W. We shall do so one by one, starting with I 1, and we shall do it greedily, trying


 Jeffrey Shaw
 2 years ago
 Views:
Transcription
1 Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering) Given set E R with λ (E) <, Vitli cover V of E, nd some ε > 0, there re disjoint I 1,..., I n V with λ ( E \ (I 1 I n ) ) < ε. Proof: Strt by picking n open set O E with λ(o) <, nd let W = {I V: I O}. It is esy to check tht W is Vitli cover of E. We now forget bout E for while, nd pick pirwise disjoint intervls I 1, I 2,... in W. We shll do so one by one, strting with I 1, nd we shll do it greedily, trying to mke ech intervl s lrge s possible. Now there my not be ny lrgest size vilble, so we settle for getting within constnt fctor: λ(i k+1 ) > 1 2 sup{ λ(j): J W nd J I j = for j = 1,...,k }. (1) In ordinry prose: At ech step, pick the next intervl disjoint from the other intervls picked this fr, so tht ny other intervl we could hve picked is less thn twice s long. It is conceivble tht this process cnnot go on forever. But if so, we hve lredy covered ll of E with finite, pirwise disjoint set of intervls from W, so we re done. If x E \ (I 1 I k ), then since I 1 I k is closed, there is some δneighbourhood of x tht does not meet I 1 I k, nd so we cn continue with the selection for one more step. If the process continues forever, then we find ( ) λ(i k ) = λ I k λ(o) <, so the sum converges. Pick n lrge enough so tht k=n+1 λ(i k ) < ε, nd let J k be the intervl with the sme center s I k, but five times the length. I clim which in turn implies n E \ I k J k, k=n+1 λ ( E \ n ) I k k=n+1 nd tht is enough to finish the proof. λ(j k ) = 5 k=n+1 λ(i k ) < 5ε, To prove the clim, let x E \ (I 1 I n ). Since W is Vitli cover for E, we cn find some I W with x I nd I I k = for k = 1,..., n. Let m be the smllest nturl number with I I m. Such n m must exist, for otherwise, I would lwys be one of the intervls we could hve picked, nd therefore λ(i k+1 ) > 1 2 λ(i ) for ll k. But this is impossible, since λ(i k ) <. There is more when k = m 1, I is still one of the J s, so we must hve λ(i m ) > 1 2 λ(i ). I I m J m This drwing illustrtes the fct tht, when I m I nd λ(i m ) > 1 2 λ(i ), then J m I. In prticulr, x J m, nd the proof is finlly complete.
2 Dini derivtives These re defined by D + f (y) f (x) f (y) f (x) f (x) = lim, D + f (x) = lim, y x y x y x y x D f (y) f (x) f (y) f (x) f (x) = lim, D f (x) = lim, y x y x y x y x nd clled the upper right, lower right, upper left, ndn lower left Dini derivtives, respectively. The Dini derivtives hve these simple properties: D ± f D ± f, D ± ( f ) = D ± f, D ± g = D f where g (x) = f ( x). Note tht ny of the Dini derivtives cn tke on either of the vlues ±. Our interest in the Dini derivtives stems from the fct tht they lwys exists, nd moreover f hs left derivtive x if nd only if D f (x) = D f (x), right derivtive if nd only if D + f (x) = D + f (x), nd (twosided) derivtive if nd only if ll four Dini derivtives re the sme.
3 Limits to growth for monotone function From now on, < b re rel numbers, nd f : (, b) R is n incresing function. 3 Lemm D + f (x) < nd D f (x) < for λ.e. x (,b). Proof: Let E = { x (,b): D + f (x) = }. The ide of the proof is tht f must hve unbounded growth on this set, if it hs positive mesure. To get this to work, we will first ssume tht f M on (,b). Pick ny (lrge) number m. It follows from the definition tht the set of intervls [x, y] where f (y) f (x) x E, y (x,b), nd > m y x is Vitli cover for E. Pick ny ε > 0, nd let [x k, y k ] be pirwise disjoint intervls of this type for k = 1,..., n with λ (E \ ([x 1, y 1 ] [x n, y n ])) < ε. It follows tht (y 1 x 1 ) + + (y n x n ) > λ (E) ε, nd therefore ( f (y1 ) f (x 1 ) ) + + ( f (y n ) f (x n ) ) > m(λ (E) ε). But becuse f is incresing, the intervls (f (x k ), f (y k )) re pirwise nonoverlpping in [ M, M], so we must finlly hve 2M > m(λ (E) ε). If λ (E) > 0, we cn choose ε < λ (E) nd m lrge enough for this to be contrdiction. If f is unbounded, we pply the bove to the restriction of f to slightly smller intervls ( + n 1,b n 1 ) insted, nd use the fct tht countble union of sets of mesure zero still hs mesure zero. We hve proved tht D + f <.e. To show the sme for D f, pply the first result to the function x f ( x) on ( b, ).
4 Limits to wiggliness for monotone function 4 Lemm D + f (x) = D + f (x) nd D f (x) = D f (x) for λ.e. x (,b). Proof: As in the previous lemm, the second.e. equlity follows from the first by pplying it to x f ( x). So we only prove the first one. Note tht D + f (x) D + f (x) lwys, so we need to show tht the set of x where D + f (x) > D + f (x) hs mesure zero. But for ny such x, we cn lwys find two rtionl numbers r nd s with D + f (x) > s > r > D + f (x). Since the number of rtionl pirs (r, s) with r < s is countble, we only need to show tht the set hs mesure zero. E = { x (,b): D + f (x) > s > r > D + f (x) } (r, s Q fixed) The proof ide is like tht for the previous lemm, only done twice: Use one inequlity to show tht the verge growth rte of f on set of intervls nerly covering E is less thn r, then use the other to find subintervls, still nerly covering E, where the verge growth rte is greter thn s. Then combine the two to get contrdiction, if λ (E) > 0. To get strted, then, we first pick some ε > 0, nd n open set O E with λ(o) < λ (E) + ε. Consider the Vitli cover for E consisting of those intervls [x, y] where x E, x < y, y O, nd f (y) f (x) < r. y x Then let ε > 0, nd use the Vitli covering lemm to pick pirwise disjoint intervls [x k, y k ] of this type, with with λ (E \ ([x 1, y 1 ] [x n, y n ])) < ε. It follows tht ( f (yk ) f (x k ) ) < r (y k x k ). Now let U = (x 1, y n ) (x n, y n ), nd crete yet nother Vitli cover, this time of E U, consisting of those intervls (u, v) where f (v) f (u) u E U, u < v, (u, v) U, nd > s. v u Use the Vitli lemm to lmost cover E U with pirwise disjoint intervls [u j, v j ] of this type, with λ (E U \ ([u 1, v 1 ] [u m, v m ])) < ε. Using resoning tht should be fmilir by now, we get ( f (v j ) f (u j ) ) > s (v j u j ). j =1 But since f is incresing, the intervls [f (u j ), f (v j )] re nonoverlpping. They re lso ech contined in some [f (x k ), f (y k )], so j =1 ( f (v j ) f (u j ) ) j =1 ( f (yk ) f (x k ) ). From the inequlities shown we get m s (v j u j ) < r (y k x k ). j =1 To estimte the right hnd side, note And for the left hnd side, s r n [x k, y k ] O, so (y k x k ) r λ(o) < r ( λ (E) + ε ). m ( m ) (v j u j ) = sλ [u j, v j ] > s ( λ (E U ) ε ) > s ( λ (E) 2ε ), j =1 so now we hve s ( λ (E) 2ε ) < r ( λ (E) + ε ). Recll tht r < s. So if λ (E) > 0, we cn pick ε > 0 smll enough tht the bove inequlity becomes contrdiction.
5 Limited number of corners From now on, < b re rel numbers, nd f : (, b) R is n rbitrry function. 5 Definition. Cll u (,b) strict locl mximum point for f if there is some δ > 0 so tht f (x) < f (u) whenever x (,b) is such tht x u < δ. 6 Lemm The set of strict locl mximum points for f is countble. Proof: Any two u vlues stisfying the definition bove for some δ > 0 must be t lest distnce δ prt, so there is only room for finite number of them in (,b). But ny strict locl mximum must stisfy the definition for some δ {1/n : n N}, so their totl number is countble. 7 Definition. A corner point for f is u (,b) so tht either D f (u) > D + f (u) or D f (u) < D + f (u). To understnd the mening of this definition, consult the illustrtion of Dini derivtives bove, which shows corner of the first kind. Turn the picture upside down to see corner of the second kind. 8 Lemm Any rel vlued function f hs only countble number of corner points. Proof: We only show tht the set of u stisfying the first inequlity is countble. The sme result for the second inequlity will then follow by replcing f by f. If the first inequlity holds, there is some q Q with D f (u) > q > D + f (u). If this holds, then x is strict locl mximum for the function g (x) = f (x) qx, since then D g (u) > 0 > D + g (u). By Lemm 6 only countble number of such u exist for ny q, nd since Q is countble, we re done.
6 Almost everywhere differentibility 9 Theorem Any monotone function is differentible lmost everywhere. Proof: We only need to consider n incresing function f : (,b) R. By Lemm 3 nd 4, f hs finite onesided derivtives lmost everywhere. If f hs onesided derivtives t some point, nd they re different, then f hs corner point there. But Lemm 8 implies tht the corner points hve mesure zero. 10 Exmple. Let f : [0,1] [0,1] be the Cntor function, nd C [0,1] the (stndrd) Cntor set. Then f is loclly constnt on the open set (0,1)\C, so f = 0 there. But λ(c ) = 0, so f = 0 lmost everywhere. In prticulr, 1 f dλ = 0, nd yet f (1) f (0) = 1. 0 This shows tht we cnnot expect the fundmentl theorem of clculus to hold for rbitrry monotone functions more is needed. Wht is hppening here is tht ll of the growth hppens within set of mesure zero.
7 An integrl (in)equlity 11 Proposition Let f : [,b] R be n incresing function. Then f is integrble, nd b f (x)dx f (b) f (). If there is some constnt L so tht f stisfies then equlity holds insted: f (y) f (x) + L(y x) whenever x < y b, b f (x)dx = f (b) f (). By incresing, I men wht some prefer to cll nondecresing. Proof: For simplicity, extend f by setting f (x) = f (b) when x > b. Define f n : [,b] R by f n (x) = n (f (x + n 1 ) f (x) ). Then f n 0, nd f n f.e. Therefore Ftou s lemm gives us b b f (x)dx lim f n (x)dx n b ( = lim n f (x + n 1 ) f (x) ) dx n = lim n n ( = lim n n ( b+n 1 +n 1 b+n 1 b f (b) f (). b f (x)dx f (x)dx n ) f (x)dx +n 1 ) f (x)dx If the stted liner growth condition holds, then f n L, so insted of using Ftou s lemm, we cn use the bounded convergence theorem (BCT), nd the first inequlity in the clcultion bove becomes equlity. Also, lim cn be replce by n ordinry limit. The finl inequlity lso becomes n equlity, since f is now continuous.
8 The first fundmentl theorem of clculus 12 Theorem If f L 1 ([,b]) nd F (x) = f (t)dt for x [,b], then F is differentible.e., nd F = f.e. in [,b]. The proof of this theorem builds on the following 13 Lemm If f L 1 ([,b]) nd f (t)dt = 0 for ll x [,b], then f = 0.e. in [,b]. Proof: Assume tht f > 0 in subset of (,b) of positive mesure. Then this subset hs compct subset K of positive mesure, nd then K f (t)dt > 0. Since b f (t)dt = 0, we must hve (,b)\k f (t)dt < 0. But (,b) \ K is n open set, nd hence is countble disjoint union of open intervls. For t lest one of those intervls, sy (c,d), we must hve (c,d) f (t)dt < 0. But (c,d) f (t)dt = ( d c) f (t)dt = 0, by ssumption nd this contrdiction shows tht f 0.e. Applying this to f, we see tht f 0.e., so f = 0.e. The bove proof is for relvlued f. The extension to complexvlued f by considering the rel nd complex prts seprtely is strightforwrd. Proof of the theorem: Assume first tht f 0. Then F is incresing, so it is differentible.e., nd s Proposition 11 shows tht F (t)dt F (x) for ll x [,b]. Furthermore, if f is bounded, then F stisfies the extr condition of Proposition 11, so the bove inequlity is ctully n equlity. But tht is equivlent to ( F (t) f (t) ) dt = 0 for ll x [,b], nd then the bove lemm shows tht F = f.e. This completes the proof in tht specil cse. If f 0 but f is unbounded, write F n (x) = ( f (t) n ) dt, nd note tht then F (x) F n (x) is the integrl of nonnegtive function, so tht F F n is n incresing function. In prticulr, F F n t ny point where both derivtives exists (i.e.,.e.). Therefore, nd using the result of the previous prgrph, F (t)dt F n (t)dt = ( ) f (t) n dt. Finlly, since f n f when n, we cn use the monotone convergence theorem to conclude tht F (t)dt f (t)dt for ll x [,b]. Just like in the bounded cse, this once more completes the proof, now for the cse when f 0. If f is relvlued integrble function, the theorem redily follows by using the result for the positive nd negtive prts of f. Similrly, if f is complex vlued, using the result for rel nd imginry prts seprtely redily yields the desired result.
9 Absolutely continuous functions Recll tht the Cntor function ψ is continuous nd differentible.e. on [0, 1], whith ψ = 0.e. yet ψ(1) ψ(0), hence it cnnot stisfy ψ(1) = ψ(0) ψ (x)dx, s the second fundmentl theorem of clculus would hve us believe. It turns out tht the remedy is to introduce stronger form of continuity. 14 Definition. A function f on n intervl [, b] is clled bsolutely continuous if for every ε > 0 there is δ > 0 so tht, whenever [ k,b k ] for k = 1,...,n re nonoverlpping intervls with n (b k k ) < δ, we hve n ( f (bk ) f ( k ) ) < ε. 15 Lemm If f L([,b]) nd F (x) = f (t)dt, then F is bsolutely continuous. Proof: Define mesure µ on M by µ(e) = f (x) dx. E [,b] I clim tht for ny ε > 0 there is some δ > 0 so tht λ(e) < δ implies µ(e) < ε. If not, there is some ε > 0 nd, for every n N, set E k M with λ(e k ) < 2 k, yet µ(e k ) ε. The lemm follows from the fct tht, for ny ε > 0, there exists δ > 0 so tht for ny mesurble E [,b] with λ(e) < δ, we hve E f (x) dx < ε. Let F n = k n E k, nd F = n N F n. Then λ(f n ) < 2 1 n, hence λ(f ) = 0. But lso µ(f n ) µ(e n ) ε, hence µ(f ) = lim n µ(f n ) ε (since F n F nd µ(f 1 ) < ). But µ(f ) > 0 while λ(f ) = 0 is clerly impossible, so the stted clim must be true. Now, given ε > 0, pick δ > 0 ccording to the clim bove. If ([ k,b k ]) n is finite sequence of pirwise nonoverlpping intervls, with n (b k k ) < δ let E = n [ k,b k ]. Then λ(e) < δ, nd so µ(e) < ε. Thus F (bk ) F ( k ) = nd the proof is complete. bk k f (t)dt bk k f (t) dt = µ(e) < ε, We shll prove the converse of the bove theorem. An importnt step long the wy is the following: 16 Lemm If f is bsolutely continuous nd differentible.e. on [,b] with f = 0.e., then f is constnt. Lter, we shll see tht bsolutely continuous functions necessrily re differentible.e. (they hve bounded vrition), but for now, we just ssume it. Proof: Let D be the set of points in (,b) in which f is differentible nd the derivtive is zero. Let ε > 0, nd let δ > 0 be s in the definition of bsolute continuity. The intervls [c,d] [,b] for which f (d) f (c) < ε d c form Vitli cover of D, so by the Vitli covering lemm nd the fct tht λ[,b] \ D = 0, there re finite, disjoint sequence of such intervls [c k,d k ] [,b] with k = 1,...,n such tht λ([,b] \ n (c k,d k )) < δ. Order these intervls so tht c 1 < d 1 < c 2 < d 2 < < c n < d n, nd write [ k,b k ] for the gps: So let [ 0,b 0 ] = [,c 1 ], [ 1,b 1 ] = [d 1,c 2 ], nd so on up to [ n,b n ] = [d n,b]. Then n k=0 (b k k ) < δ, so n k=0( f (bk ) f ( k ) ) < ε. Then nd so f (b) f () = f (b) f (d n ) + f (d n ) f (c n ) + f (c n ) f (d 1 ) + f (c 1 ) f () ( = f (bk ) f ( k ) ) ( + f (dk ) f (c k ) ), k=0 f (b) f () f (bk ) f ( k ) + f (dk ) f (c k ) k=0 < ε + ε d k c k (1 + b )ε. Since ε > 0 ws rbitrry, f (b) = f (). The sme rgument with b replced by ny x [,b] shows tht f (x) = f (), nd the proof is therefore complete.
10 The second fundmentl theorem of clculus 17 Theorem If f is bsolutely continuous on [,b], then f is differentible.e. on [,b], f L 1 ([,b]), nd for ll x [,b]. f (x) = f () + f (t)dt Proof: It is shown elsewhere tht f is differentible.e. nd tht f L 1 ([,b]) (see the textbook or perhps I will dd proof here lter). Let g (x) = f (x) f (t)dt. It follows from the first fundmentl theorem of clculus tht g = 0.e., nd from Lemm 15 tht g is bsolutely continuous. Lemm 16 then shows tht g is constnt. The vlue x = 0 shows tht then g (x) = f () for ll x, nd the proof is complete. Note tht I hve not yet lectured on bounded vrition t this point. For now, we tke these fcts for grnted: A complexvlued function hs bounded vrition if nd only if both its rel nd imginry prts do, nd relvlued function hs bounded vrition if nd only if it is the difference between two incresing functions. A function f : [,b] C is sid to hve bounded vrition (BV) if there is some M < so tht f (xk ) f (x k 1 ) M (2) whenever x 0 x 1 x n b. 18 Lemm An bsolutely continuous function on [, b] hs bounded vrition, nd is therefore differentible.e., nd its derivtive is integrble. Proof: Tke δ > 0 corresponding to ε = 1 in the definition of bsolute continuity. Let N N with Nδ > b, nd let δ = (b )/N < δ. Now ssume x 0 x 1 x n b is given. The sum in (2) will not get ny smller if we insert more points into the sequence (x k ), so we my s well ssume tht ech point in [,b] of the form + j δ for j N. Then the sum in (2) splits up into t most N sums, ech involving only x k [ + (j 1)δ, + j δ ] for some j. So in ech of the smller sums x k x k 1 < δ, nd therefore f (xk ) f (x k 1 ) < 1. Therefore, the totl sum is < N, so tht f hs bounded vrition. The sttements on differentibility nd integrbility of the derivtive follows from the fcts bout BV functions mentioned bove, nd the corresponding sttements for incresing functions.
The Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More informationLecture 3 ( ) (translated and slightly adapted from lecture notes by Martin Klazar)
Lecture 3 (5.3.2018) (trnslted nd slightly dpted from lecture notes by Mrtin Klzr) Riemnn integrl Now we define precisely the concept of the re, in prticulr, the re of figure U(, b, f) under the grph of
More informationLecture 1. Functional series. Pointwise and uniform convergence.
1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is
More informationAdvanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004
Advnced Clculus: MATH 410 Notes on Integrls nd Integrbility Professor Dvid Levermore 17 October 2004 1. Definite Integrls In this section we revisit the definite integrl tht you were introduced to when
More informationLecture 1: Introduction to integration theory and bounded variation
Lecture 1: Introduction to integrtion theory nd bounded vrition Wht is this course bout? Integrtion theory. The first question you might hve is why there is nything you need to lern bout integrtion. You
More informationPresentation Problems 5
Presenttion Problems 5 21355 A For these problems, ssume ll sets re subsets of R unless otherwise specified. 1. Let P nd Q be prtitions of [, b] such tht P Q. Then U(f, P ) U(f, Q) nd L(f, P ) L(f, Q).
More informationMath 61CM  Solutions to homework 9
Mth 61CM  Solutions to homework 9 Cédric De Groote November 30 th, 2018 Problem 1: Recll tht the left limit of function f t point c is defined s follows: lim f(x) = l x c if for ny > 0 there exists δ
More informationThe HenstockKurzweil integral
fculteit Wiskunde en Ntuurwetenschppen The HenstockKurzweil integrl Bchelorthesis Mthemtics June 2014 Student: E. vn Dijk First supervisor: Dr. A.E. Sterk Second supervisor: Prof. dr. A. vn der Schft
More information7.2 Riemann Integrable Functions
7.2 Riemnn Integrble Functions Theorem 1. If f : [, b] R is step function, then f R[, b]. Theorem 2. If f : [, b] R is continuous on [, b], then f R[, b]. Theorem 3. If f : [, b] R is bounded nd continuous
More informationMath 360: A primitive integral and elementary functions
Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:
More informationProperties of the Riemann Integral
Properties of the Riemnn Integrl Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University Februry 15, 2018 Outline 1 Some Infimum nd Supremum Properties 2
More informationMath 1B, lecture 4: Error bounds for numerical methods
Mth B, lecture 4: Error bounds for numericl methods Nthn Pflueger 4 September 0 Introduction The five numericl methods descried in the previous lecture ll operte by the sme principle: they pproximte the
More informationUNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE
UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE 1. Pointwise Convergence of Sequence Let E be set nd Y be metric spce. Consider functions f n : E Y for n = 1, 2,.... We sy tht the sequence
More informationII. Integration and Cauchy s Theorem
MTH6111 Complex Anlysis 200910 Lecture Notes c Shun Bullett QMUL 2009 II. Integrtion nd Cuchy s Theorem 1. Pths nd integrtion Wrning Different uthors hve different definitions for terms like pth nd curve.
More informationChapter 22. The Fundamental Theorem of Calculus
Version of 24.2.4 Chpter 22 The Fundmentl Theorem of Clculus In this chpter I ddress one of the most importnt properties of the Lebesgue integrl. Given n integrble function f : [,b] R, we cn form its indefinite
More informationProperties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives
Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums  1 Riemnn
More informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 200910 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some
More informationUNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3
UNIFORM CONVERGENCE Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 Suppose f n : Ω R or f n : Ω C is sequence of rel or complex functions, nd f n f s n in some sense. Furthermore,
More informationSOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL (1 + µ(f n )) f(x) =. But we don t need the exact bound.) Set
SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL 28 Nottion: N {, 2, 3,...}. (Tht is, N.. Let (X, M be mesurble spce with σfinite positive mesure µ. Prove tht there is finite positive mesure ν on (X, M such
More informationEntrance Exam, Real Analysis September 1, 2009 Solve exactly 6 out of the 8 problems. Compute the following and justify your computation: lim
1. Let n be positive integers. ntrnce xm, Rel Anlysis September 1, 29 Solve exctly 6 out of the 8 problems. Sketch the grph of the function f(x): f(x) = lim e x2n. Compute the following nd justify your
More informationRiemann is the Mann! (But Lebesgue may besgue to differ.)
Riemnn is the Mnn! (But Lebesgue my besgue to differ.) Leo Livshits My 2, 2008 1 For finite intervls in R We hve seen in clss tht every continuous function f : [, b] R hs the property tht for every ɛ >
More informationFUNDAMENTALS OF REAL ANALYSIS by. III.1. Measurable functions. f 1 (
FUNDAMNTALS OF RAL ANALYSIS by Doğn Çömez III. MASURABL FUNCTIONS AND LBSGU INTGRAL III.. Mesurble functions Hving the Lebesgue mesure define, in this chpter, we will identify the collection of functions
More informationf(x)dx . Show that there 1, 0 < x 1 does not exist a differentiable function g : [ 1, 1] R such that g (x) = f(x) for all
3 Definite Integrl 3.1 Introduction In school one comes cross the definition of the integrl of rel vlued function defined on closed nd bounded intervl [, b] between the limits nd b, i.e., f(x)dx s the
More informationMath 324 Course Notes: Brief description
Brief description These re notes for Mth 324, n introductory course in Mesure nd Integrtion. Students re dvised to go through ll sections in detil nd ttempt ll problems. These notes will be modified nd
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties
More information11 An introduction to Riemann Integration
11 An introduction to Riemnn Integrtion The PROOFS of the stndrd lemms nd theorems concerning the Riemnn Integrl re NEB, nd you will not be sked to reproduce proofs of these in full in the exmintion in
More informationNOTES AND PROBLEMS: INTEGRATION THEORY
NOTES AND PROBLEMS: INTEGRATION THEORY SAMEER CHAVAN Abstrct. These re the lecture notes prepred for prticipnts of AFSI to be conducted t Kumun University, Almor from 1st to 27th December, 2014. Contents
More informationODE: Existence and Uniqueness of a Solution
Mth 22 Fll 213 Jerry Kzdn ODE: Existence nd Uniqueness of Solution The Fundmentl Theorem of Clculus tells us how to solve the ordinry differentil eqution (ODE) du = f(t) dt with initil condition u() =
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct
More informationAppendix to Notes 8 (a)
Appendix to Notes 8 () 13 Comprison of the Riemnn nd Lebesgue integrls. Recll Let f : [, b] R be bounded. Let D be prtition of [, b] such tht Let D = { = x 0 < x 1
More informationarxiv: v1 [math.ca] 7 Mar 2012
rxiv:1203.1462v1 [mth.ca] 7 Mr 2012 A simple proof of the Fundmentl Theorem of Clculus for the Lebesgue integrl Mrch, 2012 Rodrigo López Pouso Deprtmento de Análise Mtemátic Fcultde de Mtemátics, Universidde
More informationFundamental Theorem of Calculus for Lebesgue Integration
Fundmentl Theorem of Clculus for Lebesgue Integrtion J. J. Kolih The existing proofs of the Fundmentl theorem of clculus for Lebesgue integrtion typiclly rely either on the Vitli Crthéodory theorem on
More informationThe Bochner Integral and the Weak Property (N)
Int. Journl of Mth. Anlysis, Vol. 8, 2014, no. 19, 901906 HIKARI Ltd, www.mhikri.com http://dx.doi.org/10.12988/ijm.2014.4367 The Bochner Integrl nd the Wek Property (N) Besnik Bush Memetj University
More information7.2 The Definite Integral
7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where
More informationMORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)
MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give
More informationMath 554 Integration
Mth 554 Integrtion Hndout #9 4/12/96 Defn. A collection of n + 1 distinct points of the intervl [, b] P := {x 0 = < x 1 < < x i 1 < x i < < b =: x n } is clled prtition of the intervl. In this cse, we
More informationMath 1431 Section M TH 4:00 PM 6:00 PM Susan Wheeler Office Hours: Wed 6:00 7:00 PM Online ***NOTE LABS ARE MON AND WED
Mth 43 Section 4839 M TH 4: PM 6: PM Susn Wheeler swheeler@mth.uh.edu Office Hours: Wed 6: 7: PM Online ***NOTE LABS ARE MON AND WED t :3 PM to 3: pm ONLINE Approimting the re under curve given the type
More informationHomework 11. Andrew Ma November 30, sin x (1+x) (1+x)
Homewor Andrew M November 3, 4 Problem 9 Clim: Pf: + + d = d = sin b +b + sin (+) d sin (+) d using integrtion by prts. By pplying + d = lim b sin b +b + sin (+) d. Since limits to both sides, lim b sin
More informationThe Banach algebra of functions of bounded variation and the pointwise Helly selection theorem
The Bnch lgebr of functions of bounded vrition nd the pointwise Helly selection theorem Jordn Bell jordn.bell@gmil.com Deprtment of Mthemtics, University of Toronto Jnury, 015 1 BV [, b] Let < b. For f
More information1 Sets Functions and Relations Mathematical Induction Equivalence of Sets and Countability The Real Numbers...
Contents 1 Sets 1 1.1 Functions nd Reltions....................... 3 1.2 Mthemticl Induction....................... 5 1.3 Equivlence of Sets nd Countbility................ 6 1.4 The Rel Numbers..........................
More informationPiecewise Continuous φ
Piecewise Continuous φ φ is piecewise continuous on [, b] if nd only if b in R nd φ : [, b] C There is finite set S [, b] such tht, for ll t [, b] S, φ is continuous t t: φ(t) = lim φ(u) u t u [,b] For
More informationTHE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS.
THE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrdlindeloftheorem/ This document is proof of the existenceuniqueness theorem
More information38 Riemann sums and existence of the definite integral.
38 Riemnn sums nd existence of the definite integrl. In the clcultion of the re of the region X bounded by the grph of g(x) = x 2, the xxis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These
More informationCalculus III Review Sheet
Clculus III Review Sheet 1 Definitions 1.1 Functions A function is f is incresing on n intervl if x y implies f(x) f(y), nd decresing if x y implies f(x) f(y). It is clled monotonic if it is either incresing
More informationMain topics for the First Midterm
Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 23, Sections 4.14.8, nd Sections 5.15.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the
More informationGoals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite
Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite
More informationChapter 6. Riemann Integral
Introduction to Riemnn integrl Chpter 6. Riemnn Integrl WonKwng Prk Deprtment of Mthemtics, The College of Nturl Sciences Kookmin University Second semester, 2015 1 / 41 Introduction to Riemnn integrl
More informationAdvanced Calculus: MATH 410 Uniform Convergence of Functions Professor David Levermore 11 December 2015
Advnced Clculus: MATH 410 Uniform Convergence of Functions Professor Dvid Levermore 11 December 2015 12. Sequences of Functions We now explore two notions of wht it mens for sequence of functions {f n
More informationMA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.
MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.
More informationA product convergence theorem for Henstock Kurzweil integrals
A product convergence theorem for Henstock Kurzweil integrls Prsr Mohnty Erik Tlvil 1 Deprtment of Mthemticl nd Sttisticl Sciences University of Albert Edmonton AB Cnd T6G 2G1 pmohnty@mth.ulbert.c etlvil@mth.ulbert.c
More informationA PROOF OF THE FUNDAMENTAL THEOREM OF CALCULUS USING HAUSDORFF MEASURES
INROADS Rel Anlysis Exchnge Vol. 26(1), 2000/2001, pp. 381 390 Constntin Volintiru, Deprtment of Mthemtics, University of Buchrest, Buchrest, Romni. emil: cosv@mt.cs.unibuc.ro A PROOF OF THE FUNDAMENTAL
More informationMath Solutions to homework 1
Mth 75  Solutions to homework Cédric De Groote October 5, 07 Problem, prt : This problem explores the reltionship between norms nd inner products Let X be rel vector spce ) Suppose tht is norm on X tht
More informationCMDA 4604: Intermediate Topics in Mathematical Modeling Lecture 19: Interpolation and Quadrature
CMDA 4604: Intermedite Topics in Mthemticl Modeling Lecture 19: Interpoltion nd Qudrture In this lecture we mke brief diversion into the res of interpoltion nd qudrture. Given function f C[, b], we sy
More informationPhil Wertheimer UMD Math Qualifying Exam Solutions Analysis  January, 2015
Problem 1 Let m denote the Lebesgue mesure restricted to the compct intervl [, b]. () Prove tht function f defined on the compct intervl [, b] is Lipschitz if nd only if there is constct c nd function
More informationMath 8 Winter 2015 Applications of Integration
Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl
More informationpadic Egyptian Fractions
padic Egyptin Frctions Contents 1 Introduction 1 2 Trditionl Egyptin Frctions nd Greedy Algorithm 2 3 Setup 3 4 pgreedy Algorithm 5 5 pegyptin Trditionl 10 6 Conclusion 1 Introduction An Egyptin frction
More information1 The fundamental theorems of calculus.
The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl new nme for ntiderivtive. Differentiting integrls. Tody we provide the connection
More informationMAA 4212 Improper Integrals
Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly welldefined, is too restrictive for mny purposes; there re functions which
More informationReview of Riemann Integral
1 Review of Riemnn Integrl In this chpter we review the definition of Riemnn integrl of bounded function f : [, b] R, nd point out its limittions so s to be convinced of the necessity of more generl integrl.
More informationLecture 3. Limits of Functions and Continuity
Lecture 3 Limits of Functions nd Continuity Audrey Terrs April 26, 21 1 Limits of Functions Notes I m skipping the lst section of Chpter 6 of Lng; the section bout open nd closed sets We cn probbly live
More information1 The Riemann Integral
The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re
More informationWeek 7 Riemann Stieltjes Integration: Lectures 1921
Week 7 Riemnn Stieltjes Integrtion: Lectures 1921 Lecture 19 Throughout this section α will denote monotoniclly incresing function on n intervl [, b]. Let f be bounded function on [, b]. Let P = { = 0
More informationg i fφdx dx = x i i=1 is a Hilbert space. We shall, henceforth, abuse notation and write g i f(x) = f
1. Appliction of functionl nlysis to PEs 1.1. Introduction. In this section we give little introduction to prtil differentil equtions. In prticulr we consider the problem u(x) = f(x) x, u(x) = x (1) where
More informationChapter 5 : Continuous Random Variables
STAT/MATH 395 A  PROBABILITY II UW Winter Qurter 216 Néhémy Lim Chpter 5 : Continuous Rndom Vribles Nottions. N {, 1, 2,...}, set of nturl numbers (i.e. ll nonnegtive integers); N {1, 2,...}, set of ll
More informationThe final exam will take place on Friday May 11th from 8am 11am in Evans room 60.
Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23
More informationPolynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230
Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given
More informationIMPORTANT THEOREMS CHEAT SHEET
IMPORTANT THEOREMS CHEAT SHEET BY DOUGLAS DANE Howdy, I m Bronson s dog Dougls. Bronson is still complining bout the textbook so I thought if I kept list of the importnt results for you, he might stop.
More information7 Improper Integrals, Exp, Log, Arcsin, and the Integral Test for Series
7 Improper Integrls, Exp, Log, Arcsin, nd the Integrl Test for Series We hve now ttined good level of understnding of integrtion of nice functions f over closed intervls [, b]. In prctice one often wnts
More informationBest Approximation. Chapter The General Case
Chpter 4 Best Approximtion 4.1 The Generl Cse In the previous chpter, we hve seen how n interpolting polynomil cn be used s n pproximtion to given function. We now wnt to find the best pproximtion to given
More informationReview of basic calculus
Review of bsic clculus This brief review reclls some of the most importnt concepts, definitions, nd theorems from bsic clculus. It is not intended to tech bsic clculus from scrtch. If ny of the items below
More informationFundamental Theorem of Calculus
Fundmentl Theorem of Clculus Recll tht if f is nonnegtive nd continuous on [, ], then the re under its grph etween nd is the definite integrl A= f() d Now, for in the intervl [, ], let A() e the re under
More informationDuality # Second iteration for HW problem. Recall our LP example problem we have been working on, in equality form, is given below.
Dulity #. Second itertion for HW problem Recll our LP emple problem we hve been working on, in equlity form, is given below.,,,, 8 m F which, when written in slightly different form, is 8 F Recll tht we
More informationON THE CINTEGRAL BENEDETTO BONGIORNO
ON THE CINTEGRAL BENEDETTO BONGIORNO Let F : [, b] R be differentible function nd let f be its derivtive. The problem of recovering F from f is clled problem of primitives. In 1912, the problem of primitives
More informationMA Handout 2: Notation and Background Concepts from Analysis
MA350059 Hndout 2: Nottion nd Bckground Concepts from Anlysis This hndout summrises some nottion we will use nd lso gives recp of some concepts from other units (MA20023: PDEs nd CM, MA20218: Anlysis 2A,
More informationTheoretical foundations of Gaussian quadrature
Theoreticl foundtions of Gussin qudrture 1 Inner product vector spce Definition 1. A vector spce (or liner spce) is set V = {u, v, w,...} in which the following two opertions re defined: (A) Addition of
More informationMapping the delta function and other Radon measures
Mpping the delt function nd other Rdon mesures Notes for Mth583A, Fll 2008 November 25, 2008 Rdon mesures Consider continuous function f on the rel line with sclr vlues. It is sid to hve bounded support
More informationRecitation 3: More Applications of the Derivative
Mth 1c TA: Pdric Brtlett Recittion 3: More Applictions of the Derivtive Week 3 Cltech 2012 1 Rndom Question Question 1 A grph consists of the following: A set V of vertices. A set E of edges where ech
More informationA REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007
A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus
More informationImproper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:
Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl
More informationMATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1
MATH34032: Green s Functions, Integrl Equtions nd the Clculus of Vritions 1 Section 1 Function spces nd opertors Here we gives some brief detils nd definitions, prticulrly relting to opertors. For further
More informationNumerical Analysis: Trapezoidal and Simpson s Rule
nd Simpson s Mthemticl question we re interested in numericlly nswering How to we evlute I = f (x) dx? Clculus tells us tht if F(x) is the ntiderivtive of function f (x) on the intervl [, b], then I =
More information4.4 Areas, Integrals and Antiderivatives
. res, integrls nd ntiderivtives 333. Ares, Integrls nd Antiderivtives This section explores properties of functions defined s res nd exmines some connections mong res, integrls nd ntiderivtives. In order
More informationT b a(f) [f ] +. P b a(f) = Conclude that if f is in AC then it is the difference of two monotone absolutely continuous functions.
Rel Vribles, Fll 2014 Problem set 5 Solution suggestions Exerise 1. Let f be bsolutely ontinuous on [, b] Show tht nd T b (f) P b (f) f (x) dx [f ] +. Conlude tht if f is in AC then it is the differene
More informationUniversitaireWiskundeCompetitie. Problem 2005/4A We have k=1. Show that for every q Q satisfying 0 < q < 1, there exists a finite subset K N so that
Problemen/UWC NAW 5/7 nr juni 006 47 Problemen/UWC UniversitireWiskundeCompetitie Edition 005/4 For Session 005/4 we received submissions from Peter Vndendriessche, Vldislv Frnk, Arne Smeets, Jn vn de
More informationACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019
ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH00030 SEMESTER 208/209 DR. ANTHONY BROWN 7.. Introduction to Integrtion. 7. Integrl Clculus As ws the cse with the chpter on differentil
More informationExam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH1105 Instructor: Attila Máté 1
Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixedpoint itertion to converge when solving the eqution
More informationThe area under the graph of f and above the xaxis between a and b is denoted by. f(x) dx. π O
1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the xxis etween nd is denoted y f(x) dx nd clled the
More information1 i n x i x i 1. Note that kqk kp k. In addition, if P and Q are partition of [a, b], P Q is finer than both P and Q.
Chpter 6 Integrtion In this chpter we define the integrl. Intuitively, it should be the re under curve. Not surprisingly, fter mny exmples, counter exmples, exceptions, generliztions, the concept of the
More informationAnalytical Methods Exam: Preparatory Exercises
Anlyticl Methods Exm: Preprtory Exercises Question. Wht does it men tht (X, F, µ) is mesure spce? Show tht µ is monotone, tht is: if E F re mesurble sets then µ(e) µ(f). Question. Discuss if ech of the
More information2.4 Linear Inequalities and Interval Notation
.4 Liner Inequlities nd Intervl Nottion We wnt to solve equtions tht hve n inequlity symol insted of n equl sign. There re four inequlity symols tht we will look t: Less thn , Less thn or
More informationBernoulli Numbers Jeff Morton
Bernoulli Numbers Jeff Morton. We re interested in the opertor e t k d k t k, which is to sy k tk. Applying this to some function f E to get e t f d k k tk d k f f + d k k tk dk f, we note tht since f
More informationp(t) dt + i 1 re it ireit dt =
Note: This mteril is contined in Kreyszig, Chpter 13. Complex integrtion We will define integrls of complex functions long curves in C. (This is bit similr to [relvlued] line integrls P dx + Q dy in R2.)
More informationARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac
REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b
More informationAnalysis III. Ben Green. Mathematical Institute, Oxford address:
Anlysis III Ben Green Mthemticl Institute, Oxford Emil ddress: ben.green@mths.ox.c.uk 2000 Mthemtics Subject Clssifiction. Primry Contents Prefce 1 Chpter 1. Step functions nd the Riemnn integrl 3 1.1.
More informationMATH 144: Business Calculus Final Review
MATH 144: Business Clculus Finl Review 1 Skills 1. Clculte severl limits. 2. Find verticl nd horizontl symptotes for given rtionl function. 3. Clculte derivtive by definition. 4. Clculte severl derivtives
More informationThe First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).
The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples
More informationIntegrals along Curves.
Integrls long Curves. 1. Pth integrls. Let : [, b] R n be continuous function nd let be the imge ([, b]) of. We refer to both nd s curve. If we need to distinguish between the two we cll the function the
More informationContinuous Random Variables
STAT/MATH 395 A  PROBABILITY II UW Winter Qurter 217 Néhémy Lim Continuous Rndom Vribles Nottion. The indictor function of set S is relvlued function defined by : { 1 if x S 1 S (x) if x S Suppose tht
More informationSection 5.1 #7, 10, 16, 21, 25; Section 5.2 #8, 9, 15, 20, 27, 30; Section 5.3 #4, 6, 9, 13, 16, 28, 31; Section 5.4 #7, 18, 21, 23, 25, 29, 40
Mth B Prof. Audrey Terrs HW # Solutions by Alex Eustis Due Tuesdy, Oct. 9 Section 5. #7,, 6,, 5; Section 5. #8, 9, 5,, 7, 3; Section 5.3 #4, 6, 9, 3, 6, 8, 3; Section 5.4 #7, 8,, 3, 5, 9, 4 5..7 Since
More informationEuler, Ioachimescu and the trapezium rule. G.J.O. Jameson (Math. Gazette 96 (2012), )
Euler, Iochimescu nd the trpezium rule G.J.O. Jmeson (Mth. Gzette 96 (0), 36 4) The following results were estblished in recent Gzette rticle [, Theorems, 3, 4]. Given > 0 nd 0 < s
More information