Riemann Integral Oct 31, such that

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1 Riem Itegrl Ot 31, 2007 Itegrtio of Step Futios A prtitio P of [, ] is olletio {x k } k=0 suh tht = x 0 < x 1 < < x 1 < x =. More suitly, prtitio is fiite suset of [, ] otiig d. It is helpful to thik of prtitio s dividig [, ] ito itervls I k = [x k 1, x k ], eh of whih hvig legth dx k = x k x k 1. A prtitio P is sid to e fier th P if P P. Give y two prtitios P d P, the ommo refiemet of the prtitios is P P. A step futio is futio g [, ] R suh tht there exists prtitio P for whih g is ostt o eh ope itervl (x k 1, x k ) of the prtitio. Ay prtitio tht stisfies this oditio for g will e lled step prtitio (for g). Clerly every refiemet of step prtitio for g is lso step prtitio for We use the ottio Step[, ] for the set of ll step futios o [, ]. Thikig of itegrtio s mesurig siged re uder the grph of futio, it is reltively strightforwrd to itegrte step futios. If g is step futio with step prtitio P, d if g(x) = g k o (x k 1, x k ), we defie g = g k dx k. Oe eeds to verify, however, tht this defiitio of itegrl does ot deped o the hoie of step prtitio. Tht is, if P d P re two step prtitios for g, it must hold tht g k dx k = g k dx k. (1) By reduig to ommo refiemet, it is eough to show tht (1) holds whe P is refiemet of P. Exerise 1: Estlish (1) whe P is refiemet of P. First ssume tht P = P {x }, d the dedue the result for geerl refiemet y idutio. Properties of the Itegrl Notie tht the set of step futios o [, ] is vetor suspe of B[, ]. Ideed, if f d g re step futios with step prtitios P f d P g, let P e the ommo refiemet (so P is step prtitio for f d for g). The o eh itervl (x k 1, x k ) we hve f (x) = f k d g(x) = g k d ( f + g)(x) = f k + g k. Hee P is step prtitio for f + Similrly, if f is step futio, the so is f for y R. Oe of the most importt properties of the itegrl is tht the mp tkig f to Theorem 1. Let f d g e step futios o [, ]. The f is lier. ( f + g) = f + g,

2 Riem Itegrl Ot 31, 2007 d for y α R, α α Exerise 2: Prove Theorem 1 diretly from the defiitio of the itegrl of step futios. You my hve wodered whe you were first itrodued to itegrtio why it ws defied i terms of siged re uder the grph of futio. This is doe extly to esure tht the itegrl is lier. Aother elemetry property of the itegrl is its motoiity (whih lso relies o the siged re iterprettio of the itegrl). Theorem 2. Let f d g e step futios o [, ] suh tht f (x) g(x) for every x [, ]. (Tht is, f ) The f Proof. Let P e step prtitio for f d The for eh k we hve f k g k d hee f k dx k g k dx k = Suppose f is step futio. Ay step prtitio for f is lso step prtitio for f d hee f is lso step futio. Moreover, for y x [, ] we hve d hee y Theorem 1 d Theorem 2 we hve f (x) f (x) f (x) f f We hve therefore estlished the followig estimte, whih e thought of s reltioship etwee the siged re f d the usiged re f. Theorem 3. Let f e step futio o [, ]. The f Oe fil property of the itegrl is tht it e omputed y rekig the domi up ito piees d omputig the itegrl o eh piee. Theorem 4. Suppose f is step futio o [, ] d suppose. The f + Proof. Let P e y step prtitio for Without loss of geerlity, we ssume tht x N = for some N (otherwise, we osider fier prtitio y ddig the poit ). The f k dx k = N f k dx k + k=n+1 f. f. f k dx k = f + 2

3 Riem Itegrl Ot 31, 2007 Riem Itegrle Futios We would like to exted the defiitio of the itegrl to roder lss of futios th step futios. Give futio f B[, ], we would like to defie itegrl for f tht preserves the properties of Theorems 1, 2, 3 d 4 of the itegrl for step futios. Although we will ot e le to do this for ll futios i B[, ], we will e le to do so for lrge lss of futios. Give futio f B[, ], d step futios g d G with g f G, we would wt to hve g We defie the upper Riem itegrl of f to e d the lower Reim itegrl of f to e if f G Step[,] G f G. G sup g Step[,] g f For y f B[, ], it is immedite osequee of the defiitio tht f Moreover, if f is step futio, the Hee if f is step futio, the = sup g Step[,] g f f if G Step[,] G f g G The lss of futio for whih we hve equlity of the upper d lower Riem itegrls is kow s the set of Riem itegrle futios, R[, ]. If f R[, ], the we defie f (= f ). 3

4 Riem Itegrl Ot 31, 2007 We hve just show therefore tht Step[, ] R[, ], d tht the Riem itegrl of step futio grees with the itegrl we hve lredy defied for step futios. It is perhps surprisig tht ot every futio i B[, ] is Riem itegrle. A exmple of suh futio is give y χ Q. Exerise 3: Prove tht 1 0 χ Q = 1 ut 1 0 χ Q = 0. Upper d Lower Riem Sums Sometimes it is oveiet to desrie the upper d lower Riem itegrls of futio i terms of limits of erti er optiml step futios. Let f B[, ] d let P e prtitio of [, ]. Give this prtitio, we defie M k = sup x [xk 1,x k ] f (x) d m k = if x [xk 1,x k ] f (x). We the ssoite with f d P step futios f P d f P tht re equl to M k d m k respetively o (x k 1, x k ) d re equl to f (x k ) for eh k. The upper Riem sum (for the futio f d prtitio P) is U( f, P) = f P = M k dx k d the lower Riem sum L( f, P) = f P = m k dx k. Exerise 4: Give futio f B[, ] d step futio G f with, show tht for y є > 0 there exists prtitio P suh tht є + G U( f, P ) Prove similr result for step futios g Colude tht d if P U( f, P) sup L( f, P). P Chrteriztio of Riem Itegrle Futios Give the defiitio of Riem itegrility, it is ot eessrily esy to determie whether give futio is Riem itegrle. O the fe of thigs, oe would hve to ompute the upper d lower Riem itegrls, d the verify tht they re the sme. The followig result helps idetify Riem itegrle futios without hvig to ompute upper d lower Riem itegrls. Propositio 5. Let f B[, ]. The the followig re equivlet. 1. f R[, ]. 4

5 Riem Itegrl Ot 31, For y є > 0 there exist step futios g d G with g f G d suh tht (G g) < є. 3. For y є > 0 there exists prtitio P suh tht U( f, P) < L( f, P) + є. Proof. Suppose f is Riem itegrle. Let G e step futio suh tht G f d Let g e step futio suh tht g f d G < g > f + є/2. f є/2. The So G < f + є/2 = (G g) < є. f + є/2 < g + є. Coversely, suppose f is ot Riem itegrle. Let є = g d G with g f G we hve f The for y step futios g f є G є. Hee for ll step futios G d g with g f G. (G g) є The equivlee of sttemets 2 d 3 is left for the reder. Exerise 5: Prove the equivlee of sttemets 2 d 3 i Propositio 5. Oe importt pplitio of Propositio 5 is tht it llows for esy proof tht C[, ] R[, ]. Theorem 6. C[, ] R[, ]. Proof. Let f C[, ]. Let є > 0. Sie f is uiformly otiuous, there exists δ > 0 suh tht if x z < δ, the f (x) f (z) < є/( ). Pik N N suh tht ( )/N < δ, d let P e the prtitio { + k( )/N 0 k N}. For eh k, we defie M k = sup x I k f (x) m k = if x I k f (x). 5

6 Riem Itegrl Ot 31, 2007 The Hee U( f, P) = N M k dx k N 0 M k m k є. (m k + є ) dx k = So y Propositio 5, we see tht f is Riem itegrle. N m k dx k + є = L( f, P) + є. Properties of the Itegrl We would like to exted Theorems 1, 2, 3 d 4 to ll of R[, ]. The extesio of Theorem 2 is immedite from the defiitio. Theorem 7. Suppose f, g R[, ] with f The f Proof. For y step futio H with g H we hve f H s well d hee if H Step[,] H f H if H Step[,] H g H = g = To estlish the lierity of the itegrl we eed to work little hrder. Notie tht if f is Riem itegrle, the for every there exists step futio G suh tht G f d G < f + 1 = f + 1. Hee there is sequee of step futios G, eh with G f, suh tht lim G = A similr sequee (g ) of step futios with g f lso exists. Theorem 8. Suppose f, g R[, ]. The f + g R[, ] d ( f + g) = f + (2) Also, for every α R, α f R[, ] d α α (3) 6

7 Riem Itegrl Ot 31, 2007 Proof. Let f, g R. Let є > 0 d let (h f, ) d (H f, ) e step futios suh tht h f, f H f, d suh tht lim h f, = lim H f,. Tht suh step futios exists is osequee of the Riem itegrility of Let (h g, ) d (H g, ) e similr sequees for Notie tht for eh, H f, + H g, f + Hee for eh, ( f + g) = if H Step[,] H ( f +g) Tkig the limit i we olude H (H f, + H g, ) = H f, + H g,. f + g f + A similr rgumet with the sequees (h f, ) d (h f, ) yields the iequlity f + g f + However, it is lwys true tht ( f + g) ( f + g), so we olude tht ( f + g) = ( f + g) = f + Hee ( f + g) is itegrle d equtio (2) holds. The proof tht α f is itegrle d tht (3) holds is left s exerise. Exerise 6: Suppose f R[, ] d α R. Show tht α f R[, ] d α α I order to prove the extesio of Theorem 3 to ll of R[, ], we eed to show first tht if f R[, ], the f R[, ]. This is perhps most esily doe y first showig tht the positive prt of f is Riem itegrle. Give futio f B[, ], we defie f 0 y ( f 0)(x) = mx(x, 0). Notie tht f f 0, d if if f, g B[, ] stisfy g f, the g 0 f 0. It follows tht d hee f + g f 0 + g 0 ( f + g) 0 ( f 0 + g 0) 0 = f 0 + g 0. With these fts i hd, we ow show the positive prt of f is Riem itegrle whever f is. 7

8 Riem Itegrl Ot 31, 2007 Propositio 9. Suppose f R[, ]. The f 0 R[, ]. Proof. Let f R[, ] d let є > 0 Let g d G e step futios suh tht g f G d suh tht tht (G g) < є. Now otie tht G 0 d g 0 re step futios d g 0 f 0 G 0. Moreover, G 0 = ((G g) + g) 0 (G g) 0 + g 0 = (G g) + (g 0). Hee G 0 g 0 G g d therefore G 0 g 0 Hee y Propositio 5, f 0 is Riem itegrle. Corollry 10. If f R[, ], the f R[, ] d G g < є. f f. Proof. Notie tht ( f 0) ( f 0) d f = ( f 0) + ( f 0). Sie f is sum of Riem itegrle futios, it is Riem itegrle. Moreover, ( f 0) ( f 0) ( f 0) + ( f 0) = f. The fil property to exted is the domi deompositio of the itegrl. Propositio 11. Let f B[, ] d let [, ]. The f R[, ] if d oly if f R[, ] d f R[, ]. Proof. Suppose f R[, ]. Let є > 0 d let G d g e step futios suh tht g f G d suh tht (G g) < є. The restritios of G d g to [, ] lso stisfy g f G. Moreover, sie G g 0, (G g) (G g) < є. So the restritio of f to [, ] is Riem itegrle, d similr rgumet shows tht the restritio of f to [, ] is Riem itegrle. 8

9 Riem Itegrl Ot 31, 2007 Coversely, suppose f R[, ] d f R[, ]. For ottiol oveiee, let f 1 d f 2 e the restritios of f to [, ] d [, ] respetively. Let є > 0 d let g 1 d G 1 e step futios o [, ] suh tht g 1 f 1 G 1 d (G 1 g 1 ) < є/2. Similrly, let g 2 d G 2 e step futios o [, ] suh tht g 2 f 2 G 2 d (G 2 g 2 ) < є/2. Let g e the step futio o [, ] tht is equl to g 1 o [, ), equl to g 2 o (, ] d equl to f () t. The g f d g = g + g = g 1 + Let G e similrly defied step futio with respet to G 1 d G 2, so G f d The g f G o [, ] d (G g) = Hee f is Riem itegrle o [, ]. G = (G 1 g 1 ) + G 1 + G 2. g 2. (G 2 g 2 ) < є/2 + є/2 = є. With this lst result i hd, it is ot diffiult to estlish the extesio of Theorem 4. Theorem 12. Let f R[, ]. The for y [, ] we hve f + Exerise 7: Prove Theorem 12. 9

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