Appendix to Notes 8 (a)


 Baldwin Wilcox
 2 years ago
 Views:
Transcription
1 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 <... < x n = b}. m i = inf{f(x) : x i 1 x x i } M i = sup{f(x) : x i 1 x x i }. Define the step functions (therefore, simple functions, since we hve ssumed tht f is bounded nd so M i < for ll i). nd So α D (x) = m i on [x i 1, x i ) for ll 1 i n, β D (x) = M i on [x i 1, x i ) for ll 1 i n. α D (x) f(x) β D (x) for ll x [, b]. Note tht if D D then α D (x) α D (x) nd β D (x) β D (x). Tht is, with finer prtition we get better pproximtions to f. With the nottion of integrls of simple functions we hve, with Lebesgue mesure on R, I(α D ) = n m i (x i x i 1 ) nd I(β D ) = i=1 n M i (x i x i 1 ), i=1 which re normlly known s L(D, f) nd U(D, f) in the theory of Riemnn integrtion. Then we obviously hve I(α D ) I(β D ) for ll D, nd if D D then I(α D ) I(α D ) nd I(β D ) I(β D ). Let f(x)dx = sup D I(α D ) nd Then f is Riemnn integrble if, nd only if, f(x)dx = inf D I(β D). 1
2 The common vlue is denoted by f(x)dx = R f(x)dx. f(x)dx. Theorem 1 If f is Riemnn integrble on finite intervl [, b] then it is Lebesgue integrble with the sme vlue. Proof For ech n 1 we cn find, by the definition of supremum, prtition D α n such tht when, in prticulr, 0 f(x)dx I(α D α n ) < 1 n, I(α D α n ) f(x)dx s n. Similrly choose sequence of prtitions D β n such tht I(β D β n ) Set D n = Dn α Dn β then f(x)dx s n. nd Thus I(α D α n ) I(α Dn ) I(β D β n ) I(β Dn ) f(x)dx f(x)dx. I(α Dn ) f(x)dx nd I(β Dn ) f(x)dx s n. (1) 2
3 Replcing the sequence D 1, D 2, D 3,... by D 1, D 1 D 2, D 1 D 2 D 3,...nd relbeling we cn ssume tht D n D n+1 for ll n 1 while (1) still holds. Yet D n D n+1 mens tht α Dn (x) α Dn+1 (x) nd β Dn (x) β Dn+1 (x) for ll n nd x. In prticulr {α Dn } n 1 in n incresing sequence bounded bove by f. So lim n α Dn = g exists, nd stisfies g f. Similrly {β Dn } n 1 in n decresing sequence bounded below by f. So lim n β Dn = h exists, nd stisfies h f. Now {α Dn α D1 } n 1 is n incresing sequence of nonnegtive simple Fmesurble functions tending to g α D1. So by Lebesgue s Monotone Convergence Theorem we hve L (g α D1 )dµ = lim I(α Dn α D1 ) = lim I(α Dn ) I(α D1 ) = f(x)dx I(α D1 ). Since α D1 is simple function we hve L α D 1 dµ = I(α D1 ) nd so L gdµ = Similrly, by exmining β D1 β Dn L hdµ = f(x)dx. (2) we find tht f(x)dx. So, if f is Riemnn integrble, tht is, f(x)dx = f(x)dx, then L (g h)dµ = 0. Yet h g 0, so h = g.e.(µ) on [, b]. But g f h nd so f = g.e.(µ) on [, b]. Hence L fdµ = L = = R gdµ since f = g.e.(µ) on [, b] f(x)dx by (2) f(x)dx 3 since f is Riemnn integrble
4 Let (D) = mx 1 i n (x i x i 1 ). In Theorem 1 it is possible, by dding extr points to ech of the prtitions D n, to ssume tht (D n ) 0 s n. With the nottion nd ssumptions of Theorem 1 we cn prove Lemm 1 Assume tht (D n ) 0 s n. Then for ny x / k=1 D k we hve tht f is continuous t x if, nd only if, g(x) = f(x) = h(x). Proof Recll tht f is continuous t x if, nd only if, ε > 0 δ > 0 : y if y x < δ then f(y) f(x) < ε. (3) For ech k let I k be the subintervl of D k contining x. This is unique since x / k=1 D k. Write I k = [x i 1, x i ]. ( ) Let ε > 0 be given. From (D6) we find δ > 0. Since (D n ) 0 s n there exists N such (D n ) < δ for ll n N. Then l(i k ) < δ so if y I k we hve tht y x < δ. In which cse, from (D6) we get tht f(y) f(x) < ε. In turn this mens tht we hve both inf f(y) f(x) < ε nd sup f(y) f(x) < ε. I k Yet inf Ik f(y) nd sup Ik f(y) re the vlues of α k nd β k t x. Hence, combining the inequlities, β k (x) α k (x) < 2ε. Let k to deduce h(x) g(x) < 2ε. True for ll ε > 0 gives h(x) = g(x). ( ) Assume f is not continuous t x. So ε > 0 δ > 0 : y with y x < δ nd f(y) f(x) ε. (4) For ech k 1 choose δ k = min(x x i 1, x i x) so (x δ k, x + δ k ) I k. But then by (4) we cn find y k (x δ k, x+δ k ) such tht f(y k ) f(x) ε. In prticulr, I k sup f inf f ε, I k I k in which cse β k (x) α k (x) ε nd h(x) g(x) ε. Hence h(x) g(x). This leds to Theorem 2 Assume f : [, b] R is bounded. Then f is Riemnn integrble if, nd only if, f is continuous.e.(µ) on [, b]. Proof Choose sequence of prtitions, D k, s in Lemm 1. Then. 4
5 f is continuous.e.(µ) on [, b] iff f is continuous.e.(µ) outside D k on [, b] k iff g = h.e.(µ) on [, b] by Lemm 1, iff gdµ = hdµ iff iff f(x)dx = f(x)dx f is Riemnn integrble. Mesure preserving Trnsformtions These re specil cse of mesurble functions. Definition T : (, F, µ) (, F, µ) is mesure preserving trnsformtion if (i) T 1 A F for ll A F, (ii) µ(t 1 A) = µ(a) for ll A F. Definition Let A F. A point x A is sid to be recurrent with respect to A if there exists k 1 such tht T k x A. Theorem 3 Poincre s Recurrence Theorem Assume tht µ() <. Let F be the set of points of A which re not recurrent with respect to A. Then µ(f ) = 0. (So for every A F, lmost ll points of A re recurrent.) Proof Let x F. If there exists n 1 such tht T n x F then we hve both x F A nd T n x F A, i.e. x is recurrent point with respect to A, which contrdicts the definition of F. So T n x / F for ll n 1, tht is, T n F F = for ll n 1. Now, the preimge of n empty set is empty, so given ny k, n 1 we hve = T k n (T n F F ) = T k F T (n+k) F. Hence the sets F, T 1 F, T 2 F,... re pirwise disjoint. So 5
6 ( ) > µ() µ T k F = = µ(t k F ) k=0 µ(f ) k=0 k 0 by prt (ii) of definition. Hence µ(f ) = 0. We cn sk how long it tkes point x A to wnder bck into A. To this end define n A (x) = min{n 1 : T n x A}. Assume throughout the rest of this section tht µ() = 1. Definition A mesure preserving mp T : is Ergodic if either of the following hold. (i) Whenever A F is such tht µ(t 1 A A) = 0 then either µ(a) = 0 or 1. (ii) Whenever n integrble function f stisfies f(t x) = f(x) for.e. (µ) x in then f is constnt.e.(µ) on. The first definition here mens tht if T 1 A is lmost exctly A then either µ(a) = 0 or 1. So if 0 < µ(a) < 1 then T 1 A must differ quite lot from A. We sy tht T is mixing up the spce. We do not prove here tht (i) nd (ii) re equivlent. It cn be shown tht if T is ergodic then n A dµ = 1. A Since A dµ = µ(a) we hve, tht in some sense, n A is of size 1/µ(A). This is connected with the question of how often point x will wnder into the set A F. It will be shown below tht for B F nd T ergodic, 1 n 1 lim χ n n B (T k x) = µ(b).e. (µ), (5) k=0 where χ B is the chrcteristic function of the set B, i.e. χ B (x) = 1 if x B, 0 otherwise. So if µ(b) > 0 then lmost every point of wnders into B infinitely often. 6
7 Let S n (x) = #{1 i n : T i x B} nd A n (x) = S n (x)/n. It is not obvious tht the limit lim n A n (x) will exist. We will show tht it does by looking t the limsup nd liminf of the sequence {A n (x)}. So let A(x) = lim sup A n (x) which is trivilly 1. Lemm 2 Let { n } be sequence for which lim sup n <. Let {b n } be sequence for which lim b n = 0. Then lim sup( n + b n ) = lim sup n. Proof Write A = lim sup n. Let ε > 0 be given. There exists N 1 such tht ε < b n < ε for ll n N 1 nd there exists N 2 such tht A ε < sup r < A + ε r n for ll n N 2. Choose N = mx(n 1, N 2 ), so tht for ll n N we hve which gives the result. Lemm 3 A 2ε < sup( r + b r ) < A + 2ε r n Proof A(T x) = A(x). A n (T x) = 1 n 1 i n χ B (T i (T x)) = 1 n 2 i n+1 = A n (x) + χ B(x) χ B (T n+1 x) n nd n ppliction of Lemm 2 gives the result. Theorem 4 χ B (T i x) The limit lim A n (x) exists. n Proof For given x we follow the orbit of x, nmely x, T x, T 2 x, T 3 x,.... We cll the exponent n in T n x, the time. Let ε > 0 be given. It might be tht for ll sufficiently lrge n we hve A n (x) > A(x) ε which obviously shows tht the limit exists. Otherwise the sequence {m j } defined by 7
8 m j = min{m > m j 1 : A m (x) > A(x) ε} hs infinitely mny gps. Note tht this sequence depends on x. The question must be how lrge cn these gps be? Define τ(x) = min{n : A n (x) > A(x) ε}. We first ssume tht there exists M such tht τ(x) < M.e. (µ). Let S be the exceptionl set here. Assume there is gp fter m j so A mj (x) > A(x) ε but A mj +1(x) A(x) ε. Then if T m j x / S we know there exists n < M such tht A n (T m j x) > A(T m j x) ε = A(x) ε by the lemm bove. So we hve both nd m j +1 i m j +n 1 i m j χ B (T i x) > m j (A(x) ε) χ B (T i x) = 1 i n Adding these two inequlities gives tht is 1 i m j +n χ B (T i (T m j x)) > n(a(x) ε). χ B (T i x) > (m j + n)(a(x) ε), A mj +n(x) > A(x) ε. Thus m j+1 m j + n < m j + M. So if T m j x / S, the gp m j+1 m j is less thn M. So if x / T k S k=1 set of mesure zero, ll the gps re less thn M. Thus given x / k=1 T k S nd given N choose j (which will depend on x s well N since the sequence of m j depends on x) such tht m j N < m j+1, then for lmost ll x we hve 8
9 S N (x) S mj (x) > m j (A(x) ε) > (N M)(A(x) ε). Since this inequlity is true for lmost ll x we cn integrte to get (A(x) ε)dµ = 1 N M 1 N M = Nµ(B) N M, n N χ T n Bdµ µ(t n B) since T is mesure preserving. Let N nd then ε 0 to deduce A(x)dµ µ(b). (6) The ssumption bove concerning M my not hold. Define S M = {x : τ(x) > M}, the collection of which is nested sequence of sets, S 1 S 2 S 3... From the definition of lim sup A n (x) we know tht given ε > 0 nd x there exists (infinitely mny) N such tht A N (x) > A(x) ε in which cse τ(x) N nd so x / S N. Let S = M 1 S M, then we hve seen tht x / S for ll x, tht is, S =. Since the S N re nested we cn thus find n M such tht µ(s M ) < ε. Let B = B S M. We pply the rguments bove with B replced by B, so S n(x) = #{1 i n : T i x B }, A n(x) = S n(x)/n nd A (x) = lim sup A n(x). We follow the method bove looking t the gps in the sequence of {m j }. If fter some m j we hve gp then A m j (x) > A (x) ε but A m j +1(x) A (x) ε, in prticulr A m j +1(x) < A m j (x). If it were the cse tht T m j+1 x B then n N A m j +1(x) = S m j (x) + χ B (T mj+1 x) m j + 1 > S m j (x) = A m m j (x), j = S m j (x) + 1 m j + 1 hving used the observtion tht D > C > 0 implies C+1 > C. Hence D+1 D we must hve T mj+1 x / B. In prticulr T mj+1 x / S M, in which cse 9
10 τ(t mj+1 x) M nd so, s in the rgument bove, the gp fter m j is bounded by M. Thus following the rgument tht led to (6) will led to A (x)dµ µ(b ). But A n(x) A n (x) for ll n in which cse A (x) A(x), while µ(b ) < µ(b) + ε. Let ε 0 to deduce A(x)dµ µ(b). Hence this inequlity holds whtever we ssume bout M. A similr rgument gives µ(b) A(x)dµ where A(x) = lim inf A n (x). Hence A(x)dµ = A(x)dµ = µ(b) (7) nd so A(x) = A(x) for.e. (µ) x. Hence lim n A n (x) exists for lmost every (µ) x (nd is integrble). The remining question must be wht is the vlue of this limit? From the definition of ergodic bove the pproprite one for the present sitution sttes tht whenever n integrble function f stisfies f(t x) = f(x) for.e. (µ) x in then f is constnt.e.(µ) on. From the lemm bove we hve tht A(T x) = A(x) for ll x nd so for the points t which the limit exists we hve lim n A n (T x) = lim n A n (x), i.e. this holds.e.(µ) on. Hence if T is ergodic we hve tht lim n A n (x) = c, constnt,.e.(µ). For the vlue of c use (7) tht shows tht µ(b) = lim A n(x)dµ = cdµ = cµ() = c, n since µ() = 1. Hence.e.(µ). #{1 i n : T i x B} lim n n = µ(b) 10
FUNDAMENTALS 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 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 informationThe 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 informationMATH 409 Advanced Calculus I Lecture 19: Riemann sums. Properties of integrals.
MATH 409 Advnced Clculus I Lecture 19: Riemnn sums. Properties of integrls. Drboux sums Let P = {x 0,x 1,...,x n } be prtition of n intervl [,b], where x 0 = < x 1 < < x n = b. Let f : [,b] R be bounded
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 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 informationW. We shall do so one by one, starting with I 1, and we shall do it greedily, trying
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)
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 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 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 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 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 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 informationa n = 1 58 a n+1 1 = 57a n + 1 a n = 56(a n 1) 57 so 0 a n+1 1, and the required result is true, by induction.
MAS221(21617) Exm Solutions 1. (i) A is () bounded bove if there exists K R so tht K for ll A ; (b) it is bounded below if there exists L R so tht L for ll A. e.g. the set { n; n N} is bounded bove (by
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 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 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 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 informationMATH 409 Advanced Calculus I Lecture 18: Darboux sums. The Riemann integral.
MATH 409 Advnced Clculus I Lecture 18: Drboux sums. The Riemnn integrl. Prtitions of n intervl Definition. A prtition of closed bounded intervl [, b] is finite subset P [,b] tht includes the endpoints
More informationChapter 4. Lebesgue Integration
4.2. Lebesgue Integrtion 1 Chpter 4. Lebesgue Integrtion Section 4.2. Lebesgue Integrtion Note. Simple functions ply the sme role to Lebesgue integrls s step functions ply to Riemnn integrtion. Definition.
More informationCalculus in R. Chapter Di erentiation
Chpter 3 Clculus in R 3.1 Di erentition Definition 3.1. Suppose U R is open. A function f : U! R is di erentible t x 2 U if there exists number m such tht lim y!0 pple f(x + y) f(x) my y =0. If f is di
More informationMAT612REAL ANALYSIS RIEMANN STIELTJES INTEGRAL
MAT612REAL ANALYSIS RIEMANN STIELTJES INTEGRAL DR. RITU AGARWAL MALVIYA NATIONAL INSTITUTE OF TECHNOLOGY, JAIPUR, INDIA302017 Tble of Contents Contents Tble of Contents 1 1. Introduction 1 2. Prtition
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 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 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 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 informationPrinciples of Real Analysis I Fall VI. Riemann Integration
21355 Principles of Rel Anlysis I Fll 2004 A. Definitions VI. Riemnn Integrtion Let, b R with < b be given. By prtition of [, b] we men finite set P [, b] with, b P. The set of ll prtitions of [, b] will
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 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 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 informationFor a continuous function f : [a; b]! R we wish to define the Riemann integral
Supplementry Notes for MM509 Topology II 2. The Riemnn Integrl Andrew Swnn For continuous function f : [; b]! R we wish to define the Riemnn integrl R b f (x) dx nd estblish some of its properties. This
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 information2 Definitions and Basic Properties of Extended Riemann Stieltjes Integrals
2 Definitions nd Bsic Properties of Extended Riemnn Stieltjes Integrls 2.1 Regulted nd Intervl Functions Regulted functions Let X be Bnch spce, nd let J be nonempty intervl in R, which my be bounded or
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 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 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 informationMATH1050 CauchySchwarz Inequality and Triangle Inequality
MATH050 CuchySchwrz Inequlity nd Tringle Inequlity 0 Refer to the Hndout Qudrtic polynomils Definition (Asolute extrem for relvlued functions of one rel vrile) Let I e n intervl, nd h : D R e relvlued
More informationSTUDY GUIDE FOR BASIC EXAM
STUDY GUIDE FOR BASIC EXAM BRYON ARAGAM This is prtil list of theorems tht frequently show up on the bsic exm. In mny cses, you my be sked to directly prove one of these theorems or these vrints. There
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 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 informationChapter 0. What is the Lebesgue integral about?
Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous
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 informationDefinite integral. Mathematics FRDIS MENDELU
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the
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 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 informationDefinite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function
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 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 informationChapter 6. Infinite series
Chpter 6 Infinite series We briefly review this chpter in order to study series of functions in chpter 7. We cover from the beginning to Theorem 6.7 in the text excluding Theorem 6.6 nd Rbbe s test (Theorem
More informationDEFINITE INTEGRALS. f(x)dx exists. Note that, together with the definition of definite integrals, definitions (2) and (3) define b
DEFINITE INTEGRALS JOHN D. MCCARTHY Astrct. These re lecture notes for Sections 5.3 nd 5.4. 1. Section 5.3 Definition 1. f is integrle on [, ] if f(x)dx exists. Definition 2. If f() is defined, then f(x)dx.
More informationHomework 4. (1) If f R[a, b], show that f 3 R[a, b]. If f + (x) = max{f(x), 0}, is f + R[a, b]? Justify your answer.
Homework 4 (1) If f R[, b], show tht f 3 R[, b]. If f + (x) = mx{f(x), 0}, is f + R[, b]? Justify your nswer. (2) Let f be continuous function on [, b] tht is strictly positive except finitely mny points
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 informationCzechoslovak Mathematical Journal, 55 (130) (2005), , Abbotsford. 1. Introduction
Czechoslovk Mthemticl Journl, 55 (130) (2005), 933 940 ESTIMATES OF THE REMAINDER IN TAYLOR S THEOREM USING THE HENSTOCKKURZWEIL INTEGRAL, Abbotsford (Received Jnury 22, 2003) Abstrct. When relvlued
More informationa n+2 a n+1 M n a 2 a 1. (2)
Rel Anlysis Fll 004 Tke Home Finl Key 1. Suppose tht f is uniformly continuous on set S R nd {x n } is Cuchy sequence in S. Prove tht {f(x n )} is Cuchy sequence. (f is not ssumed to be continuous outside
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 informationIntroduction to Real Analysis (Math 315) Martin Bohner
ntroduction to Rel Anlysis (Mth 315) Spring 2005 Lecture Notes Mrtin Bohner Author ddress: Version from April 20, 2005 Deprtment of Mthemtics nd Sttistics, University of Missouri Roll, Roll, Missouri 654090020
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. On some properties of definite integrals. We prove
This short collection of notes is intended to complement the textbook Anlisi Mtemtic 2 by Crl Mdern, published by Città Studi Editore, [M]. We refer to [M] for nottion nd the logicl stremline of the rguments.
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 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 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 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 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 informationThis is a short summary of Lebesgue integration theory, which will be used in the course.
3 Chpter 0 ntegrtion theory This is short summry of Lebesgue integrtion theory, which will be used in the course. Fct 0.1. Some subsets (= delmängder E R = (, re mesurble (= mätbr in the Lebesgue sense,
More informationINTRODUCTION TO INTEGRATION
INTRODUCTION TO INTEGRATION 5.1 Ares nd Distnces Assume f(x) 0 on the intervl [, b]. Let A be the re under the grph of f(x). b We will obtin n pproximtion of A in the following three steps. STEP 1: Divide
More informationarxiv: v1 [math.ca] 11 Jul 2011
rxiv:1107.1996v1 [mth.ca] 11 Jul 2011 Existence nd computtion of Riemnn Stieltjes integrls through Riemnn integrls July, 2011 Rodrigo López Pouso Deprtmento de Análise Mtemátic Fcultde de Mtemátics, Universidde
More information7.2 The Definition of the Riemann Integral. Outline
7.2 The Definition of the Riemnn Integrl Tom Lewis Fll Semester 2014 Upper nd lower sums Some importnt theorems Upper nd lower integrls The integrl Two importnt theorems on integrbility Outline Upper nd
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 informationMath 118: Honours Calculus II Winter, 2005 List of Theorems. L(P, f) U(Q, f). f exists for each ǫ > 0 there exists a partition P of [a, b] such that
Mth 118: Honours Clculus II Winter, 2005 List of Theorems Lemm 5.1 (Prtition Refinement): If P nd Q re prtitions of [, b] such tht Q P, then L(P, f) L(Q, f) U(Q, f) U(P, f). Lemm 5.2 (Upper Sums Bound
More informationAdvanced Calculus I (Math 4209) Martin Bohner
Advnced Clculus I (Mth 4209) Spring 2018 Lecture Notes Mrtin Bohner Version from My 4, 2018 Author ddress: Deprtment of Mthemtics nd Sttistics, Missouri University of Science nd Technology, Roll, Missouri
More informationProf. Girardi, Math 703, Fall 2012 Homework Solutions: 1 8. Homework 1. in R, prove that. c k. sup. k n. sup. c k R = inf
Knpp, Chpter, Section, # 4, p. 78 Homework For ny two sequences { n } nd {b n} in R, prove tht lim sup ( n + b n ) lim sup n + lim sup b n, () provided the two terms on the right side re not + nd in some
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 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 informationA HELLY THEOREM FOR FUNCTIONS WITH VALUES IN METRIC SPACES. 1. Introduction
Ttr Mt. Mth. Publ. 44 (29), 159 168 DOI: 1.2478/v1127956z t m Mthemticl Publictions A HELLY THEOREM FOR FUNCTIONS WITH VALUES IN METRIC SPACES Miloslv Duchoň Peter Mličký ABSTRACT. We present Helly
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 informationQuestion 1. Question 3. Question 4. Graduate Analysis I Chapter 5
Grdute Anlysis I Chpter 5 Question If f is simple mesurle function (not necessrily positive) tking vlues j on j, j =,,..., N, show tht f = N j= j j. Proof. We ssume j disjoint nd,, J e nonnegtive ut J+,,
More informationThe Riemann Integral
Deprtment of Mthemtics King Sud University 20172018 Tble of contents 1 Antiderivtive Function nd Indefinite Integrls 2 3 4 5 Indefinite Integrls & Antiderivtive Function Definition Let f : I R be function
More informationIntegration Techniques
Integrtion Techniques. Integrtion of Trigonometric Functions Exmple. Evlute cos x. Recll tht cos x = cos x. Hence, cos x Exmple. Evlute = ( + cos x) = (x + sin x) + C = x + 4 sin x + C. cos 3 x. Let u
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 informationSections 5.2: The Definite Integral
Sections 5.2: The Definite Integrl In this section we shll formlize the ides from the lst section to functions in generl. We strt with forml definition.. The Definite Integrl Definition.. Suppose f(x)
More informationRiemann Stieltjes Integration  Definition and Existence of Integral
 Definition nd Existence of Integrl Dr. Adity Kushik Directorte of Distnce Eduction Kurukshetr University, Kurukshetr Hryn 136119 Indi. Prtition Riemnn Stieltjes Sums Refinement Definition Given closed
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 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 informationWeek 10: Riemann integral and its properties
Clculus nd Liner Algebr for Biomedicl Engineering Week 10: Riemnn integrl nd its properties H. Führ, Lehrstuhl A für Mthemtik, RWTH Achen, WS 07 Motivtion: Computing flow from flow rtes 1 We observe the
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 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 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 informationGeneralized Riemann Integral
Generlized Riemnn Integrl Krel Hrbcek The City College of New York, New York khrbcek@sci.ccny.cuny.edu July 27, 2014 These notes present the theory of generlized Riemnn integrl, due to R. Henstock nd J.
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 informationLECTURE. INTEGRATION AND ANTIDERIVATIVE.
ANALYSIS FOR HIGH SCHOOL TEACHERS LECTURE. INTEGRATION AND ANTIDERIVATIVE. ROTHSCHILD CAESARIA COURSE, 2015/6 1. Integrtion Historiclly, it ws the problem of computing res nd volumes, tht triggered development
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 informationMath 301 Integration I
Mth 301 Integrtion I Lecture notes of Prof. Hichm Gebrn hichm.gebrn@yhoo.com Lebnese University, Fnr, Fll 20162017 http://fs2.ul.edu.lb/mth.htm http://hichmgebrn.wordpress.com 2 Introduction nd orienttion
More informationSTURMLIOUVILLE BOUNDARY VALUE PROBLEMS
STURMLIOUVILLE BOUNDARY VALUE PROBLEMS Throughout, we let [, b] be bounded intervl in R. C 2 ([, b]) denotes the spce of functions with derivtives of second order continuous up to the endpoints. Cc 2
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 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 informationf(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral
Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one
More informationA4 Integration (2017 HT)
A4 Integrtion (217 HT) Zhongmin Qin (mil: qinz@mths.ox.c.uk) Mthemticl Institute, Oxford December 3, 216 Contents 1 Introduction 3 2 xtended rel line, upper nd lower limits 5 3 Mesure spces 6 4 The Lebesgue
More informationProblem Set 4: Solutions Math 201A: Fall 2016
Problem Set 4: s Mth 20A: Fll 206 Problem. Let f : X Y be onetoone, onto mp between metric spces X, Y. () If f is continuous nd X is compct, prove tht f is homeomorphism. Does this result remin true
More informationarxiv:math/ v2 [math.ho] 16 Dec 2003
rxiv:mth/0312293v2 [mth.ho] 16 Dec 2003 Clssicl Lebesgue Integrtion Theorems for the Riemnn Integrl Josh Isrlowitz 244 Ridge Rd. Rutherford, NJ 07070 jbi2@njit.edu Februry 1, 2008 Abstrct In this pper,
More information