The Riemnn nd the Generlised Riemnn Integrl Clvin 17 July 14 Contents 1 The Riemnn Integrl 1.1 Riemnn Integrl............................................ 1. Properties o Riemnn Integrble Funtions............................. 5 1.3 The Fundmentl Theorem o Clulus................................ 9 1.4 Lebesgue s Integrbility Criterion................................... 11 1.5 Substitution Theorem nd Integrtion by Prts........................... 13 The Generlised Riemnn Integrl/Henstok- Kurzweil Integrl 16.1 Deinition nd Min Properties.................................... 16. Improper nd Lebesgue Integrl.................................... 3.3 Ininite Integrl............................................. 6.3.1 Integrl on [,........................................ 6.3. Integrls on, b]....................................... 9.3.3 Integrl on,....................................... 9.4 Convergene Theorems......................................... 31 3 Reerene 34
Riemnn Integrl 1 The Riemnn Integrl 1.1 Riemnn Integrl Here, we will not provide motivtion o the integrl, disuss its interprettion s the re under the grph, or its pplitions. We will ous only on the mthemtil spets o the integrl. Deinition 1.1.1 Prtition. A prtition P o [, b] is inite set o points x, x 1,..., x n suh tht = x < x 1 <... < x n 1 < x n = b We denote prtition P by P := {[x i 1, x i ]} n nd [x i 1, x i ] is lled the i th intervl o P. Deinition 1.1. Mesh/norm o Prtition P. We deine the norm o P to be P := mx {x 1 x, x x 1,..., x n x n 1 } 1.1 Thereore, we n sy tht the norm o P is the length o the lrgest subintervl in the Prtition P. Deinition 1.1.3 Tgged Prtition. For prtition P = {[x i 1, x i ]} n in [, b]. Let us hoose point t i [x i 1, x i ] or every i < n, we ll t i s tgs o subintervl [x i 1, x i ]. The prtition P together with its tgs re lled tgged prtition o intervl [, b] denoted by Remrks: P := {[x i 1, x i ], t i } n 1. Tgs n be hosen in ny wyeg. let endpoint, right endpoint, midpoint, et. An endpoint o subintervl n be used s tg or onseutive subintervls Deinition 1.1.4 Riemnn Sum. Let : [, b] R be untion with tgged prtition P = {[x i 1, x i ], t i } n. Then we deine the Riemnn Sum o to be S; P = t i x i x i 1 1. Using this knowledge o the Prtition, now we n deine the Riemnn Integrl Deinition 1.1.5 Riemnn Integrl. A untion : [, b] R is sid to be Riemnn Integrble on [, b] i there exists L R suh tht or every > there exists δ > suh tht i P is tgged prtition o [, b] with P < δ, then S; P L < We usully ll L s the Riemnn Integrl o over [, b] denoted by L = or xdx We denote the set o ll Riemnn Integrble untions on [, b] s R[, b]. Remrk: 1. Although the deinition o the Riemnn Integrl is similr to the deinition o the limit o the Riemnn Sum s the norm goes to, lim S; P = L P type o limit rom wht we hve studied beore. Theorem 1.1.6 Uniqueness Theorem. I R[, b], then its integrl is unique, S; P is not untion o P. So this is dierent
Riemnn Integrl 3 Proo. Assume insted tht L 1 nd L both stisy Deinition 1.1.5 with L 1 L, without loss o generlity, ssume tht L > L 1. Then, or ll >, there exists δ 1 > suh tht i tgged prtition P 1 with norm P 1 < δ 1, then S ; P 1 L 1 < Also, there exists δ > suh tht i tgged prtition P with norm P < δ, then S ; P L < Choose δ = min{δ 1, δ } >. Let P be tgged prtition with norm P < δ. Sine P < δ 1 nd P < δ, then S ; P L 1 < 1.3 nd S ; P L < 1.4 We then ix = L L1 >. Then rom Eqution 1.3, we get + L 1 < S ; P Similrly, Eqution 1.4 gives us + L < S ; P < + L 1 = 3L 1 L < + L = L 1 + L < S ; P < S ; P < L 1 + L < 3L L 1 So we hve 3L 1 L < S ; P < L 1 + L < S ; P < 3L L 1 whih is impossible so we hve ontrdition tht L 1 L. Thereore, the vlue o the integrl is unique. The ollowing theorem llows us to orm lgebri ombintions o integrble untion Theorem 1.1.7. Suppose, g R[, b], then:. I k R, then k R[, b] nd b. + g R[, b] nd k = k + g = + g. I x gx or ll x [, b] then Proo. Consider P = {[x i 1, x i ], t i } n on [, b], then Sk; P = kt i x i x i 1 = k Using similr methods, we n onlude tht S + g; P = S; P + Sg; P g t i x i x i 1 = ks; P nd S; P Sg; P
4 Riemnn Integrl For prt, s R[, b], then or ll >, there exists δ > suh tht i tgged prtition P with norm P < δ, then S ; P < k So S k; P k = ks ; P k = k S ; P < k = k Thereore, the Riemnn Integrl o the untion k is equl to k For prt b, s, g R[, b], then or ll >, there exists δ 1 > suh tht i tgged prtition P 1 with norm P 1 < δ 1, then S ; P 1 < Similrly, there exists δ > suh tht i tgged prtition P with norm P < δ, then S g; P g < We let δ = min{δ 1, δ } >. Sine P < δ 1 nd P < δ, then S ; P < nd S g; P So, by Tringle Inequlity S + g; P + S ; P g < g = S ; P + S g; P + S g; P g < + = Thereore, the Riemnn Integrl o the untion + g is equl to + For prt, we use Eqution 1.5. S ; P < = < S ; P = S g; P b b g < = S g; P As we know tht S; P Sg; P, we hve So, we n onlude tht g < g + = < b b g g < g; = S P g + g ; < S P Theorem 1.1.8. Let h : [, b] R suh tht hx = exept or inite number o points in [, b], then h R[, b] nd h = Proo. We just need to proo or the se o only 1 point i.e. hx = exept or point, then we n use indution to prove or inite number o points. For ll >, let δ < S h > suh tht i tgged prtition P < δ, then h; P = ht i x i x i 1 h P < h δ = Thereore, the untion h is integrble with h = < g + 1.5
Properties o Riemnn Integrble Funtions 5 Corollry 1.1.9. I g R[, b] nd i x = gx exept or inite number o points in [, b], then is Riemnn integrble nd = g. Proo. Consider h : [, b] R suh tht hx = x gx then we n use Theorem 1.1.8 to show tht h = Thereore, g = h = = g = = = g Theorem 1.1.1 Boundedness Theorem on Integrls. I R[, b], then is bounded on [, b]. Remrk: 1. The onverse All bounded untions re Riemnn Integrble is not true, nd exmple is shown in Exmple 1.. Proo. Assume by ontrdition tht R[, b] is n unbounded untion with there exists δ > suh tht i tgged prtition P with P < δ, then S ; P L < 1 = S = L. By hoosing = 1, ; P < L + 1 1.6 Then we let Q = {[x i 1, x i ]} n be prtition with Q < δ. Sine is not bounded on [, b], then there exists subintervl [x k 1, x k ] in whih is not bounded. Sine is not bounded or [x k 1, x k ], or ll M >, there exists x [, b] suh tht x > M. Here we ix M suh tht M = L + 1 + i k t ix i x i 1 To mke the ontrdition with Eqution 1.6, we need to hoose pproprite tgs. For subintervl [x i 1, x i ] suh tht i k, pik t i = x i. For i = k we hoose t k [x k 1, x k ] suh tht t k x k x k 1 > M = L + 1 + t i x i x i 1 By using the Tringle Inequlity x + y x y x, y R, Q S ; tk x k x k 1 t i x i x i 1 > L + 1 This ontrdits Eqution 1.6. 1. Properties o Riemnn Integrble Funtions Theorem 1..1 Cuhy Criterion. Let : [, b] R. R[, b] i nd only i or every >, there exists δ > suh tht i tgged prtitions P nd Q o [, b] with P < δ nd Q < δ, then S ; P Q S ; < Proo. = As R[, b], let Q < δ, then S So, we hve S ; P S And so we re done. ; Q S = ; P i k i k = L. I we hoose tgged prtitions P nd Q suh tht P < δ nd ; P L < S ; nd Q L < L + L S ; Q S ; P L + S ; Q L < + =
6 Properties o Riemnn Integrble Funtions = For every >, let δ > suh tht i tgged prtitions P nd Q with P < δ nd Q < δ, then < S ; P S ; Q Let P n be sequene o tgged prtitions with P n < δ or ll n N. So we hoose m, n N nd without loss o generlity, we ssume m > n. Clerly, this mens tht P n < δ nd P m < δ whih in turn implies S ; P n S ; P m < This shows tht S ; P n is uhy sequene nd in turn, onverges to some number L R. Or in other words, >, N N : n > N = S, P n L < Finlly, to show tht R[, b] nd tht = L, given >, we let δ > nd n > N suh tht i tgged prtition Q with Q < δ, then S ; Q L S ; Q S ; P S n + ; P n L < + = Exmple 1... Show tht the Dirihlet untion, { 1 i x Q x = i x R \ Q is not Riemnn integrble on [, b] or ll, b R Proo. We re going to use the onverse o the Cuhy Criterion. Thereore we need to prove tht: There exists > or every δ >, there exists tgged prtitions P nd Q with P < δ nd Q < δ suh tht S ; P Q S ;. Tking = b, we let P nd Q be prtitions on [, b] For ll n N, P = {[x i 1, x i ], t i } n suh tht x =, x n = b nd t i Q or ll i < n nd S ; P = While Then, Q = {[x i 1, x i ], t i } n suh tht x =, x n = b nd t i R \ Q or ll i < n t i x i x i 1 = x i x i 1 = x n x n 1 +x n 1 x n + +x 1 x = x n x = b S S ; Q = ; P t i x i x i 1 = S ; Q b = b > = Thereore, the Dirihlet untion is not Riemnn integrble on [, b] or ll, b R Theorem 1..3 Squeeze Theorem. : [, b] R, then R[, b] i nd only i or ll >, there exists untions g, h R[, b] with gx x hx x [, b] suh tht h g <
Properties o Riemnn Integrble Funtions 7 Proo. = We n let x = gx = hx or ll x [, b] nd let > = Given >, we hoose untions g, h R[, b] suh tht suh tht gx x hx x [, b] 1.7 h g < 3 < 1.8 As g, h R[, b], there exists δ > suh tht i ny tgged prtition P with P < δ, then S g; P g < 3 nd S h; P h < 3 It ollows tht g g; 3 < S P Due to Eqution 1.7, we know tht S g; P Then So, S nd S h; P ; P S g ; 3 < S P < < h; P h + 3. Thereore, h + 3 I we hoose nother tgged prtition Q with Q < δ, we lso hve S h g + < S 3 ; P g ; 3 < S Q < S ; Q ; P < h + 3 S ; Q < h g + 3 h g + 3 < 3 + 3 = This stisies the Cuhy Criterion Theorem 1..1, thereore R[, b] Theorem 1..4. I : [, b] R is step untion, then R[, b] Proo. Consider the untion ϕ A : [, b] R suh tht ϕ A x = { 1 i x A i x / A where A = [, d] or some, d R with < < d < b Given >, let δ > suh tht i tgged prtition with P < δ, S ϕ A ; P = ϕt i x i x i 1 = x A x i x i 1 = d We lim tht the integrl o ϕ A x is d. So, S ϕ A ; P d = d d = < Thereore, untion ϕ A R[, b]. To omplete the proo, we write ny step untion s nd by using Theorem 1.1.7, we re done. k x = i ϕ Ai x
8 Properties o Riemnn Integrble Funtions Theorem 1..5. I : [, b] R is ontinuous on [, b], then R[, b] Proo. As is ontinuous on [, b], then is uniormly ontinuous over [, b], i.e. >, δ >, x, y [, b] : x y < δ = x y < b Let P be prtition nd so s is ontinuous on [, b], or eh subintervl [x i 1, x i ], there exists u i, v i [x i 1, x i ] suh tht or eh i th subintervl, u i nd v i hs minimum vlue nd mximum vlue respetively on [x i 1, x i ] Then, we deine step untion φ nd Φ where φx = u i nd Φx = v i or ll x [x i 1, x i ] Below is n illustrtion o the untion, φ, nd Φ where the urve is the untion, the step untion bove is Φ nd the step untion below is φ Clerly, we hve φx x Φx or ll x [, b] So, or ll >, there exists δ >, suh tht i ny prtition P with P < δ, then Then, Φ φ = v i u i x i x i 1 < x i x i 1 < δ or ll i < n = v i u i < δ b x i x i 1 = n b x i x i 1 = b b = Finlly, we just pply the squeeze theorem Theorem 1..3 to omplete the proo. Theorem 1..6. I : [, b] R is monotone on [, b], then R[, b] As the proo is very similr to Theorem 1..5. The proo is omitted. Theorem 1..7 Additivity Theorem. Let : [, b] R nd [, b], then is Riemnn Integrble on [, b] i nd only i is Riemnn Integrble on [, ] nd is Riemnn Integrble on [, b] nd so i the ondition is true, = +
The Fundmentl Theorem o Clulus 9 Proo. = As R[, b] with = L 1, then or ll >, there exists δ 1 > suh tht i ny tgged prtition P 1 with P 1 < δ 1, then S ; P 1 L 1 < Also, s R[, b] with = L, then or ll >, there exists δ > suh tht i ny tgged prtition P with P < δ, then S ; P L <. I M is bound or, hoose δ = min { } δ 1, δ, M, suh tht i tgged prtition P with P < δ, we will prove tht S ; P L 1 + L < i Cse 1: Let P = {[x i 1, x i ], t i } n. I is prtition point o P i.e. = x i or some i < n, we split the prtition to P 1 nd P on [, ] nd [, b] respetively. Then, sine we know tht P 1 < δ 1 nd P < δ nd S ; P = S ; P 1 + S ; P, S ; P L 1 + L = S ; P 1 L 1 + S ; P L S ; P S 1 L 1 + ; P L < + = ii Cse : I is not prtition is point o P, then there exists k < n suh tht [x k 1, x k ]. We onsider tgged prtition P with one more prtition point thn P t, i.e. P = {[x, x 1 ], t 1,..., [x k 1, ],, [, x k ],,..., [x n 1, x n ], t n } So, or ll >, there exists δ > suh tht i ny prtition P with P < δ < M, then S ; P S ; P = tk x k x k 1 x k x k 1 = t k x k x k 1 < M x k x k 1 < M This stisies the Cuhy Criterion Theorem 1..1 nd so, R[, b] = M = As R[, b], given >, let δ > suh tht i tgged prtitions P nd Q with P < δ nd Q < δ, then S ; P Q S ; < We onsider P nd Q on [, ] with P < δ nd Q < δ. By dding the sme dditionl prtition points nd tgs in [, b] to P nd Q, we get P nd Q suh tht P < δ nd Q < δ nd so S ; P S ; Q S = ; P Q S ; < This shows tht it ulills the Cuhy Criterion Theorem 1..1. So is Riemnn Integrble on [, ] The sme pplies to [, b] nd so, we re done. 1.3 The Fundmentl Theorem o Clulus In this subsetion, we re going to disuss the onnetion o the derivtive n the integrl. This subsetion will disuss minly the two orms o the Fundmentl Theorem o Clulus. Deinition 1.3.1. Let : [, b] R be untion. A untion F : [, b] R is lled n ntiderivtive o i F x = x or ll x [, b] Deinition 1.3.. I R[, b], then the untion F : [, b] R deined by F x = is lled the indeinite integrl o with bsepoint x or x [, b] Using this deinition, we n deine the Fundmentl Theorem o Clulus.
1 The Fundmentl Theorem o Clulus Theorem 1.3.3 Fundmentl Theorem o Clulus First Form. Let : [, b] R nd is Riemnn Integrble on [, b]. Suppose tht there is untion F : [, b] R suh tht: F is ontinuous on [, b], b F x = x or most x [, b] exept or inite set o points, Then, we hve = F b F Remrks: 1. The untion F suh tht F x = x or ll x [, b] is lled the ntiderivtive o x. We n permit inite number o points where F does not exists or F Proo. As R[, b], or ll >, there exists δ > suh tht i ny tgged prtition P with P < δ then S ; P < We let [x i 1, x i ] be the i th subintervl o P. Then, by pplying the Men Vlue Theorem on F, or eh i < n, there exists u i x i 1, x i suh tht By dding ll the i th terms, we get F b F = F x i F x i 1 x i x i 1 = F u i = u i F x i F x i 1 = u i x i x i 1 F x i F x i 1 = u i x i x i 1 1.9 Then, letting P u = {[x i 1, x i ], u i } n ;, Eqution 1.9 is equl to S P u So, we n onlude tht, or ll >. But, sine we n mke > s smll s possible, we iner tht F b F = F b F b < Exmple 1.3.4. Let, b R suh tht < < b. Let untion F : [, b] R suh tht F x = x. Then let : [, b] R suh tht 1 i x > x = i x = 1 i x < Proo. We n see tht F x = x or ll [, b] \ {} s F x is not dierentible t. However, the Fundmentl Theorem o Clulus still pplies. b
Lebesgue s Integrbility Criterion 11 We n see tht the untion x is step untion nd so = b + nd F b F = b = b + nd thereore, = F b F Theorem 1.3.5 Fundmentl Theorem o Clulus Seond Form. Let R[, b], nd let be ontinuous t [, b]. Then the indeinite integrl is dierentible t nd F =. F = or [, b] Proo. Suppose, b onsider the right hnd derivtive o F t, sine is ontinuous t, given >, there exists δ > suh tht i x + δ, then x < = < x < + 1.1 Let h stisy < h < δ, the Additivity Theorem Theorem 1..7 implies tht is Riemnn Integrble on [, ], [, + h] nd [, + h] nd tht +h = +h Now on [, + h], stisy Eqution 1.1. So we hve h = +h F + h F = = = = F + h F +h F + h F h F + h F h But, sine we n hoose > to be very smll, we n iner tht +h F + h F lim = F = h + h + = + h By proving using the similr method or the let hnd side, we onlude tht F =. 1.4 Lebesgue s Integrbility Criterion In this subsetion, neessry nd suiient ondition or untion to be Riemnn Integrble nd some pplition will be given. Beore strting with the Lebesgue s Integrbility Criterion, we strt by deining the null set Deinition 1.4.1 Lebesgue Outer Mesure. Let A R nd let l, b = b or ll open intervl, b. Then the Lebesgue Outer Meusure o A µ A is deined to be { } µ A = in li k : {I k } is sequene o open intervls in whih A I k Remrks: k=1 1. We n see here tht µ A or ll A R Deinition 1.4. Null Set. A set A R is null set i or ll >, µ A Exmple 1.4.3. All ountble sets re null sets. k=1
1 Lebesgue s Integrbility Criterion Proo. Let set A = {x k : k N} be ountbly ininite set. Then, we deine the sequene o open intervl {I k } to be I k = x k k+1, x k + k+1 or ll k N Then, li k = k = As µ A is the inimum, we hve µ A < k=1 k=1 Exmple 1.4.4. The set o rtionl numbers Q is null set. Proo. We strt by denoting tht Q = {r k : k Z}. Then we n use similr proo with the proo in Exmple 1.4.3 Deinition 1.4.5. I P x is sttement bout the point x [, b], we sy tht P x holds lmost everywhere on I i there exists null set Z I suh tht P x holds or ll x I \ Z or we my write P x x I Now, we re redy to introdue the Lebesgue s Integrbility Criterion Theorem 1.4.6 Lebesgue s Integrbility Criterion. A bounded untion : [, b] R is Riemnn Integrble i nd only i is ontinuous lmost everywhere on [, b] Proo. The proo to this theorem is omitted. Exmple 1.4.7 Thome Funtion. The Thome Funtion is integrble on [, 1]. The Thome Funtion is deined s { 1 x = q i x Q suh tht x = p q in lowest terms nd q > i x R \ Q Proo. Here x is ontinuous or ll x R \ Q nd disontinuous or ll x Q x is disontinuous t x Q. Let Q, let { n } be sequene suh tht n R \ Q or ll n N nd { n }, then lim n n = x is ontinuous t x R \ Q. By Arhimeden Property, let b R \ Q, then there exists n suh tht 1 n <. Consider the intervl b 1, b + 1, we n sy tht there is only inite number o rtionls with denomintor less thn n. So we n hoose δ > smll enough suh tht b δ, b + δ does not ontin rtionl number with denomintor less thn n. It ollows tht x b < δ where x A. We hve x b = x 1 n <. So x is ontinuous or ll x R \ Q. So the set o ll disontinuous points is S = Q [, 1] nd so s the set is ountbly ininite, then S is null set nd by the Lebesgue Integrbility Criterion, the Thome untion is Riemnn Integrble on [, 1]. Theorem 1.4.8 Compostition Theorem. Let R[, b] with [, b] [, d] nd let g : [, d] R be ontinuous. Then, g R[, b] Remrks: 1. The ondition g is ontinuous n not be dropped Proo. We n split this proo into two ses i Cse 1. I is ontinuous t [, b], then g is lso ontinuous t [, b]. This implies tht g is Riemnn integrble on [, b] ii Cse. I hs inite number o disontinuous points in [, b], then s stiies the Lebesgue s Integrbility Theorem Theorem 1.4.6, then the set S = {u : is disontinuous t u} is null set s it is ountble By exmple 1.4.3. It ollows tht the set S S where S = {u : g is disontinuous t u}. Thereore, by Lebesque s Integrbility Theorem Theorem 1.4.6, g is Riemnn Integrble on [, b]
Substitution Theorem nd Integrtion by Prts 13 Both ses implies tht g is Riemnn integrble on [, b] Exmple 1.4.9. Let nd 1 i x > x = i x = 1 i x < { 1 gx = q i x Q suh tht x = p q in lowest terms nd q > i x R \ Q We lredy know tht, g R[, 1] Show tht gx is not Riemnn Integrble on [, 1] Proo. We n see tht gx > or ll x Q nd gx = or ll x R \ Q. So { 1 i x Q gx = x R \ Q whih we know tht it is not Riemnn Integrble Theorem 1.4.1 Produt Theorem. I nd g is Riemnn integrble on [, b], then the produt g is lso Riemnn integrble on [, b]. Proo. For eh untion : [, b] R, we onsider the untion hx = x or x [ M, M] where [ M, M] is the bounds o the untion. Then by the omposition theorem Theorem 1.4.8, we know tht = h is Riemnn Integrble on [, b]. Thereore, the untions, g nd + g re Riemnn Integrble on [, b]. To omplete the proo, we simply write g s g = 1 whih shows tht g is Riemnn Integrble on [, b] [ + g g ] 1.5 Substitution Theorem nd Integrtion by Prts Theorem 1.5.1 Substitution Theorem. Let α : [, d] R be dierentible t [, d] with its derivtive α being ontinuous on [, d]. I : [, b] R with d b nd α [, d] [, b], then Proo. Consider d α x α xdx = F u = u α αd α αdα xdx or u [, b] By the Fundmentl Theorem o Clulus, F u = u Let Hx = F αx Also by the Fundmentl Theorem o Clulus, Then, αd α Hd = d H ddx αdα = F αd = Hd To omplete the proo, we just need to ind H d. As nd α re both dierentible on [, b] nd [, d] respetively, then H d = F αx α x = αx α x Thereore, d αx α xdx = αd α αdα
14 Substitution Theorem nd Integrtion by Prts Theorem 1.5. Integrtion by Prts. Let F nd G be dierentible on [, b], let = F nd g = G with, g R[, b]. Then G = [F G] b F g Proo. By Produt Rule o Dierentition, s F nd G re dierentible, F G = F G + F G = G + F g As ll F, G,, g R[, b], then G, F g R[, b]. So, by using the Fundmentl Theorem o Clulus, Thereore, [F G] b = F G = G + G = [F G] b F g F g Theorem 1.5.3 Equivlene Theorem. A untion : [.b] R is Drboux Integrble i nd only i it is Riemnn Integrble Proo. = Assume is Drboux Integrble on [, b]. Then, or ll >, let P be prtition on [, b] suh tht U; P L; P <. We then deine the step untion φ nd Φ in whih nd So we hve φx = m k or x [x k 1, x k k < n Φx = M k or x [x k 1, x k k < n φx x Φx x [, b] As φ, Φ re step untions, it is Riemnn Integrble nd moreover, Thereore, we hve Φ = φ = M k x k x k 1 = U ; P k=1 m k x k x k 1 = L ; P k=1 Φ φ = U; P L; P < nd so by squeeze theorem, R[, b]. As = L; P = U; P nd so i tgged prtition Q with Q < δ, then S; Q U; P < U; P L; P < = Assume is Riemnn Integrble on [, b]. By Cuhy riterion, or ll >, there exists δ > suh tht or ll tgged prtitions P M nd P m, i P M < δ nd P m < δ, then S ; P M S ; P m < 3 We then ix prtition P = {[x i 1, x i ]} n. For ll i < n, onsider set A i = {x : x [x i 1, x i ]}. Let M i be sup A i, s M i is supremum, >, hoose tgs u i [x i 1, x i ] suh tht M i < u i + 3nx i x i 1
Substitution Theorem nd Integrtion by Prts 15 This implies, nd so, M i x i x i 1 < u i x i x i 1 + 3n M i x i x i 1 < u i x i x i 1 + 3n U; P < S; P M + 3 where P M = {[x i 1, x i ], u i } n i 1 Similrly, let m i be in A i, >, hoose tgs v i [x i 1, x i ] suh tht m i > v i 3nx i x i 1 This implies nd so, m i x i x i 1 > v i x i x i 1 3n m i x i x i 1 > v i x i x i 1 3n L; P > S; P m 3 where P M = {[x i 1, x i ], v i } n i 1 Thereore, U; P L; P < S ; P M S ; P m + < S ; P M S ; P m + 3 < 3 + 3 =
16 Deinition nd Min Properties The Generlised Riemnn Integrl/Henstok- Kurzweil Integrl The Riemnn Integrl is powerul tool nd hs mny pplitions. However, it ws ound out lter tht it ws not indequte. For exmple, the Fundmentl Theorem o Clulus F = F b F does not hold or ll dierentible untion. To tkle with these indequies, we use slightly dierent pproh, to rive t n integrl lled the Generlised Riemnn Integrl..1 Deinition nd Min Properties Deinition.1.1. A guge δ on [, b] is stritly positive untion δ : [, b], Deinition.1.. A tgged prtition P = {[x i 1, x i ], t i } n o [, b] is sid to be δ-ine i t i [x i 1, x i ] [t i δt i, t i + δt i ] or ll i N, i < n t i δt i x i 1 t i x i t i + δt i Remrks: 1. δt i is dependent on t i, so or every t i, it is possible to hve dierent vlue o δ The deinition o the Generlised Riemnn Integrl is very similr to the ordinry Riemnn Integrl nd mny o the proos re similr. However, the slight hnge results in muh lrger set o integrble untion. Deinition.1.3. A untion : [.b] R is sid to be Generlised Riemnn Integrble on [, b] i there exists number L R suh tht or every >, there exists guge δ on [, b] suh tht i P is δ-ine prtition o [, b], then S ; P L < Remrks: 1 The use o guge llow more ontrol on the length o the subintervl thn using the norm we n put smller subintervl when the untion is rpidly inresing nd lrger subintervl when the untion is nerly onstnt. The set o ll Generlised Riemnn Integrble untion is denoted by R [, b] 3 The Generlised Riemnn Integrl o R[, b] is lso denoted by or xdx Theorem.1.4 Uniqueness Theorem. I R [, b], then the vlue o the integrl is uniquely deined. Remrks: 1 The proo below is very similr to the Uniqueness Theorem o the Riemnn Integrl
Deinition nd Min Properties 17 Proo. Assume L 1 nd L both stisy Deinition.1.3 with L 1 L nd without loss o generlity, ssume L > L 1. Let >, then there exists guge δ 1 suh tht i P 1 is δ 1 -ine prtition, then S ; P 1 L 1 < Similry, there exists δ suh tht i P is δ -ine prtition, then S ; P L < Deine δt = min {δ 1 t, δ t} or eh t [, b] so tht δ guge on [, b]. I P is δ-ine prtition, the P is both δ 1 -ine nd δ -ine. So we hve S ; P L 1 <.1 S ; P L <. We then let = L L1 >. Then rom Eqution.1, we hve nd rom Eqution., we hve 3L 1 L L 1 + L < S ; P < S ; P < L 1 + L < 3L L 1 We then get 3L 1 L < S ; P < L 1 + L < S ; P < 3L L 1 whih is impossible thereore we hve ontrdition tht L 1 L nd so the vlue o the integrl is unique. Theorem.1.5 Consisteny Theorem. Let : [, b] R. I is Riemnn Integrble on [, b], then is Generlised Riemnn Integrble on [, b] nd their integrls re the sme, i.e. Riemnn = Generlised Riemnn Proo. Given >, we re going to ind n pproprite guge on [, b]. Sine R[, b], there exists δ > suh tht i ny tgged prtition P with P, then S ; P L < We let untion δ : [, b], suh tht δ t = 1 4 δ or ll t [, b] so tht δ n be guge on [, b]. I P = {[x i 1, x i ], t i } n is δ -ine prtition, then I i [t i δ t i, t i + δ t i ] = [t i 14 δ, t i + 14 ] δ As < x i x i 1 < t i + 1 4 δ t i + 1 4 δ = 1 δ < δ or ll i N, i < n We n see tht P lso stisies P < δ nd s is Riemnn Integrble, the sttement S ; P L < must be true. So, we n see tht is lso Generlised Riemnn Integrble. Thereore, by hoosing onstnt untion s guge, we n see tht ll Riemnn Integrble untion re Generlised Riemnn Integrble. We know tht rom Exmple 1.. tht the Dirihlet untion is not Riemnn Integrble, however we re going to prove shortly tht it is Generlised Riemnn Integrble.
18 Deinition nd Min Properties Exmple.1.6. The Dirihlet untion x = is Generlised Riemnn Integrble on [, 1] nd 1 = { 1 i x Q i x R \ Q Proo. We deine the set Q [, 1] s Q = {r 1, r,..., r n } or ll n N. Given >, deine δr n = nd n+ δx = 1 when x R \ Q. So δ is guge on [, 1] nd i ny prtition P = {[x i 1, x i ], t i } n is δ-ine, then we hve x i x i 1 < δt i. As x = i x is irrtionl, we only onsider the sum o ll the rtionl tgs to ind S ; Q. < r n x i x i 1 = x i x i 1 < n+ = n+1 Sine t most subintervl n hve the sme tgs, we hve S ; Q < n=1 S ; Q < S ; Q < Thereore, is Generlised Riemnn Integrble nd 1 = Theorem.1.7. Suppose, g R [, b] n+1 = n = n=1. I k R, then k R [, b] nd b. + g R [, b] nd k = k + g = + g. I x gx or ll x [, b], then Proo. As the proo the similr to Theorem 1.1.7, the proo is omitted Below is the Cuhy Criterion or Generlised Riemnn Integrl, the proo is similr with the Cuhy Criterion or Riemnn Integrl Theorem 1..1 Theorem.1.8 Cuhy Criterion. A untion : [, b] R is Generlised Riemnn Integrble i nd only i or every >, there exists guge δ on [, b] suh tht i ny tgged δ-ine prtition P, Q, then S ; P Q S ; < Proo. = As R [, b], let. I we hoose guge δ suh tht P nd Q re δ -ine prtition o [, b], then S ; P L < S ; nd Q L < We set δt = δ t or ll t [, b], so i P nd Q re δ-ine, then S ; P S ; Q S ; P L + S ; Q L < + = And so we re done. g
Deinition nd Min Properties 19 = For every >, there exists guge δ on [, b] suh tht i ny tgged prtition P, Q re δ-ine, then S ; P S ; Q < Let P n be sequene o δ-ine prtitions. So, we hoose m, n N nd WLOG, we ssume m > n. Clerly, P n nd P m re δ-ine, whih in turn implies S ; P n S ; P m < This shows tht S; P n is uhy sequene nd in turn, onverges to some number L R. Or in other words, >, N N : n > N = S ; P n L < Finlly, to show tht R [, b] nd tht = L, given >, we let δ be guge nd n > N suh tht i tgged prtition Q is δ-ine, then S ; Q L < S ; Q S ; P S n + ; P n L < + = Thereore, the untion is Generlised Riemnn Integrble on [, b] with integrl = L. Theorem.1.9 Squeeze Theorem. Let : [, b] R. Then R [, b] i nd only i or every >, there exists untions g, h R [, b] with nd suh tht h g. gx x hx or ll x [, b] Proo. The proo is similr to the proo in Theorem 1..3 nd so will be omitted. Theorem.1.1 Additivity Theorem. Let : [, b] R, let [, b]. Then R [, b] i nd only i is Generlised Riemnn Integrble on both [, ] nd [, b] nd = Proo. = Suppose = L 1 nd = L. Then given >, there exists guge δ 1 on [, ] suh tht i tgged prtition P 1 is δ 1 -ine on [, ], then S 1 ; P 1 L 1 < Similrly, there exists guge δ on [, b] suh tht i tgged prtition P is δ -ine on [, b], then S ; P L < Deine guge δ on [, b] by + min { δ 1 t, 1 t} i t [, δt = min {δ 1, δ } i t = min { δ t, 1 t} i t, b] Using this guge will ore the δ-ine prtition to hve hs tg or ny subintervl ontining. We will show tht i Q is δ-ine prtition o [, b], then there exists δ 1 -ine prtition Q 1 on [, ] nd δ -ine prtition Q on [, b] suh tht S ; Q = S 1 ; Q 1 + S ; Q.3
Deinition nd Min Properties i Cse 1. I is prtition point on Q, then it belongs to two subintervl o Q nd is the tg or both subintervl. Let Q 1 be the prt o Q in [, ] nd so Q 1 is δ 1 -ine. Similrly, let Q be the prt o Q in [, b] nd so Q is δ -ine. So then S ; Q = t i x i x i 1 = x i [,] t i x i x i 1 + = S ; Q 1 + S ; Q x i [,b] t i x i x i 1 ii Cse. I is not prtition point in Q = {[x i 1, x i ], t i } n, then s Q is δ-ine, is ored to hve s its tgs in one o its intervls, sy [x k 1, x k ]. Then s x k x k 1 = x k 1 + x k, we n split subintervl [x k 1, x k ], to [x k 1, ], nd [, x k ], nd so using similr methods with se i, we split Q to Q 1, δ 1 -ine prtition in [, ] nd Q, δ -ine prtition in [, b] nd so S ; Q = S ; Q 1 + S ; Q Using Eqution.3, nd the tringle inequlity, S ; Q L 1 + L = S ; Q 1 L 1 + S ; Q L S ; Q S 1 L 1 + ; Q L < + = Thereore, we onlude tht R [, b] nd = = Let R [, b] nd given >, there exists guge δ suh tht i ny tgged prtitions P, Q re δ-ine, then S ; P + Q S ; < Then let P, Q be δ-ine prtitions on [, ]. By dding the sme prtition points nd tgs on [, b] to both P nd Q, we n extend both prtitions to δ-ine prtitions P nd Q o [, b] nd S ; P S ; Q = S ; P S ; Q So S ; P S ; S Q = ; P Q S ; < nd so the prtitions P nd Q o [, ] stisies the Cuhy Criterion. We then repet or the intervl [, b] nd so we hve proved tht R [, ] nd R [, b]. To prove tht = we use the t tht R [, ] with integrl L 1 nd tht R [, b] with integrl L nd we pply the = prt o the proo tht = L 1 + L = + We will now begin with the Fundmentl Theorem o Clulus or Generlised Riemnn Integrl. Notie tht the irst orm is stronger thn tht o the ordinry Riemnn Integrl. Theorem.1.11 Fundmentl Theorem o ClulusFirst Form. Suppose there exists ountble set E [, b], untion, F : [, b] R suh tht +
Deinition nd Min Properties 1 1 F is ontinuous on [, b]. F x = x or ll x [, b] \ E i.e. F x = x or lmost ll x [, b] exept or inite number o points ontined in set E. Then R [, b] nd Remrks: = F b F 1. Here, ompred to the Fundmentl Theorem o Clulus or ordinry Riemnn Integrl, R [, b] is onlusion insted o ondition. Proo. We will proo or the se E =, so we ssume holds or ll x [, b]. Given >, we use guge δt > or t [, b] suh tht i < z t δt or z [, b], then F z F t t z t < b Thereore, we hve F z F t tz t z t b or z [t δt, t + δt] [, b]. Now, or ll subintervl [x i 1, x i ], t i o δt-ine prtition, we hve t i [x i 1, x i ] [t i δt i, t i + δt i ]. I we onsider F x i F x i 1 t i x i x i 1, then we get F x i F t i t i x i t i + F t i F x i 1 + t i t i x i 1 F x i F t i t i x i t i + F x i 1 F t i t i x i 1 t i x i t i b + x i 1 t i = x i t i b b + t i x i 1 = x i x i 1 b b So i t i [x i 1, x i ] [t i δt i, t i + δt i ], then F x i F x i 1 t i x i x i 1 x i x i 1 b Let prtition P = {[x i 1, x i ], t i } n to be δ-ine, then t i [x i 1, x i ] [t i δt i, t i + δt i ] or ll i n, i N. Then F x i F x i 1 = F b F So, S ; P F b F = F b F S ; P n = F x i F x i 1 t i x i x i 1 F x i F x i 1 t i x i x i 1 Thereore, we onlude tht R [, b] nd = F b F. 1 b x b i x i 1 < b = Theorem.1.1 Fundmentl Theorem o ClulusSeond Form. Let R [, b]. Let F be the indeinite integrl o suh tht Then we hve F x = x tdt or x [, b]
Deinition nd Min Properties 1 F is ontinuous on [, b]. There exists null set E suh tht i x [, b] \ E, then F is dierentible t x nd F x = x i.e. F is dierentible t x or lmost ll x [, b] with F x = x exept or inite number o points ontined in set E. 3 I is ontinuous t [, b], then F = Proo. Proo omitted Theorem.1.13 Substitution Theorem. Let I = [, b] nd J = [, d]. Let F : I R nd α : J R be ontinuous untion with αj I. Suppose there exists E I nd E α J suh tht x = F x or x I \ E nd α t exists or t J \ E α, nd tht E = α 1 E E α is ountble. I we ix x = or ll x E nd α t = or t E α. We n onlude tht R αj, tht α α R J nd tht d αd αα = F α d = Proo. Sine α is ontinuous on J, then the rnge o α, αj, is losed intervl in I. Also, α 1E is ountble, so E αj = αα 1 E is lso ountble. As x = F x or ll x αj \ E, then the Fundmentl Theorem o Clulus implies tht R αj nd αd α α = F αd α = F αd F α I t J \ E, then t J \ E α nd αt I \ E, then the Chin Rule implies tht F α t = F αt α t = αt α t or t J \ E Sine E is ountble, the Fundmentl Theorem o Clulus implies tht α α t R J nd d α α = F α d = F αd F α Theorem.1.14 Multiplition Theorem. I R [, b] nd i g : [, b] R is monotone, then g R [, b] Remrks: 1. Note tht the produt o Generlised Riemnn Integrble untions is not neessrily Generlised Riemnn Integrble. Proo. Proo omitted. Theorem.1.15 Integrtion By Prts. Let F nd G be dierentible on [, b], then F G R [, b] i nd only i F G R [, b]. We hve F G = F G b F G Proo. Proo omitted.
Improper nd Lebesgue Integrl 3. Improper nd Lebesgue Integrl Consider h : [, 1] R suh tht hx = { x 1 i < x 1 i x = We n see tht it is unbounded on the let end point. But, h is still belongs to R[γ, 1] or every γ [, 1]. So, we deine the improper Riemnn Integrl o h on [,1] to be 1 1 1 1 dx = lim dx x γ + γ x We lso hve the sme sitution with k : [, 1] R with { sin 1 kx = x i < x 1 i x = Note tht here we re not neessrily deling with Generlised Riemnn Integrl in these ses. Theorem..1 Hke s Theorem. I : [, b] R, then R [, b] i nd only i or ll γ [, b], R [, γ] nd lim = A R.4 γ b In this se, = A. Remrks: 1. The theorem implies tht the Generlised Riemnn Integrl nnot be extended by tking limits. So i Eqution.4 holds, lredy belongs to R [, b].. We n know whether untion is integrble on [, b] just by looking on subintervl [, γ] with γ < b. So this is just nother tool to show tht untion is Generlised Riemnn Integrl on [, b]. 3. We n use Eqution.4 to evlute the integrl o untion. Exmple... Let k=1 k suh k R or ll k N be series onverging to A R. Let : [, 1] R suh tht 1 = k = A nd then R [, 1]. x = { k k i k 1 x < k i x = 1 k=1 where k = 1 1 or ll k N, k Proo. Clerly, we n see tht on [, γ] or γ [, 1] is step untion nd so is integrble. So, by summing the res, = k k k k 1 + r γ = k=1 k=1 k k 1 k + 1 k 1 + r γ = k k 1 k + r γ = k=1 k + r γ where r γ n+1. But s the series is onvergent, lim n = nd so r γ s n beomes lrge enough nd n so, lim = lim k = A γ 1 n So R [, b]. Exmple..3. We will use k=1 k nd rom Exmple... I k=1 k onverges k=1 k onverge bsolutely, then onverge. k=1 k=1
4 Improper nd Lebesgue Integrl b I k=1 k does not onverges k=1 k onverge onditionly, then does not onverge. Proo. We just pply Exmple.. to k=1 k nd. Below, we re going to give the deinition o lss o Riemnn Integrble untion - The Lebesgue Integrble Funtion. Deinition..4 Lebesgue Integrble Funtions. Let R [, b] suh tht R [, b]. Then is sid to be Lebesgue Integrble on [, b]. The olletion o ll Lebesgue Integrble untions on [, b] is denoted by L[, b] Remrk: 1. Note tht this is not the stndrd deinition o Lebesgue Integrble Funtions. However, by using the set o Generlised Riemnn Integrble untions, it is esier to prove i untion is Lebesgue Integrble. Riemnn/Generlised Riemnn Integrl Lebesgue Integrl Theorem..5 Comprison Test. I, g R [, b] nd x gx or ll x [, b], then L[, b] nd g.5 Proo. We re going to omit the proo or L[, b] nd we re just going to prove Eqution.5. So we note tht. Theorem.1.7 implies tht nd so Similrly, s x gx, g Theorem..6. I, g L[, b] nd i R, then, + g L[, b] nd = nd + g Proo. x = x or ll x [, b]. I R [, b], then, R [, b]. So L[, b]. By Tringle Inequlity, x + gx x + gx or ll x [, b]. Sine + g R [, b], by omprison test Theorem..5, we hve + g L[, b] nd + g + g = + + g g
Improper nd Lebesgue Integrl 5 Theorem..7. Let R [, b]. Then the ollowing sttement re equivlent. L[, b] b There exists g L[, b] suh tht x gx or ll x [, b] There exists h L[, b] suh tht hx x or ll x [, b] Proo. = b Let g = b = Consider untion g. As g or ll x [, b] nd sine g R [, b], then g = g R [, b] nd so by Deinition..4, g L[, b]. Then just pply Theorem..6 to g = + g. Proo or uses similr method just opposite nd so will be omitted. Theorem..8. I, g L[, b], then mx {, g}, min {, g} L[, b] where { x i x gx or ll x [, b] mx {, g} = gx i gx > x or ll x [, b] nd min {, g} = { x i x gx or ll x [, b] gx i gx < x or ll x [, b] Proo. We n write mx{, g} nd min{, g} s mx{, g} = 1 x + gx + x gx nd min{, g} = 1 x + gx x gx So by Theorem..6, x gx L[, b] nd lso x gx L[, b]. So lso by Theorem..6, mx{, g} = 1 x + gx + x gx nd min{, g} = 1 x + gx x gx lso belong to L[, b]. Theorem..9. Let, g, α, β R [, b]. I α, g α or i β, β g, then mx{, g} nd min{, g} lso belong to R [, b] respetively. Proo. Let α nd g α, then mx{, g} α, then g = mx{, g} g α g As α g nd α g R [, b], α g L[, b]. By omprison test Theorem..5, g L[, b] nd so by Theorem..6, mx{, g} L[, b]. The seond prt o the proo is similr with the proo bove nd so will be omitted. Deinition..1 Seminorm nd Distne. I L[, b], then we deine seminorm o to be = I, g L[, b], then we deine the distne between nd g to be Theorem..11 Properties o Seminorm. dist, g =, g = g i. or ll L[, b]. ii. I x = or ll x [, b], then =.
6 Ininite Integrl iii. I L[, b], R, then =. iv. Let, g L[, b], then + g + g. Proo. The proo is trivil nd so will be omitted. For prt iv. onsider the Tringle Inequlity. Theorem..1 Properties o the Distne Funtion. i. dist, g or ll, g L[, b] ii. I x = gx or ll x [, b], then dist, g = iii. dist, g = distg, or ll, g L[, b] iv. dist, h dist, g + distg, h or ll, g, h L[, b] Proo. Proo is trivil nd so is omitted Theorem..13 Completeness Theorem. The sequene { n } o untion in L[, b] onverge to untion L[, b] i nd only i or ll >, there exists M R suh tht i m, n M then m n = dist m n < Proo. = As { n } onverges, then it is Cuhy Sequene. So, given >, there exists M R, suh tht i m, n M, then m n < b u where u is the mximum vlue m n n hieve or eh m, n. Then, by Theorem..6, m n L[, b]. So And so we re done. = Proo Omitted. dist m, n = m n < u b b u =.3 Ininite Integrl In the elementry lulus ourse, the usul pproh to ind ininite integrls is to deine the integrl on [, s imporoper integrl nd so = lim γ.3.1 Integrl on [, We n deine prtition nd P = {[x, x 1 ], t 1,..., [x n 1, x n ], t n, [x n, ], t n+1 } S; Q = t i x i x i 1 + t n+1 x n As the inl term t n+1 x n is not meningul, we only onsider the irst n turns in whih n is lrge enough. So, or the integrl over [, only del with the irst n subprtition. P = {[x, x 1 ], t i,..., [x n 1, x n ], t n } Note tht the prtition P only lk the subintervl [x n, ], t n+1 suh tht [, = n [x i 1, x i ] [x n,
Ininite Integrl 7 i nd For P to be meningul, we put ondition on x n in whih we deine number d suh tht P is δ, d -ine [x i 1, x i ] [t i δt i, t i + δt i ] or ll t N, t n [x n, [ 1 d, Deinition.3.1. Let : [, R. is Generlised Riemnn Integrble on [, i there exsits A R suh tht or ll > there exists guge δ [, suh tht i P is δ, d -ine tgged prtition in [,, then S ; P A I suie the ondition bove, then R [, nd = A Deinition.3.. Let : [, R, is Lebesgue Integrble is, R [,. I suie the ondition, then L[,. Theorem.3.3 Hke s Theorem. I : [, R, then R [, i nd only i or ll γ,, : [, γ] is Generlised Riemnn Integrble on [, γ] nd In this se, = A Proo. Proo Omitted. lim = A R γ Theorem.3.4. I, g R [,, then + g R [, nd + g = Proo. Given >, there exists guge δ on [, suh tht i tgged prtition P is δ -ine, then S ; P < Similrly, there exist guge δ g on [, suh tht i tgged prtition P is δ g -ine, then S g; P g < So, we use the guge δt = min {δ t, δ g t} or t [, suh tht i tgged prtition P is δ-ine, then S + g; P + g S ; P + S g; P g < + = + g Theorem.3.5. Let : [, R nd,, then R [, i nd only i is Generlised Riemnn Integrl on [, ] nd [,. In this se, = Proo. = As R [,, using Hke s Theorem Theorem.3.3 we get or ll γ,, is integrble on [, γ] nd = lim γ +
8 Ininite Integrl We then pply the dditivity theorem Theorem.1.1, to [, γ] = [, ] [, γ] nd so So we hve = lim = γ = + + lim = γ = We n get the proo just by working bkwrds rom the irst prt o the proo. Exmple.3.6. Let α > 1 nd α x = 1 x α + or x [1,. We show tht α R [1,. Proo. Let y 1,, then α is ontinuous on [1, γ] nd so R [1, γ]. By Fundmentl Theorem o Clulus, we lso hve 1 x α dx = 1 γ 1 α x1 α = 1 [ 1 1 ] α 1 γ α 1 Then, Hke s Theorem implies tht 1 1 dx = lim xα γ 1 1 1 1 1 dx = lim 1 1 xα γ α 1 γ α 1 = 1 α 1 when α > 1 Then by Fundmentl Theorem o Clulus Theorem.1.1, we n show tht R [1, Exmple.3.7. Let x = { sin x x i x, 1 i x = Consider the Generlised Riemnn Integrbility o xdx. Proo. Sine x is ontinuous on [, γ] or ll γ,, R [, γ]. To prove tht xdx hs limit s γ rehes, we let < β < γ nd use integrtion by prts. β xdx xdx = β sin x x dx = os x γ x β β os x x dx Sine os x 1 nd sin x 1, we re going to prove tht the bove terms pproh to s β < γ goes to. Consider the Cuhy sequene { 1 n}. As it is Cuhy sequene, we know tht For ll >, there exists M R suh tht i M < β < γ, then 1 γ 1 β < Then, or ll >, there exists M R suh tht i M < β < γ, then os x γ γ os x x β x dx = os x γ γ os x + x β x dx γ 1 x + β β β β 1 x dx 1 γ 1 β < = Thereore, the Cuhy Criterion pplies nd nd Hke s Theorem implies tht R [,. Theorem.3.8 Fundmentl Theorem. Let E [, suh tht E is ountble nd tht, F : [, R suh tht 1 F is ontinuous on [, nd lim F x exists. x F x = x or ll x,, x E Then R [, nd = lim F x F.6 x Proo. I γ [,, we pply the Fundmentl Theorem to intervl [, γ] to onlude tht R [, γ] nd = F γ F. We then use Hke s Theorem to show tht R [, nd we hve Eqution.6.
Ininite Integrl 9.3. Integrls on, b] In similr wy, we just onsider tgged prtition signiint enough so tht we don t hve to del with the let end point. Let b R nd :, b] R suh tht R, b] then given > there exists guge on, b] onsisting o number d > nd stritly positive untion, b] So tgged prtition P is d, δ-ine i, b] =, x n [x i 1, x i ] So tht nd [x i 1, x i ] [t i δt i, t i + δt i ] or ll i n, i N ], x ], 1d.3.3 Integrl on, Let :, R suh tht R,. Given >, there exists guge on, ontining stritly positive untion on, nd d, d >. So tgged subprtition P = {[x i 1, x i ], t i } n is d, δ, d -ine i n, =, x ] [x i 1, x i ] [x n, suh tht nd [x i 1, x i ] [t i δt i, t i + δt i ] or ll i N, i n ], x ], 1d nd [x n, [ 1 d, Theorem.3.9 Hke s Theorem. I :, R, then R, i nd only i or ll β < γ,, R [β, γ] nd lim = C R β β γ + Theorem.3.1 Fundmentl Theorem. Let E, nd E is ountble nd, F :, R suh tht F is ontinuous, nd lim F x exists. x ± b F x = x or ll x, with x / E Then R, nd = lim F x lim F y x y Theorem.3.11 Comprison Test or Improper Integrl. Let g be non-negtive untion ontinuous on, b, let be untion ontinuous on, b. Suppose I improper integrl g exists, then lso exists Remrk: 1. We llow b = nd/or = x gx or llx, b. This theorem is nlogous to omprison test o series
3 Ininite Integrl Proo. Proo Omitted. Exmple.3.1 The Gmm Funtion. The gmm untion Γ :, R in whih Γt = x t 1 e x dx.7 is n generliztion o the toril untion suh tht it extends to the rel number. Show tht it is Generlised Riemnn Integrble, tht the gmm untion stisies the eqution : nd so dedue tht Γt + 1 = t! or ll t N Γt + 1 = tγt.8 Proo. We irst strt by splitting the integrl into two prts by Additivity Theorem Γt = x t 1 e x dx = For the irst integrl 1 xt 1 e x dx, we note tht 1 x t 1 e x dx + x t 1 e x x t 1 or ll x, t R We lso know tht x t 1 R [, 1] using Hke s Theorem. 1 1 [ x x t 1 dx = lim x t 1 t dx = lim γ + γ + t ] 1 γ 1 x t 1 e x dx [ ] 1 = lim γ + t γt = 1 t t i t > Thereore, by the Comprison Test or Improper Integrls Theorem.3.11, 1 xt 1 e x dx exists. For the seond integrl x t 1 e x dx, onsider gx = x, x = x t+1 e x. We note tht 1 lim gx = lim x x xt+1 e x = As the untion gx onverges, it is bounded. So there exists number M R suh tht whih implies tht gx = x t+1 e x M or ll x 1 x = x t 1 e x Mx It is lso esy to prove tht the untion Mx R [1, nd so the rest o the proo ollows rom the irst integrl nd Hke s Theorem. By Hke s Theorem, s the gmm untion is Generlised Riemnn Integrble, given >, the gmm untion is Generlised Riemnn Integrble on [, γ] nd Using Integrtion by Prts, lim γ Γt + 1 = x t e x dx = lim γ Applying Hke s Theorem gin, we get Γt + 1 = x t e x dx = lim γ x t e x γ + t x t e x dx = t x t e x dx x t 1 e x dx = lim t x t 1 e x dx γ x t 1 e x dx = tγt nd so we re done with the irst prt. To prove the seond prt, we irst ompute Γ1 using Hke s Theorem nd the Fundmentl Theorem o Clulus Γ1 = e x dx = lim e x [ dx = lim e x ] γ = lim γ γ e γ + e = 1 γ Thereore, Γt + 1 = tt 1... Γ1 = t!
Convergene Theorems 31.4 Convergene Theorems We now turn rom untions to sequene o untions in terms o Generlised Riemnn Integrl. Theorem.4.1 Uniorm Convergene Theorem. Let { n } be sequene in R [, b] nd suppose { n } onverges uniormly on [, b] to. Then R [, b] nd Remrks: = lim k k 1. A sequene o bounded untion { n } suh tht n : A R where A R or ll n N onverges uniormly on A A to : A R is or ll >, there exists N N suh tht i n > N, then n x x < or ll x A. Sequene {x n } is Cuhy sequene i or ll >, there exists N N suh tht i n > m > N, then 3. The theorem does not pply or ininite intervl n m < Proo. As { k } onverge uniormly, given >, there exists N N suh tht i n > N, then n x < or ll x [, b] b Similrly, i m > N, then So we get m x x < b or ll x [, b] n x m x = n x x + x m x < n x x + m x x < b This implies Then by Theorem.1.7, b < Thereore, the sequene b < nx m x < b or ll x [, b] or ll x [, b] b b n m < = < n m < = n m < b { } b n is Cuhy Sequene in R nd so onverges to some number A R So, now we just need to show tht R [, b] with integrl = A. For ll >, we use N N s bove. I P = {[x i 1, x i ], t i } n is tgged prtition o [, b] nd i n N, then S n ; P S ; P n = n t i t i x i x i 1 n t i t i x i x i 1 < x i x i 1 = b Fix m N suh tht m A <, let δ be guge on [, b] suh tht i P is δ-ine, then m S m ; P < Then we hve S ; P A S ; P S m ; P + S ; P m + As we n set to be s smll s we wnt, we onlude tht R [, b] nd = A m A < b + + = b +
3 Convergene Theorems Another wy o obtining lim k n is using the Equi-integrbility Theorem disussed below Deinition.4. Equi-integrbility Hypothesis. The sequene { k } in R [, b] is sid to be equi-integrble i or ll >, there exists guuge δ [, b] suh tht i P is δ-ine prtition on [, b] nd n N, then S n ; P b n < Theorem.4.3 Equi-integrbility Theorem. I { n } R [, b] is equi-integrble on [, b] nd i then R [, b] nd x = lim n nx or ll x [, b], = lim n n Proo. For ll >, by Equi-integrbility Hypothesis Deinition.4., there exists guge δ on [, b] suh tht i P = {[x i 1, x i ], t i } n on [, b] is δ-ine, we hve S n ; P b n < or ll n [, b] Sine P hs inite number o tgs nd t = m, n N, then S n ; P S lim nt or ll t [, b], there exists N N suh tht i n m ; P n n t i m t i x i x i 1 b.9 I we tke the limit s m in Eqution.9, we get S n ; P S ; P b or ll n N Also, using m, n N, Deinition.4. nd Eqution.9, we get b n m = n S n ; P + S n ; P S m ; P + S S n ; P n + S n ; P S m ; P + S m ; P m ; P m m + b + = + b { } b n is Cuhy sequene onverging to some number Sine we n set > to be s smll s we wnt, { } b A R. As n onverges, we write or ll >, there exists N N suh tht i n > N, then n A As we know tht sequene { n} onverge to some number A R, we just need to show tht R [, b] with its integrl equl to the sme number A. So, given >, there exists guge δ on [, b] suh tht i P is δ-ine on [, b] nd n N, then S ; P A = S ; P S n ; P + S n ; P n + n A S ; P S n ; P + S n ; P n + n A b + + = + b As we n set to be s smll s we wnt, we onlude tht R [, b] with = A
Convergene Theorems 33 Theorem.4.4 Monotone Convergene Theorem. Let { k } be monotone sequene o untion in R [, b] suh tht x = lim nx lmost everywhere in [, b], then R [, b] i nd only i { n n} is bounded in R nd i the onditition is stisied, Proo. Proo Omitted = lim n n Theorem.4.5 Dominted Convergene Theorem. Let { n } be sequene in R [, b]. Let x = lim nx n lmost everywhere on [, b]. I there exists g, h R [, b] suh tht gx n x hx or lmost every x [, b], then R [, b] nd = lim n n Moreover, i g, h L[, b], then n, L[, b] nd k = k Proo. Proo Omitted.
34 3 REFERENCE 3 Reerene 1. Brtle, Robert G. & Sherbert, Donld R.198. Introdution to Rel Anlysis - 4th edition