Chpter 7 The Riemnn Integrl 7.1 Discussion: How Should Integrtion be Defined? The Fundmentl Theorem of Clculus is sttement bout the inverse reltionship between differentition nd integrtion. It comes in two prts, depending on whether we re differentiting n integrl or integrting derivtive. Under suitble hypotheses on the functions f nd F, the Fundmentl Theorem of Clculus sttes tht (i) (ii) if G(x) = F (x) dx = F (b) F () nd x f(t) dt, then G (x) =f(x). Before we cn undertke ny type of rigorous investigtion of these sttements, we need to settle on definition for f. Historiclly, the concept of integrtion ws defined s the inverse process of differentition. In other words, the integrl of function f ws understood to be function F tht stisfied F = f. Newton, Leibniz, Fermt, nd the other founders of clculus then went on to explore the reltionship between ntiderivtives nd the problem of computing res. This pproch is ultimtely unstisfying from the point of view of nlysis becuse it results in very limited number of functions tht cn be integrted. Recll tht every derivtive stisfies the intermedite vlue property (Drboux s Theorem, Theorem 5.2.7). This mens tht ny function with jump discontinuity cnnot be derivtive. If we wnt to define integrtion vi ntidifferentition, then we must ccept the consequence tht function s simple s { 1 for 0 x<1 h(x) = 2 for 1 x 2 is not integrble on the intervl [0, 2]. 183
184 Chpter 7. The Riemnn Integrl c 1 c 2 c 3 c n x 0= x 1 x 2 x 3... x n 1 x n=b Figure 7.1: A Riemnn Sum. A very interesting shift in emphsis occurred round 1850 in the work of Cuchy, nd soon fter in the work of Bernhrd Riemnn. The ide ws to completely divorce integrtion from the derivtive nd insted use the notion of re under the curve s strting point for building rigorous definition of the integrl. The resons for this were complicted. As we hve mentioned erlier (Section 1.2), the concept of function ws undergoing trnsformtion. The trditionl understnding of function s holistic formul such s f(x) = x 2 ws being replced with more liberl interprettion, which included such bizrre constructions s Dirichlet s function discussed in Section 4.1. Serving s ctlyst to this evolution ws the budding theory of Fourier series (discussed in Section 8.3), which required, mong other things, the need to be ble to integrte these more unruly objects. The Riemnn integrl, s it is clled tody, is the one usully discussed in introductory clculus. Strting with function f on [, b], we prtition the domin into smll subintervls. On ech subintervl [x k 1,x k ], we pick some point c k [x k 1,x k ] nd use the y-vlue f(c k ) s n pproximtion for f on [x k 1,x k ]. Grphiclly speking, the result is row of thin rectngles constructed to pproximte the re between f nd the x-xis. The re of ech rectngle is f(c k )(x k x k 1 ), nd so the totl re of ll of the rectngles is given by the Riemnn sum (Fig. 7.1) n f(c k )(x k x k 1 ). k=1 Note tht re here comes with the understnding tht res below the x-xis re ssigned negtive vlue. Wht should be evident from the grph is tht the ccurcy of the Riemnnsum pproximtion seems to improve s the rectngles get thinner. In some
7.1. Discussion: How Should Integrtion be Defined? 185 sense, we tke the limit of these pproximting Riemnn sums s the width of the individul subintervls of the prtitions tends to zero. This limit, if it exists, is Riemnn s definition of f. This brings us to hndful of questions. Creting rigorous mening for the limit just referred to is not too difficult. Wht will be of most interest to us nd ws lso to Riemnn is deciding wht types of functions cn be integrted using this procedure. Specificlly, wht conditions on f gurntee tht this limit exists? The theory of the Riemnn integrl turns on the observtion tht smller subintervls produce better pproximtions to the function f. On ech subintervl [x k 1,x k ], the function f is pproximted by its vlue t some point c k [x k 1,x k ]. The qulity of the pproximtion is directly relted to the difference f(x) f(c k ) s x rnges over the subintervl. Becuse the subintervls cn be chosen to hve rbitrrily smll width, this mens tht we wnt f(x) to be close to f(c k ) whenever x is close to c k. But this sounds like discussion of continuity! We will soon see tht the continuity of f is intimtely relted to the existence of the Riemnn integrl f. Is continuity sufficient to prove tht the Riemnn sums converge to welldefined limit? Is it necessry, or cn the Riemnn integrl hndle discontinuous function such s h(x) mentioned erlier? Relying on the intuitive notion of re, it would seem tht 2 h = 3, but does the Riemnn integrl rech this 0 conclusion? If so, how discontinuous cn function be before it fils to be integrble? Cn the Riemnn integrl mke sense out of something s pthologicl s Dirichlet s function on the intervl [0, 1]? A function such s g(x) = { x 2 sin( 1 x ) for x 0 0 for x =0 rises nother interesting question. Here is n exmple of differentible function, studied in Section 5.1, where the derivtive g (x) isnot continuous. As we explore the clss of integrble functions, some ttempt must be mde to reunite the integrl with the derivtive. Hving defined integrtion independently of differentition, we would like to come bck nd investigte the conditions under which equtions (i) nd (ii) from the Fundmentl Theorem of Clculus stted erlier hold. If we re mking wish list for the types of functions tht we wnt to be integrble, then in light of eqution (i) it seems desirble to expect this set to t lest contin the set of derivtives. The fct tht derivtives re not lwys continuous is further motivtion not to content ourselves with n integrl tht cnnot hndle some discontinuities.
186 Chpter 7. The Riemnn Integrl m k M k =x 0 x k 1 x k b=x n Figure 7.2: Upper nd Lower Sums. 7.2 The Definition of the Riemnn Integrl Although it hs the benefit of some modern polish, the development of the integrl presented in this chpter is closely relted to the procedure just discussed. In plce of Riemnn sums, we will construct upper sums nd lower sums (Fig. 7.2), nd in plce of limit we will use supremum nd n infimum. Throughout this section, it is ssumed tht we re working with bounded function f on closed intervl [, b], mening tht there exists n M>0 such tht f(x) M for ll x [, b]. Prtitions, Upper Sums, nd Lower Sums Definition 7.2.1. A prtition P of [, b] is finite, ordered set P = { = x 0 <x 1 <x 2 < <x n = b}. For ech subintervl [x k 1,x k ]ofp, let m k = inf{f(x) :x [x k 1,x k ]} nd M k = sup{f(x) :x [x k 1,x k ]}. The lower sum of f with respect to P is given by n L(f,P) = m k (x k x k 1 ). k=1 Likewise, we define the upper sum of f with respect to P by n U(f,P) = M k (x k x k 1 ). k=1
7.2. The Definition of the Riemnn Integrl 187 For prticulr prtition P, it is cler tht U(f,P) L(f,P). The fct tht this sme inequlity holds if the upper nd lower sums re computed with respect to different prtitions is the content of the next two lemms. Definition 7.2.2. A prtition Q is refinement of prtition P if Q contins ll of the points of P. In this cse, we write P Q. Lemm 7.2.3. If P Q, then L(f,P) L(f,Q), nd U(f,P) U(f,Q). Proof. Consider wht hppens when we refine P by dding single point z to some subintervl [x k 1,x k ]ofp. m k m k =m k x k 1 z x k Focusing on the lower sum for moment, we hve where m k (x x x k 1 ) = m k (x k z)+m k (z x k 1 ) m k(x k z)+m k(z x k 1 ), m k = inf {f(x) :x [z,x k ]} nd m k = inf {f(x) :x [x k 1,z]} re ech necessrily s lrge or lrger thn m k. By induction, we hve L(f,P) L(f,Q), nd n nlogous rgument holds for the upper sums. Lemm 7.2.4. If P 1 nd P 2 re ny two prtitions of [, b], then L(f,P 1 ) U(f,P 2 ). Proof. Let Q = P 1 P 2 be the so-clled common refinement of P 1 nd P 2. Becuse Q P 1 nd Q P 2, it follows tht L(f,P 1 ) L(f,Q) U(f,Q) U(f,P 2 ).
188 Chpter 7. The Riemnn Integrl Integrbility Intuitively, it helps to visulize prticulr upper sum s n overestimte for the vlue of the integrl nd lower sum s n underestimte. As the prtitions get more refined, the upper sums get potentilly smller while the lower sums get potentilly lrger. A function is integrble if the upper nd lower sums meet t some common vlue in the middle. Rther thn tking limit of these sums, we will insted mke use of the Axiom of Completeness nd consider the infimum of the upper sums nd the supremum of the lower sums. Definition 7.2.5. Let P be the collection of ll possible prtitions of the intervl [, b]. The upper integrl of f is defined to be U(f) = inf{u(f,p) :P P}. In similr wy, define the lower integrl of f by The following fct is not surprising. L(f) = sup{u(f,p) :P P}. Lemm 7.2.6. For ny bounded function f on [, b], it is lwys the cse tht U(f) L(f). Proof. Exercise 7.2.1. Definition 7.2.7 (Riemnn Integrbility). A bounded function f defined on the intervl [, b] isriemnn-integrble if U(f) =L(f). In this cse, we define f or f(x) dx to be this common vlue; nmely, f = U(f) =L(f). The modifier Riemnn in front of integrble ccurtely suggests tht there re other wys to define the integrl. In fct, our work in this chpter will expose the need for different pproch, one of which is discussed in Section 8.1. In this chpter, the Riemnn integrl is the only method under considertion, so it will usully be convenient to drop the modifier Riemnn nd simply refer to function s being integrble. Criteri for Integrbility To summrize the sitution thus fr, it is lwys the cse for bounded function f on [, b] tht sup{l(f,p) :P P}= L(f) U(f) = inf{u(f,p) :P P}. The function f is integrble if the inequlity is n equlity. The mjor thrust of our investigtion of the integrl is to describe, s best we cn, the clss
7.2. The Definition of the Riemnn Integrl 189 of integrble functions. The preceding inequlity revels tht integrbility is relly equivlent to the existence of prtitions whose upper nd lower sums re rbitrrily close together. Theorem 7.2.8. A bounded function f is integrble on [, b] if nd only if, for every ɛ>0, there exists prtition P ɛ of [, b] such tht U(f,P ɛ ) L(f,P ɛ ) <ɛ. Proof. Let ɛ>0. If such prtition P ɛ exists, then U(f) L(f) U(f,P ɛ ) L(f,P ɛ ) <ɛ. Becuse ɛ is rbitrry, it must be tht U(f) =L(f), so f is integrble. (To be bsolutely precise here, we could throw in reference to Theorem 1.2.6.) The proof of the converse sttement is fmilir tringle inequlity rgument with prentheses in plce of bsolute vlue brs becuse, in ech cse, we know which quntity is lrger. Becuse U(f) is the gretest lower bound of the upper sums, we know tht, given some ɛ>0, there must exist prtition P 1 such tht U(f,P 1 ) <U(f)+ ɛ 2. Likewise, there exists prtition P 2 stisfying L(f,P 2 ) >L(f) ɛ 2. Now, let P ɛ = P 1 P 2 be the common refinement. Keeping in mind tht the integrbility of f mens U(f) =L(f), we cn write U(f,P ɛ ) L(f,P ɛ ) U(f,P 1 ) L(f,P 2 ) = (U(f,P 1 ) U(f))+(L(f) L(f,P 2 )) < ɛ 2 + ɛ 2 = ɛ. In the discussion t the beginning of this chpter, it becme cler tht integrbility is closely tied to the concept of continuity. To mke this observtion more precise, let P = { = x 0 <x 1 <x 2 < <x n = b} be n rbitrry prtition of [, b], nd define x k = x k x k 1. Then, U(f,P) L(f,P) = n (M k m k ) x k, where M k nd m k re the supremum nd infimum of the function on the intervl [x k 1,x k ] respectively. Our bility to control the size of U(f,P) L(f,P) hinges on the differences M k m k, which we cn interpret s the vrition in the rnge of the function over the intervl [x k 1,x k ]. Restricting the vrition of f over rbitrrily smll intervls in [, b] is precisely wht it mens to sy tht f is uniformly continuous on this set. k=1
190 Chpter 7. The Riemnn Integrl Theorem 7.2.9. If f is continuous on [, b], then it is integrble. Proof. The first crucil observtion is tht becuse f is continuous on compct set, it is uniformly continuous. This mens tht, given ɛ>0, there exists δ>0 so tht x y <δgurntees f(x) f(y) < ɛ b. Now, let P be prtition of [, b] where x k = x k x k 1 is less thn δ for every subintervl of P. M k =f(z k ) m k =f(y k ) x k 1 z k y k x k }{{} x k x k 1 <δ Given prticulr subintervl [x k 1,x k ]ofp, we know from the Extreme Vlue Theorem (Theorem 4.4.3) tht the supremum M k = f(z k ) for some z k [x k 1,x k ]. In ddition, the infimum m k is ttined t some point y k lso in the intervl [x k 1,x k ]. But this mens z k y k <δ,so Finlly, U(f,P) L(f,P) = M k m k = f(z k ) f(y k ) < ɛ b. n (M k m k ) x k < ɛ b k=1 nd f is integrble by the criterion given in Theorem 7.2.8. Exercises n x k = ɛ, Exercise 7.2.1. Let f be bounded function on [, b], nd let P be n rbitrry prtition of [, b]. First, explin why U(f) L(f,P). Now, prove Lemm 7.2.6. Exercise 7.2.2. Consider f(x) =2x + 1 over the intervl [1, 3]. Let P be the prtition consisting of the points {1, 3/2, 2, 3}. () Compute L(f,P), U(f,P), nd U(f,P) L(f,P). (b) Wht hppens to the vlue of U(f,P) L(f,P) when we dd the point 5/2 to the prtition? (c) Find prtition P of [1, 3] for which U(f,P ) L(f,P ) < 2. k=1
7.3. Integrting Functions with Discontinuities 191 Exercise 7.2.3. Show directly (without ppeling to Theorem 7.2) tht the constnt function f(x) =k is integrble over ny closed intervl [, b]. Wht is f? Exercise 7.2.4. () Prove tht bounded function f is integrble on [, b] if nd only if there exists sequence of prtitions (P n ) n=1 stisfying lim [U(f,P n) L(f,P n )] = 0. n (b) For ech n, let P n be the prtition of [0, 1] into n equl subintervls. Find formuls for U(f,P n ) nd L(f,P n )iff(x) =x. The formul 1+2+3+ +n = n(n +1)/2 will be useful. (c) Use the sequentil criterion for integrbility from () to show directly tht f(x) =x is integrble on [0, 1]. Exercise 7.2.5. Assume tht, for ech n, f n is n integrble function on [, b]. If (f n ) f uniformly on [, b], prove tht f is lso integrble on this set. (We will see tht this conclusion does not necessrily follow if the convergence is pointwise.) Exercise 7.2.6. Let f :[, b] R be incresing on the set [, b] (i.e., f(x) f(y) whenever x<y). Show tht f is integrble on [, b]. 7.3 Integrting Functions with Discontinuities The fct tht continuous functions re integrble is not so much fortunte discovery s it is evidence for well-designed integrl. Riemnn s integrl is modifiction of Cuchy s definition of the integrl, which ws crfted specificlly to work on continuous functions. The interesting issue is discovering just how dependent the Riemnn integrl is on the continuity of the integrnd. Exmple 7.3.1. Consider the function { 1 for x 1 f(x) = 0 for x =1 on the intervl [0, 2]. If P is ny prtition of [0, 2], quick clcultion revels tht U(f,P) = 2. The lower sum L(f,P) will be less thn 2 becuse ny subintervl of P tht contins x = 1 will contribute zero to the vlue of the lower sum. The wy to show tht f is integrble is to construct prtition tht minimizes the effect of the discontinuity by embedding x = 1 into very smll subintervl. Let ɛ>0, nd consider the prtition P ɛ = {0, 1 ɛ/3, 1+ɛ/3, 2}. Then, ( L(f,P ɛ ) = 1 1 ɛ ) ( +0(ɛ)+1 1 ɛ ) 3 3 = 2 2 3 ɛ.
192 Chpter 7. The Riemnn Integrl Becuse U(f,P ɛ )=2,wehve U(f,P ɛ ) L(f,P ɛ )= 2 3 ɛ<ɛ. We cn now use Theorem 7.2.8 to conclude tht f is integrble. Although the function in Exmple 7.3.1 is extremely simple, the method used to show it is integrble is relly the sme one used to prove tht ny bounded function with single discontinuity is integrble. The nottion in the following proof is more cumbersome, but the essence of the rgument is tht the misbehvior of the function t its discontinuity is isolted inside prticulrly smll subintervl of the prtition. Theorem 7.3.2. If f :[, b] R is bounded, nd f is integrble on [c, b] for ll c (, b), then f is integrble on [, b]. An nlogous result holds t the other endpoint. Proof. Let M be bound for f so tht f(x) M for ll x [, b]. If is prtition of [, b], then U(f,P) L(f,P) = P = { = x 0 <x 1 <x 2 < x n = b} n (M k m k ) x k k=1 = (M 1 m 1 )(x 1 )+ n (M k m k ) x k k=2 = (M 1 m 1 )(x 1 )+(U(f,P [x1,b]) L(f,P [x1,b])), where P [x1,b] = {x 1 <x 2 < <x n = b} is the prtition of [x 1,b] obtined by deleting from P. Given ɛ>0, the first step is to choose x 1 close enough to so tht (M 1 m 1 )(x 1 ) < ɛ 2. This is not too difficult. Becuse M 1 m 1 2M, we cn pick x 1 so tht x 1 ɛ 4M. Now, by hypothesis, f is integrble on [x 1,b] so there exists prtition P 1 of [x 1,b] for which U(f,P 1 ) L(f,P 1 ) < ɛ 2. Finlly, we let P 2 = {} P 1 be prtition of [, b], from which it follows tht U(f,P 2 ) L(f,P 2 ) (2M)(x 1 )+(U(f,P 1 ) L(f,P 1 )) < ɛ 2 + ɛ 2 = ɛ.
7.3. Integrting Functions with Discontinuities 193 Theorem 7.3.2 only llows for discontinuity t the endpoint of n intervl, but tht is esily remedied. In the next section, we will prove tht integrbility on the intervls [, b] nd [b, d] is equivlent to integrbility on [, d]. This property, together with n induction rgument, leds to the conclusion tht ny function with finite number of discontinuities is still integrble. Wht if the number of discontinuities is infinite? Exmple 7.3.3. Recll Dirichlet s function g(x) = { 1 for x rtionl 0 for x irrtionl from Section 4.1. If P is some prtition of [0, 1], then the density of the rtionls in R implies tht every subintervl of P will contin point where g(x) =1. It follows tht U(g, P) = 1. On the other hnd, L(g, P) = 0 becuse the irrtionls re lso dense in R. Becuse this is the cse for every prtition P, we see tht the upper integrl U(f) = 1 nd the lower integrl L(f) = 0. The two re not equl, so we conclude tht Dirichlet s function is not integrble. How discontinuous cn function be before it fils to be integrble? Before jumping to the hsty (nd incorrect) conclusion tht the Riemnn integrl fils for functions with more thn finite number of discontinuities, we should relize tht Dirichlet s function is discontinuous t every pointin[0, 1]. It would be useful to investigte function where the discontinuities re infinite in number but do not necessrily mke up ll of [0, 1]. Thome s function, lso defined in Section 4.1, is one such exmple. The discontinuous points of this function re precisely the rtionl numbers in [0, 1]. In Section 7.6, we will see tht Thome s function is Riemnn-integrble, rising the br for llowble discontinuous points to include potentilly infinite sets. The conclusion of this story is contined in the doctorl disserttion of Henri Lebesgue, who presented his work in 1901. Lebesgue s elegnt criterion for Riemnn integrbility is explored in gret detil in Section 7.6. For the moment, though, we will tke short detour from questions of integrbility nd construct proof of the celebrted Fundmentl Theorem of Clculus. Exercises Exercise 7.3.1. Consider the function { 1 for 0 x<1 h(x) = 2 for x =1 over the intervl [0, 1]. () Show tht L(f,P) = 1 for every prtition P of [0, 1]. (b) Construct prtition P for which U(f,P) < 1+1/10. (c) Given ɛ>0, construct prtition P ɛ for which U(f,P ɛ ) < 1+ɛ.
194 Chpter 7. The Riemnn Integrl Exercise 7.3.2. In Exmple 7.3.3, we lerned tht Dirichlet s function g(x) is not Riemnn-integrble. Construct sequence g n (x) of integrble functions with g n g pointwise on [0, 1]. This demonstrtes tht the pointwise limit of integrble functions need not be integrble. Compre this exmple to the result requested in Exercise 7.2.5. Exercise 7.3.3. Here is n lternte explntion for why function f on [, b] with finite number of discontinuities is integrble. Supply the missing detils. Embed ech discontinuity in sufficiently smll open intervl nd let O be the union of these intervls. Explin why f is uniformly continuous on [, b]\o, nd use this to finish the rgument. Exercise 7.3.4. Assume f :[, b] R is integrble. () Show tht if one vlue of f(x) is chnged t some point x [, b], then f is still integrble nd integrtes to the sme vlue s before. (b) Show tht the observtion in () holds if finite number of vlues of f re chnged. (c) Find n exmple to show tht by ltering countble number of vlues, f my fil to be integrble. Exercise 7.3.5. Let { 1 if x =1/n for some n N f(x) = 0 otherwise. Show tht f is integrble on [0, 1] nd compute 1 0 f. Exercise 7.3.6. A set A [, b] hs content zero if for every ɛ>0 there exists finite collection of open intervls {O 1,O 2,...,O N } tht contin A in their union nd whose lengths sum to ɛ or less. Using O n to refer to the length of ech intervl, we hve N A n=1 O n nd N O n ɛ. k=1 () Let f be bounded on [, b]. Show tht if the set of discontinuous points of f hs content zero, then f is integrble. (b) Show tht ny finite set hs content zero. (c) Content zero sets do not hve to be finite. They do not hve to be countble. Show tht the Cntor set C defined in Section 3.1 hs content zero. (d) Prove tht h(x) = { 1 if x C 0 if x/ C. is integrble, nd find the vlue of the integrl.
7.4. Properties of the Integrl 195 7.4 Properties of the Integrl Before embrking on the proof of the Fundmentl Theorem of Clculus, we need to verify wht re probbly some very fmilir properties of the integrl. The discussion in the previous section hs lredy mde use of the following fct. Theorem 7.4.1. Assume f :[, b] R is bounded, nd let c (, b). Then, f is integrble on [, b] ifnd only iff is integrble on [, c] nd [c, b]. In this cse, we hve c f = f + f. Proof. If f is integrble on [, b], then for ɛ>0 there exists prtition P such tht U(f,P) L(f,P) <ɛ. Becuse refining prtition cn only potentilly bring the upper nd lower sums closer together, we cn simply dd c to P if it is not lredy there. Then, let P 1 = P [, c] be prtition of [, c], nd P 2 = P [c, b] be prtition of [c, b]. It follows tht U(f,P 1 ) L(f,P 1 ) <ɛ nd U(f,P 2 ) L(f,P 2 ) <ɛ, implying tht f is integrble on [, c] nd [c, b]. Conversely, if we re given tht f is integrble on the two smller intervls [, c] nd [c, b], then given n ɛ>0 we cn produce prtitions P 1 nd P 2 of [, c] nd [c, b], respectively, such tht c U(f,P 1 ) L(f,P 1 ) < ɛ 2 nd U(f,P) L(f,P) < ɛ 2. Letting P = P 1 P 2 produces prtition of [, b] for which U(f,P) L(f,P) <ɛ. Thus, f is integrble on [, b]. Continuing to let P = P 1 P 2 s erlier, we hve which implies f c f + c c f U(f,P) < L(f,P)+ɛ f + c = L(f,P 1 )+L(f,P 2 )+ɛ c f + c f + ɛ, f. To get the other inequlity, observe tht f U(f,P 1 )+U(f,P 2 ) < L(f,P 1 )+L(f,P 2 )+ɛ = L(f,P)+ɛ f + ɛ.
196 Chpter 7. The Riemnn Integrl Becuse ɛ>0 is rbitrry, we must hve c f + c f f,so s desired. c f + c The proof of Theorem 7.4.1 demonstrtes some of the stndrd techniques involved for proving fcts bout the Riemnn integrl. Admittedly, mnipulting prtitions does not lend itself to gret del of elegnce. The next result ctlogs the reminder of the bsic properties of the integrl tht we will need in our upcoming rguments. Theorem 7.4.2. Assume f nd g re integrble functions on the intervl [, b]. f = (i) The function f + g is integrble on [, b] with (f + g) = f + g. (ii) For k R, the function kf is integrble with kf = k f. (iii) If m f M, then m(b ) f M(b ). (iv) If f g, then f g. (v) The function f is integrble nd f f. Proof. Properties (i) nd (ii) re reminiscent of the Algebric Limit Theorem nd its mny descendnts (Theorems 2.3.3, 2.7.1, 4.2.4, nd 5.2.4). In fct, there is wy to use the Algebric Limit Theorem for this rgument s well. An immedite corollry to Theorem 7.2.8 is tht function f is integrble on [, b] if nd only if there exists sequence of prtitions (P n ) stisfying (1) lim n [U(f,P n) L(f,P n )] = 0, nd in this cse f = lim U(f,P n) = lim L(f,P n ). (A proof for this ws requested s Exercise 7.2.4.) To prove (ii) for the cse k 0, first verify tht for ny prtition P we hve U(kf, P) =ku(f,p) nd L(kf, P) =kl(f,p). Exercise 1.3.5 is used here. Becuse f is integrble, there exist prtitions (P n ) stisfying (1). Turning our ttention to the function (kf), we see tht lim n [U(kf, P n) L(kf, P n )] = lim n k [U(f,P n) L(f,P n )] = 0, nd the formul in (ii) follows. The cse where k<0 is similr except tht we hve U(kf, P n )=kl(f,p n ) nd L(kf, P n )=ku(f,p n ). f,
7.4. Properties of the Integrl 197 A proof for (i) cn be constructed using similr methods nd is requested in Exercise 7.4.5. To prove (iii), observe tht U(f,P) f L(f,P) for ny prtition P. Sttement (iii) follows if we tke P to be the trivil prtition consisting of only the endpoints nd b. For (iv), let h = g f 0 nd use (i) nd (iii). Becuse f f f, sttement (v) will follow from (iv) provided tht we cn show tht f is ctully integrble. The proof of this fct is outlined in Exercise 7.4.1. To this point, the quntity f is only defined in the cse where <b. Definition 7.4.3. If f is integrble on the intervl [, b], define Also, define b f = c c f =0. Definition 7.4.3 is nturl convention to simplify the lgebr of integrls. If f is n integrble function on some intervl I, then it is strightforwrd to verify tht the eqution c f = f + f from Theorem 7.4.1 remins vlid for ny three points, b, nd c chosen in ny order from I. Uniform Convergence nd Integrtion If (f n ) is sequence of integrble functions on [, b], nd if f n f, then we re inevitbly going to wnt to know whether f. c (2) f n f. This is n rchetypicl instnce of one of the mjor themes of nlysis: When does mthemticl mnipultion such s integrtion respect the limiting process? If the convergence is pointwise, then ny number of things cn go wrong. It is possible for ech f n to be integrble but for the limit f not to be integrble
198 Chpter 7. The Riemnn Integrl (Exercise 7.3.2). Even if the limit function f is integrble, eqution (2) my fil to hold. As n exmple of this, let { n if 0 <x<1/n f n (x) = 0 if x =0orx 1/n. Ech f n hs two discontinuities on [0, 1] nd so is integrble with 1 0 f n =1. For ech x [0, 1], we hve lim f n (x) = 0 so tht f n 0 pointwise on [0, 1]. But now observe tht the limit function f = 0 certinly integrtes to 0, nd 0 lim n As finl remrk on wht cn go wrong in (2), we should point out tht it is possible to modify this exmple to produce sitution where lim 1 0 f n does not even exist. One wy to resolve ll of these problems is to dd the ssumption of uniform convergence. Theorem 7.4.4. Assume tht f n f uniformly on [, b] nd tht ech f n is integrble. Then, f is integrble nd lim n 1 0 f n = f n. f. Proof. The proof tht f is integrble ws requested s Exercise 7.2.5. reminder of this rgument is sked for in Exercise 7.4.3. The Exercises Exercise 7.4.1. () Let f be bounded function on set A, nd set M = sup{f(x) :x A}, m = inf{f(x) :x A}, M = sup{ f(x) : x A}, nd m = inf{ f(x) : x A}. Show tht M m M m. (b) Show tht if f is integrble on the intervl [, b], then f is lso integrble on this intervl. (c) Provide the detils for the rgument tht in this cse we hve f b f. Exercise 7.4.2. Review Definition 7.4.3. Show tht if c b nd f is integrble on the intervl [c, b], then it is still the cse tht f = c f + c f. Exercise 7.4.3. Prove Theorem 7.4.4 including n rgument for Exercise 7.2.5 if it is not lredy done.
7.5. The Fundmentl Theorem of Clculus 199 Exercise 7.4.4. Decide which of the following conjectures is true nd supply short proof. For those tht re not true, give counterexmple. () If f is integrble on [, b] then f is lso integrble on this set. (b) Assume g is integrble nd g 0on[, b]. If g(x) > 0 for n infinite number of points x [, b], then g>0. (c) If g is continuous on [, b] nd g 0 with g(x 0 ) > 0 for t lest one point x 0 [, b], then g>0. (d) If f > 0, there is n intervl [c, d] [, b] nd δ>0such tht f(x) δ for ll x [c, d]. Exercise 7.4.5. Let f nd g be integrble functions on [, b]. () Show tht if P is ny prtition of [, b], then U(f + g, P) U(f,P)+U(g, P). Provide specific exmple where the inequlity is strict. Wht does the corresponding inequlity for lower sums look like? (b) Review the proof of Theorem 7.4.2 (ii), nd provide n rgument for prt (i) of this theorem. Exercise 7.4.6. Review the discussion immeditely preceding Theorem 7.4.4. () Produce n exmple of sequence f n 0 pointwise on [0, 1] where 1 lim n 0 f n does not exist. (b) Produce nother exmple (if necessry) where f n 0 nd the sequence 1 0 f n is unbounded. (c) Is it possible to construct ech f n to be continuous in the exmples of prts () nd (b)? (d) Does it seem possible to construct the sequence (f n ) to be uniformly bounded? (Uniformly bounded mens tht there exists single M>0stisfying f n M for ll n N. Exercise 7.4.7. Assume tht g n nd g re bounded integrble functions with g n g on [0, 1]. The convergence is not uniform; however, the convergence is uniform on ny set of the form [δ, 1] where 0 <δ<1. Show tht lim n 1 0 g n = 1 0 g. 7.5 The Fundmentl Theorem of Clculus The derivtive nd the integrl hve been independently defined, ech in its own rigorous mthemticl terms. The definition of the derivtive is motivted by the problem of finding tngent lines nd is given in terms of functionl limits of difference quotients. The definition of the integrl grows out of the desire to describe res under nonconstnt functions nd is given in terms of supremums nd infimums of finite sums. The Fundmentl Theorem of Clculus revels the remrkble inverse reltionship between the two processes.
200 Chpter 7. The Riemnn Integrl The result is stted in two prts. The first is computtionl sttement tht describes how n ntiderivtive cn be used to evlute n integrl over prticulr intervl. The second sttement is more theoreticl in nture, expressing the fct tht every continuous function is the derivtive of its indefinite integrl. Theorem 7.5.1 (Fundmentl Theorem of Clculus). (i) If f :[, b] R is integrble, nd F :[, b] R stisfies F (x) =f(x) for ll x [, b], then f = F (b) F (). (ii) Let g :[, b] R be integrble, nd define G(x) = for ll x [, b]. Then, G is continuous on [, b]. If g is continuous t some point c [, b], then G is differentible t c nd G (c) =g(c). Proof. (i) Let P be prtition of [, b] nd pply the Men Vlue Theorem to F on typicl subintervl [x k 1,x k ]ofp. This yields point t k (x k 1,x k ) where x F (x k ) F (x k 1 ) = F (t k )(x k x k 1 ) = f(t k )(x k x k 1 ). Now, consider the upper nd lower sums U(f,P) nd L(f,P). Becuse m k f(t k ) M k (where m k is the infimum on [x k 1,x k ] nd M k is the supremum), it follows tht n L(f,P) [F (x k ) F (x k 1 )] U(f,P). k=1 But notice tht the sum in the middle telescopes so tht n [F (x k ) F (x k 1 )] = F (b) F (), k=1 which is independent of the prtition P. Thus we hve L(f) F (b) F () U(f). Becuse L(f) =U(f) = f, we conclude tht f = F (b) F (). (ii) To prove the second sttement, tke x, y [, b] nd observe tht x y x G(x) G(y) = g g = g g y x y g M x y,
7.5. The Fundmentl Theorem of Clculus 201 where M > 0 is bound on g. This shows tht G is Lipschitz nd so is uniformly continuous on [, b] (Exercise 4.4.9). Now, let s ssume tht g is continuous t c [, b]. In order to show tht G (c) =g(c), we rewrite the limit for G (c) s ( G(x) G(c) 1 x c ) lim = lim g(t) dt g(t) dt x c x c x c x c ( 1 x ) = lim g(t) dt. x c x c c We would like to show tht this limit equls g(c). Thus, given n ɛ>0, we must produce δ>0 such tht if x c <δthen ( (1) 1 x ) g(t) dt g(c) x c <ɛ. c The ssumption of continuity of g gives us control over the difference g(t) g(c). In prticulr, we know tht there exists δ>0 such tht t c <δimplies g(t) g(c) <ɛ. To tke dvntge of this, we cleverly write the constnt g(c) s g(c) = 1 x g(c) dt x c c nd combine the two terms in eqution (1) into single integrl. Keeping in mind tht x c t c, we hve tht for ll x c <δ, ( 1 x ) g(t) dt g(c) x c = 1 x [g(t) g(c)] dt c x c c 1 x g(t) g(c) dt (x c) c 1 x < ɛdt= ɛ. (x c) c Exercises Exercise 7.5.1. We hve seen tht not every derivtive is continuous, but explin how we t lest know tht every continuous function is derivtive. Exercise 7.5.2. () Let f(x) = x nd define F (x) = x f. Find formul 1 for F (x) for ll x. Where is F continuous? Where is F differentible? Where does F (x) =f(x)? (b) Repet prt () for the function { 1 if x<0 f(x) = 2 if x 0.
202 Chpter 7. The Riemnn Integrl Exercise 7.5.3. The hypothesis in Theorem 7.5.1 (i) tht F (x) =f(x) for ll x [, b] is slightly stronger thn it needs to be. Crefully red the proof nd stte exctly wht needs to be ssumed with regrd to the reltionship between f nd F for the proof to be vlid. Exercise 7.5.4 (Nturl Logrithm). Let H(x) = x 1 1 t dt, where we consider only x>0. () Wht is H(1)? Find H (x). (b) Show tht H is strictly incresing; tht is, show tht if 0 <x<y, then H(x) <H(y). (c) Show tht H(cx) = H(c) + H(x). (Think of c s constnt nd differentite g(x) = H(cx).) Exercise 7.5.5. The Fundmentl Theorem of Clculus cn be used to supply shorter rgument for Theorem 6.3.1 under the dditionl ssumption tht the sequence of derivtives is continuous. Assume f n f pointwise nd f n g uniformly on [, b]. Assuming ech f n is continuous, we cn pply Theorem 7.5.1 (i) to get x for ll x [, b]. Show tht g(x) =f (x). f n = f n (x) f n () Exercise 7.5.6. Use prt (ii) of Theorem 7.5.1 to construct nother proof of prt (i) of Theorem 7.5.1 using the following strtegy. Given f nd F s in prt (i), set G(x) = x f. Wht is the reltionship between F nd G? Exercise 7.5.7 (Averge Vlue). If g is continuous on [, b], show tht there exists point c (, b) where g(c) = 1 b Exercise 7.5.8. Given function f on [, b], define the totl vrition of f to be { n } Vf = sup f(x k ) f(x k 1 ), k=1 where the supremum is tken over ll prtitions P of [, b]. () If f is continuously differentible (f exists s continuous function), use the Fundmentl Theorem of Clculus to show Vf f. (b) Use the Men Vlue Theorem to estblish the reverse inequlity nd conclude tht Vf = f. g.
7.6. Lebesgue s Criterion for Riemnn Integrbility 203 Exercise 7.5.9. Let { 1 if x<1orx>1 h(x) = 0 if x =1, nd define H(x) = x h. Show tht even though h is not continuous t x =1, 0 H(x) is still differentible t x =1. Exercise 7.5.10. Assume f is integrble on [, b] nd hs jump discontinuity t c (, b). This mens tht both one-sided limits exist s x pproches c from the left nd from the right, but tht lim x c f(x) lim x c + f(x). (This phenomenon is discussed in more detil in Section 4.6.) Show tht F (x) = x f is not differentible t x = c. Exercise 7.5.11. The Epilogue to Chpter 5 mentions the existence of continuous monotone function tht fils to be differentible on dense subset of R. Combine the results of Exercise 7.5.10 nd Exercise 6.4.8 to show how to construct such function. 7.6 Lebesgue s Criterion for Riemnn Integrbility We now return to our investigtion of the reltionship between continuity nd the Riemnn integrl. We hve proved tht continuous functions re integrble nd tht the integrl lso exists for functions with only finite number of discontinuities. At the opposite end of the spectrum, we sw tht Dirichlet s function, which is discontinuous t every point on [0, 1], fils to be Riemnn-integrble. The next exmples show tht the set of discontinuities of n integrble function cn be infinite nd even uncountble. Riemnn-integrble Functions with Infinite Discontinuities Recll from Section 4.1 tht Thome s function 1 if x =0 t(x) = 1/n if x = m/n Q\{0} is in lowest terms with n>0 0 if x/ Q is continuous on the set of irrtionls nd hs discontinuities t every rtionl point. Let s prove tht Thome s function is integrble on [0, 1] with 1 0 t =0. Let ɛ>0. The strtegy, s usul, is to construct prtition P ɛ of [0, 1] for which U(t, P ɛ ) L(t, P ɛ ) <ɛ.
204 Chpter 7. The Riemnn Integrl Exercise 7.6.1. ) First, rgue tht L(t, P ) = 0 for ny prtition P of [0, 1]. b) Consider the set of points D ɛ/2 = {x : t(x) ɛ/2}. How big is D ɛ/2? c) To complete the rgument, explin how to construct prtition P ɛ of [0, 1] so tht U(t, P ɛ ) <ɛ. We first met the Cntor set C in Section 3.1. We hve since lerned tht C is compct, uncountble subset of the intervl [0, 1]. The request of Exercise 4.3.12 is to prove tht the function { 1 if x C g(x) = 0 if x/ C is continuous t every point of the complement of C nd hs discontinuities t ech point of C. Thus, g is not continuous on n uncountbly infinite set. Exercise 7.6.2. Using the fct tht C = n=0 C n, where ech C n consists of finite union of closed intervls, rgue tht g is Riemnn-integrble on [0, 1]. Sets of Mesure Zero Thome s function fils to be continuous t ech rtionl number in [0, 1]. Although this set is infinite, we hve seen tht ny subset of Q is countble. Countbly infinite sets re the smllest type of infinite set. The Cntor set is uncountble, but it is lso smll in sense tht we re now redy to mke precise. In the introduction to Chpter 3, we presented n rgument tht the Cntor set hs zero length. The term length is wkwrd here becuse it relly should only be pplied to intervls or unions of intervls, which the Cntor set is not. There is generliztion of the concept of length to more generl sets clled the mesure of set. Of interest to our discussion re subsets tht hve mesure zero. Definition 7.6.1. A set A R hs mesure zero if, for ll ɛ>0, there exists countble collection of open intervls O n with the property tht A is contined in the union of ll of the intervls O n nd the sum of the lengths of ll of the intervls is less thn or equl to ɛ. More precisely, if O n refers to the length of the intervl O n, then we hve A nd O n ɛ. n=1 O n Exmple 7.6.2. Consider finite set A = { 1, 2,..., N }. To show tht A hs mesure zero, let ɛ>0 be rbitrry. For ech 1 n N, construct the intervl ( G n = n ɛ 2N, n + ɛ ). 2N Clerly, A is contined in the union of these intervls, nd N G n = n=1 N n=1 n=1 ɛ N = ɛ.
7.6. Lebesgue s Criterion for Riemnn Integrbility 205 Exercise 7.6.3. Show tht ny countble set hs mesure zero. Exercise 7.6.4. Prove tht the Cntor set (which is uncountble) hs mesure zero. Exercise 7.6.5. Show tht if two sets A nd B ech hve mesure zero, then A B hs mesure zero s well. In ddition, discuss the proof of the stronger sttement tht the countble union of sets of mesure zero lso hs mesure zero. (This second sttement is true, but completely rigorous proof requires result bout double summtions discussed in Section 2.8.) α-continuity Definition 7.6.3. Let f be defined on [, b], nd let α>0. The function f is α-continuous t x [, b] if there exists δ>0 such tht for ll y, z (x δ, x+δ) it follows tht f(y) f(z) <α. Let f be bounded function on [, b]. For ech α>0, define D α to be the set of points in [, b] where the function f fils to be α-continuous; tht is, (1) D α = {x [, b] :f is not α-continuous t x.}. The concept of α-continuity ws previously introduced in Section 4.6. Severl of the ensuing exercises ppered s exercises in this section s well. Exercise 7.6.6. If α 1 <α 2, show tht D α2 D α1. Now, let (2) D = {x [, b] :f is not continuous t x }. Exercise 7.6.7. () Let α > 0 be given. Show tht if f is continuous t x [, b], then it is α-continuous t x s well. Explin how it follows tht D α D. (b) Show tht if f is not continuous t x, then f is not α-continuous for some α>0. Now, explin why this gurntees tht D = D 1/n. n=1 Exercise 7.6.8. Prove tht for fixed α>0, the set D α is closed. Exercise 7.6.9. By imitting the proof of Theorem 4.4.8, show tht if, for some fixed α>0, f is α-continuous t every point on some compct set K, then f is uniformly α-continuous on K. By uniformly α-continuous, we men tht there exists δ>0 such tht whenever x nd y re points in K stisfying x y <δ, it follows tht f(x) f(y) <α.
206 Chpter 7. The Riemnn Integrl Compctness Revisited Compctness of subsets of the rel line cn be described in three equivlent wys. The following theorem ppers towrd the end of Section 3.3. Theorem 7.6.4. Let K R. The following three sttements re ll equivlent, in the sense tht ifny one is true, then so re the two others. (i) Every sequence contined in K hs convergent subsequence tht converges to limit in K. (ii) K is closed nd bounded. (iii) Given collection ofopen intervls {G α : α Λ} tht covers K; tht is, K α Λ G α, there exists finite subcollection {G α1,g α2,g α3,...,g αn } ofthe originl set tht lso covers K. The equivlence of (i) nd (ii) hs been used throughout the core mteril in the text. Chrcteriztion (iii) hs been less centrl but is essentil to the upcoming rgument. So tht the mteril in this section is self-contined, we quickly outline proof tht (i) nd (ii) imply (iii). (This lso ppers s Exercise 3.3.8.) Proof. Assume K stisfies (i) nd (ii), nd let {G α : α Λ} be n open cover of K. For contrdiction, let s ssume tht no finite subcover exists. Let I 0 be closed intervl contining K, nd then bisect I 0 into two closed intervls A 1 nd B 1. It must be tht either A 1 K or B 1 K (or both) hs no finite subcover consisting of sets from {G α : α Λ}. Let I 1 be hlf of I 0 contining prt of K tht cnnot be finitely covered. Repeting this construction results in nested sequence of closed intervls I 0 I 1 I 2 with the property tht, for ny n, I n K cnnot be finitely covered nd lim n I n =0. Exercise 7.6.10. () Show tht there exists n x K such tht x I n for ll n. (b) Becuse x K, there must exist n open set G α0 from the originl collection tht contins x s n element. Explin why this furnishes us with the desired contrdiction. Lebesgue s Theorem We re now prepred to completely ctegorize the collection of Riemnn-integrble functions in terms of continuity. Theorem 7.6.5 (Lebesgue s Theorem). Let f be bounded function defined on the intervl [, b]. Then, f is Riemnn-integrble ifnd only ifthe set ofpoints where f is not continuous hs mesure zero.
7.6. Lebesgue s Criterion for Riemnn Integrbility 207 Proof. Let M>0stisfy f(x) M for ll x [, b], nd let D nd D α be defined s in the preceding equtions (1) nd (2). Let s first ssume tht D hs mesure zero nd prove tht our function is integrble. ( ) Set ɛ α = 2(b ). Exercise 7.6.11. Show tht there exists finite collection of disjoint open intervls {G 1,G 2,...,G N } whose union contins D α nd tht stisfies N n=1 G n < ɛ 4M. Exercise 7.6.12. Let K be wht remins of the intervl [, b] fter the open intervls G n re ll removed; tht is, K =[, b]\ N n=1 G n. Argue tht f is uniformly α-continuous on K. Exercise 7.6.13. Finish the proof in this direction by explining how to construct prtition P ɛ of [, b] such tht U(f,P ɛ ) L(f,P ɛ ) ɛ. It will be helpful to brek the sum U(f,P ɛ ) L(f,P ɛ )= n (M k m k ) x k into two prts, one over those subintervls tht contin points of D α nd the other over subintervls tht do not. ( ) For the other direction, ssume f is Riemnn-integrble. We must rgue tht the set D of discontinuities of f hs mesure zero. Fix α>0, nd let ɛ>0 be rbitrry. Becuse f is Riemnn-integrble, there exists prtition P ɛ of [, b] such tht U(f,P ɛ ) L(f,P ɛ ) <αɛ. Exercise 7.6.14. () Use the subintervls of the prtition P ɛ to prove tht D α hs mesure zero. Point out tht it is possible to choose cover for D α tht consists of finite number of open intervls. (Sets for which this is possible re sometimes clled content zero. See Exercise 7.3.6.) (b) Show how this implies tht D hs mesure zero. k=1 A Nonintegrble Derivtive To this point, our one exmple of nonintegrble function is Dirichlet s nowherecontinuous function. We close this section with nother exmple tht hs specil significnce. The content of the Fundmentl Theorem of Clculus is tht integrtion nd differentition re inverse processes of ech other. This led us to sk (in the finl prgrph of the discussion in Section 7.1) whether we could integrte every derivtive. For the Riemnn integrl, the nswer is resounding
208 Chpter 7. The Riemnn Integrl 1 Figure 7.3: A preliminry sketch of f 1 (x). no. Wht follows is the construction of differentible function whose derivtive cnnot be integrted with the Riemnn integrl. We will once gin be interested in the Cntor set C = C n, n=0 defined in Section 3.1. As n initil step, let s crete function f(x) tht is differentible on [0, 1] nd whose derivtive f (x) hs discontinuities t every point of C. The key ingredient for this construction is the function g(x) = { x 2 sin(1/x) if x>0 0 if x 0. Exercise 7.6.15. () Find g (0). (b) Use the stndrd rules of differentition to compute g (x) for x 0. (c) Explin why, for every δ>0, g (x) ttins every vlue between 1 nd 1 s x rnges over the set ( δ, δ). Conclude tht g is not continuous t x =0. Now, we wnt to trnsport the behvior of g round zero to ech of the endpoints of the closed intervls tht mke up the sets C n used in the definition of the Cntor set. The formuls re wkwrd but the bsic ide is strightforwrd. Strt by setting f 0 (x) =0 on C 0 =[0, 1]. To define f 1 on [0, 1], first ssign f 1 (x) = 0 for ll x C 1 = [ 0, 1 ] [ ] 2 3 3, 1.
7.6. Lebesgue s Criterion for Riemnn Integrbility 209 1 Figure 7.4: A grph of f 2 (x). In the remining open middle third, put trnslted copies of g oscillting towrd the two endpoints (Fig. 7.3). In terms of formul, we hve 0 if x [0, 1/3] g(x 1/3) if x is just to the right of 1/3 f 1 (x) = g( x +1/3) if x is just to the left of 2/3 0 if x [2/3, 1]. Finlly, we splice the two oscillting pieces of f 1 together in such wy tht mkes f 1 differentible. This is no gret fet, nd we will skip the detils so s to keep our ttention focused on the two endpoints 1/3 nd 2/3. These re the points where f 1(x) fils to be continuous. To define f 2 (x), we strt with f 1 (x) nd do the sme trick s before, this time in the two open intervls (1/9, 2/9) nd (7/9, 8/9). The result (Fig. 7.4) is differentible function tht is zero on C 2 nd hs derivtive tht is not continuous on the set {1/9, 2/9, 1/3, 2/3, 7/9, 8/9}. Continuing in this fshion yields sequence of functions f 0,f 1,f 2,... defined on [0, 1]. Exercise 7.6.16. () If c C, wht is lim n f n (c)? (b) Why does lim n f n (x) exist for x/ C? Now, set f(x) = lim f n(x). n Exercise 7.6.17. () Explin why f (x) exists for ll x/ C. (b) If c C, rgue tht f(x) (x c) 2 for ll x [0, 1]. Show how this implies f (c) =0.
210 Chpter 7. The Riemnn Integrl (c) Give creful rgument for why f (x) fils to be continuous on C. Remember tht C contins mny points besides the endpoints of the intervls tht mke up C 1,C 2,C 3,.... Let s tke inventory of the sitution. Our gol is to crete nonintegrble derivtive. Our function f(x) is differentible, nd f fils to be continuous on C. We re not quite done. Exercise 7.6.18. Why is f (x) Riemnn-integrble on [0, 1]? The reson the Cntor set hs mesure zero is tht, t ech stge, 2 n 1 open intervls of length 1/3 n re removed from C n 1. The resulting sum ( ) 1 2 n 1 3 n n=1 converges to one, which mens tht the pproximting sets C 1,C 2,C 3,... hve totl lengths tending to zero. Insted of removing open intervls of length 1/3 n t ech stge, let s see wht hppens when we remove intervls of length 1/3 n+1. Exercise 7.6.19. Show tht, under these circumstnces, the sum of the lengths of the intervls mking up ech C n no longer tends to zero s n. Wht is this limit? If we gin tke the intersection n=0 C n, the result is Cntor-type set with the sme topologicl properties it is closed, compct nd perfect. But consequence of the previous exercise is tht it no longer hs mesure zero. This is just wht we need to define our desired function. By repeting the preceding construction of f(x) on this new Cntor-type set of positive mesure, we get differentible function whose derivtive hs too mny points of discontinuity. By Lebesgue s Theorem, this derivtive cnnot be integrted using the Riemnn integrl. 7.7 Epilogue Riemnn s definition of the integrl ws modifiction of Cuchy s integrl, which ws originlly designed for the purpose of integrting continuous functions. In this gol, the Riemnn integrl ws complete success. For continuous functions t lest, the process of integrtion now stood on its own rigorous footing, defined independently of differentition. As nlysis progressed, however, the dependence of integrbility on continuity becme problemtic. The lst exmple of Section 7.6 highlights one type of wekness: not every derivtive cn be integrted. Another limittion of the Riemnn integrl rises in ssocition with limits of sequences of functions. To get sense of this, let s once gin consider Dirichlet s function g(x) introduced in Section 4.1. Recll tht g(x) = 1 whenever x is rtionl, nd g(x) = 0 t every irrtionl point. Focusing on the intervl [0, 1] for moment, let {r 1,r 2,r 3,r 4...}
7.7. Epilogue 211 be n enumertion of the countble number of rtionl points in this intervl. Now, let g 1 (x) =1ifx = r 1 nd define g 1 (x) = 0 otherwise. Next, define g 2 (x) =1ifx is either r 1 or r 2, nd let g 2 (x) = 0 t ll other points. In generl, for ech n N, define { 1 if x {r1,r g n (x) = 2,...,r n } 0 otherwise. Notice tht ech g n hs only finite number of discontinuities nd so is Riemnnintegrble with 1 0 g n = 0. But we lso hve g n g pointwise on the intervl [0, 1]. The problem rises when we remember tht Dirichlet s nowherecontinuous function is not Riemnn-integrble. Thus, the eqution (1) lim n 1 0 g n = fils to hold, not becuse the vlues on ech side of the equl sign re different but becuse the vlue on the right-hnd side does not exist. The content of Theorem 7.4.4 is tht this eqution does hold whenever we hve g n g uniformly. This is resonble wy to resolve the sitution, but it is bit unstisfying becuse the deficiency in this cse is not entirely with the type of convergence but lies in the strength of the Riemnn integrl. If we could mke sense of the right-hnd side vi some other definition of integrtion, then mybe eqution (1) would ctully be true. Such definition ws introduced by Henri Lebesque in 1901. Generlly speking, Lebesgue s integrl is constructed using generliztion of length clled the mesure of set. In the previous section, we studied sets of mesure zero. In prticulr, we showed tht the rtionl numbers in [0,1] (becuse they re countble) hve mesure zero. The irrtionl numbers in [0,1] hve mesure one. This should not be too surprising becuse we now hve tht the mesures of these two disjoint sets dd up to the length of the intervl [0, 1]. Rther thn chopping up the x-xis to pproximte the re under the curve, Lebesgue suggested prtitioning the y-xis. In the cse of Dirichlet s function g, there re only two rnge vlues zero nd one. The integrl, ccording to Lebesgue, could be defined vi 1 0 g = 1 [mesure of set where g =1]+0 [mesure of set where g =0] = 1 0+0 1=0. With this interprettion of 1 g, eqution (1) is now vlid! 0 The Lebesgue integrl is presently the stndrd integrl in dvnced mthemtics. The theory is tught to ll grdute students, s well s to mny dvnced undergrdutes, nd it is the integrl used in most reserch ppers where integrtion is required. The Lebesgue integrl generlizes the Riemnn integrl in the sense tht ny function tht is Riemnn-integrble is Lebesgueintegrble nd integrtes to the sme vlue. The rel strength of the Lebesgue 1 0 g
212 Chpter 7. The Riemnn Integrl integrl is tht the clss of integrble functions is much lrger. Most importntly, this clss includes the limits of different types of Cuchy sequences of integrble functions. This leds to group of extremely importnt convergence theorems relted to eqution (1) with hypotheses much weker thn the uniform convergence ssumed in Theorem 7.4.4. Despite its prevlence, the Lebesgue integrl does hve few drwbcks. There re functions whose improper Riemnn integrls exist but tht re not Lebesgue-integrble. Another disppointment rises from the reltionship between integrtion nd differentition. Even with the Lebesgue integrl, it is still not possible to prove f = f(b) f() without some dditionl ssumptions on f. Around 1960, new integrl ws proposed tht cn integrte lrger clss of functions thn either the Riemnn integrl or the Lebesgue integrl nd suffers from neither of the preceding weknesses. Remrkbly, this integrl is ctully return to Riemnn s originl technique for defining integrtion, with some smll modifictions in how we describe the fineness of the prtitions. An introduction to the generlized Riemnn integrl is the topic of Section 8.1.
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