On the fundamental theorem of calculus. Jesper Singh

Transcription

1 On the fundmentl theorem of clculus Jesper Singh VT 2015 Exmensrbete, 30hp Mgisterexmen i mtemtik, 240hp Institutionen för mtemtik och mtemtisk sttistik

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3 Abstrct The Riemnn integrl hve mny flws, some tht becomes visible in the fundmentl theorem of clculus. The min point of this essy is to introduce the guge integrl, nd prove much more suitble version of tht theorem. Smmnfttning Riemnnintegrlen hr mång brister. Viss utv dess ser mn i integrlklkylens huvudsts. Huvudmålet med denn uppsts är tt introducer guge integrlen och vis en mer lämplig version v huvudstsen.

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5 Contents Abstrct Smmnfttning 1. Introduction 1 2. The Riemnn integrl The forml definition nd some exmples Properties of the Riemnn integrl nd integrble functions The fundmentl theorem of clculus The guge integrl Guges nd δ-fine prtitions The guge integrl nd the reltion to the Riemnn integrl A little tste of exceptionl sets, null sets nd null functions Properties of the guge integrl The fundmentl theorems of clculus for the guge integrl 36 Appendix A. An open letter to uthors of clculus books 49 Acknowledgements 55 References 57

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7 1. Introduction The fundmentl theorem of clculus is historiclly mjor mthemticl brekthrough, nd is bsolutely essentil for evluting integrls. In tody s modern society it is simply difficult to imgine life without it. The history goes wy bck to sir Isc Newton long before Riemnn mde the first sound foundtion of the Riemnn integrl itself. It is rgued in [1] tht the first published proof of the fundmentl theorem of clculus is due to Jmes Gregory. The book titled Geometrie Prs Universlis tht ws published in 1668 by the Scottish mthemticin nd stronomer contins first version of the fundmentl theorem of clculus. Lter in history it is considered tht Newton himself discovered this theorem, even though tht version ws published t lter dte. For further informtion on the history of the fundmentl theorem of clculus we refer to [1]. The min point of this essy is the fundmentl theorem of clculus, nd in modern nottions it is stted s follows. The fundmentl theorem of clculus for the Riemnn integrl: Prt I Let f, F : [, b] R be two functions tht stisfies the following ) f is Riemnn integrble on [, b]; b) F is continuous on [, b], nd c) F (x) = f(x) for ll x [, b]. Then we hve tht f = F (b) F (). Prt II Let f : [, b] R be continuous function, nd define G(x) = Then G is differentible on [, b], nd G (x) = f(x) for ll x [, b]. In Theorem 2.11 we give proof of Prt I, nd in Corollry 2.1 we hve Prt II. The Riemnn integrl hve mny flws, some tht becomes visible in the bove theorem. The min points re tht there re too few Riemnn integrble functions, nd of course it is nnoying with ll the ssumptions. As seen in Exmple 2.7 there re functions f such tht there exists primitive function F, i.e. F = f on [0, 1], but f fils to be Riemnn integrble. The French mthemticin Arnud Denjoy ws interested in integrting such functions. In 1912, he published pper [5] were he defined n integrl hndling problems like this. Lter in 1914, Oskr Perron defined n integrl seemingly different to tht of Denjoy ([11]). Both these definitions re very technicl nd difficult to grsp. Surprisingly, these definitions re equivlent. It ws not until 1957, tht Jroslv Kurzweil found out n elementry definition of this integrl, nd lter developed by Henstock ([8]). As we shll see we only need to mke some minor chnges to the Riemnn integrl, nd we get something tht is surprisingly superior to the integrls of Riemnn nd Lebesgue ([2]). To put things in perspective it cn be interesting to know tht function f is Lebesgue integrble if, nd only if, f nd f re integrble in the sense of Denjoy, Perron, Kurzweil, nd Henstock ([6]). There re mny nmes for the things we love. This integrl goes under nmes like for exmple Denjoy-Perron integrl, Henstock Kurzweil integrl, nd the generlized Riemnn integrl. We shll simply cll it the guge integrl. One cn only speculte why this integrl is still in the 1 ˆ x f.

8 shdows, nd we shll not do tht here. For people interested in the history of the Riemnn, nd Lebesgue integrl we refer to [7], nd for those interested in the movement in trying to use the guge integrl in freshmen clculus we refer to n open letter ttched in Appendix A. Before we stte the fundmentl theorem of clculus for the guge integrl let us recll the definition of the Riemnn integrl, nd wht chnges tht needs to be mde to get the guge integrl. A prtition of the intervl [, b] R, is set P = {x 0, x 1,..., x n 1, x n }, x j I, tht stisfies The norm of prtition P is defined by = x 0 < x 1 <... < x n 1 < x n = b. P = mx{x 1 x 0, x 2 x 1,..., x n x n 1 }. If point t i [x i 1, x i ] hs been chosen from ech subintervl for ech i = 1, 2,..., n, then the points t i re clled tgs of the subintervl [x i 1, x i ], nd the ordered pirs P = {([x i 1, x i ], t i )} n of subintervls nd corresponding tgs is clled tgged prtition of [, b]. Now, for function f : [, b] R nd tgged prtition P = {([x i 1, x i ], t i )} n, then the Riemnn sum of f corresponding to P is defined by S(f, P ) = f(t i )(x i x i 1 ) Furthermore, we sy tht the function f is Riemnn integrble on [, b] if A R for every ε > 0 there exists number δ ε > 0 such tht if P = {[x i 1, x i ], t i } n is ny tgged prtition of [, b] with P < δ ε, then we hve tht S(f, P ) A < ε. We then sy tht f = A. We shll for exmple see tht ll continuous functions re Riemnn integrble (Theorem 2.8), s well s ll monotone functions (Theorem 2.9). The trnsition to guge integrl is surprisingly trivil. We only need to replce the constnt δ ε > 0 bove with function δ ε : [, b] (0, ). Now to the fundmentl theorem of clculus for the guge integrl. We now see clerly tht the generlity of this theorem is fr greter thn the corresponding one for the Riemnn integrl. The fundmentl theorem of clculus for the guge integrl: Prt I Let f, F : [, b] R be two functions tht stisfies the following ) F is continuous on [, b], nd b) F (x) = f(x) for ll x [, b]\e. Here E [, b] is countble set or the empty set. Then we hve tht f is guge integrble, nd f = F (b) F (). 2

9 Prt II Let f : [, b] R be function tht is guge integrble, nd define G(x) = ˆ x Then G is continuous on [, b], nd G (x) = f(x) lmost everywhere w.r.t. the Lebesgue mesure. Prt I in the bove theorem is Theorem 3.15, nd Prt II is Theorem Section 2 is ll bout the Riemnn integrl, nd this section is bsed on [3], nd section 3 is devoted to the guge integrl nd there we hve used [2]. We shll follow pure mthemticl pth, however the reder with n undergrdute bckground in mthemticl nlysis should hve no problem to follow the mteril tht is presented. A good prerequisite nd informtion bout bsic nlysis cn be found in [10]. f. 3

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11 2. The Riemnn integrl The structure of this section bout the Riemnn integrl is s follows. In subsection 2.1, we introduce the Riemnn integrl, prove tht it is sound definition, nd of course give some elementry exmples. Lter in subsection 2.2 we prove ll bsic properties tht is needed to give proof of the im of this section, nmely the fundmentl theorem of clculus. This is done in subsection The forml definition nd some exmples. In this subsection we shll introduce the Riemnn sum (Definition 2.2), nd then the concept of Riemnn integrl (Definition 2.3). We then prove tht the definition of Riemnn integrl is well-posed (Theorem 2.1). We end this subsection with some exmples. Definition 2.1. If I = [, b] is closed bounded intervl tht belongs to R, then prtition of I is finite, ordered set P = {x 0, x 1,..., x n 1, x n } of points in I such tht = x 0 < x 1 <... < x n 1, x n = b. The points of P is used to divide I = [, b] into non-overlpping subintervls I 1 = [x 0, x 1 ], I 2 = [x 1, x 2 ],..., I n = [x n 1, x n ]. More often thn not we will denote the prtition P by the nottion P = {[x i 1, x i ]} n. The norm of P is defined by the number P = mx{x 1 x 0, x 2 x 1,..., x n x n 1 }. Hence the norm is the length of the lrgest subintervl of I. Mny prtitions hve the sme norm, therefore the prtition is not function of the norm. If point t i I i hs been chosen from ech subintervl I i = [x i 1, x i ], for i = 1, 2,..., n, then the points t i re clled tgs of the subintervl I i. A set of ordered pirs P = {([x i 1, x i ], t i )} n of subintervls nd corresponding tgs is clled tgged prtition of I. The dot over the P indictes tht tg hs been selected from ech subintervl. The tgs cn be chosen in infinitely mny wys, for instnce we cn choose the left endpoints, or the right endpoints, or the midpoints of the subintervls. Note nd endpoint of subintervl cn be used s the tg for t most two consecutive subintervls. The norm of tgged prtition is defined in the sme mtter s for n ordinry prtition, so the tgs wont depend on the choice of tgs. Definition 2.2. let f : I R nd let P = {([x i 1, x i ], t i )} n is tgged prtition of I, then the sum S(f, P ) = f(t i )(x i x i 1 ), is clled the Riemnn sum of f corresponding to P. As we know from clculus the Riemnn sum under positive function f on [, b] is the sum of the res of n rectngles whose bse re the subintervls I i = [x i 1, x i ] with heights f(t i ). Definition 2.3. A function f : I R is sid to be Riemnn-integrble on I if there exists number A R such tht for every ε > 0 there exists number δ ε such tht if P = {I i, t i } n is ny tgged prtition of I with P < δ ε, then S(f, P ) A < ε, nd we write lim P 0 S(f, P ) = A. The set of ll Riemnn integrble functions on [, b] will be denoted by R[, b]. 5

12 Nottion 2.1. The interprettion of lim P 0 S(f, P ) = A is tht the integrl A is the limit of the Riemnn sums S(f, P ) when the norm P 0. This is different thn the limit of function since S(f, P ) is not function of P. Mny different prtitions cn shre the sme norm. Insted of A we often use the clculus nottion A = f or f(x)dx. The letter x is just dummy vrible nd cn be replced with ny other letter s long it does not become mbiguity. Theorem 2.1. If f R[, b], then the vlue of the integrl is uniquely determined. Proof. Assume tht A nd A stisfies Definition 2.3 nd let ε > 0 be given. Then there exists δ ε/2 > 0 such tht if P 1 is ny tgged prtition with P 1 < δ ε/2, we hve S(f, P 1 ) A < ε/2. Similrly there exists δ ε/2 > 0 such tht if P 2 is ny tgged prtition with P 2 < δ ε/2, then S(f, P 2 ) A < ε/2. Put δ ε = min{δ ε/2, δ ε/2 } > 0 nd let P be tgged prtition with P < δ ε. Since P < δ ε/2 nd P < δ ε/2, we get S(f, P ) A < ε/2 nd S(f, P ) A < ε/2. Hence by the Tringle Inequlity it follows tht A A = L S(f, P ) S(f, P ) L L S(f, P ) + S(f, P ) L < ε/2 + ε/2 = ε. Since ε > 0 is rbitrry, it follows tht L = L. If Definition 2.3 is used to show tht function is Riemnn integrble we must know (or guess) the vlue A of the integrl, lso we must construct δ ε tht will do for n rbitrry ε > 0. The determintion of the vlue A cn be done by clculting Riemnn sums nd guessing the vlue A, the construction of sufficient δ ε will often give us hrd time. Lter on we will prove some theorems tht will be very hndy when it comes to determine if f R[, b]. For now lets hve look t some instructive exmples. Exmple 2.1. If f : [, b] R nd f = c, where c is constnt, then f R[, b] nd b f = c(b ). Let f(c) = c for ll x [, b]. If P = {([x i 1, x i ], t i )} n is ny tgged prtition of [, b], then S(f, P ) = f(t i )(x i x i 1 ) = c(x i x i 1 ) = c(b ). Thus, for ny ε > 0 we cn choose δ ε with ese, lets put δ ε = 1 so tht if P < 1, then S(f, P ) A = S(f, P ) c(b ) = 0 < ε. Since ε > 0 is rbitrry, the conclusion is tht f R[, b] nd b f = c(b ). 6

13 Exmple 2.2. Let g : [0, 1] R be defined by g(x) = x for ll x [0, 1]. Then g R[0, 1] with 1 0 g = 1 2. Now the trick of using prticulr points s tgs will be good guess of the integrl vlue. If Q = {I i } n is ny prtition of [0, 1], choose the tg t i I i = [x i 1, x i ] to be the midpoint q i = 1 2 (x i 1 + x i ) we now hve tgged prtition Q = {(I i, q i )} n. The Riemnn sum becomes S(g, Q) = g(q i )(x i x i 1 = 1 2 (x i + x i 1 )(x i x i 1 ) = 1 2 (x2 i x 2 i 1) = 1 2 ( ) = 1 2. Now let P = {(I i, t i )} n be n rbitrry tgged prtition of [0, 1] with P < δ so tht x i x i 1 < δ for i = 1,..., n. Also let Q hve the sme prtition points, but choose the tg q i to be the midpoint of the intervl I i. The tgs of P cn be nywhere between x i 1 nd x i. Since t i, q i [x i 1, x i ], we hve t i q i < δ. Using the Tringle Inequlity, we deduce S(g, P ) S(g, Q) = t i (x i x i 1 ) q i (x i x i 1 ) = (t i q i )(x i x i 1 ) n t i q i (x i x i 1 ) < δ (x i x i 1 ) = δ(x n x 0 ) = δ(1 0) = δ. Since S(g, Q) = 1 2, we infer tht if P is ny tgged prtition with P < δ, then S(g, P ) 1 2 < δ. Therefore ny δ ε ε will do. If we choose δ ε = ε nd works bckwrds we conclude tht g R[0, 1] nd 1 0 g = 1 0 xdx = 1 2. Chnging function t finite number of points does not ffect its integrbility nor the vlue of the integrl. Theorem 2.2. Let g : [, b] R be integrble on [, b] nd if g(x) = f(x) except for finite number of points in [, b], then f is Riemnn integrble nd b f = b g. Proof. Lets prove the cse of one exceptionl point, then the extension to finite number of points cn be done by mthemticl induction. Assume c [, b] nd f(x) = g(x) for ll x c nd let A = b g. For ny tgged prtition P the Riemnn sums S(f, P ) nd S(g, P ) re identicl with exception of t most two terms, tht is the cse when c = x i or c = x i 1. Therefore S(f, P ) S(g, P ) = (f(x i ) g(x i )) (x i x i 1 ) = [f(c) g(c)] (x i x i 1 ) ( f(c) + g(c) ) (x i x i 1 ) ( f(c) + g(c) ) P. Now, given ε > 0, let δ 1 > 0 stisfy δ 1 < ε/2( f(c) + g(c) ), then S(f, P ) S(g, P ) < ε/2. Since g R[, b] there exists δ 2 > 0 such tht when P < δ 2 we hve S(g, P ) A < ε/2. Now let δ = min{δ 1, δ 2 }. Then if P < δ we obtin S(f, P ) A = S(f, P ) S(g, P )+S(g, P ) A S(f, P ) S(g, P ) + S(g, P ) A < ε/2+ε/2 = ε. Hence f R[, b] with b f = A. 7

14 The interprettion of Theorem 2.2 is tht singleton set hs no length so the re under the curve is zero. Exmple 2.3. Let g : [, b] R be defined by { c if x (, b) g(x) = 0 if x {, b}. Since the function f in Exmple 2.1 is integrble with integrl c(b ), nd g = f except t the finite points or b, then g R[, b] with b g = c(b ) by Theorem Properties of the Riemnn integrl nd integrble functions. In this subsection we strt by proving some bsic properties of the Riemnn integrl, like the linerity in Theorem 2.3, tht every Riemnn integrble function is bounded on [, b] (Theorem 2.4), but the most importnt for the rest of this section on the Riemnn integrl is tht continuous functions defined on compct intervl [, b] is Riemnn integrble (Theorem 2.8). The obstcle involved when determining the vlue of the integrl nd of δ ε suggest tht it would be very nice nd useful to hve some generl theorem. The next result enbles us to perform certin lgebric mnipultions of functions tht belong to R[, b]. Theorem 2.3. Suppose f, g R[, b] nd k R. Then ) kf R[, b] nd b kf = k b f; b) (f + g) R[, b] nd b (f + g) = b f + b g; c) if f(x) g(x) for ll x [, b], then b f b g. Proof. ) For k = 0 there is nothing to prove so we ssume k 0. Let ε > 0 be given. If P = {([x i 1, x i ], t i )} n is tgged prtition of [, b] nd since f R[, b] there exists δ ε > 0 such tht S(f, P ) f < ε whenever P k < δ ε. Thus, if P < δ ε then S(kf, P ) k f = ks(f, P ) k f = k S(f, P ) Hence lim P 0 S(kf, P ) = k b f, therefore kf R[, b] nd b kf = k b f < k ε k = ε. f by Definition 2.3. b) Let ɛ > 0 be given. Since f, g R[, b] there exists δ ε/2 > 0 such tht if P 1 is ny tgged prtition with P 1 < δ ε/2, then S(f, P ) f < ε 2, nd there exists δ ε/2 > 0 such tht if P 2 is ny tgged prtition with P 2 < δ ε/2, then S(g, P ) g < ε 2. 8

15 Let δ ε = min{δ ε/2, δ ε/2 } > 0 nd let P be tgged prtition with P < δ ε. Since both P < δ ε/2 nd P < δ, we get ε/2 ( S(f + g, P ) f + g) = S(f, P ) + S(g, P ) S(f, P ) f g f + S(g, P ) g < ε 2 + ε 2 = ε. Hence lim P 0 S(f + g, P ) = b f + b g, so (f + g) R[, b] nd b f + g = b f + b g by Definition 2.3. c) Let ε > 0 be given. Since f, g R[, b] then for ny tgged prtition P with P < δ ε we hve S(f, P ) f < ε nd 2 S(g, P ) g < ε 2. Removing the bsolute vlues nd using the fct tht S(f, P ) S(g, P ) gives us dding ε/2 yields f ε 2 < S(f, P ) S(g, P ) < f < S(f, P ) + ε 2 S(g, P ) + ε 2 < g + ε 2, g + ε. Since b fz b g + ε for every ε > 0 the conclusion is tht b f b g. Now when some lgebric combintions been estblished lets look t certin demnds on the function f to be required for the Riemnn integrl to exist. An unbounded function for sure is not Riemnn integrble. Let B[, b] denote the set of ll bounded functions on [, b]. Theorem 2.4. If f R[, b], then f B[, b]. Proof. The proof is by contrdiction. Assume f / B[, b] nd b f = A. Then there exists δ > 0 nd ε > 0 such tht if P is ny tgged prtition of [, b] with P < δ, we hve S(f, P ) A < ε, which gives A ε < S(f, P ) < A + ε, hence S(f, P ) < A + ε A + ε. (2.1) Now let Q = {[x i 1, x i ]} n be prtition of [, b] with Q < δ. Since f is unbounded on [, b], there exists t lest one subintervl in Q, sy [x k 1, x k ], where f is not bounded. If f where bounded on ech [x i 1, x i ] [, b] by sy M i, then f would be bounded on [, b] by mx{m 1,..., M n }. The strtegy now is to tg Q in such wy tht eqution 2.1 is contrdicted. To do so tg Q by t i = x i for i k nd t k [x k 1, x k ] such tht f(t k )(x k x k 1 ) > A + ε + f(t i )(x i x i 1 ), this is lwys possible since f is not bounded, thus f(t k )(x k x k 1 ) f(t i )(x i x i 1 ) > A + ε. i k 9 i k

16 By the inequlity form A + B A B we find S(f, Q) = f(t k)(x k x k 1 ) + f(t i )(x i x i 1 ) i k f(t k )(x k x k 1 ) f(t i )(x i x i 1 ) > A + ε. Hence we hve found tgged prtition of [, b] which contrdicts eqution 2.1, therefore n unbounded function cnnot be integrble. Since f R[, b] the conclusion is tht f B[, b]. Note tht there is no gurntee tht bounded function is Riemnn integrble(it my or my not be). Lets hve look t nother importnt steppingstone tht will be needed to estblish the integrbility of severl clsses of functions such s step, monotone nd continuous functions. This result is known s the Cuchy criterion. The mjor gin by the Cuchy criterion is tht it mkes it possible to prove tht function is Riemnn integrble without the need to know the integrls vlue. There is price to be pid thou, we need to consider two Riemnn sums insted of just one. Theorem 2.5 (Cuchy criterion). A function f : [, b] R belongs to R[, b] if nd only if for every ε > 0 there exists η ε > 0 such tht if P nd Q re ny tgged prtitions of [, b] with P < η ε nd Q < η ε, then S(f, P ) S(f, Q) < ε. Proof. Let f R[, b] with b f = A, let η ε = δ ε/2 > 0 be such tht if P, Q re tgged prtitions of [, b] such tht P < η ε nd Q < η ε, then S(f, P ) A < ε/2 nd S(f, Q) A < ε/2. Hence S(f, P ) S(f, Q) = S(f, P ) A + A S(f, Q) i k S(f, P ) A + A S(f, P ) < ε/2 + ε/2 = ε. Conversely for ech n N let δ n > 0 be such tht if P nd Q be tgged prtitions of [, b] with P < δ n nd Q < δ n, then S(f, P ) S(f, Q) < 1/n. We cn ssume tht δ n δ n+1 for ech n N, if not just substitute δ n with δ n = min{δ 1,..., δ n }. For ech n N, let P n be tgged prtition of [, b] with P n < δ n. Hence, if m > n then both P n < δ n nd P m < δ n, therefore S(f, P n ) S(f, P m ) < 1/n if m > n, consequently {S(f, P m )} m=1 is Cuchy sequence in R nd therefore this sequence converges in R to sy A, thus sending m to the limit gives us lim m S(f, P m ) = A, nd S(f, P n ) L 1/n for ll n N. To be convinced tht b f = A, given ε > 0, let K N stisfy K > 2/ε. If Q prtition of [, b] with Q < δ K, then S(f, Q) A = S(f, Q) S(f, P K ) + S(f, P K ) A Since ε > 0 is rbitrry, f R[, b] nd b f = A. is ny tgged S(f, Q) S(f, P K ) + S(f, P K ) A 1/K + 1/K = 2/K < ε. 10

17 Theorem 2.5 is prcticl tool to determine if either f R[, b] or f / R[, b]. Exmple 2.4. Let f : [0, 1] R be defined by { 1 if x Q [0, 1] f(x) = 0 if x (R\Q) [0, 1]. Then f / R[, b]. Hence we hve to show tht there exists ε 0 > 0 such tht for ny δ ε0 > 0 there exists prtitions P nd Q with P < ε 0 nd Q < ε 0 such tht S(f, P ) S(f, Q) ε 0. In order to do so we choose ε 0 = 1/2. If P is ny prtition with tgs t i Q [0, 1] then S(f, P ) = 1, on the other hnd if Q is ny tgged prtition with tgs t k (R\Q) [0, 1] then S(f, P ) = 0. Hence S(f, P ) S(f, Q) = 1 > 1/2, since P, Q re tgged prtitions with rbitrry smll norms, the conclusion is tht f / R[, b]. The definition of the Riemnn integrl gives us two types of obstcles. The first one is tht there is infinitely wys to choose tgs t i. Secondly there is infinitely mny prtitions with norm less thn δ ε. Below we will estblish useful theorem tht will ese us from some of the difficulties when proving tht function f is integrble, the theorem is clled the squeeze theorem. The bsic ide is tht function cn be squeezed between two functions known to be integrble. It ws this ide tht led the French mthemticin Gston Drboux to develop different pproch to integrtion by introducing upper nd lower integrls. For fully tretment of the Drboux integrl we refer to [10]. Theorem 2.6. Let f : [, b] R. Then f R[, b] if nd only if for every ε > 0 there exists functions α ε, ω ε R[, b] with such tht α ε (x) f(x) ω ε (x) for ll x [, b], (ω ε α ε ) < ε. Proof. Choose α ε = ω ε = f for ll ε > 0, then b (ω ε α ε ) = 0 < ε. Conversely, let ε > 0 be given. Since α ε, ω ε R[, b], there exists δ ε > 0 such tht if P is ny tgged prtition with P < δ ε, then ˆ b S(α ε, P ˆ b ) α ε < ε nd S(ω ε, P ) ω ε. These inequlities gives α ε ε < S(α ε, P ) nd S(ω ε, P ) < Since α ε f ω ε we hve S(α ε, P ) S(ω ε, P ), hence α ε ε < S(f, P ) < ω ε + ε. If Q is nother tgged prtition with Q < δ ε, then we lso hve α ε ε < S(f, Q) < 11 ω ε + ε, ω ε + ε.

18 by subtrcting these two inequlities nd using the ssumtion b (ω ε α ε ) < ε, we conclude tht S(f, P ) S(f, Q) ( < ω ε + ε α ε ε) = ω ε Since ε > 0 is rbitrry f R[, b] by Theorem 2.5. α ε + 2ε = (ω ε α ε ) + 2ε < ε + 2ε = 3ε. Step functions re importnt in the nture of Riemnn integrbility, nd will be used to prove tht every continuous function over nd intervl [, b] is lso Riemnn integrble over the sme intervl. First we prove specil cse in the following lemm, then we prove the generl sttement in Theorem 2.7. Lemm 2.1. If J [, b] hving endpoints c < d nd if ϕ J (x) = 1 for x J nd ϕ J (x) = 0 for x [, b] \ J, then ϕ J R[, b] nd b ϕ J = d c. Proof. With c = nd d = b it is just Exmple 2.1 ll over gin so ssume < c nd d < b. If P = {([x i 1, x i ], t i )} n is ny tgged prtition of [, b], then it is cler tht S(ϕ J, P ) = ϕ J (t i )(x i x i 1 ) = ϕ J (t 1 )( c) + ϕ J (t 2 )(d c) + ϕ J (t 3 )(b d) = ϕ J (t 2 )(d c) = d c. Hence, for ny ε > 0, we cn choose δ ε = 1 so if P < δ ε, then S(ϕ J, P ) (d c) = 0 < ε. Since ε > 0 is rbitrry, the conclusion is tht ϕ J R[, b] nd b ϕ J = d c. Theorem 2.7. If ϕ : [, b] R is step function, then ϕ R[, b]. Proof. Step functions like in Lemm 2.1 re clled elementry step functions, for further studies see [2]. Any step function ϕ cn be written s liner combintion of elementry step functions: m ϕ = k j ϕ Jj, j=1 where J j hs endpoints c j < d j. Lemm 2.1 nd Theorem 2.3 implies tht ϕ R[, b] nd tht ϕ = (k 1 ϕ J k m ϕ Jm ) = k 1 ϕ J k m ϕ Jm = k 1 (d 1 c 1 ) k m (d m c m ) = m k j (d j c j ). Some exmples regrding step functions nd the squeeze theorem would be pproprite. Exmple 2.5. Let g : [0, 1] R be defined by { 2 if x [0, 1] g(x) = 3 if x (1, 3]. 12 j=1

19 To show tht g R[0, 3] using Definition 2.3 would be hlf complicted nd require some tricks, see [2] Exmple 7.1.4, pge 202. However since m 2 g = k j ϕ Jj = k j ϕ Jj, is step function nd therefore by Theorem 2.7 we hve ˆ 1 2 g = k j ϕ Jj = 2(1 0) + 3(3 1) = = 8. 0 j=1 j=1 The next exmple combines step functions with the useful squeeze theorem. Exmple 2.6. Let h(x) = x on [0, 1] nd let P n = {0, 1/n, 2/n,..., (n 1)/n, n/n = 1}. Define α n : [0, 1] R by { ( h k 1 ) n = k 1 k 1 n, if x [ n α n (x) =, k n ), k = 1, 2,..., n 1 h ( ) n 1 n = n 1 n 1 n, if x [ n, 1], k = n. Hence α n h on [0, 1/n), [1/n, 2/n),..., [(n 2)/n, (n 1)/n), [(n 1)/n, 1]. Similrly we define ω n : [0, 1] R by { ( h k ) n = k n ω n (x) =, k 1 if x [ n, k n ), k = 1, 2,..., n 1 h ( n n ) = 1) = n 1 n 1 n, if x [ n, 1], k = n. Hence ω n h on [0, 1/n), [1/n, 2/n),..., [(n 2)/n, (n 1)/n), [(n 1)/n, 1]. By Theorem 2.7 we hve ˆ 1 (k 1) α n = 1 0 n n = 1 n 2 k 1 = 1 ( n 1) n2 k=1 k=1 = 1 n(n 1) = 1 ( 1 1 ). n2 2 2 n Also ˆ 1 0 ω n = k=1 k n 1 n = 1 n 2 k=1 Thus we hve α n (x) h(x) ω n (x) for x [0, 1] nd j=1 k = 1 ( n 1 + n) n2 = 1 n(n + 1) = 1 ( ). n2 2 2 n ˆ 1 0 ω n ˆ 1 0 α n = ˆ 1 0 (ω n α n ) = 1 n. Therefore for given ε > 0 we cn choose n N so lrge tht 1 n < ε, Theorem 2.6 tells us tht n h R[0, 1] nd since lim n k=1 α n n 1/n = lim n k=1 ω n 1/n = 1/2 we lso know tht 1 0 h = 1/2 since 1 0 α n 1 0 h 1 0 ω n. We re now equipped to prove tht ll continuous functions over [, b] belongs to R[, b], to do so we will use Theorem 2.6 nd some properties bout continuous functions. Let C[, b] denote ll continuous functions on [, b]. Theorem 2.8. If f : [, b] R nd f C[, b], then f R[, b]. 13

20 Proof. f is continuous on [, b] nd therefore f is uniformly continuous on [, b]. For further explntion bout uniformly continuity see [10]. Hence, for given ε > 0 there exists δ ε such tht if u, v [, b] nd u v < δ ε, then we hve f(u) f(v) < ε/(b ). Let P = {I i } n be prtition such tht P < δ ε. Since f C[, b], there exists points u, v [, b] such tht f(u) f(x) f(v) see [10] Theorem 4.5, pge 104. Let u i I i be the point such tht f(u i ) f(x) for ll x I i, nd let v i be the point such tht f(x) f(v i ) for x I i. Define step function α ε : [, b] R by { f(ui ) for x [x i 1, x i ), i = 1,..., n 1 α ε (x) = f(u n ) for x [x n 1, x n ]. Similrly define ω ε : [, b] R by { f(vi ) for x [x i 1, x i ), i = 1,..., n 1 ω ε (x) = f(v n ) for x [x n 1, x n ]. This gives us Hence 0 (ω ε α ε ) = α ε (x) f(x) ω ε (x) for ll x [, b]. (f(v i ) f(u i )) (x i x i 1 ) < Therefore it follows from Theorem 2.6 tht f R[, b]. ( ) ( ) ε ε (x i x i 1 ) = (x n x o ) = ε. b b Monotone functions re not lwys continuous over n intervl [, b], however they re lwys integrble on [, b]. Theorem 2.9. If f : [, b] R is monotone on [, b], then f R[, b]. Proof. Assume f(x i 1 ) f(x i ) on [, b] (the cse when f is decresing is hndled similrly). Prtition [, b] into n subintervls I k = [x k 1, x k ] with equl lenght, thus x k x k 1 = (b )/n, k = 1,..., n. Since f is incresing on I k, the minimum vlue is ttined t the left endpoint x k 1 nd the mximum vlue is ttined t the right endpoint x k. Define α ε : [, b] R by { f(xk 1 ) for x [x k 1, x k ), k = 1,..., n 1 α ε (x) = f(x n 1 ) for x [x n 1, x n ]. Similrly define ω ε : [, b] R by { f(xk ) for x [x k 1, x k ), i = 1,..., n 1 ω ε (x) = f(x n ) for x [x n 1, x n ]. Then we hve Theorem 2.7 gives nd α ε = k=1 ω ε = α ε (x) f(x) ω ε for ll x [, b]. f(x k 1 )(x k x k 1 ) = b n (f(x 0) + f(x 1 ) f(x n 1 )), k=1 f(x k )(x k x k 1 ) = b n (f(x 1) + f(x 2 ) f(x n )). 14

21 Subtrcting nd cncelltion of terms yields (ω ε α ε ) = (f(x k ) f(x k 1 )) (x k x k 1 ) k=1 = b n (f(x n) f(x 0 )) = b (f(b) f()). n Hence for given ε > 0, we choose n N so lrge tht n > (b )(f(b) f())/ε. Consequently b (ω ε α ε ) < ε nd therefore f R[, b] by Theorem 2.6. Lets return to rbitrry functions tht belong to R[, b]. nevertheless it requires proof. The next result is quite obvious, Theorem Let f : [, b] R nd let c (, b). Then f R[, b] if nd only if its restriction f 1 of f to [, c] is in R[, c] nd f 2 of f to [c, b] is in R[, b]. In this cse f = ˆ c Proof. Suppose tht f 1 R[, c] with c f 1 = A 1 nd f 2 R[c, b] with b c f 1 = A 2. Then for given ε > 0 there exists δ > 0 such tht if P 1 is tgged prtitions of [, c] with P 1 < δ, then S(f 1, P 1 ) A 1 < ε/3. Also there exist δ > 0 such tht if P 2 is ny tgged prtition of [c, b] with P 2 < δ, then S(f 2, P 2 ) A 2 < ε/3. If M is bound for f, define δ ε = min{δ, δ, ε/6m} nd let Q be tgged prtition of [, b] with Q < δ. We hve to prove tht f + c f. S(f, Q) (A 1 + A 2 ) < ε. There re two cses to consider i) If c Q nd we split Q into prtition Q 1 of [, c] nd prtition Q 2 of [c, b]. Since S(f, Q) = S(f, Q 1 ) + S(f, Q 2 ) nd since Q 1 < δ nd Q 2 < δ, we hve S(f, Q) (A 1 + A 2 ) = S(f, Q 1 ) + S(f, Q 2 ) (A 1 + A 2 ) S(f, Q 1 ) A 1 + S(f, Q 2 ) A 2 < ε 3 + ε 3 = 2ε 3 < ε. ii) If c / Q = {(I k, t k )} m k=1, then there exists k m such tht c (x k 1, x k ). Just dd c to Q nd let Q 1 be the tgged prtition of [, c] defined by similrly let Q 1 = {(I 1, t 1 ),..., (I k 1, t k 1 ), ([x k 1, c], c)}. Q 2 be the tgged prtition of [c, b] defined by Subtrcting mny terms shows tht Q 2 = {([c, x k ], c), (I k+1, t k+1 ),..., (I m, t m )}. S(f, Q) S(f, Q 1 ) S(f, Q 2 ) m k 1 = f(t k )(x k x k 1 ) f(t k )(x k x k 1 ) k=1 k=1 m k=k 1 = f(t k )(x k x k 1 ) f(c)(c x k 1 ) f(c)(x k c) f(t k )(x k x k 1 ) = f(t k )(x k x k 1 ) f(c)(x k x k 1 ) = (f(t k ) f(c)) (x k x k 1 ). Also since f(t k ) f(c) 2M nd x k x k 1 < ε/6m it follows tht S(f, Q) S(f, Q 1 ) S(f, Q 2 ) 2M(x k x k 1 ) < 2M 15 ε 6M = ε 3.

22 Now since Q 1 < δ ε δ nd Q 2 < δ ε δ, it follows tht S(f, Q) (A 1 + A 2 ) = S(f, Q) S(f, Q 1 ) S(f, Q 2 + S(f, Q 1 ) + S(f, Q 2 (A 1 + A 2 ) S(f, Q) S(f, Q 1 ) S(f, Q 2 + S(f, Q 1 A 1 + S(f, Q 2 A 2 < ε 3 + ε 3 + ε 3 = ε. Since ε > 0 is rbitrry, we conclude tht f R[, b] nd b f = c f + b c f. Conversely, suppose tht f R[, b]. For given ε > 0, let η ε > 0, by Theorem 2.5 we hve S(f, P ) S(f, Q < ε, for ny tgged prtitions P nd Q with norm less thn η ε. Let f 1 be the restriction of f to [, c]. Also let P 1, Q 1 be tgged prtitions of [, c] with P 1, Q 1 < η ε. By dding dditionl prtition points nd tgs from [c, b], we cn extend P 1 nd Q 1 to tgged prtitions P nd Q of [, b] tht stisfy P, Q < η ε. If we dd the sme dditionl points nd tgs in [c, b] for both P nd Q, then S(f 1, P 1 ) S(f 1, Q 1 ) = S(f, P ) S(f, Q). Since P, Q < η ε, we hve S(f 1, P 1 ) S(f 1, Q 1 ) < ε. Therefore the restriction f 1 of f to [, c] belongs to R[, c] by Theorem 2.5. A repetition of the rgument bove shows tht the restriction f 2 of f belongs to R[c, d], only difference is tht the prtition we re borrowing points from is [, c], insted of [c, b]. Now since since f R[, c] nd f R[c, b] hs been estblished the equlity c f + b c f = b f follows from the first prt of the theorem The fundmentl theorem of clculus. This is the finl prt of the Riemnn integrl. We shll stte nd prove the fundmentl theorem of clculus. Furthermore, we give one exmple to show some of its flws. Theorem Let f, F : [, b] R such tht Then ) F C[, b]; b) F (x) = f(x) for ll x [, b]; c) f R[, b]. f = F (b) F (). Proof. Since f R[, b], given ε > 0, there exists δ ε > 0 such tht if P is ny tgged prtition with P < δ ε, then S(f, P ) f < ε. Since F is continuous on [, b] nd if [x i 1, x i ] P then the men-vlue theorem pplied to F on [x i 1, x i ] implies tht there exists u i (x i 1, x i ), such tht F (x i ) F (x i 1 ) = F (u i ) (x i x i 1 ) for i = 1,..., n. 16

23 For more informtion on the Men-vlue theorem of differentil cclulus see [10] Theorem 5.6, pge 134. Adding, using the fct tht F (u i ) = f(u i ) nd cncelltion of terms gives F (b) F () = (F (x i ) F (x i 1 )) = f(u i )(x i x i 1 ). Let P u = {([x i 1, x i ], u i )} n, thus S(f, P u ) = n f(u i)(x i x i 1 ), therefore S(f, P u ) f = F (b) F () f < ε. Since ε > 0 is rbitrry, we conclude tht b f = F (b) F (). Even if F exist for ll x [, b] it is not lwys true tht f R[, b]. Exmple 2.7. Let F : [0, 1] R be defined by { x 2 cos(1/x 2 ) if x (0, 1] F (x) = 0 if x = 0. Differentiting yields F (x) = { 2x cos(1/x 2 ) + (2/x) sin(1/x 2 ) for x (0, 1], 0 for x = 0. For x (0, 1] the chin nd product rules pplies nd for x = 0 we hve F (x) F (0) F (0) x 0 = x 2 cos(1/x 2 ) x = x cos(1/x 2 ) x < ε. Hence for x 0 < δ choose ε = δ, this proves tht F (0) = 0. Therefore F is continuous nd differentible t every point x [0, 1], but since F fils to be bounded on [0, 1] it follows tht F / R[0, 1]. So condition c) cnnot be omitted in Theorem In section 3 we will see tht f G[0, 1] nd Theorem 3.15 will even provide us with ntiderivtive of f in some sense. We now wish to differentite n integrl with fluctuting upper limit, the following will be needed. Definition 2.4. If f R[, b], then the function F : [, b] R defined by F (x) = ˆ x is clled the indefinite integrl of f with bse point. f for x [, b], Definition 2.5. A function f : [, b] R is sid to be Lipschitz function (or stisfy Lipschitz condition) on [, b] if there exists constnt K > 0 such tht f(x 1 ) f(x 2 ) K x 1 x 2 for ll x 1, x 2 [, b]. Every Lipschitz function is uniformly continuous. Theorem If f : [, b] R stisfies Lipschitz condition, then f is uniformly continuous on [, b]. Proof. Let f be s in Definition 2.5. For given ε > 0, we choose δ = ε/k. Now if x 1 x 2 < δ for ll x 1, x 2 [, b], then f(x 1 ) f(x 2 ) K x 1 x 2 < K ε K = ε. 17

24 The indefinite integrl F of Riemnn integrble function f stisfies Lipschitz condition nd is therefore continuous. Theorem The indefinite integrl F in Definition 2.4 is continuous on [, b]. If f(x) M for ll x [, b], then F (x) F (y) M x y for ll x, y [, b]. Proof. Theorem 2.10 implies tht if x, y [, b] nd y x, then from this we hve F (x) = ˆ x f = ˆ y f + ˆ x y F (x) F (y) = f = F (y) + By hypothesis if M f(x) M for ll x [, b], then Theorem 2.3 implies tht since y x it follows tht M(x y) ˆ x F (x) F (y) = y ˆ x y ˆ x y f. f M(x y), f M x y. Therefore F is uniformly continuous on [, b] by Theorem The derivtive of the indefinite integrl F exists t every point where f i continuous. Theorem Let f R[, b] nd let lim x c+ f(x) = f(c) for c [, b). Then the indefinite integrl hs right hnd derivtive t c equl to f(c). F (x) = Proof. Assume c [, b). Since f is continuous t c, given ε > 0 there exists δ ε > 0 such tht if c x < c + δ ε, then f(c) ε < f(x) < f(c) + ε. (2.2) Let h stisfy 0 < h < δ ε. Since f R[, b] Theorem 2.10 implies tht f (R[, c], R[, c + h], R[c, c + h]) nd ˆ x F (c + h) F (c) = f, ˆ c+h On the intervl [c, c + h] inequlity 2.2 holds for f, therefore Hence ˆ c+h c f(c) ε < ˆ c+h c c f(x) < (f(c) ε) h < F (c + h) F (c) = divide by h > 0 nd subtrct f(c) everywhere, we obtin thus f. ˆ c+h c ˆ c+h F (c + h) F (c) ε < f(c) < ε, h F (c + h) F (c) f(c) h < ε, 18 c ˆ x y f, f(c) + ε. f < (f(c) + ε) h,

25 since ε > 0 is rbitrry we conclude tht F (c + h) F (c) lim = f(c). h 0+ h Approching the limit from the left gives the following result. Theorem Let f R[, b] nd let lim x c f(x) = f(c) for c [, b). Then the indefinite integrl hs left hnd derivtive t c equl to f(c). F (x) = Proof. Assume c (, b]. Since f is continuous t c, given ε > 0 there exists δ ε > 0 such tht if c δ ε < x c, then f(c) ε < f(x) < f(c) + ε. (2.3) Let h stisfy 0 < h < δ ε. Since f R[, b] Theorem 2.10 implies tht f (R[, c h], R[, c], R[c h, c]) nd ˆ x F (c) F (c h) = f, ˆ h c h On the intervl [c h, c] inequlity 2.3 holds for f, therefore Hence ˆ c c h f(c) ε < ˆ c c h f(x) < (f(c) ε) h < F (c) F (c h) = divide by h > 0 nd subtrct f(c) everywhere, we obtin ˆ c ˆ c f. c h c h f(c) + ε. F (c) F (c h) ε < f(c) < ε, h thus F (c) F (c h) f(c) h < ε, since ε > 0 is rbitrry we conclude tht If f C[, b], then we hve fmilir result. F (c) F (c h) lim = f(c). h 0 h f < (f(c) + ε) h, Corollry 2.1. If f C[, b], then the indefinite integrl F (x) = x f is differentible on the whole intervl [, b] nd F (x) = f(x) for ll x [, b]. Proof. Let c (, b). Since f is continuous t c we hve lim x c+ f(x) = lim x c f(x) = f(c). Therefore ccording to Theorems 2.14 nd 2.15 bove we hve F F (c + h) F (c) F (c) F (c h) F (c + h) F (c) (c) = lim = lim = lim = f(c). h 0+ h h 0 h h 0 h If c = or c = b we get F +() F ( + h) F () = lim = f(), h 0+ h 19

26 respectively F (b) F (b) F (b h) lim = f(b). h 0 h The indefinite integrl of f need not to be n ntiderivtive of f, mybe the derivtive does not exist t c or F (c) f(c), however if f C[, b], then its indefinite integrl F is gurnteed to be n ntiderivtive of f by Corollry

27 3. The guge integrl In this section we shll generlize the Riemnn integrl. First in subsection 3.1 we introduce some nottions, nd then in subsection 3.2 we define the guge integrl. There we lso prove tht every Riemnn integrble function is lso guge integrble, nd give n exmple tht there is function tht is guge integrble but not Riemnn integrble. To be ble to formulte the dvnced version of the fundmentl theorem of clculus for the guge integrl we need some bsic mesure theory, this is done in subsection 3.3. Much in the sme vein s for the Riemnn integrl we prove in subsection 3.4 some bsic properties. Finlly in subsection 3.5 we obtin the im of this essy the fundmentl theorem of clculus for the guge integrl Guges nd δ-fine prtitions. In this subsection we shll introduce some necessry nottion nd bckground for the guge integrl. Definition 3.1. The closed neighborhood of x with rdius r > 0 is the set B[x, r] = {y R : x y r}, which is lso clled the closed bll with center x nd rdius r. The open neighborhood of x with rdius r > 0 is the set B(x, r) = {y R : x y < r}, which is lso clled the open bll with center x nd rdius r. If I = [, b], nd b, we define the length of I to be l(i) = b. It is esy to see tht l(i) 0, nd tht l(i) = 0 if nd only if = b. Similrly, the length of ny intervl of the forms (, b), [, b), (, b], is lso defined to be b. The length of the empty set is defined s l( ) = 0, this lengths coincide with the Lebesgue mesure. We sy tht n intervl I R is degenerte if it contins t most one point, nd tht it is non degenerte if it contins t lest two points, in which cse it contins infinitely mny points. We sy tht two intervls I 1, I 2 R re disjoint if I 1 I 2 =. We sy tht two intervls I 1, I 2 R re non overlpping if I 1 I 2 = or I 1 I 2 = {p}, where {p} is singleton set, which is n endpoint of both intervls. Exmple 3.1. If J R is bounded open intervl nd suppose tht I 1,..., I m re nonoverlpping closed nd bounded intervls contined in J. Then m l(i i ) l(j). If I 1,..., I m re degenerte then m l(i i) = 0 l(j). Suppose tht the intervls I i re nondegenerte, so i < b i. If necessry we relbel such tht m. Since the intervls re nonoverlpping, we hve α < 1 < b 1 2 < b 2... m < b m < β, where J = (α, β). It follows tht b i i i+1 i for i = 1,..., m 1 nd b m m < β m. Adding the inequlities, we get m m (b 1 1 ) + (b 2 2 ) (b m m ) = (b i i ) = l(i i ) < β 1 < β α = l(j). 21

28 Definition 3.2. If I = [, b] R, then the function δ : I R is sid to be guge on I if δ(t) > 0 for ll t I. The intervl round t I controlled by the guge δ is the intervl B[t, δ(t)] = [t δ(t), t + δ(t)]. Definition 3.3. Let I = [, b] nd let P = {I i, t i } n be tgged prtition of I. If δ is guge on I, then we sy tht P is δ-fine if I i [t i δ(t i ), t i + δ(t i )] for ll i = 1,..., n. Thus ech subintervl I i is subset of the intervl B[t i, δ(t i )] controlled by the point t i. When tgged prtition P is δ-fine we sometimes write P δ. As lwys some exmples will be needed to clrify the theory. Exmple 3.2. If δ > 0 nd δ R, define the guge δ : I R by letting δ(t) = δ for ll x I. This guge is clled constnt guge, nd ws used in Definition 2.3. A prtition P = {(I i, t i )} n for this constnt guge is δ-fine if nd only if I i [t i δ, t i + δ] = B[t i, δ] for ll i = 1,..., n. Since P is δ-fine the length l(i i ) = x i x i 1 2δ for ll i. Exmple 3.3. If δ 1 nd δ 2 re guges on I = [, b]. Define δ(t) = min{δ 1, δ 2 } for t I, then δ is guge on [, b]. Every P = {I i, t i )} n tht is δ-fine will utomticlly be both δ 1-fine nd δ 2 -fine. This is esy to relize since if P δ then for ech tgged point t i we hve t i δ 2 (t i ) t i δ 1 (t i ) x i 1 t i x i t i + δ 1 (t i ) t i + δ 2 (t i ) for ll i = 1,..., n. ssuming tht δ(t) = min{δ 1 (t), δ 2 (t)} = δ 1 (t). Thus I i B[t i, δ(t i )] B[t i, δ 1 (t i )] B[t i, δ 2 (t i )], nd similrly if δ(t) = min{δ 1 (t), δ 2 (t)} = δ 2 (t). This cn be extended to ny finite number of guges on I. Exmple 3.4. If δ 1 nd δ 2 re guges on I = [, b] nd if δ 1 (t) δ 2 (t) for ll t I nd if P = {I i, t i } n is δ 1-fine, then it is lso δ 2 -fine. Since P δ 1, then for ll i = 1,..., n t i δ 1 (t i ) x i 1 x i t i + δ 1 (t i ), or I i = [x i 1, x i ] B[t i, δ 1 (t i )] = [t i δ 1 (t i ), t i + δ 1 (t i )]. Using the fct tht δ 1 (t i ) δ 2 (t i ) for i = 1,..., n we get I i = [x i 1, x i ] B[t i, δ 1 (t i )] = [t i δ 1 (t i ), t i + δ 1 (t i )] [t i δ 2 (t i ), t i + δ 2 (t i )] = B[t i, δ 2 (t i )]. Hence I B[t i, δ 2 (t i )] so P δ 2. Exmple 3.5. If δ 1,..., δ m re guges on I = [, b], then δ defined by: δ (t) = min{δ 1 (t),..., δ m (t)}, for t I, is guge on I nd ech prtition P of I is δ -fine if nd only if P is δ k -fine for ech k = 1,..., m. Assume P δ, then δ > 0 nd I i B[t i, δ (t i )] = [t i δ (t i ), t i + δ (t i )] [t i δ k (t i ), t i + δ k (t i )] m k=1 for ll i = 1,..., n. 22

29 Conversely, ssume tht P = {I i, t i } n δ k for ll k = 1,..., m then I i [t i δ k (t i ), t i + δ k (t i )] for ech k = 1,..., m, since δ (t i ) δ k (t i ) for ll k = 1,..., m we get thus P δ. I i [t i δ (t i ), t i + δ (t i )] [t i δ k (t i ), t i + δ k (t i )] m k=1 B[t i, δ k (t i )] m k=1, Exmple 3.6. Let I = [0, 1] nd let δ(t) = { 1 2 t, 0 < t 1 1 4, t = 0. ) The prtition P = {I i, t i } 3 = {([0, 1 4 ], 0), ([ 1 4, 1 2 ], 1 2 ), ([ 1 2, 1], 3 4 )} is δ-fine, hence [ I 1 = 0, 1 ] [ , ] [ = 1 4 4, 1 ] [ = B [t 1, δ(t 1 )] = B 0, 1 ], 4 4 [ 1 I 2 = 4, 1 ] [ , ] [ 1 = B [t 2, δ(t 2 )] = B 2 2, 1 ] 2 t 2, [ ] [ 1 3 I 3 = 2, , ] [ 3 = B [t 3, δ(t 3 )] = B 4 4, 1 ] 2 t 3. Therefore P δ. b) The prtition P = {I i, t i } 3 = {([0, 1 4 ], 0), ([ 1 4, 1 2, 1 2 ), ([ 1 2, 1], 6 10 )} is not δ-fine, since [ I 1 = 0, 1 ] [ 1 4 4, 1 ] 4 [ 1 I 2 = 4, 1 ] [ 0, 1 ] 2 2 [ = B [ 1 = B 2, 1 4 0, 1 ] = B [t 1, δ(t 1 )], 4 ] [ 1 = B 2, 1 ] 2 t 2 = B [t 2, δ(t 2 )]. But [ 6 B [t 3, δ(t 3 )] = B 10, 3 ] [ 6 = , so I 3 = [ 1 2, 1] [ 3 10, 9 10 ] = B[ 6 10, 3 10 ] = B[t 3, δ(t 3 )] nd thus P δ. ] [ 6 3 = 10 10, 9 ], 10 In this prticulr cse every δ-fine prtition must hve tg t 1 = 0. Since P is δ-fine [0, x 1 ] [t 1 δ(t 1 ), t 2 + δ(t 2 )] = B[t 1, δ(t 1 )] must hold. This implies tht t 1 δ(t 1 ) 0. Assume tht t 1 > 0, then δ(t 1 ) = 1 2 t 1 so t 1 δ(t 1 ) = t t 1 = 1 2 t 1 > 0, this contrdicts the inequlity t 1 δ(t 1 ) 0. Therefore we must hve t 1 = 0 if P δ. Exmple 3.7. Let < c < b nd let δ be guge on [, b]. If P is δ-fine prtition of [, c] nd if P is prtition of [c, b] tht is δ-fine, then P = P P is δ-fine prtition of [, b]. We cn modify the guge in Exmple 3.6 b) δ-fine for both prtitions. { 3 4 Exmple 3.8. Let I = [0, 1] nd let δ 1 (t) = t, 0 < t 1 1 4, t = 0. 23

30 ) The prtition will be δ 1 -fine ccording to Exmple 3.4 since δ(t) δ 1 (t). b) Since δ(t) δ 1 (t) the only intervl thts need to be checked is I 3, [ ] [ 1 6 I 3 = 2, , ] [ = 10 20, 21 ] [ 6 = B 20 10, 3 ] 4 t 3, thus I 3 = [ 1 2, 1] [ 3 20, [ 20] 21 = B 6 10, 3 4 t ] 3 nd therefore P δ 2. The question rises why do guges work? Before the development of the integrl we would like to nswer this question: Why do non constnt guges work better thn constnt guges? In the Riemnn integrl, the mesure of fineness of prtition P is given by the mximum length P of the subintervls I i. In the pproch of the guge integrl, with δ-fine prtitions, more vrition of the lengths of the subintervls I i is llowed s long s the subintervls where the function quickly chnges hve smll length. There is no need to mke the subintervls length smll where function is roughly constnt. Another importnt dvntge with non constnt guges is tht we cn force prticulr point to be tg, this cn come in hndy when prticulr point is the origin of difficulty, by choosing the problem point s tg we cn control the difficulty t times. If I is compct intervl nd δ is guge on I we cn picture tht every point t I restrin every point in the intervl B[t, δ(t)] = [t δ(t), t + δ(t)]. Some points in I controls lrge intervls, nd other points controls intervls tht re very smll. This mkes us inquisitive, for n rbitrry guge δ when cn we find tgged prtition P = {(I i, t i )} n, where ech tg t i controls its ssocited subintervl I i. Theorem 3.3 will nswer this question nd gurntee tht given ny guge δ we cn lwys find tgged prtitions tht re δ-fine. This result ws first discovered by Pierre Cousin, in order to prove Cousin s theorem we need the Archimeden property nd the Nested intervl property. Lemm 3.1. Let S R be non empty set, M = sup S nd y < M then there exist nd x 0 S with x 0 > y. Proof. By contrdiction, suppose tht x y for every x S. Then y is upper bound for S, nd y < M contrdicts tht M = sup S. Hence there must exist n x 0 S with x 0 > y. Theorem 3.1 (Archimeden property). If α, β R with α > 0 nd b > 0 then there exist n n N nd nα > β. Proof. The proof is by contrdiction. Suppose tht nα β for every n N. Let S = {n : n N}. Now S is bounded bove by b, therefore M = sup S exists. Now M α < M, so by Lemm 3.1 there exists n n 0 N such tht n 0 α > M α. But then n 0 α + α > M, nd so (n 0 + 1)α > M. Since n N, we re led to the contrdiction tht (n 0 + 1)α S nd (n 0 + 1)α > sup S. Hence nα b cnnot be true for every n N. Therefore there exist n n N with nα > b Corollry 3.1. Given rbitrry ε > 0 there exists n n N such tht 1/n < ε. Proof. Just put ε = α nd b = 1 in Theorem 3.1. Theorem 3.2 (Nested intervl property). Suppose tht I 1 = [ 1, b 1 ],..., I n = [ n, b n ] with I n... I 2 I 1 nd lim n (b n n ) = 0. Then there is exctly one number A R common to ll the intervls I n. Proof. The sequence { n } n=1 is monotone incresing nd bounded bove by b 1. Therefore lim n = A n 24

31 exists nd A = sup n n b k for ech k N. Thus k A b k for every nturl number k. Tht is A is contined in ech intervl I k. Suppose B I n for every n N. Then n B b n for ll n, nd so 0 B n b n n. Since lim n (b n n ) = 0 we lso hve lim n (B n ) = 0. Hence B = lim n n = A, so A R is the only common number to ll the intervls I n. Theorem 3.3. If I = [, b] R is non degenerte compct intervl nd δ is guge on I, then there exists tgged prtition P of I such tht P δ. Proof. By contrdiction we suppose tht I does not hve δ-fine prtition. Let c = 1 2 ( + b) nd bisect I to [, c] nd [c, b]. Since we ssumed tht I hd no δ-fine prtition t lest one of these subintervls cnnot hve δ-fine prtition. Becuse if they both were δ-fine then their union would be δ-fine prtition s we sw in Exmple 3.7. Let I 1 = [, c] if this is the subintervl with no δ-fine prtition, otherwise let I 1 = [c, b]. Renme I 1 s [ 1, b 1 ], let c 1 = 1 2 nd bisect I 1 to [ 1, c 1 ] nd [c 1, b 1 ]. As bove, t lest one of these subintervls does not hve δ-fine prtition. Let I 2 = [ 1, c 1 ] if it does not hve δ-fine prtition, otherwise let I 2 = [c 1, b 1 ]. Renme I 2 s [ 2, b 2 ] nd bisect gin. Continue this bisection on nd on gives us sequence {I n } n=1 of compct subintervls of I = [, b] tht is nested in the mening tht [, b] = I I 1... I n I n+1... Theorem 3.2 implies tht there is exctly one number ξ common to ll of the intervls I n. Since δ(ξ) > 0, Theorem 3.1 implies tht there exist k N such tht l(i k ) = (b )/2 k < δ(ξ), where l p [ξ δ(ξ), ξ + δ(ξ)]. Thus the intervl round ξ is controlled by the guge δ, therefore the pir (l p, ξ) is δ-fine prtition of l p. But we hve constructed subintervls I n of I tht hve no δ-fine prtitions. Hence for every guge δ on I, there exists δ-fine prtition of I The guge integrl nd the reltion to the Riemnn integrl. In Definition 3.6, we shll define the concept of guge integrl. Then in Theorem 3.6 it is proved tht every Riemnn integrble function is guge integrble, but s shown in Exmple 3.13 there re functions tht re guge integrble but not Riemnn integrble. Lets restte the definition for function f to be Riemnn integrble. The purpose in doing so is comprison tht little modifiction to the definition will give us huge dvntges nd widely broden the clss of integrble functions. Definition 3.4. A function f : I R is sid to be Riemnn-integrble on I if there exists number A R such tht for every ε > 0 there exists number δ ε such tht if P = {I i, t i } n is ny tgged prtition of I such tht l(i i ) δ ε for i = 1,..., n then nd we write lim P 0 S(f, P ) = A. S(f, P ) A < ε, Definition 3.5. A function f : I R is sid to be guge integrble on I if there exists number B R such tht for every ε > 0 there exists guge γ ε on I such tht if P = {I i, t i } n is ny tgged prtition of I such tht l(i i ) γ ε (t i ) for i = 1,..., n then nd we write lim P 0 S(f, P ) = B. S(f, P ) B < ε, (3.1) The only difference from Definition 3.4 is tht the constnt δ ε is rplced by guge on I, tht is, by function δ ε : I (0, ). The following definition is more prcticl thou. 25

32 Definition 3.6. A function f : I R is sid to be guge-integrble on I if there exists number C R such tht for every ε > 0 there exists guge δ ε on I such tht if P = {I i, t i } n is ny tgged prtition of I tht is δ ɛ -fine, then S(f, P ) C < ε, (3.2) nd we write lim P 0 S(f, P ) = C. The collection of ll guge integrble functions on I = [, b] will be denoted by G[, b] or G(I). Theorem 3.4. Definitions 3.5 nd 3.6 led to equivlent integrls. Proof. Let ε > 0 nd f G(I) be given in the sense of Definition 3.5. Then there exists number B R nd guge γ ε (t) on I such tht for ny tgged prtition P = {I i, t i } n we hve l(i i ) γ ε (t i ) for i = 1,..., n. Define δ ε (t) = 1 2 γ ε(t) for t I, so tht δ ε is guge on I, then for ny δ ε -fine prtition of I we hve [ I i [t i δ ε (t i ), t i + δ ε (t i )] = t i 1 2 γ ε(t i ), t i + 1 ] 2 γ ε(t i ). Everything in Definition 3.5 is stisfied so inequlity 3.1 holds. Hence if P is ny δ ε -fine prtition of I, then S(f, P ) B < ε. Since ε > 0 is rbitrry f G(I) in the sense of Definition 3.6 with C = B. Conversely, suppose tht f G(I) in the sense of Definition 3.6. Then there exists number C R such tht given ε > 0 there exists guge δ ε. Define γ ε (t) = δ ε, so γ ε is guge on I. If the prtition P = {(I i, t i )} n is δ ε-fine then I i [t i δ ε, t i + δ ε ] for ll i = 1,..., n nd so l(i i ) γ ε (t i ) = δ ε (t i ). Hence Definition 3.6 is stisfied so inequlity 3.2 holds. Thus if P is ny prtition of I with l(i i ) γ ε (t i ) for ll i = 1,..., n then S(f, P ) C < ε. Since ε > 0 is rbitrry, then f G(I) in the sense of defintion 3.5 with B = C. The number B from Definition 3.5 nd the number C from Definition 3.6 is unique. Theorem 3.5. There is only one number C tht stisfies Definition 3.6. Proof. The proof is by contrdiction. Suppose C Ĉ nd let ε = 1 4 C Ĉ > 0. If C stisfies Definition 3.6, then there exists guge δ ε on I such tht if P is ny δ ε -fine prtition of I then S(f, P ) < ε. Similrly if Ĉ stisfies Definition 3.6, there exists guge ˆδ ε on I such tht if prtition of I then S(f, P ) < ε. Let δ ε = min{ δ ε, ˆδ ε }, then δ ε is guge on I(see Exmple 3.5) nd let P ˆδ ε. By the tringle inequlity we get C Ĉ = C S(f, P ) + S(f, P ) Ĉ P is ny ˆδ ε -fine P δ ε. Then P δ ε nd C S(f, P ) + S(f, P ) Ĉ ε + ε = 1 2 C Ĉ < C Ĉ. Ofcourse C Ĉ C Ĉ, therefore the ssumption C Ĉ is flse nd C = Ĉ = C. We will now see tht the Riemnn integrl is subset of the guge integrl. Theorem 3.6. A function f tht is Riemnn integrble is lso guge integrble. 26