Riemann Integration. Chapter 1

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1 Mesure, Itegrtio & Rel Alysis. Prelimiry editio. 8 July Sheldo Axler 1 Chpter 1 Riem Itegrtio This chpter reviews Riem itegrtio. Riem itegrtio uses rectgles to pproximte res uder grphs. This chpter begis by crefully presetig the defiitios ledig to the Riem itegrl. The big result i the first sectio sttes tht cotiuous rel-vlued fuctio o closed bouded itervl is Riem itegrble. The proof depeds upo the theorem tht cotiuous fuctios o closed bouded itervls re uiformly cotiuous. The secod sectio of this chpter focuses o severl deficiecies of Riem itegrtio. As we will see, Riem itegrtio does ot do everythig tht we would like itegrl to do. These deficiecies will provide motivtio i future chpters for the developmet of mesures d itegrtio with respect to mesures. Digitl sculpture of Berhrd Riem ( ), the Germ mthemtici whose method of itegrtio is tught i clculus courses.

2 Mesure, Itegrtio & Rel Alysis. Prelimiry editio. 8 July Sheldo Axler 2 Chpter 1 Riem Itegrtio 1A Review: Riem Itegrl Let R deote the complete ordered field of rel umbers. See the Appedix for importt properties of R, especilly relted to the cocepts of ifimum d supremum, tht you should kow before redig other prts of this book. 1.1 Defiitio prtitio Suppose, b R with < b. A prtitio of [, b] is fiite list of the form x 0, x 1,..., x, where = x 0 < x 1 < < x = b. We use prtitio x 0, x 1,..., x of [, b] to thik of [, b] s uio of closed subitervls, with y two such subitervls itersectig t most i edpoit: [, b] = [x 0, x 1 ] [x 1, x 2 ] [x 1, x ]. The ext defiitio itroduces cle ottio for the ifimum d supremum of the vlues of fuctio o some subset of its domi. 1.2 Defiitio ottio for ifimum d supremum of fuctio If f is rel-vlued fuctio d A is subset of the domi of f, the if A f = if{ f (x) : x A} d sup A f = sup{ f (x) : x A}. The lower d upper Riem sums, which we ow defie, pproximte the re uder the grph of oegtive fuctio. 1.3 Defiitio lower d upper Riem sums Suppose f : [, b] R is bouded fuctio d P is prtitio x 0,..., x of [, b]. The lower Riem sum L( f, P, [, b]) d the upper Riem sum U( f, P, [, b]) re defied by L( f, P, [, b]) = j=1 (x j x j 1 ) if [x j 1,x j ] f d U( f, P, [, b]) = j=1 (x j x j 1 ) sup [x j 1,x j ] f. Our ituitio suggests tht for prtitio with oly smll gp betwee cosecutive poits, the lower Riem sum should be bit less th the re uder the grph, d the upper Riem sum should be bit more th the re uder the grph.

3 Mesure, Itegrtio & Rel Alysis. Prelimiry editio. 8 July Sheldo Axler Sectio 1A Review: Riem Itegrl 3 The pictures i the ext exmple help covey the ide of these pproximtios. The bse of the j th rectgle hs legth x j x j 1 d hs height if f for the [x j 1,x j ] lower Riem sum d height sup [x j 1,x j ] f for the upper Riem sum. 1.4 Exmple lower d upper Riem sums Defie f : [0, 1] R by f (x) = x 2. Let P deote the prtitio 0, 1, 2,..., 1 of [0, 1]. The two figures here show the grph of f i red. The ifimum of this fuctio f is ttied t the left edpoit of ech subitervl [ j 1, j ]; the supremum is ttied t the right edpoit. L(x 2, P 16, [0, 1]) equls the sum of the res of these rectgles. U(x 2, P 16, [0, 1]) equls the sum of the res of these rectgles. d For the prtitio P, we hve x j x j 1 = 1 for ech j = 1,...,. Thus L(x 2, P, [0, 1]) = 1 U(x 2, P, [0, 1]) = 1 (j 1) 2 j=1 2 = j=1 j 2 2 = , s you should verify [use the formul = ( ) 6 ]. The ext result sttes tht djoiig more poits to prtitio icreses the lower Riem sum d decreses the upper Riem sum. 1.5 Iequlities with Riem sums Suppose f : [, b] R is bouded fuctio d P, P re prtitios of [, b] such tht the list defiig P is sublist of the list defiig P. The L( f, P, [, b]) L( f, P, [, b]) U( f, P, [, b]) U( f, P, [, b]). Proof To prove the first iequlity, suppose P is the prtitio x 0,..., x d P is the prtitio x 0,..., x N of [, b]. For ech j = 1,...,, there exist k {0,..., N 1} d positive iteger m such tht x j 1 = x k < x k+1 <... < x k+m = x j. We hve

4 Mesure, Itegrtio & Rel Alysis. Prelimiry editio. 8 July Sheldo Axler 4 Chpter 1 Riem Itegrtio m (x j x j 1 ) if f = [x j 1,x j ] (x k+i x k+i 1 ) if f i=1 [x j 1,x j ] m i=1 (x k+i x k+i 1 ) if f. [x k+i 1,x k+i ] The iequlity bove implies tht L( f, P, [, b]) L( f, P, [, b]). The middle iequlity i this result follows from the observtio tht the ifimum of ech set of rel umbers is less th or equl to the supremum of tht set. The proof of the lst iequlity i this result is similr to the proof of the first iequlity d is left to the reder. The followig result sttes tht if the fuctio is fixed, the ech lower Riem sum is less th or equl to ech upper Riem sum. 1.6 Lower Riem sums upper Riem sums Suppose f : [, b] R is bouded fuctio d P, P re prtitios of [, b]. The L( f, P, [, b]) U( f, P, [, b]). Proof Let P be the prtitio of [, b] obtied by mergig the lists tht defie P d P. The L( f, P, [, b]) L( f, P, [, b]) U( f, P, [, b]) U( f, P, [, b]), where ll three iequlities bove come from 1.5. We hve bee workig with lower d upper Riem sums. Now we defie the lower d upper Riem itegrls. 1.7 Defiitio lower d upper Riem itegrls Suppose f : [, b] R is bouded fuctio. The lower Riem itegrl L( f, [, b]) d the upper Riem itegrl U( f, [, b]) of f re defied by d L( f, [, b]) = sup L( f, P, [, b]) P U( f, [, b]) = if U( f, P, [, b]), P where the supremum d ifimum bove re tke over ll prtitios P of [, b].

5 Sectio 1A Review: Riem Itegrl 5 I the defiitio bove, we tke the supremum (over ll prtitios) of the lower Riem sums becuse djoiig more poits to prtitio icreses the lower Riem sum (by 1.5) d should provide more ccurte estimte of the re uder the grph. Similrly, i the defiitio bove, we tke the ifimum (over ll prtitios) of the upper Riem sums becuse djoiig more poits to prtitio decreses the upper Riem sum (by 1.5) d should provide more ccurte estimte of the re uder the grph. Our first result bout the lower d upper Riem itegrls is esy iequlity. 1.8 Lower Riem itegrl upper Riem itegrl Suppose f : [, b] R is bouded fuctio. The L( f, [, b]) U( f, [, b]). Proof The desired iequlity follows from the defiitios d 1.6. The lower Riem itegrl d the upper Riem itegrl c both be resobly cosidered to be the re uder the grph of fuctio. Which oe should we use? The pictures i Exmple 1.4 suggest tht these two qutities re the sme for the fuctio i tht exmple; we will soo verify this suspicio. However, s we will see i the ext sectio, there re fuctios for which the lower Riem itegrl does ot equl the upper Riem itegrl. Isted of choosig betwee the lower Riem itegrl d the upper Riem itegrl, the stdrd procedure i Riem itegrtio is to cosider oly fuctios for which those two qutities re equl. This decisio hs the huge dvtge of mkig the Riem itegrl behve s we wish with respect to the sum of two fuctios (see Exercise 5 i this sectio). 1.9 Defiitio Riem itegrble; Riem itegrl A bouded fuctio o closed bouded itervl is clled Riem itegrble if its lower Riem itegrl equls its upper Riem itegrl. If f : [, b] R is Riem itegrble, the the Riem itegrl defied by f = L( f, [, b]) = U( f, [, b]). Let Z deote the set of itegers d Z + deote the set of positive itegers Exmple computig Riem itegrl Defie f : [0, 1] R by f (x) = x 2. The 2 U( f, [0, 1]) if Z = 1 3 = sup Z L( f, [0, 1]), f is Mesure, Itegrtio & Rel Alysis. Prelimiry editio. 8 July Sheldo Axler

6 Mesure, Itegrtio & Rel Alysis. Prelimiry editio. 8 July Sheldo Axler 6 Chpter 1 Riem Itegrtio where the two iequlities bove come from Exmple 1.4 d the two equlities esily follow from dividig the umertors d deomitors of both frctios bove by 2. The prgrph bove shows tht U( f, [0, 1]) 3 1 L( f, [0, 1]). Whe combied with 1.8, this shows tht L( f, [0, 1]) = U( f, [0, 1]) = 1 3. Thus f is Riem itegrble d 1 0 f = 1 3. Now we come to key result bout Riem itegrtio. Uiform cotiuity provides the mjor tool tht mkes the proof work Cotiuous fuctios re Riem itegrble Every cotiuous rel-vlued fuctio o ech closed bouded itervl is Riem itegrble. Proof Suppose, b R with < b d f : [, b] R is cotiuous fuctio (thus f is bouded, by 0.79 i the Appedix). Let ε > 0. Becuse f is uiformly cotiuous (by 0.78), there exists δ > 0 such tht 1.12 f (s) f (t) < ε for ll s, t [, b] with s t < δ. Let Z + be such tht b < δ. Let P be the eqully-spced prtitio = x 0, x 1,..., x = b of [, b] with for ech j = 1,..... The x j x j 1 = b U( f, [, b]) L( f, [, b]) U( f, P, [, b]) L( f, P, [, b]) = b ( ) sup f if f [x j 1,x j ] [x j 1,x j ] j=1 (b )ε, where the first equlity follows from the defiitios of U( f, [, b]) d L( f, [, b]) d the lst iequlity follows from We hve show tht U( f, [, b]) L( f, [, b]) b ε for ll ε > 0. Thus 1.8 implies tht L( f, [, b]) = U( f, [, b]). Hece f is Riem itegrble. A ltertive ottio for we could lso write f is f (x) dx. Here x is dummy vrible, so f (t) dt or use other vrible. This ottio becomes useful whe we wt to write somethig like 1 0 x2 dx isted of usig fuctio ottio. The ext result gives frequetly-used estimte for Riem itegrl.

7 Mesure, Itegrtio & Rel Alysis. Prelimiry editio. 8 July Sheldo Axler Sectio 1A Review: Riem Itegrl Bouds o Riem itegrl Suppose f : [, b] R is Riem itegrble. The (b ) if f f (b ) sup f [,b] [,b] Proof Let P equl the trivil prtitio = x 0, x 1 = b. The (b ) if f = L( f, P, [, b]) L( f, [, b]) = f, [,b] provig the first iequlity i the result. The secod iequlity i the result is proved similrly d is left to the reder. EXERCISES 1A 1 Suppose f : [, b] R is bouded fuctio such tht L( f, P, [, b]) = U( f, P, [, b]) for some prtitio P of [, b]. Prove tht f is costt fuctio o [, b]. 2 Suppose s < t b. Defie f : [, b] {0, 1} by { 1 if s < x < t, f (x) = 0 otherwise. Prove tht f is Riem itegrble o [, b] d tht f = t s. 3 Suppose f : [, b] R is bouded fuctio. Prove tht f is Riem itegrble if d oly if for ech ε > 0, there exists prtitio P of [, b] such tht U( f, P, [, b]) L( f, P, [, b]) < ε. 4 Suppose f, g : [, b] R re bouded fuctios. Prove tht d L( f, [, b]) + L(g, [, b]) L( f + g, [, b]) U( f + g, [, b]) U( f, [, b]) + U(g, [, b]). 5 Suppose f, g : [, b] R re Riem itegrble. Prove tht f + g is Riem itegrble o [, b] d ( f + g) = f + g.

8 Mesure, Itegrtio & Rel Alysis. Prelimiry editio. 8 July Sheldo Axler 8 Chpter 1 Riem Itegrtio 6 Suppose f : [, b] R is Riem itegrble. Prove tht the fuctio f is Riem itegrble o [, b] d ( f ) = f. 7 Suppose f : [, b] R is Riem itegrble. Suppose g : [, b] R is fuctio such tht g(x) = f (x) for ll except fiitely my x [, b]. Prove tht g is Riem itegrble o [, b] d g = 8 Suppose f : [, b] R is Riem itegrble. Prove tht f = lim b j=1 f. f ( + j(b ) ). 9 Suppose f : [, b] R is bouded fuctio. For Z +, let P deote the prtitio tht divides [, b] ito 2 itervls of equl size. Prove tht L( f, [, b]) = lim L( f, P, [, b]) d U( f, [, b]) = lim U( f, P, [, b]). 10 Suppose f : [, b] R is Riem itegrble. Prove tht if c, d R d c < d b, the f is Riem itegrble o [c, d]. [To sy tht f is Riem itegrble o [c, d] mes tht f with its domi restricted to [c, d] is Riem itegrble.] 11 Suppose f : [, b] R is bouded fuctio d c (, b). Prove tht f is Riem itegrble o [, b] if d oly if f is Riem itegrble o [, c] d f is Riem itegrble o [c, b]. Furthermore, prove tht if these coditios hold, the c f = f + f. c 12 Suppose f : [, b] R is Riem itegrble. Defie F : [, b] R by 0 if t =, F(t) = t f if t (, b]. Prove tht F is cotiuous o [, b]. 13 Suppose f : [, b] R is Riem itegrble. Prove tht f is Riem itegrble d tht f f. 14 Suppose f : [, b] R is fuctio such tht f (c) f (d) for ll c, d [, b] with c < d. Prove tht f is Riem itegrble o [, b].

9 Mesure, Itegrtio & Rel Alysis. Prelimiry editio. 8 July Sheldo Axler Sectio 1B Why the Riem Itegrl is Not Good Eough 9 1B Why the Riem Itegrl is Not Good Eough The Riem itegrl works well eough to be tught to millios of clculus studets roud the world ech yer. However, the Riem itegrl hs severl deficiecies. I this sectio, we will discuss the followig three issues: Riem itegrtio does ot hdle fuctios with my discotiuities; Riem itegrtio does ot hdle ubouded fuctios; Riem itegrtio does ot work well with limits. I Chpter 2, we will strt to costruct theory to remedy these problems. We begi with the followig exmple of fuctio tht is ot Riem itegrble Exmple fuctio tht is ot Riem itegrble Defie f : [0, 1] {0, 1} by f (x) = If [, b] [0, 1] with < b, the { 1 if x is rtiol, 0 if x is irrtiol. if f = 0 d sup f = 1 [,b] [,b] becuse [, b] cotis irrtiol umber (by 0.38) d rtiol umber (by 0.30). Thus L( f, P, [0, 1]) = 0 d U( f, P, [0, 1]) = 1 for every prtitio P of [0, 1]. Hece L( f, [0, 1]) = 0 d U( f, [0, 1]) = 1. Becuse L( f, [0, 1]) = U( f, [0, 1]), we coclude tht f is ot Riem itegrble. This exmple is disturbig becuse (s we will see lter), there re fr fewer rtiol umbers th irrtiol umbers. Thus f should, i some sese, hve itegrl 0. However, the Riem itegrl of f is ot defied. Tryig to pply the defiitio of the Riem itegrl to ubouded fuctios would led to udesirble results, s show i the ext exmple Exmple Riem itegrtio does ot work with ubouded fuctios Defie f : [0, 1] [0, ) by f (x) = { 1 x if 0 < x 1, 0 if x = 0. If x 0, x 1,..., x is prtitio of [0, 1], the sup f =. Thus if we tried to pply [x 0,x 1 ] the defiitio of the upper Riem sum to f, we would hve U( f, P, [0, 1]) = for every prtitio P of [0, 1]. However, we should cosider the re uder the grph of f to be 2, ot, becuse lim 0 1 f = lim 0 (2 2 ) = 2.

10 Mesure, Itegrtio & Rel Alysis. Prelimiry editio. 8 July Sheldo Axler 10 Chpter 1 Riem Itegrtio Clculus courses del with the previous exmple by defiig 1 1 lim 0 be 1 x dx. If usig this pproch, we would defie 1 0 1/2 lim 0 1 dx + lim x b 1 1/2 1 1 x dx. 1 0 x dx to be ( ) x dx to 1 x However, the ide of tkig Riem itegrls over subdomis d the tkig limits c fil with more complicted fuctios, s show i the ext exmple Exmple re seems to mke sese, but Riem itegrl is ot defied Let r 1, r 2,... be sequece tht icludes ech rtiol umber i (0, 1) exctly oce d tht icludes o other umbers (0.56 implies tht such sequece exists). For Z +, defie f : [0, 1] [0, ) by 1 x r if x > r, f (x) = 0 if x r. Defie f : [0, 1] [0, ] by f (x) = f (x) 2 =1. Becuse every subitervl of [0, 1] with more th oe elemet cotis ifiitely my rtiol umbers (s follows from 0.30), f is ubouded o every such subitervl. Thus the Riem itegrl of f is udefied o every subitervl of [0, 1] with more th oe elemet. However, the re uder the grph of ech f is less th 2. The formul defiig f the shows tht we should expect the re uder the grph of f to be less th 2 rther th udefied. The ext exmple shows tht the poitwise limit of sequece of Riem itegrble fuctios bouded by 1 eed ot be Riem itegrble Exmple Riem itegrtio does ot work well with poitwise limits Let r 1, r 2,... be sequece tht icludes ech rtiol umber i [0, 1] exctly oce d tht icludes o other umbers (0.56 implies tht such sequece exists). For Z +, defie f : [0, 1] {0, 1} by 1 if x = r j for some j {1,... }, f (x) = 0 otherwise. Defie f : [0, 1] {0, 1} by f (x) = { 1 if x is rtiol, 0 if x is irrtiol.

11 Mesure, Itegrtio & Rel Alysis. Prelimiry editio. 8 July Sheldo Axler Sectio 1B Why the Riem Itegrl is Not Good Eough 11 The ech f is Riem itegrble d 1 0 f = 0, s you should verify. Clerly lim f (x) = f (x) for ech x [0, 1]. However, f is ot Riem itegrble (see Exmple 1.14) eve though f is the poitwise limit of sequece of itegrble fuctios bouded by 1. Becuse lysis relies hevily upo limits, good theory of itegrtio should llow for iterchge of limits d itegrls, t lest whe the fuctios re ppropritely bouded. Thus the previous exmple poits out serious deficiecy i Riem itegrtio. Now we come to positive result, but s we will see, eve this result idictes tht Riem itegrtio hs some problems Iterchgig Riem itegrl d limit Suppose, b, M R with < b. Suppose f 1, f 2,... is sequece of Riem itegrble fuctios o [, b] such tht f (x) M for ll Z + d ll x [, b]. Suppose lim f (x) exists for ech x [, b]. Defie f : [, b] R by f (x) = lim f (x). If f is Riem itegrble o [, b], the f = lim f. The result bove suffers from two problems. The first problem is the udesirble hypothesis tht the limit fuctio f is Riem itegrble. Idelly, tht property would follow from the other hypotheses, but Exmple 1.17 shows tht we must explicitly iclude the ssumptio tht f is Riem itegrble. The secod problem with the result bove is tht it does ot seem to hve resoble proof usig just the tools of Riem itegrtio. Thus proof of the result bove will ot be give here. A proof of stroger result will be give lter, usig the tools of mesure theory tht we will develop strtig with the ext chpter. The lck of good Riem-itegrtio-bsed proof of the result bove idictes tht Riem itegrtio is ot the idel theory of itegrtio. We hve ot discussed differetitio (but we will do so i Chpter 4). However, you should recll from your clculus clss the followig versio of the Fudmetl Theorem of Clculus: if f is differetible o ope itervl cotiig [, b] d f is cotiuous o [, b], the f (b) f () = f.

12 Mesure, Itegrtio & Rel Alysis. Prelimiry editio. 8 July Sheldo Axler 12 Chpter 1 Riem Itegrtio Note the hypothesis bove tht f is cotiuous o [, b]. It would be ice ot to hve tht hypothesis. However, tht hypothesis (or somethig close to it) is eeded becuse there exist fuctios f tht re differetible everywhere o ope itervl cotiig [, b] but f is ot Riem itegrble o [, b]. I other words, if we use Riem itegrtio, the the right side of the equtio bove eed ot mke sese eve if f is defied everywhere. EXERCISES 1B 1 Defie f : [0, 1] R s follows: 0 if is irrtiol, f () = 1 if is rtiol d is the smllest positive iteger such tht = m for some iteger m. Show tht f is Riem itegrble d compute 2 Suppose f : [, b] R is bouded fuctio. Prove tht f is Riem itegrble if d oly if 1 0 f. L( f, [, b]) = L( f, [, b]). 3 Give exmple of rel-vlued fuctios f, g defied o [0, 1] such tht L( f + g, [0, 1]) = L( f, [0, 1]) + L(g, [0, 1]). d U( f + g, [0, 1]) = U( f, [0, 1]) + U(g, [0, 1]). 4 Give exmple of sequece of cotiuous rel-vlued fuctios f 1, f 2,... o [0, 1] d cotiuous rel-vlued fuctio f o [0, 1] such tht for ech x [0, 1] but 5 Show tht f (x) = lim f (x) f = lim f. 0 ( lim lim (cos(m!πx) ) = m for every x R. [This exmple is due to Heri Lebesgue.] { 1 if x is rtiol, 0 if x is irrtiol