x) d M(x) := lim d#0 M d (x). (4.63) These limits exist becuse (s is esy to check) d 7! m d (x) is o-decresig while d 7! M d (x) is o-icresig d The coectio with cotiuity of f is ow provided by: m d (x) pple f (x) pple M d (x), d > 0. (4.64) Lemm 4.23 Let f :! R be bouded fuctio d let m(x) d M(x) be defied s bove. The we hve: (1) m(x) pple f (x) pple M(x) for ll x 2, (2) f is cotiuous t x 2 if d oly if m(x)=m(x), (3) M is upper-semicotiuous d m is lower-semicotiuous o, d thus (4) both m d M re Lebesgue mesurble. We ote tht fuctio g is lower-semicotiuous if {g > } is ope for ll, d is uppersemicotiuous if {g < } is ope for ll. Altertively, g is lower-semicotiuous if x! x implies g(x) pple limif! g(x ), etc. Proof of Lemm (1) is immedite from (4.64). For (2) we ote tht f is cotiuous t x if d oly if for ech e > 0 there is d > 0 such tht (the oscilltio of f obeys) osc Ad (x)( f ) < e for A d (x) := y 2 : y x < d. (4.65) But osc Ad (x)( f )=M d (x) m d (x) d so the clim follows. For (3) we ote tht, M(x) < implies M 2d (x) < for some d > 0 smll. But the lso M d (y) < for y such tht y x < d. Hece {M < } is ope d so M is upper-semicotiuous. The lower-semicotiuity of m is proved logously. Filly, to get (4) we ote tht lower, resp., upper-semicotiuous fuctio preimges itervls (, ), resp., (, ) ito ope sets which re ex defiitio Lebesgue mesurble. Ivokig Lemm 4.10 d lso the ltertive coditios ( ) for mesurbility fterwrds we coclude tht both M d m re Lebesgue mesurble. We ow defie other cocept:

12 44 Defiitio 4.24 A step fuctio f o is fuctio of the form f(x) := i 1 Ii (4.66) where 1,..., 2 R d {I i } re disjoit itervls such tht S I i =. Note tht every step fuctio is utomticlly simple. A key poit is tht, for f :! R bouded, the Drboux-Riem lower/upper itegrls re give by b o f (x)dx = sup f dl : f step fuctio, f pple f (4.67) d b f (x)dx = if f dl : f step fuctio, f o f, (4.68) where we otice tht step fuctio is Lebesgue mesurble d the Lebesgue itegrl R f dl is exctly the vlue of the Riem sum ssocited with f. From this we immeditely coclude: Lemm 4.25 (Riem itegrbility implies Lebesgue itegrbility) b For f : [, b]! R bouded, b f (x)dx pple ( f 1 )dl pple ( f 1 )dl pple f (x)dx. (4.69) I prticulr, if f is Riem itegrble o, the it is Lebesgue itegrble o. Proof. Every step fuctio is simple fuctio d so we get (4.69) by comprig ( ) with ( ). If f is Riem itegrble o, the extreme eds of (4.69) coicide d so ll iequlities re i fct equlities. I prticulr, f 1 is Lebesgue itegrble. The coectio to the bove fuctios m d M is ow provided by: Lemm 4.26 Let f :! R be bouded. If f,f 0 re step fuctios such tht f pple f pple f 0, the lso f pple m pple M pple f 0 t the cotiuity poits of both f d f 0. I prticulr, we hve b f (x)dx = mdl d b f (x)dx = M dl (4.70) Proof. If f is step fuctio such tht f pple f d x is cotiuity poit of f, the f is costt o d-eighborhood of x. But the f lies below the ifimum of f o this eighborhood d so f(x) pple m d (x) pple m(x). This implies the first cluse of the lemm. Sice l ssigs zero mss to sigletos, f pple f pple f 0 thus implies f dl pple mdl d f 0 dl M dl. (4.71) The we hve b f (x)dx pple mdl by direct pplictio of ( ). d b f (x)dx M dl. (4.72)

13 To prove tht equlities tke plce, cosider the collectio {I i : i = 1,...,2 } of the dydic itervls of the form, + 2 (b ), + 2 (b ), (b ),..., + k2 (b ), +(k + 1)2 (b ),..., b 2 (b ),b. (4.73) (Notice tht the first itervl is closed; ll others hlf-ope.). The edpoits of these itervls lie i the set D := {k2 : k = 0,...,2 }. Note tht S 2 I i = d so f := 2 if f (x) 1 Ii d f 0 := x2i i 2 45 sup f (x) 1 Ii (4.74) x2i i re step fuctios. Let D := S 2N D. Sice f pple f pple f 0 re cotiuous o D c, the first cluse i the lemm d the defiitio of m d d M d show f pple m d f! m f 0 M d f 0! M o D c. (4.75) Sice f is bouded, the fuctios {f },{f},m,m 0 re uiformly bouded. Moreover, D is coutble d so the covergece f! m d f 0! M tkes plce Lebesgue lmost everywhere, d thus i mesure. Ivokig gi ( ) we get b f (x) limsup f dl = mdl (4.76)! d similrly b f (x) pple limif f dl = M dl, (4.77)! where the limit is cosequece of the Bouded Covergece Theorem. The clim follows. We re ow filly redy to prove the desired result: Proof of Theorem Suppose f is Riem itegrble. By Lemm 4.26, the Lebesgue itegrls of M d m with respect to the Lebesgue mesure o re equl. Hece, by Mrkov s iequlity (Lemm 4.14), for ech e > 0 we hve µ M m > e pple 1 (M m)dl = 0. (4.78) e Tkig e # 0, the Upwrd Mootoe Covergece Theorem for sets shows µ(m m > 0) =0. Lemm 4.23 implies tht f is cotiuous.e. O the other hd, if f is bouded d cotiuous.e., the lso M = m.e. by Lemm The the Lebesgue itegrls of (bouded fuctios) m d M with respect to (fiite) Lebesgue mesure o re equl. By Lemm 4.26 gi, f is Riem itegrble.