# Review of Riemann Integral

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1 1 Review of Riemnn Integrl In this chpter we review the definition of Riemnn integrl of bounded function f : [, b] R, nd point out its limittions so s to be convinced of the necessity of more generl integrl. 1.1 Definition nd Some Chrcteriztions Let f : [, b] R be bounded function. The ide of Riemnn integrl of f is to ssocite unique number γ to f such tht, in cse f(x) 0 for x [, b], then γ cn be thought of s the the re of the region R f bounded by the grph of f, x-xis, nd the ordintes t nd b. For this purpose, first we consider prtition P of [, b], tht is, finite set P := {x i : i = 0, 1,..., n} such tht nd consider the sums = x 0 < x 1 < x 2 <... < x k = b, L(P, f) := m i x i, U(P, f) := M i x i, where, for i = 1,..., k, x i = x i x i 1, m i = inf{f(x) : x i 1 x x i } nd M i = sup{f(x) : x i 1 x x i }. The quntities L(P, f) nd U(P, f) re clled the lower sum nd upper sum ssocited with (P, f). Note tht if f(x) 0 for ll x [, b], then L(P, f) is the totl re of the rectngles with lengths m i nd widths x i x i 1, nd U(P, f) is the totl re of the rectngles with lengths M i nd widths x i x i 1, for i = 1,..., k. Thus, it is intuitively cler tht the required re, sy γ, under the grph of f must stisfy the reltion: L(P, f) γ U(P, f) 1

2 2 Review of Riemnn Integrl for ll prtitions P of [, b]. With this requirement in mind, we hve the following definition. Definition A bounded function f : [, b] R is sid to be Riemnn integrble on [, b] if there exists unique γ R such tht L(P, f) γ U(P, f) for ll prtitions P of [, b], nd in tht cse γ is clled the Riemnn integrl of f nd it is denoted by f(x)dx. Let P be the set of ll prtitions of [, b]. Clerly L(P, f) U(P, f) P P. Denoting m = inf{f(x) : x b}, M = sup{f(x) : x b}, for ny prtition P = {x i }, we hve nd L(P, f) = L(P, f) = m i x i M i x i m x i = m(b ) M x i = M(b ). Thus, the set {L(P, f) : P P} is bounded bove by M(b ), nd the set {U(Q, f) : Q P} is bounded below by m(b ). Hence, α(f) := sup{l(p, f) : P P} M(b ), β(f) := inf{u(p, f) : P P} m(b ) exist s rel numbers. Using the quntities α(f) nd β(f), we hve the following chrcteriztion of Riemnn integrbility. Theorem A bounded function f : [, b] R is Riemnn integrble on [, b] if nd only if α(f) = β(f). Proof. Suppose f : [, b] R is Riemnn integrble on [, b], nd let γ be the Riemnn integrl of f, i.e., L(P, f) γ U(P, f) P P. (1)

3 Definition nd Some Chrcteriztions 3 Thus, α(f) γ β(f). Consequently, L(P, f) α(f) γ β(f) U(P, f) P P. Now, since γ is the only number stisfying (1), we hve α(f) = γ = β(f). Conversely, suppose α(f) = β(f). Then we hve L(P, f) α(f) = β(f) U(P, f) P P. Thus, γ := α(f) = β(f) stisfies (1). Suppose there is nother number, sy γ R stisfying L(P, f) γ U(P, f) P P. (2) Without loss of generlity, ssume tht γ < γ. Then, from (2), we hve α(f) γ β(f). It then follows tht γ = γ. Thus, we hve proved tht there exists unique γ R stisfying (1). Remrk As you would hve observed, the proof of Theorem ws very esy. We gve its proof in detil, minly becuse of the fct tht in stndrd books, the Riemnn integrbility of bounded function f : [, b] R defined by requiring it to stisfy α(f) = β(f). Given two prtitions P 1 nd P 2 of [, b], we cn consider new prtition P by using ll the prtition points of P 1 nd P 2, tking repeted points only once. Such prtition will be clled the prtition obtined by combining P 1 nd P 2, nd is usully denoted by P 1 P 2. The chrcteriztion given in Theorem below is useful in deducing mny properties of Riemnn integrl. For its proof, we shll use the observtion tht if P nd Q re ny two prtitions of [, b], then L(P, f) U(Q, f), nd consequently, α(f) β(f). Indeed, given ny two prtitions P nd Q of [, b], we hve so tht α(f) β(f). L(P, f) L(P Q, f) U(P Q, f) U(Q, f),

4 4 Review of Riemnn Integrl Theorem Let f : [, b] R be bounded function. Then f is Riemnn integrble if nd only if for every ε > 0, there exists prtition P of [, b] such tht U(P, f) L(P, f) < ε. Proof. Suppose f is Riemnn integrble nd let ε > 0 be given. By the definition of α(f) nd β(f), there exists prtition P 1 nd P 2 of [, b] such tht α(f) ε/2 < L(P 1, f) nd U(P 2, f) β(f) + ε/2. Let P be the prtition obtined by combining P 1 nd P 2. Then, we hve α(f) ε/2 < L(P 1, f) L(P, f) U(P, f) U(P 2, f) β(f) + ε/2. Since α(f) = β(f) (cf. Theorem 1.1.1), it follows tht U(P, f) L(P, f) [β(f) + ε/2] [α(f) ε/2] = ε. Conversely, suppose tht for every ε > 0, there exists prtition P of [, b] such tht U(P, f) L(P, f) < ε. Since we hve L(P, f) α(f) β(f) U(P, f), β(f) α(f) U(P, f) L(P, f) < ε. This is true for every ε > 0. Hence, α(f) = β(f), nd hence f is Riemnn integrble. Here is n n immedite consequence of the bove theorem. Corollry A bounded function f : [, b] R is Riemnn integrble if nd only if there exists sequence (P n ) of prtitions of [, b] such tht U(P n, f) L(P n, f) 0 s n, nd in tht cse the sequences (U(P n, f)) nd (L(P n, f)) converge to the sme limit f(x)dx. Remrk In Theorem we hve used the following fct: If, b re rel numbers such tht < b + ε for ll ε > 0, then b. Indeed, if the conclusion is not true, then would hve > b, nd in tht cse we cn find n ε > 0 such tht b + ε, which would led to contrdiction to the hypothesis tht < b + ε for ll ε > 0. Thus, in order to show tht b, one my show tht b + ε for ll ε > 0. This procedure will be dopted t mny occsion in the due course.

5 Definition nd Some Chrcteriztions 5 Next we give nother chrcteriztion of Riemnn integrbility. For tht purpose we introduce the following definition. Definition Let P = {x i } n i=0 be prtition of [, b] nd let T P := {ξ i } n with ξ i [x i 1, x i ], i = 1,..., n. The set T P is clled tg set corresponding to the prtition P nd the sum S(P, T P, f) := n f(ξ i ) x i is clled the Riemnn sum of f ssocited with the prtition P nd the tg set T P. The quntity P := mx{x i x i 1 : i = 1,..., n} is clled the mesh of the prtition P. It is obvious tht L(P, f) S(P, T P, f) U(P, f) for ny prtition P nd the tg set T P. Therefore, we hve the following result. Theorem If bounded function f : [, b] R is Riemnn integrble, then for every ε > 0, there there exists prtition P such tht S(P, T P, f) f(x)dx < ε for every tg set T P. In fct, the following chrcteriztion of the Riemnn integrbility holds (see Ghorpde nd Limye [2]) Theorem A function f : [, b] R is Riemnn integrble if nd only if there exists γ R stisfying the following condition: For every ε > 0, there exists δ > 0 such tht for every prtition P of [, b] with with P < δ nd for every tg set T P of P, we hve S(P, T P, f) γ < ε Corollry Suppose f : [, b] R is Riemnn integrble function. If (P n ) is prtition of [, b] such tht P n 0 s n nd if T n is tg on P n for ech n N, then S(P n, T n, f) f(x)dx s n.

6 6 Review of Riemnn Integrl 1.2 Advntges nd Some Disdvntges We my observe, in view of Corollry 1.1.3, tht if (P n ) is sequence of prtitions of [, b] such tht (U(P n, f)) nd (L(P n, f)) converge to the sme limit sy γ, then f is Riemnn integrble, nd γ = f(x)dx. If f : [, b] R is continuous function, then it cn be shown tht for ny sequence (P n ) of prtitions of [, b] stisfying P n 0 s n, we hve U(P n, f) L(P n, f) 0 s n. Thus, by Corollry 1.1.3, we cn conclude the following: Every continuous function f : [, b] R is Riemnn integrble. The following results re lso true (See Ghorpde nd Limye [2] or Rudin [4]): 1. Every bounded function f : [, b] R hving tmost finite number of discontinuities is Riemnn integrble. 2. Every monotonic function f : [, b] R is Riemnn integrble. Thus, the set of ll Riemnn integrble functions is very lrge. In fct we hve the following theorem, known s Lebesgue s criterion for Riemnn integrbility, whose proof depends on some techniques involving the concepts of oscilltion of function on subset nd t point. For the proof one my refer Delninger [1]. Lebesgue s criterion for Riemnn integrbility: A bounded function f : [, b] R is Riemnn integrble if nd only if the set of points t which f is discontinuous is of mesure zero. Here, the concept of set of mesure zero is used in the sense of the following definition. Definition A set E R is sid to be of mesure zero if for every ε > 0, there exists countble fmily {I n } of open intervls such tht E n I n nd n l(i n) < ε, where l(i n ) is the length of the intervl I n. Recll tht, if I is n intervl with endpoints, b R, then length of I, denoted by l(i), is b.

7 Advntges nd Some Disdvntges 7 Exmple It cn be esily verified tht every finite subset of R is of mesure zero. In fct, every countble subset of R is of mesure zero. To see this, consider countble set E = { n : n Λ}, where Λ is {1,..., k} for some n N or Λ = N. For ε > 0, let I n := ( n ε/2 n+1, n + ε/2 n+1 ), n Λ. Then E n Λ I n nd n Λ l(i n ) = n Λ (ε/2 n ) ε. Cn n uncountble set be of mesure zero? question ffirmtively in the next chpter. We shll nswer this Functions with only finite number of discontinuity in [, b] re plenty. Here is n exmple with infinite number of discontinuities. Exmple Let I = [, b], S := { n : n N} I nd let f : I R be defined by f( n ) = 1/n for ll n N nd f(x) = 0 for x S. Clerly, this function is not continuous t every x S. We show tht f is continuous t every x I \ S. Let x 0 I \ S. Then f(x 0 ) = 0. For ε > 0, we hve to find δ > 0 such tht x x 0 < δ implies f(x) < ε. For δ > 0, let J δ := (x 0 δ, x 0 + δ) I. For ε > 0, let k N be such tht 1/k < ε. Choose δ > 0 such tht 1, 2,..., k J δ. For instnce, we my choose 0 < δ < min{ x 0 i : i = 1,..., k}. Then we hve J δ { 1, 2,...} = { k+1, k+2,...}. Hence, for x J δ, we hve either f9x) = 0 or f(x) = 1/n for some n > k. Thus, f(x) 1 k < ε. Thus we hve proved tht f is continuous t x 0. In the bove, if we tke S to be the set of ll rtionl numbers in I, then the corresponding function f is continuous t every irrtionl number in I nd discontinuous t every rtionl number in I; nd consequently, f is continuous except on set of mesure zero. Although the set set of Riemnn integrble functions on [, b] is quite lrge, this clss is short of some desirble properties. For exmple observe the following drw bcks of Riemnn integrbility nd Riemnn integrtion:

8 8 Review of Riemnn Integrl If (f n ) is sequence of Riemnn integrble functions on [, b] nd if f n (x) f(x) s n for every x [, b], then it is not necessry tht f is Riemnn integrble. Even if the function f in the bove is Riemnn integrble, it is not necessry tht f n(x)dx f(x)dx s n. To illustrte the lst two sttements consider the following two exmples: Exmple Let {r 1, r 2,...} be n enumertion of the set rtionl numbers in [0, 1]. For ech n N, let { 0 if x {r1,..., r f n (x) = n } 1 if x {r 1,..., r n }. Then ech f n is Riemnn integrble, s it is continuous except t finite number of points. Note tht f n (x) f(x) s n for every x [0, 1], where f : [0, 1] R is the Dirichlet s function function defined by { 0 if x rtionl f(x) = 1 if x irrtionl. Note tht this function is not Riemnn integrble since for ny prtition P of [, b], we hve L(P, f) = 0 nd U(P, f) = 1, so tht α(f) = 0 nd β(f) = 1. Hence, f is not Riemnn integrble. Thus, though 1 0 f n(x)dx = 1 for every n N nd f n (x) f(x) for every x [0, 1], we cnnot even tlk bout the integrl of f. Exercise 1.1 Cn we ssert the fct tht f in Exmple is not Riemnn integrble using the Lebesgue s criterion of Riemnn integrbility? Exmple For n N, let f n : [0, 1] R be defined by f n (x) = nχ (0,1/n] (x), x [0, 1], where χ E denotes the chrcteristic function of E, tht is, { 1 if x E, χ E (x) = 0 if x E. Then we see tht f n (x) f(x) 0 s n, but 1 0 f n(x) dx = 1 for every n N. Hence f n(x)dx f(x)dx. Exmple Consider f n : [0, 1] R defined by { ne nx if x (0, 1] f n (x) = 0 if x = 0. Then we hve f n (x) f(x) = 0 s n for every x [0, 1]. Note tht 1 0 f n(x)dx = 1 e n 1 s n. Thus, 1 0 f n(x)dx 1 0 f(x)dx.

9 Advntges nd Some Disdvntges 9 In Exmples nd 1.2.5, we see tht, lthough the sequence (f n (x)) converges for ech x [0, 1], it is not uniformly bounded, tht is, there does not exist n M > 0 such tht f n (x) M for ll x [, b] nd for ll n N. In fct, in both the exmples, the sequence is (f n (1/n)) is unbounded. As consequence of our generl theory, we shll deduce the following known result in the theory of Riemnn integrtion (cf. Rudin [4]): If (f n ) is uniformly bounded sequence of Riemnn integrble functions defined on [, b] such tht f n (x) f(x) s n for ech x [, b] for some Riemnn integrble function f, then f n(x)dx f(x)dx s n. So, the the crux of the mtter is the following: If (f n ) is uniformly bounded sequence of Riemnn integrble functions defined on [, b] such tht f n (x) f(x) s n for ech x [, b], cn we hve new type of integrl which generlizes the Riemnn integrl, sy g S(g), such tht S(f) exists nd S(f n ) S(f)? This is wht we re going to do in this course. The new type of integrl will be clled the Lebesgue integrl. In order to do this we hve to introduce the concept of mesure of subset of R which is generliztion of the concept of length of n intervl. We end this chpter with remrk on how to compute the Riemnn integrl in certin specil cse. Recll tht, in school, integrtion is defined s inverse of differentition. More precisely, if f : [, b] R is such tht there is continuous function g : [, b] R which is differentible on (, b), nd if f(x) = g (x) for ll x [, b], then f(x)dx := g(b) g(). In the bove, it is tcitly ssumed tht f is Riemnn integrble. But, derivtive of function need not be Riemnn integrble s the following exmple shows. Exmple Let g(x) = { x 2 cos π x 2 if 0 < x 1 0 if x = 0. Then we see tht g is differentible nd { g 2x cos π + 2π (x) = x 2 x sin π if 0 < x 1 x 2 0 if x = 0. ( )

10 10 Review of Riemnn Integrl Note tht g is not bounded, nd hence it is not Riemnn integrble. However, if f is Riemnn integrble, then ( ) provides formul to compute the Riemnn integrl. This fct is known s the fundmentl theorem of Riemnn integrtion. Let us estblish this fct. Theorem Let f : [, b] R be Riemnn integrble function such tht there is continuous function g : [, b] R which is differentible on (, b), nd f(x) = g (x) for ll x [, b]. Then, f is Riemnn integrble nd f(x)dx = g(b) g(). Proof. Consider prtition P : = x 0 < x 1 <... < x n = b of [, b]. Then, by Men Vlue Theorem (cf. Ghorpde nd Limye [2]), there exists ξ (x i 1, x i ) such tht g(x i ) g(x i 1 ) = g (ξ)(x i x i 1 ) = f(ξ i )(x i x i 1 ) for i = 1,..., n. Hence, with := {ξ i }, S(f, P, ) = n f(ξ i )(x i x i 1 ) = n [g(x i ) g(x i 1 )] = g(b) g(). Since f is Riemnn integrble, by Theorem 1.1.5, the Riemnn integrl of f is g(b) g(). Remrk Observe tht in the proof of Theorem 1.2.1, the Riemnn sum S(f, P, ) is independent of the prtition P. Therefore, one my think, in view of Theorem 1.1.5, tht we need not impose the Riemnn integrbility of f s hypothesis in Theorem But, if you look t the line rguments in the proof, you see tht the prtition P is rbitrry, but the set of tgs on P is not rbitrry, wheres in the sttement of Theorem the set of tgs lso must be rbitrry.

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