The Riemann Integral. Chapter Definition of the Riemann integral

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1 Chpter The Riemnn Integrl now of some universities in Englnd where the Lebesgue integrl is tught in the first yer of mthemtics degree insted of the Riemnn integrl, but now of no universities in Englnd where students lern the Lebesgue integrl in the first yer of mthemtics degree. (Approximte quottion ttributed to T. W. Körner) Let f : [, b] R be bounded (not necessrily continuous) function on compct (closed, bounded) intervl. We will define wht it mens for f to be Riemnn integrble on [, b] nd, in tht cse, define its Riemnn integrl b f(x) dx. The integrl of f on [, b] is rel number whose geometricl interprettion is the signed re under the grph y = f(x). The Riemnn integrl is the simplest integrl to define, nd it llows one to integrte every continuous function s well s some not-too-bdly discontinuous functions. It isn t the only integrl nd there re mny others, the most importnt of which is the Lebesgue integrl. The Lebesgue integrl llows one to integrte unbounded or highly discontinuous functions whose Riemnn integrls do not exist, nd it hs better mthemticl properties thn the Riemnn integrl. The definition of the Lebesgue integrl, however, requires the use of mesure theory, which we will not describe here. In ny event, the Riemnn integrl is dequte for mny purposes, nd even if one needs the Lebesgue integrl, it s better to understnd the Riemnn integrl first... Definition of the Riemnn integrl We sy tht two intervls re lmost disjoint if they re disjoint or intersect only t common endpoint. For exmple, the intervls [0, ] nd [, 3] re lmost disjoint, wheres the intervls [0, 2] nd [, 3] re not.

2 2. The Riemnn Integrl Definition.. Let I be compct intervl of nonzero length. A prtition of I is finite collection { : k =,..., n} of lmost disjoint, compct subintervls of nonzero length whose union is I. A prtition of [, b] with subintervls = [x k, x k ] is determined by the set of endpoints of the intervls = x 0 < x < < x n = b. Abusing nottion, we will denote prtition P either by its intervls or by its endpoints P = {I, I 2,..., I n } P = {x 0, x,..., x n }. We ll dopt either nottion s convenient; the context should mke it cler which one is being used. Exmple.2. The set of intervls {[0, /5], [/5, /4], [/4, /3], [/3, /2], [/2, ]} is prtition of [0, ]. The corresponding set of endpoints is {0, /5, /4, /3, /2, }. We denote the length of n intervl I = [, b] by I = b. Note tht the sum of the lengths = x k x k of the lmost disjoint subintervls in prtition { : k =,..., n} of n intervl I is equl to length of the whole intervl. This is obvious geometriclly; lgebriclly, it follows from the telescoping series = (x k x k ) = x n x n + x n x n x 2 x + x x 0 = x n x 0 = I. Suppose tht f : [, b] R is bounded function on the compct intervl I = [, b] with M = sup f, m = inf f. I I If P = { : k =,..., n} is prtition of I with endpoints {x 0, x,..., x n }, let M k = sup f, m k = inf f. These suprem nd infim re well-defined nd finite since f is bounded. Moreover, m m k M k M. If f is continuous on I, then it is bounded nd ttins its mximum nd minimum vlues on ech intervl, but bounded discontinuous function need not ttin its supremum or infimum.

3 .. Definition of the Riemnn integrl 3 We define the upper Riemnn sum of f with respect to the prtition P by U(f; P ) = M k = M k (x k x k ) nd the lower Riemnn sum of f with respect to the prtition P by L(f; P ) = m k = m k (x k x k ). Geometriclly, U(f; P ) is the sum of the res of rectngles bsed on the intervls tht lie bove the grph of f, nd L(f; P ) is the sum of the res of rectngles tht lie below the grph of f. Note tht m(b ) L(f; P ) U(f; P ) M(b ). Let Π(, b), or Π for short, denote the collection of ll prtitions of [, b]. The set of ll upper Riemnn sums {U(f; P ) : P Π} is bounded from below by m(b ), nd we cn therefore define the upper Riemnn integrl of f on [, b] by U(f) = inf U(f; P ). P Π Similrly, the set of ll lower Riemnn sums {L(f; P ) : P Π} is bounded from bove by M(b ), nd we define the lower Riemnn integrl of f on [, b] by L(f) = sup L(f; P ). P Π Note the use of lower-upper nd upper-lower pproximtions for the integrls: we tke the infimum of the upper sums nd the supremum of the lower sums. We will show in Corollry.3 below tht we lwys hve L(f) U(f), but they need not be equl. Definition.3. A bounded function f : [, b] R is Riemnn integrble on [, b] if its upper nd lower Riemnn integrls re equl, mening tht U(f) = L(f). In tht cse, the Riemnn integrl of f on [, b], denoted by b b f(x) dx, f, f [,b] or similr nottions, is the common vlue of U(f) nd L(f). Let us illustrte the definition of Riemnn integrbility with number of exmples. Exmple.4. Define f : [0, ] R by { /x if 0 < x, f(x) = 0 if x = 0.

4 4. The Riemnn Integrl Then 0 x dx isn t defined s Riemnn integrl becuse f is unbounded. In fct, if is prtition of [0, ], then 0 < x < x 2 < < x n < sup f =, [0,x ] so the Riemnn sums of f re not well-defined. An integrl with n unbounded intervl of integrtion, such s x dx, lso isn t defined s Riemnn integrl. In this cse, Riemnn sum ssocited with prtition of [, ) into intervls of finite length (for exmple, = [k, k + ] with k N) is n infinite series rther thn finite sum, leding to questions of convergence. One cn interpret these integrls s limits of Riemnn integrls, or improper Riemnn integrls, 0 dx = lim x 0 + x dx, dx = lim x b x dx. Such improper Riemnn integrls involve two limits limit of Riemnn sums to define the Riemnn integrls, followed by limit of Riemnn integrls nd they re not proper Riemnn integrls in the sense of Definition.3. Both of the improper integrls in this exmple diverge to infinity. Next, we consider some exmples of bounded functions on compct intervls, where Definition.3 does pply. Exmple.5. The constnt function f(x) = on [0, ] is Riemnn integrble, nd 0 dx =. To show this, let P = {I,..., I n } be ny prtition of [0, ] with endpoints Since f is constnt, nd therefore {x 0, x,..., x n }. M k = sup f =, m k = inf f = for k =,..., n, U(f; P ) = L(f; P ) = (x k x k ) = x n x 0 =. Geometriclly, this eqution is the obvious fct tht the sum of the res of the rectngles over (or, equivlently, under) the grph of constnt function is lwys

5 .. Definition of the Riemnn integrl 5 exctly equl to the re under the grph. Thus, every upper nd lower sum of f on [0, ] is equl to, which implies tht the upper nd lower integrls U(f) = inf U(f; P ) = inf{} =, P Π re equl, nd the integrl of f is. L(f) = sup L(f; P ) = sup{} = P Π More generlly, the sme rgument shows tht every constnt function f(x) = c is integrble nd b c dx = c(b ). The following is n exmple of discontinuous function tht is Riemnn integrble. Exmple.6. The function is Riemnn integrble, nd f(x) = { 0 if 0 < x if x = 0 0 f dx = 0. To show this, let P = {I,..., I n } be prtition of [0, ]. Then, since f(x) = 0 for x > 0, M k = sup f = 0, m k = inf f = 0 for k = 2,..., n. The first intervl in the prtition is I = [0, x ], where 0 < x, nd M =, m = 0, since f(0) = nd f(x) = 0 for 0 < x x. It follows tht Thus, L(f) = 0 nd U(f; P ) = x, L(f; P ) = 0. U(f) = inf{x : 0 < x } = 0, so U(f) = L(f) = 0 re equl, nd the integrl of f is 0. Note tht in this exmple, the infimum of the upper Riemnn sums is not ttined nd U(f; P ) > U(f) for every prtition P. A similr rgument shows tht if function f : [, b] R is zero except t finitely mny points in [, b], then it is Riemnn integrble nd its integrl is 0. The next exmple shows tht the Riemnn integrl cn fil to exist even for bounded functions on compct intervl. Exmple.7. The Dirichlet function f : [0, ] R defined by { if x Q [0, ] f(x) = 0 if x [0, ] \ Q is not Riemnn integrble. If P = { : k n} is prtition of [0, ], then M k = sup f =, m k = inf = 0,

6 6. The Riemnn Integrl since every intervl of non-zero length contins both rtionl nd irrtionl numbers. It follows tht U(f; P ) =, L(f; P ) = 0 for every prtition P of [0, ], so U(f) = nd L(f) = 0 re not equl. We remrk tht the Dirichlet function is Lebesgue integrble. integrl is given by 0 f = A + 0 B Its Lebesgue where A = Q [0, ] is the set of rtionl numbers in [0, ] nd B = [0, ] \ Q is the set of irrtionl numbers. Here, E denotes the Lebesgue mesure of the set E, which is countbly dditive extension of the length of n intervl to more generl sets. It turns out tht A = 0 (s is true for ny countble set of rel numbers) nd B =, so the Lebesgue integrl of the Dirichlet function is 0. The morl of the previous exmple is tht the Riemnn integrl of highly discontinuous function need not exist. A precise sttement bout Riemnn integrbility cn be given in terms of Lebesgue mesure. One cn show tht set E R hs Lebesgue mesure zero if nd only if for every ɛ > 0 there is countble collection of open intervls {( k, b k ) : k N} such tht E ( k, b k ), (b k k ) < ɛ. A bounded function on compct intervl is Riemnn integrble if nd only if the set of points t which it is discontinuous hs Lebesgue mesure zero. For exmple, the set of discontinuities of the function in Exmple.6 consists of single point {0}, which hs Lebesgue mesure zero. On the other hnd, the Dirichlet function in Exmple.7 is discontinuous t every point of [0, ], nd its set of discontinuities hs Lebesgue mesure one. From now on, we ll only consider the Riemnn integrl; integrble will men Riemnn integrble, nd integrl will men Riemnn integrl..2. Refinements of prtitions As the previous exmples illustrte, direct verifiction of integrbility from Definition.3 is unwieldy even for the simplest functions becuse we hve to consider ll possible prtitions of the intervl of integrtion. We cn obtin simpler nd more esily verified conditions for Riemnn integrbility by considering refinements of prtitions. Definition.8. A prtition Q = {J l : l =,..., m} is refinement of prtition P = { : k =,..., n} if every intervl in P is n lmost disjoint union of one or more intervls J l in Q Equivlently, if we represent prtitions by their endpoints, then Q is refinement of P if Q P, mening tht every endpoint of P is n endpoint of Q.

7 .2. Refinements of prtitions 7 Exmple.9. Consider the prtitions of [0, ] with endpoints P = {0, /2, }, Q = {0, /3, 2/3, }, R = {0, /4, /2, 3/4, }. Then Q is not refinement of P but R is refinement of P. Given two prtitions, neither need be refinement of the the other. However, two prtitions P, Q lwys hve common refinement; the smllest one is R = P Q, mening tht the endpoints of R re exctly the endpoints of P or Q (or both). Exmple.0. Let P = {0, /2, } nd Q = {0, /3, 2/3, }, s in Exmple.9. Then Q isn t refinement of P nd P isn t refinement of Q. The prtition S = P Q, or S = {0, /3, /2, 2/3, }, is refinement of both P nd Q. The prtition S is not refinement of R, but T = R S, or T = {0, /4, /3, /2, 2/3, 3/4, }, is common refinement of ll of the prtitions {P, Q, R, S}. As we show next, refining prtitions decreses upper sums nd increses lower sums. (The proof is esier to understnd thn it is to write out drw picture!) Proposition.. If f : [, b] R is bounded nd P, Q re prtitions of [, b] such tht Q is refinement of P, then Proof. Let U(f; Q) U(f; P ), L(f; P ) L(f; Q). P = { : k =,..., n}, Q = {J l : l =,..., m} be prtitions of [, b], where the intervls re listed in incresing order of their endpoints. Define M k = sup f, m k = inf f, M l = sup J l f, m l = inf J l f. Since Q is refinement of P, ech intervl in P is n lmost disjoint union of intervls {J l : p k l q k } in Q for some p k, q k, where = p q < p 2 q 2 < p 3 q 3 < < p n q n = n. In prticulr, J l if p k l q k, so M l M k, m k m l for p k l q k. Using the fct tht the sum of the lengths of the J-intervls is the length of the I-intervl, we get tht It follows tht q k U(f; Q) = l=p k M l J l q k m M l J l = l= q k M k J l = M k J l = M k. l=p k l=p k q k l=p k M l J l M k U(f; P )

8 8. The Riemnn Integrl Similrly, nd q k q k m l J l m k J l = m k, l=p k l=p k L(f; Q) == q k l=p k m l J l m k L(f; P ). Using this result, we cn prove tht ll lower sums re less thn or equl to ll upper sums, not just the lower nd upper sums ssocited with the sme prtition. Proposition.2. If f : [, b] R is bounded nd P, Q re ny prtitions of [, b], then L(f; P ) U(f; Q). Proof. Let R be refinement of both P nd Q. Then, by Proposition., It follows tht L(f; P ) L(f; R), U(f; R) U(f; Q). L(f; P ) L(f; R) U(f; R) U(f; Q). An immedite consequence of this resultis tht the lower integrl is lwys less thn or equl to the upper integrl. Corollry.3. If f : [, b] R is bounded, then L(f) U(f). Here we use the following lemm, whose conclusion is obvious but requires proof. Lemm.4. Suppose tht A, B re nonempty sets of rel numbers such tht b for ll A nd b B. Then sup A inf B. Proof. The condition implies tht every b B is n upper bound of A, so sup A b. Hence, sup A is lower bound of B, so sup A inf B..3. Existence of the Riemnn integrl The following theorem gives simple criterion for integrbility. Theorem.5. A bounded function f : [, b] R is Riemnn integrble if nd only if for every ɛ > 0 there exists prtition P of [, b] (depending on ɛ) such tht U(f; P ) L(f; P ) < ɛ.

9 .3. Existence of the Riemnn integrl 9 Proof. First, suppose tht the condition holds. Let ɛ > 0 nd choose prtition P tht stisfies the condition. Then, since U(f) U(f; P ) nd L(f; P ) L(f), we hve 0 U(f) L(f) U(f; P ) L(f; P ) < ɛ. Since this inequlity holds for ll ɛ > 0, we must hve U(f) = L(f), nd f is integrble. Conversely, suppose tht f is integrble, mening tht U(f) = L(f). Given ny ɛ > 0, there re prtitions Q, R such tht U(f; Q) < U(f) + ɛ 2, L(f; R) > L(f) ɛ 2. Let P be common refinement of Q nd R. Then, by Proposition., U(f; P ) L(f; P ) U(f; Q) L(f; R) < U(f) L(f) + ɛ. Since U(f) = L(f), the condition follows. Furthermore, it is sufficient to find sequence of prtitions whose upper nd lower sums pproch ech other. Theorem.6. A bounded function f : [, b] R is Riemnn integrble if nd only if there is sequence (P n ) of prtitions such tht In tht cse, b lim [U(f; P n) L(f; P n )] = 0. n f = lim n U(f; P n) = lim n L(f; P n). Proof. First, suppose tht the condition holds. Then, given ɛ > 0, there is n n N such tht U(f; P n ) L(f; P n ) < ɛ, so Theorem.5 implies tht f is integrble nd U(f) = L(f). Moreover, since U(f) U(f; P n ) nd L(f; P n ) L(f), 0 U(f; P n ) U(f) = U(f; P n ) L(f) U(f; P n ) L(f; P n ). Since the limit of the right-hnd side is zero, the squeeze theorem implies tht It then lso follows tht lim U(f; P n) = U(f). n lim L(f; P n) = lim U(f; P n) lim [U(f; P n) L(f; P n )] = U(f). n n n Conversely, if f is integrble then, by Theorem.5, for every n N there exists prtition P n such tht 0 U(f; P n ) L(f; P n ) < n nd then [U(f; P n ) L(f; P n )] 0 s n. Note tht, since lim [U(f; P n) L(f; P n )] = lim U(f; P n) lim L(f; P n), n n n the conditions of the theorem re stisfied if the limits U(f; P n ) nd L(f; P n ) exist nd re equl. Conversely, the proof of the theorem shows tht if the limit of

10 0. The Riemnn Integrl U(f; P n ) L(f; P n ) is zero, then the limits of U(f; P n ) nd L(f; P n ) exist nd re equl. This isn t true for generl sequences, where one my hve lim( n b n ) = 0 even though lim n nd lim b n do not exist. Exmple.7. Consider f(x) = x 2 on [0, ]. Let P n be prtion of [0, ] into n-intervls of equl length /n, with x k = k/n for 0 k n. Then, using the formul for the sum of squres we get nd It follows tht U(f; P n ) = L(f; P n ) = k 2 = n(n + )(2n + ), 6 x 2 k n = n 3 k 2 = 6 x 2 k n = n n 3 k 2 = 6 lim U(f; P n) = lim L(f; P n) = n n 3, ( + ) ( 2 + ) n n ( ) ( 2 ). n n nd Theorem.6 implies tht so x 2 is integrble on [0, ] with 0 x 2 dx = 3. The min ppliction of Theorems.5 is the following fundmentl result tht every continuous function is Riemnn integrble. Theorem.8. A continuous function f : [, b] R on compct intervl is Riemnn integrble. Proof. A continuous function on compct set is bounded, so its upper nd lower sums re well-defined, nd we just need to verify the condition in Theorems.5. Let ɛ > 0. A continuous function on compct set is uniformly continuous, so there exists δ > 0 such tht f(x) f(y) < ɛ b for ll x, y [, b] such tht x y < δ. Choose prtition P = { : k =,..., n} of [, b] such tht < δ for every k; for exmple, we cn tke n intervls of equl length (b )/n with n > (b )/δ. Since f is continuous, it ttin its mximum nd minimum vlues M k nd m k on the compct intervl t points x k nd y k in, which stisfy x k y k < δ. It follows tht M k m k = f(x k ) f(y k ) < ɛ b.

11 .3. Existence of the Riemnn integrl The upper nd lower sums of f therefore stisfy U(f; P ) L(f; P ) = M k = m k (M k m k ) < ɛ b < ɛ, nd Theorems.5 implies tht f is integrble.