Class Notes for Math 921/922: Real Analysis, Instructor Mikil Foss
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University of Nebraska - Lincoln DigitalCommons@University of Nebraska - Lincoln Math Department: Class Notes and Learning Materials Mathematics, Department of 2010 Class Notes for Math 921/922: Real Analysis, Instructor Mikil Foss Laura Lynch University of Nebraska-Lincoln, [email protected] Follow this and additional works at: https://digitalcommons.unl.edu/mathclass Part of the Science and Mathematics Education Commons Lynch, Laura, "Class Notes for Math 921/922: Real Analysis, Instructor Mikil Foss" (2010). Math Department: Class Notes and Learning Materials. 4. https://digitalcommons.unl.edu/mathclass/4 This Article is brought to you for free and open access by the Mathematics, Department of at DigitalCommons@University of Nebraska - Lincoln. It has been accepted for inclusion in Math Department: Class Notes and Learning Materials by an authorized administrator of DigitalCommons@University of Nebraska - Lincoln. Class Notes for Math 921/922: Real Analysis, Instructor Mikil Foss Topics include: Semicontinuity, equicontinuity, absolute continuity, metric spaces, compact spaces, Ascoli’s theorem, Stone Weierstrass theorem, Borel and Lebesque measures, measurable functions, Lebesque integration, convergence theorems, Lp spaces, general measure and integration theory, Radon- Nikodyn theorem, Fubini theorem, Lebesque-Stieltjes integration, Semicontinuity, equicontinuity, absolute continuity, metric spaces, compact spaces, Ascoli’s theorem, Stone Weierstrass theorem, Borel and Lebesque measures, measurable functions, Lebesque integration, convergence theorems, Lp spaces, general measure and integration theory, Radon-Nikodyn theorem, Fubini theorem, Lebesque-Stieltjes integration. Prepared by Laura Lynch, University of Nebraska-Lincoln August 2010 Course Information O±ce Hours: 11:30-1 MWF Optional Meeting Time: Thursdays at 5pm Assessment Homework 6 34pts Midterm 33pts Final 33pts (Dec 13 : 1 ¡ 3pm) 1 Chapter 1 Drawbacks to Riemann Integration 1. Not all bounded functions are Riemann Integrable. 2. All Riemann Integrable functions are bounded. 3. To use the following theorem, we must have f 2 R[a; b]: 1 Theorem 1 (Dominated Convergence Theorem for Riemann Integrals, Arzela). Let ffngn=1 ½ R[a; b] and f 2 R[a; b] be given. Suppose there exists g 2 R[a; b] such that jfn(x)j < g(x) for all x 2 [a; b] and lim fn(x) = f(x) for all n!1 Z b Z b x 2 [a; b]; then lim fn(x)dx = f(x)dx: n!1 a a 8 < p 1 if x = q in lowest terms with 1 · q · n on [0; 1]; Example. De¯ne fn(x) = :0 otherwise: Then fn(x) ! ÂQ\[0;1] =: f: Notice here jfn(x)j < 2 for all x 2 [0; 1] but f is not Riemann Integrable. Thus we can not use the theorem. 4. The space R[a; b] is not complete with respect to many useful metrics. Good Properties of Riemann Integration 1. R[a; b] is a vector space R b 2. The functional f 7! a f(x)dx is linear on R[a; b]: R b 3. a f(x)dx ¸ 0 when f(x) ¸ 0 for all x 2 [a; b]: 4. Theorem 1 holds. 1.1 Measurable and Topological Spaces De¯nition (p 113,119). Let X be a nonempty set. 1. A collection T of subsets of X is called a topology of X if it possesses the following properties: (a) ; 2 T and X 2 T : n n (b) If fUjgj=1 ⊆ T ; then \1 Uj 2 T : (c) If fU®g®2A ⊆ T , then [®2AU® 2 T : 2. If T is a topology on X; then (X; T ) is called a topological space. If T is understood, we may just call X itself a topological space. The members of T are called open sets in X: The complements of open sets in X are called closed sets. 3. If X and Y are topological spaces and f :X ! Y; then f is called continuous if f ¡1(V) is open in X for all V that are open in Y: Examples. 1. If X = R; then f;; Rg is a topology on R: 2. If X = R; then the power set fP(R)g is a topology on R: 3. f;; [a>0f(¡a; a)g; Rg is a topology on R: 4. f;; [a<b2Rf(a; b)g; [a2Rf(¡1; a); (a; 1)g; Rg is a topology on R = R [ f§1g: De¯nition (pg21,25,43). 1. A collection M of subsets of X is called a σ¡algebra if (a) ; 2 M and X 2 M: (b) If E 2 M; then EC 2 M: 1 1 (c) If fEjgj=1 ⊆ M; then [j=1Ej 2 M: 2. If M is a σ¡algebra on X; then (X; M) is called a measurable space. If M is understood, then we may just call X itself a measurable space. The members of M are called measurable sets. 3. If (X; M) and (Y; N ) are measurable spaces, then f :X ! Y is called (M; N )¡measurable or measurable if f ¡1(E) 2 M whenever E 2 N : Examples. 1. (R; f;; Rg) is a σ¡algebra 2. (R; f;; f1g; R n f1g; Rg) is a σ¡algebra 3. (R; fE ⊆ RjE or EC is countableg) is a σ¡algebra 1 1 j¡1 Lemma 1. If fEjgj=1 ½ P(X); then fFjgj=1 ⊆ P(X) de¯ned by Fj = Ej n [k=1Ek is a sequence of mutually disjoint sets 1 1 and [j=1Ej = [j=1Fj: Note. As a result of Lemma 1, we can actually modify part (c) of our de¯nition of a σ¡algebra to say 1 1 (c) If fEjg1 ½ M is a sequence of mutually disjoint sets, then [1 Ej 2 M: Remarks. 1. Property (1a) of our de¯nition for σ¡ algebra could be replaced with \; 2 M or X 2 M:" 1 C 1 1 1 C C 2. If fEjgj=1 ⊆ M; then fEj gj=1 ⊆ M and \j=1Ej = [[j=1Ej ] ⊆ M: 3. If E; F 2 M; then E n F = E \ F C 2 M: Theorem 2. If E is a collection of subsets of X; then there exists a unique smallest σ¡algebra M(E) that contains the members of E: Note: By smallest, we mean any other σ¡algebra will contain all the sets in M(E): Proof. Let be the family of all σ¡algebras containing E: Note 6= ; as P(X) 2 : De¯ne M(E) = \M2M: We want to show M(E) is a σ¡algebra. 1. ;; X 2 M(E) since ;; X 2 M for all M 2 : 2. Let E 2 M(E): Then E 2 M for all M 2 which implies EC 2 M for all M 2 which implies EC 2 M(E): 1 1 1 3. If fEjgj=1 ½ M(E); then fEjgj=1 ½ M for all M 2 which implies [1 Ej 2 M for all M 2 which implies 1 [1 Ej 2 M(E): Remark. M(E) is called the σ¡algebra generated by E: De¯nition. Let (X; T ) be a topological space. The σ¡algebra generated by T is called the Borel σ¡algebra on X and is denoted BX: The members of a Borel σ¡algebra are called Borel sets. Proposition 1 (p 22). BR is generated by each of the following: 1. E1 = f(a; b): a < bg 2. E2 = f[a; b]: a < bg 3. E3 = f(a; b]: a < bg or E4 = f[a; b): a < bg 4. E5 = f(a; 1): a 2 Rg or E6 = f(¡1; a): a 2 Rg 5. E7 = f[a; 1): a 2 Rg or E8 = f(¡1; a]: a 2 Rg Proof. In text. Remark. The Borel σ¡algebra on R is BR = fE ⊆ R : E \ R 2 BRg: It can be generated by E = f(a; 1]: a 2 Rg: Proposition 2. If (X; M) is a measurable space and f :X ! R; then TFAE 1. f is measurable 2. f ¡1((a; 1)) 2 M for all a 2 R 3. f ¡1([a; 1)) 2 M for all a 2 R 4. f ¡1((¡1; a)) 2 M for all a 2 R 5. f ¡1((¡1; a]) 2 M for all a 2 R Proposition 3 (p 43). Let (X; M) and (Y; N ) be measurable spaces. If N is generated by E ⊆ P(Y); then f :X ! Y is (M; N )¡measurable if and only if f ¡1(E) 2 M for all E 2 E: Proof. ()) Since E ⊆ N ; if f is measurable then f ¡1(E) 2 M for all E 2 E by de¯nition. (() De¯ne O = fE 2 Y: f ¡1(E) 2 Mg: Want to show O is a σ¡algebra. Then since E ⊆ O and N is generated by E; ¡1 C ¡1 C ¡1 1 1 ¡1 we will be able to conclude N ⊆ O: Recall (p4 of text) f (E ) = [f (E)] and f ([1 Ej) = [1 f (Ej): Claim: O is a σ¡algebra. Proof: 1. Since f ¡1(;) = ; and f ¡1(Y) = f ¡1(;C ) = (f ¡1(;))C = (;)C = X 2 M; we have ;; Y 2 O: 2. Suppose F 2 O: Then f ¡1(F ) 2 M by de¯nition of O and f ¡1(F C ) = [f ¡1(F )]C 2 M since M is a σ¡algebra. 1 ¡1 1 ¡1 1 1 ¡1 3. Suppose fFjgj=1 2 O: Then ff (Fj)gj=1 2 M and f ([j=1Fj) = [j=1f (Fj) 2 M as M is a σ¡algebra. 1 Thus [j=1Fj 2 O: Hence O is a σ¡algebra on Y; which contains E: Then N 2 O implies f is measurable.