4th Preparation Sheet - Solutions
|
|
- Jordan Barker
- 5 years ago
- Views:
Transcription
1 Prof. Dr. Rainer Dahlhaus Probability Theory Summer term 017 4th Preparation Sheet - Solutions Remark: Throughout the exercise sheet we use the two equivalent definitions of separability of a metric space (X, d): The topology of (X, d) is generated by countably many open sets There exists a countable dense subset A X of (X, d). Task P14 (Separability). Solve the following tasks. (a) Let L 1 (R) := {f : R R f is measurable and f(x) dx < }, (b) Let and let f 1 := f(x) dx be the L 1 -norm. Show that (L 1 (R), 1 ) is separable. Hint: Approximate f by simple functions. Use Theorem 3.4(i) of the lecture and the separability of R. BV (R) := {f : R R f has bounded variation}, where f BV := sup n N sup <t0 <...<t n< n i=1 f(t i) f(t i 1 ) is the variation of f. Show that (BV (R), BV ) is not separable. (c) Let l 0 (R) := {a = (a n ) n N R (a n ) n N is convergent to 0}, and define a := sup n N a n. Show that (l 0 (R), ) is separable. (d) Let l (R) := {a = (a n ) n N R (a n ) n N is a bounded sequence}, and define a := sup n N a n. Show that (l (R), ) is not separable. Hint: You may use without proof that M := {0, 1} N := {a = (a n ) n N : a n {0, 1}} is uncountable which directly follows from the fact that this set can be identified with the binary representation of all real numbers in [0, 1]. Solution: (a) Let (D n ) n N be a countable basis of the topology of (R, ). Let M := n N i 1,...,i { n n N D i k } which is still a countable basis of the topology. This modification allows us to assume that finite unions of elements of M are in M again. We propose the set of primitive functions A := { n y k 1 Bk : n N, y k Q, B k M} = n N 1 y 1,...,y n Q B 1,...,B n Q { n y k 1 Bk }
2 which is countable as a union of countable sets. Reduction to nonnegative functions: By the decomposition f = f + f into positive and negative parts and the inequality f g 1 f + g f g 1, we can assume w.l.o.g. that f 0. Let > 0. Reduction to primitive functions: By results from Probability Theory 1 we can find a sequence of nonnegative primitive functions φ n : R R such that φ n f. By definition of the measure integral, we have (λ is the Lebesgue measure) 0 φ n dλ f dλ, i.e. f φ n dλ 0. Thus we can find a nonnegative primitive function φ N = M N x k1 Ak with A k A (w.l.o.g. λ(a k ) > 0), x k [0, ) (w.l.o.g. x k > 0) such that f φ N 1. 4 Reduction to rational primitive functions: Since Q is dense in R we can find y k Q (0, ) such that x k y k, K = 1,..., M 4M N λ(a k ) N. Put ψ N := M N y k1 Ak, then M N φ N ψ N 1 x k y k λ(a k ) 4. Reduction to functions out of A: By Theorem 3.4(i) of the lecture applied to (R,, λ) there exist open sets O k A k such that λ(o k \A k ) 4M N y k. Since (R, ) is separable, for each k N we can find (C k,i ) i N M (M is the countable basis of the topology of (R, )) such that O k = i N C k,i. Defining the disjoint sets D k,i := C k,i \ i 1 j=1 C k,j, we see that λ(o k ) = i N λ(d k,i ), therefore there have to exist m k N such that Define M N ψ N := 4M N y k > λ(o k ) m k i=1 λ(d k,i) = λ ( O k \ m k i=1 C k,i). y k 1 m k i=1 C k,i. Since m k i=1 C k,i M (see the motivation at the beginning), we have ψ N A. Furthermore, which shows in total that ψ N ψ M N N 1 y k 1 Ak 1 m k i=1 C dλ < k,i 4 + 4, λ(o k \A k )+λ(o k \ m k i=1 C k,i) f ψ N 1 f φ N 1 + φ N ψ N 1 + ψ N ψ N 1 <. (b) For x R, define f x : R R, f x (y) := 1 (,x] (y). Obviously it holds that f x BV = 1 (f x has only one jump of height 1). For x 1, x [0, 1], x 1 < x we have f x f x1 = 1 (x1,x ], thus f x1 f x BV = (since f x f x1 has two jumps of height 1). Suppose that there exists a countably dense subset A BV (R) of (BV (R), BV ). Because
3 A is dense in (BV (R), BV ), for each x R we can find an a x A with f x a x < 1. In total, with x 1, x R, x 1 x : = f x f x1 f x1 a x1 + a x1 a x + a x f x 1 = 1 1 < 1 f x 1 a x1 a x f x a x1 a x, which shows a x1 a x. This is a contradiction to the property of A to be countable, because is uncountable. (c) Define {a x : x R} A A := m N{b = (b n ) n N : b n Q for 1 n m, b n = 0 for n > m} = m N Q m {(0) n N } which is countable as a countable union of countable sets. Obviously A consists of sequences converging to 0, thus A l 0 (R). We now show that A is dense in (l 0 (R), ). Let (a n ) n N l 0 (R) be arbitrary. Let > 0. Since a n 0, there exists N N such that for all n > N, a n <. Since Q is dense in R, for each 1 n N there exists b n Q such that a n b n. We now choose b = (b n) n N A, where b n = 0 for n > N. Then {, n = 1,..., N, a n b n a n, n > N, thus sup n N a n b n <. (d) Note that for all m M (see the hint for the definition of M), m = 1 <, thus m l (N). For m, m M, m 1 m there has to exist at least one index n N with m n m n, and thus m m = sup n N m n m n 1. Suppose that there exists a countably dense subset A l (N) of (l (N), ). Because A is dense in (l (N), ), for each m M we can find a m A such that m a m < 1 4. In total, with m 1, m M, m 1 m : 1 m 1 m m 1 a m1 + a m1 a m + a m m 1 = < 1 m 1 a m1 a m m a m1 a m, which shows a m1 a m. This is a contradiction to the property of A to be countable, because is uncountable. {a m : m M} A Task P15 (Polish spaces). Solve the following tasks. (a) Let (X, d) be a Polish space and U X an open, nonempty subset of X. Show that (U, d) is again Polish. 1 Hint: Define the metric ρ(x, y) := d(x, y)+ f(x) f(y) on U, where f(x) :=. inf{d(x,z):z X\U} For separability, show that ρ induces the same topology as d on U. 3
4 (b) Let C[0, ) := {f : [0, ) R f is continuous}. Define the metric d(f, g) := k sup f(x) g(x) x [0,k] on C[0, ). Assume that it is already known that (C[0, k], ) is Polish for k N. Show that (C[0, ), d) is Polish. Solution: (a) Note that the topology of (U, d) is the subspace topology {A U : A X open in (X, d)}. In the following we will show that g : (U, d) ((0, ), ), x inf{d(x, z) : z X\U} is continuous. g is well-defined since for each fixed x U, we can find > 0 such that B (x) = {z X : d(x, z) < } U which implies g(x) /. Since d(x, z) d(x, y) + d(y, z) inf d(x, z) inf d(y, z) d(x, y) z X\U z X\U it follows that (swapping x, y in the above inequality for the same result with interchanged signs) inf d(x, z) inf d(y, z) d(x, y). z X\U z X\U This finishes the proof of the continuity of g. Since 1 : (0, ) (0, ) is continuous, we x conclude that f : (U, d) ((0, ), ) is continuous as a composition (*). We first show that ρ generates the same topology: Let A U be open in (U, d), and let x A. Then there exists > 0 such that {z U : d(x, z) < } A. Since d ρ, we have {z U : ρ(x, z) < } {z U : d(x, z) < } A, thus A is open in (U, ρ). Let A U be open in (U, ρ), and let x A. Then there exists > 0 such that {z U : ρ(x, z) < } A. Since f : (U, d) ((0, ), ) is continuous (cf. (*)), there exists δ > 0 such that d(x, z) < δ implies f(x) f(z) /. Choose := min{δ, /}. Then {z U : d(x, z) < } {z U : ρ(x, z) < } A, thus A is open in (U, d). Separability: Let A X be a countable dense subset of (X, d). Then A := A U is still countable. Let x U be arbitrary. Since U is open, there exists > 0 such that {z X : d(x, z) < } U. Since A is dense in (X, d), we find some y A with d(x, y) < /. This implies that y U, i.e. y A U = A. This shows that A is a countable dense subset of U. Completeness: Let (x n ) n N U be a Cauchy sequence in (U, ρ). Since d ρ, it follows that (x n ) n N is also Cauchy in (X, d). Since (X, d) is complete, there exists a limit x X such that d(x n, x) 0. Obviously, it holds that x U U (where U is the boundary in (X, d)). Assume that x U X\U. Then inf{d(x n, z) : z X\U} 0. This yields f(x n ) as well as f(x m ) (n, m ) and is therefore a contradiction the fact that (x n ) n N is a Cauchy sequence in (U, ρ) (since ρ(x, y) = d(x, y) + f(x) f(y) ). Thus x U. Since f is continuous on U (see (*)) and d(x n, x) 0 with x U, we obtain that f(x n ) f(x) 0. Thus ρ(x n, x) 0, i.e. (x n ) n N is convergent in (U, ρ). (b) Separability: For k N, let A k be the countable dense set in (C[0, k], ). Define A := {f C[0, ) : f [0,k] A k, f(x) = f(k) for all x > k}. k N =:B k 4
5 Note that there exists a one-to-one mapping from B k to A k since the domain of definition of the functions of A k is only extended to [0, ) glueing a constant line on these functions. Thus A is countable as a countable union of countable sets. Fix some f C[0, ). Let > 0. For N N large enough it holds that k=n+1 k <. Since f [0,N] C[0, N], there exists some g A N C[0, N] such that f [0,N] g <. The N corresponding element g B N A then fulfills d(f, g) = k sup f(x) g(x) x [0,k] N sup x [0,N] f(x) g(x) + k=n+1 k < N N + =. Remark: A direct proof of separability in the manner of task 14 from exercise sheet 4 without using prior information on C[0, k] is also possible. Completeness: Let (f n ) n N be an arbitrary Cauchy sequence in (C[0, ), d). Since d(f n, f m ) = k sup x [0,k] f n (x) f m (x) 0 for m n, each summand has to converge to 0, i.e. for all k N it holds that sup f n (x) f m (x) 0 (m n ). x [0,k] This shows that for all k N, (f n [0,k] ) n N is a Cauchy sequence in (C[0, k], ). Since (C[0, k], ) is complete, there exists a limit f k C[0, k] with f n [0,k] f k 0. Define f(x) := f k (x) for x [0, k]. Since the uniform limits are unique, this definition is welldefined. We conclude that f C[0, ). Furthermore, f n [0,k] f [0,k] 0. Now let > 0 be arbitrarily chosen. Let K N be large enough such that k=k+1 k <. Choose N N large enough such that for all n N, sup x [0,K] f n (x) f(x) <. Then we have for n N: K d(f n, f) K sup x [0,K] f n (x) f(x) + k=k+1 k < K K + =, showing that d(f n, f) 0. Task P16 (The set of cylinder sets). Let (S, A) be a measurable space and T an arbitrary index set. For index sets J I T, π I,J : S I S J, (x i ) i I (x j ) j J are the projections. Recall that B(A) T := σ(π T,{t} : t T ) is the smallest σ-algebra which makes the one-dimensional projections measurable. Put C := (a) Show that C is an algebra over S T. (b) Show that σ(c) = A T. I T finite π 1 T,I (AI ) Remark: In the lecture, one would often use (S, A) = (R, B(R)) and T = [0, ). Then C = n N 0 t 1 <t <...<t n π T,{t1,...,t n}(b(r) n ). Solution: (a) In the following we show the three properties of X to be an algebra: (i) Choose I = {t} T. Because S A, we have S T π 1 T,{t} (A), thus ST C. (ii) Let B 1, B C. Then there exist finite I 1, I T and A 1 A I 1, A A I such that B 1 = π 1 T,I 1 (A 1 ), B = π 1 T,I (A ). 5
6 Define I := I 1 I and A := π 1 I,I 1 (A 1 ) π 1 I,I (A ). Then we have (by the stability of the preimage w.r.t. unions) B 1 B = (π I,I1 π T,I ) 1 (A 1 ) (π I,I π T,I ) 1 (A ) = π 1 T,I (A). Because the projections π I,I1 : (S I, A I ) (S I 1, A I 1 ), π I,I : (S I, A I ) (S I, A I) are measurable (*), we have A A I and thus B 1 B π 1 T,I (AI ) C. Proof of (*): It is well-known that E := { i I 1 A i : i I 1 : A i A i } generates the σ- algebra A I 1, i.e. σ(e) = A I 1. For i I 1 A i = E E we have π 1 I,I 1 (E) = i I 1 π 1 I,{i} (A i) A I (since A I is the smallest σ-algebra which makes the projections to one component measurable). Therefore, π 1 I,I 1 (E) A I. In total, π 1 I,I 1 (A I 1 ) A I. (iii) Let B C. Then there exists a finite I T and A A I such that B = π 1 T,I (A). Because A I is a σ-algebra, we have A c A I. Thus B c = π 1 T,I (Ac ) π 1 T,I (AI ) C. (b) Because A T = σ(π 1 T,{t} (A) : t T ) and π 1 T,{t} (A) C, we have AT σ(c). Let I T be a finite index set. It is well-known that the set E := { i I A i i I : A i A} generates the product σ-algebra A I. It is therefore enough to show that π 1 T,I (E) AT (*), because then we obtain π 1 T,I (AI ) = π 1 T,I (σ(e)) well-known rule = σ(π 1 T,I (E)) AT σ-algebra A T. As a consequence, C A T (note that I T finite is a possibly uncountable union, but we do not unite sets, but collections of sets!). Proof of (*): If A = i I A i with A i A i, then π 1 T,I (A) = i I π 1 T,{i} (A i) A T, because it is the finite intersection of sets that are in A T. Task P17 (Measurability and continuity). Let X = (X t ) t 0 be a real-values stochastic process on (Ω, A, P), i.e. for each t 0, X t : Ω R is (A, B(R))-measurable. (a) Show that the above measurability condition is equivalent to the fact that X : Ω R [0, ) is A-B(R) [0, ) -measurable (see the definition in Task P16). From now on assume that X is right-continuous (i.e. t X t (ω) is continuous for all ω Ω). (b) Show that X is product measurable, i.e. X : Ω [0, ) R is ( A B([0, )) ) -B(R)- measurable. (c) Show that the following subsets of Ω are in A: M 1 := { t 0 : X t 0}, M := { lim X t = 0}, t M 3 := {t X t is Lipschitz continuous with constant L}, L > 0. Solution: (a) We have to show the equivalence [ ] t 0 : Xt 1 (B(R)) A X 1 (B(R) [0, ) ) A. 6
7 Since B(R) [0, ) = σ(e) with E = t 0 {π 1 [0, ),{t}(b(r))}, we have X 1 (B(R) [0, ) ) = X 1 (σ(e)) ( = σ (X 1 ( = σ t 0 t 0 well-known rule = σ(x 1 (E)) )) ( π 1 [0, ),{t} (B(R)) = σ Xt 1 (B(R)) ) = σ(x t : t 0). t 0 X 1 (π 1 [0, ),{t} (B(R))) ) Now, " "is obvious. For " ", note that by assumption t 0 X 1 t (B(R)) A (note that we build the union of sets of sets and thus this is still a subset of A), thus X 1 (B(R) [0, ) ) = σ( t 0 X 1 t (B(R))) A. (b) Define the stochastic process X (n) t := X nt /n for t 0, where denotes flooring. Since nt /n t for n and t X t is continuous, it holds that X (n) X on Ω [0, ) (indeed, for each t 0 and ω Ω we have X (n) t (ω) = X nt /n (ω) X t (ω)). We now show that X (n) is A B([0, ))-B(R)-measurable, then it follows that X is also measurable with respect to the same σ-algebras as limit of X (n). We only have to show measurability on the generating system {[0, c] : c > 0} of B(R). It holds that = = {(ω, t) Ω [0, ) : X (n) t (ω) c} {(ω, t) : X nt /n (ω) c, t [ k n, k + 1 n )} k=0 ( {ω Ω : X k/n (ω) c} [ k n, k + 1 ) n ) A B([0, )), k=0 A B([0, )) since {A B : A A, B B([0, ))} A B([0, )) and σ-algebras are closed with respect to countable unions. (c) M 1 : By continuity we have that M 1 ( ) = {t [0, ) Q : X t 0} = t [0, ) Q {X t 0} A A as a countable union of elements of A. (*) holds since for arbitrary t [0, ) we can find a sequence t n [0, ) Q with t n t. Since X has continuous paths, 0 X tn X t, thus X t 0 (this implies {t [0, ) Q : X t 0} M 1, the other relation is obvious). For the other two cases M and M 3, these facts are similarly proved and we will therefore abbreviate the argumentations. M : It holds that M = { k N : N N : t [N, ) : X t 1 k } X = cont. k N N N t [N, ) Q { X t 1 k } A, A since countable unions and intersections of elements of σ-algebras are again elements of the σ-algebra. M 3 : Here we have M 3 = { s, t [0, ) : X t X s L s t } X cont. = 7 s,t [0, ) Q { X t X s L s t } A. A
8 Submission: No submission! This sheet will (partly) be discussed in the exercise groups. Homepage of the lecture: 8
3 (Due ). Let A X consist of points (x, y) such that either x or y is a rational number. Is A measurable? What is its Lebesgue measure?
MA 645-4A (Real Analysis), Dr. Chernov Homework assignment 1 (Due ). Show that the open disk x 2 + y 2 < 1 is a countable union of planar elementary sets. Show that the closed disk x 2 + y 2 1 is a countable
More informationFunctional Analysis Exercise Class
Functional Analysis Exercise Class Week 2 November 6 November Deadline to hand in the homeworks: your exercise class on week 9 November 13 November Exercises (1) Let X be the following space of piecewise
More informationProblem Set 2: Solutions Math 201A: Fall 2016
Problem Set 2: s Math 201A: Fall 2016 Problem 1. (a) Prove that a closed subset of a complete metric space is complete. (b) Prove that a closed subset of a compact metric space is compact. (c) Prove that
More information2 (Bonus). Let A X consist of points (x, y) such that either x or y is a rational number. Is A measurable? What is its Lebesgue measure?
MA 645-4A (Real Analysis), Dr. Chernov Homework assignment 1 (Due 9/5). Prove that every countable set A is measurable and µ(a) = 0. 2 (Bonus). Let A consist of points (x, y) such that either x or y is
More informationMetric Spaces Math 413 Honors Project
Metric Spaces Math 413 Honors Project 1 Metric Spaces Definition 1.1 Let X be a set. A metric on X is a function d : X X R such that for all x, y, z X: i) d(x, y) = d(y, x); ii) d(x, y) = 0 if and only
More informationMAT 570 REAL ANALYSIS LECTURE NOTES. Contents. 1. Sets Functions Countability Axiom of choice Equivalence relations 9
MAT 570 REAL ANALYSIS LECTURE NOTES PROFESSOR: JOHN QUIGG SEMESTER: FALL 204 Contents. Sets 2 2. Functions 5 3. Countability 7 4. Axiom of choice 8 5. Equivalence relations 9 6. Real numbers 9 7. Extended
More informationMetric Spaces. Exercises Fall 2017 Lecturer: Viveka Erlandsson. Written by M.van den Berg
Metric Spaces Exercises Fall 2017 Lecturer: Viveka Erlandsson Written by M.van den Berg School of Mathematics University of Bristol BS8 1TW Bristol, UK 1 Exercises. 1. Let X be a non-empty set, and suppose
More informationMetric Spaces and Topology
Chapter 2 Metric Spaces and Topology From an engineering perspective, the most important way to construct a topology on a set is to define the topology in terms of a metric on the set. This approach underlies
More informationReal Analysis Problems
Real Analysis Problems Cristian E. Gutiérrez September 14, 29 1 1 CONTINUITY 1 Continuity Problem 1.1 Let r n be the sequence of rational numbers and Prove that f(x) = 1. f is continuous on the irrationals.
More informationTHEOREMS, ETC., FOR MATH 515
THEOREMS, ETC., FOR MATH 515 Proposition 1 (=comment on page 17). If A is an algebra, then any finite union or finite intersection of sets in A is also in A. Proposition 2 (=Proposition 1.1). For every
More informationMetric Spaces Math 413 Honors Project
Metric Spaces Math 413 Honors Project 1 Metric Spaces Definition 1.1 Let X be a set. A metric on X is a function d : X X R such that for all x, y, z X: i) d(x, y) = d(y, x); ii) d(x, y) = 0 if and only
More information(1) Consider the space S consisting of all continuous real-valued functions on the closed interval [0, 1]. For f, g S, define
Homework, Real Analysis I, Fall, 2010. (1) Consider the space S consisting of all continuous real-valued functions on the closed interval [0, 1]. For f, g S, define ρ(f, g) = 1 0 f(x) g(x) dx. Show that
More informationREAL AND COMPLEX ANALYSIS
REAL AND COMPLE ANALYSIS Third Edition Walter Rudin Professor of Mathematics University of Wisconsin, Madison Version 1.1 No rights reserved. Any part of this work can be reproduced or transmitted in any
More informationBounded uniformly continuous functions
Bounded uniformly continuous functions Objectives. To study the basic properties of the C -algebra of the bounded uniformly continuous functions on some metric space. Requirements. Basic concepts of analysis:
More informationBootcamp. Christoph Thiele. Summer As in the case of separability we have the following two observations: Lemma 1 Finite sets are compact.
Bootcamp Christoph Thiele Summer 212.1 Compactness Definition 1 A metric space is called compact, if every cover of the space has a finite subcover. As in the case of separability we have the following
More informationEconomics 204 Summer/Fall 2011 Lecture 5 Friday July 29, 2011
Economics 204 Summer/Fall 2011 Lecture 5 Friday July 29, 2011 Section 2.6 (cont.) Properties of Real Functions Here we first study properties of functions from R to R, making use of the additional structure
More informationMATS113 ADVANCED MEASURE THEORY SPRING 2016
MATS113 ADVANCED MEASURE THEORY SPRING 2016 Foreword These are the lecture notes for the course Advanced Measure Theory given at the University of Jyväskylä in the Spring of 2016. The lecture notes can
More informationLebesgue Integration on R n
Lebesgue Integration on R n The treatment here is based loosely on that of Jones, Lebesgue Integration on Euclidean Space We give an overview from the perspective of a user of the theory Riemann integration
More informationn [ F (b j ) F (a j ) ], n j=1(a j, b j ] E (4.1)
1.4. CONSTRUCTION OF LEBESGUE-STIELTJES MEASURES In this section we shall put to use the Carathéodory-Hahn theory, in order to construct measures with certain desirable properties first on the real line
More informationMeasure and integration
Chapter 5 Measure and integration In calculus you have learned how to calculate the size of different kinds of sets: the length of a curve, the area of a region or a surface, the volume or mass of a solid.
More informationMATH 51H Section 4. October 16, Recall what it means for a function between metric spaces to be continuous:
MATH 51H Section 4 October 16, 2015 1 Continuity Recall what it means for a function between metric spaces to be continuous: Definition. Let (X, d X ), (Y, d Y ) be metric spaces. A function f : X Y is
More informationLecture Notes in Advanced Calculus 1 (80315) Raz Kupferman Institute of Mathematics The Hebrew University
Lecture Notes in Advanced Calculus 1 (80315) Raz Kupferman Institute of Mathematics The Hebrew University February 7, 2007 2 Contents 1 Metric Spaces 1 1.1 Basic definitions...........................
More informationREVIEW OF ESSENTIAL MATH 346 TOPICS
REVIEW OF ESSENTIAL MATH 346 TOPICS 1. AXIOMATIC STRUCTURE OF R Doğan Çömez The real number system is a complete ordered field, i.e., it is a set R which is endowed with addition and multiplication operations
More informationMATH 202B - Problem Set 5
MATH 202B - Problem Set 5 Walid Krichene (23265217) March 6, 2013 (5.1) Show that there exists a continuous function F : [0, 1] R which is monotonic on no interval of positive length. proof We know there
More informationTools from Lebesgue integration
Tools from Lebesgue integration E.P. van den Ban Fall 2005 Introduction In these notes we describe some of the basic tools from the theory of Lebesgue integration. Definitions and results will be given
More informationReal Analysis Math 131AH Rudin, Chapter #1. Dominique Abdi
Real Analysis Math 3AH Rudin, Chapter # Dominique Abdi.. If r is rational (r 0) and x is irrational, prove that r + x and rx are irrational. Solution. Assume the contrary, that r+x and rx are rational.
More informationProblem set 1, Real Analysis I, Spring, 2015.
Problem set 1, Real Analysis I, Spring, 015. (1) Let f n : D R be a sequence of functions with domain D R n. Recall that f n f uniformly if and only if for all ɛ > 0, there is an N = N(ɛ) so that if n
More informationF 1 =. Setting F 1 = F i0 we have that. j=1 F i j
Topology Exercise Sheet 5 Prof. Dr. Alessandro Sisto Due to 28 March Question 1: Let T be the following topology on the real line R: T ; for each finite set F R, we declare R F T. (a) Check that T is a
More informationABSTRACT INTEGRATION CHAPTER ONE
CHAPTER ONE ABSTRACT INTEGRATION Version 1.1 No rights reserved. Any part of this work can be reproduced or transmitted in any form or by any means. Suggestions and errors are invited and can be mailed
More information1 Topology Definition of a topology Basis (Base) of a topology The subspace topology & the product topology on X Y 3
Index Page 1 Topology 2 1.1 Definition of a topology 2 1.2 Basis (Base) of a topology 2 1.3 The subspace topology & the product topology on X Y 3 1.4 Basic topology concepts: limit points, closed sets,
More informationChapter 2 Metric Spaces
Chapter 2 Metric Spaces The purpose of this chapter is to present a summary of some basic properties of metric and topological spaces that play an important role in the main body of the book. 2.1 Metrics
More informationANALYSIS WORKSHEET II: METRIC SPACES
ANALYSIS WORKSHEET II: METRIC SPACES Definition 1. A metric space (X, d) is a space X of objects (called points), together with a distance function or metric d : X X [0, ), which associates to each pair
More informationMAT 544 Problem Set 2 Solutions
MAT 544 Problem Set 2 Solutions Problems. Problem 1 A metric space is separable if it contains a dense subset which is finite or countably infinite. Prove that every totally bounded metric space X is separable.
More informationFunctional Analysis Winter 2018/2019
Functional Analysis Winter 2018/2019 Peer Christian Kunstmann Karlsruher Institut für Technologie (KIT) Institut für Analysis Englerstr. 2, 76131 Karlsruhe e-mail: peer.kunstmann@kit.edu These lecture
More informationIntroduction to Functional Analysis
Introduction to Functional Analysis Carnegie Mellon University, 21-640, Spring 2014 Acknowledgements These notes are based on the lecture course given by Irene Fonseca but may differ from the exact lecture
More informationReal Analysis Chapter 4 Solutions Jonathan Conder
2. Let x, y X and suppose that x y. Then {x} c is open in the cofinite topology and contains y but not x. The cofinite topology on X is therefore T 1. Since X is infinite it contains two distinct points
More informationReal Analysis Notes. Thomas Goller
Real Analysis Notes Thomas Goller September 4, 2011 Contents 1 Abstract Measure Spaces 2 1.1 Basic Definitions........................... 2 1.2 Measurable Functions........................ 2 1.3 Integration..............................
More informationIntroduction and Preliminaries
Chapter 1 Introduction and Preliminaries This chapter serves two purposes. The first purpose is to prepare the readers for the more systematic development in later chapters of methods of real analysis
More informationMeasure Theory on Topological Spaces. Course: Prof. Tony Dorlas 2010 Typset: Cathal Ormond
Measure Theory on Topological Spaces Course: Prof. Tony Dorlas 2010 Typset: Cathal Ormond May 22, 2011 Contents 1 Introduction 2 1.1 The Riemann Integral........................................ 2 1.2 Measurable..............................................
More informationMath 209B Homework 2
Math 29B Homework 2 Edward Burkard Note: All vector spaces are over the field F = R or C 4.6. Two Compactness Theorems. 4. Point Set Topology Exercise 6 The product of countably many sequentally compact
More informationLecture 2: Random Variables and Expectation
Econ 514: Probability and Statistics Lecture 2: Random Variables and Expectation Definition of function: Given sets X and Y, a function f with domain X and image Y is a rule that assigns to every x X one
More informationconverges as well if x < 1. 1 x n x n 1 1 = 2 a nx n
Solve the following 6 problems. 1. Prove that if series n=1 a nx n converges for all x such that x < 1, then the series n=1 a n xn 1 x converges as well if x < 1. n For x < 1, x n 0 as n, so there exists
More informationAnalysis Qualifying Exam
Analysis Qualifying Exam Spring 2017 Problem 1: Let f be differentiable on R. Suppose that there exists M > 0 such that f(k) M for each integer k, and f (x) M for all x R. Show that f is bounded, i.e.,
More informationMH 7500 THEOREMS. (iii) A = A; (iv) A B = A B. Theorem 5. If {A α : α Λ} is any collection of subsets of a space X, then
MH 7500 THEOREMS Definition. A topological space is an ordered pair (X, T ), where X is a set and T is a collection of subsets of X such that (i) T and X T ; (ii) U V T whenever U, V T ; (iii) U T whenever
More informationMetric Space Topology (Spring 2016) Selected Homework Solutions. HW1 Q1.2. Suppose that d is a metric on a set X. Prove that the inequality d(x, y)
Metric Space Topology (Spring 2016) Selected Homework Solutions HW1 Q1.2. Suppose that d is a metric on a set X. Prove that the inequality d(x, y) d(z, w) d(x, z) + d(y, w) holds for all w, x, y, z X.
More informationSummer Jump-Start Program for Analysis, 2012 Song-Ying Li
Summer Jump-Start Program for Analysis, 01 Song-Ying Li 1 Lecture 6: Uniformly continuity and sequence of functions 1.1 Uniform Continuity Definition 1.1 Let (X, d 1 ) and (Y, d ) are metric spaces and
More informationNotes 1 : Measure-theoretic foundations I
Notes 1 : Measure-theoretic foundations I Math 733-734: Theory of Probability Lecturer: Sebastien Roch References: [Wil91, Section 1.0-1.8, 2.1-2.3, 3.1-3.11], [Fel68, Sections 7.2, 8.1, 9.6], [Dur10,
More informationvan Rooij, Schikhof: A Second Course on Real Functions
vanrooijschikhofproblems.tex December 5, 2017 http://thales.doa.fmph.uniba.sk/sleziak/texty/rozne/pozn/books/ van Rooij, Schikhof: A Second Course on Real Functions Some notes made when reading [vrs].
More informationMATH5011 Real Analysis I. Exercise 1 Suggested Solution
MATH5011 Real Analysis I Exercise 1 Suggested Solution Notations in the notes are used. (1) Show that every open set in R can be written as a countable union of mutually disjoint open intervals. Hint:
More informationThus, X is connected by Problem 4. Case 3: X = (a, b]. This case is analogous to Case 2. Case 4: X = (a, b). Choose ε < b a
Solutions to Homework #6 1. Complete the proof of the backwards direction of Theorem 12.2 from class (which asserts the any interval in R is connected). Solution: Let X R be a closed interval. Case 1:
More informationg 2 (x) (1/3)M 1 = (1/3)(2/3)M.
COMPACTNESS If C R n is closed and bounded, then by B-W it is sequentially compact: any sequence of points in C has a subsequence converging to a point in C Conversely, any sequentially compact C R n is
More informationMath 426 Homework 4 Due 3 November 2017
Math 46 Homework 4 Due 3 November 017 1. Given a metric space X,d) and two subsets A,B, we define the distance between them, dista,b), as the infimum inf a A, b B da,b). a) Prove that if A is compact and
More informationChapter 1. Measure Spaces. 1.1 Algebras and σ algebras of sets Notation and preliminaries
Chapter 1 Measure Spaces 1.1 Algebras and σ algebras of sets 1.1.1 Notation and preliminaries We shall denote by X a nonempty set, by P(X) the set of all parts (i.e., subsets) of X, and by the empty set.
More informationAnalysis Comprehensive Exam Questions Fall F(x) = 1 x. f(t)dt. t 1 2. tf 2 (t)dt. and g(t, x) = 2 t. 2 t
Analysis Comprehensive Exam Questions Fall 2. Let f L 2 (, ) be given. (a) Prove that ( x 2 f(t) dt) 2 x x t f(t) 2 dt. (b) Given part (a), prove that F L 2 (, ) 2 f L 2 (, ), where F(x) = x (a) Using
More informationMath 201 Topology I. Lecture notes of Prof. Hicham Gebran
Math 201 Topology I Lecture notes of Prof. Hicham Gebran hicham.gebran@yahoo.com Lebanese University, Fanar, Fall 2015-2016 http://fs2.ul.edu.lb/math http://hichamgebran.wordpress.com 2 Introduction and
More informationNAME: Mathematics 205A, Fall 2008, Final Examination. Answer Key
NAME: Mathematics 205A, Fall 2008, Final Examination Answer Key 1 1. [25 points] Let X be a set with 2 or more elements. Show that there are topologies U and V on X such that the identity map J : (X, U)
More informationFrom now on, we will represent a metric space with (X, d). Here are some examples: i=1 (x i y i ) p ) 1 p, p 1.
Chapter 1 Metric spaces 1.1 Metric and convergence We will begin with some basic concepts. Definition 1.1. (Metric space) Metric space is a set X, with a metric satisfying: 1. d(x, y) 0, d(x, y) = 0 x
More informationl(y j ) = 0 for all y j (1)
Problem 1. The closed linear span of a subset {y j } of a normed vector space is defined as the intersection of all closed subspaces containing all y j and thus the smallest such subspace. 1 Show that
More informationSome Background Material
Chapter 1 Some Background Material In the first chapter, we present a quick review of elementary - but important - material as a way of dipping our toes in the water. This chapter also introduces important
More informationReal Analysis, 2nd Edition, G.B.Folland Elements of Functional Analysis
Real Analysis, 2nd Edition, G.B.Folland Chapter 5 Elements of Functional Analysis Yung-Hsiang Huang 5.1 Normed Vector Spaces 1. Note for any x, y X and a, b K, x+y x + y and by ax b y x + b a x. 2. It
More informationSpring 2014 Advanced Probability Overview. Lecture Notes Set 1: Course Overview, σ-fields, and Measures
36-752 Spring 2014 Advanced Probability Overview Lecture Notes Set 1: Course Overview, σ-fields, and Measures Instructor: Jing Lei Associated reading: Sec 1.1-1.4 of Ash and Doléans-Dade; Sec 1.1 and A.1
More information1/12/05: sec 3.1 and my article: How good is the Lebesgue measure?, Math. Intelligencer 11(2) (1989),
Real Analysis 2, Math 651, Spring 2005 April 26, 2005 1 Real Analysis 2, Math 651, Spring 2005 Krzysztof Chris Ciesielski 1/12/05: sec 3.1 and my article: How good is the Lebesgue measure?, Math. Intelligencer
More information2. Function spaces and approximation
2.1 2. Function spaces and approximation 2.1. The space of test functions. Notation and prerequisites are collected in Appendix A. Let Ω be an open subset of R n. The space C0 (Ω), consisting of the C
More informationExercises from other sources REAL NUMBERS 2,...,
Exercises from other sources REAL NUMBERS 1. Find the supremum and infimum of the following sets: a) {1, b) c) 12, 13, 14, }, { 1 3, 4 9, 13 27, 40 } 81,, { 2, 2 + 2, 2 + 2 + } 2,..., d) {n N : n 2 < 10},
More informationHW 4 SOLUTIONS. , x + x x 1 ) 2
HW 4 SOLUTIONS The Way of Analysis p. 98: 1.) Suppose that A is open. Show that A minus a finite set is still open. This follows by induction as long as A minus one point x is still open. To see that A
More information3. (a) What is a simple function? What is an integrable function? How is f dµ defined? Define it first
Math 632/6321: Theory of Functions of a Real Variable Sample Preinary Exam Questions 1. Let (, M, µ) be a measure space. (a) Prove that if µ() < and if 1 p < q
More informationPROBABILITY THEORY II
Ruprecht-Karls-Universität Heidelberg Institut für Angewandte Mathematik Prof. Dr. Jan JOHANNES Outline of the lecture course PROBABILITY THEORY II Summer semester 2016 Preliminary version: April 21, 2016
More informationMidterm 1. Every element of the set of functions is continuous
Econ 200 Mathematics for Economists Midterm Question.- Consider the set of functions F C(0, ) dened by { } F = f C(0, ) f(x) = ax b, a A R and b B R That is, F is a subset of the set of continuous functions
More informationMATHS 730 FC Lecture Notes March 5, Introduction
1 INTRODUCTION MATHS 730 FC Lecture Notes March 5, 2014 1 Introduction Definition. If A, B are sets and there exists a bijection A B, they have the same cardinality, which we write as A, #A. If there exists
More informationBrownian Motion and Conditional Probability
Math 561: Theory of Probability (Spring 2018) Week 10 Brownian Motion and Conditional Probability 10.1 Standard Brownian Motion (SBM) Brownian motion is a stochastic process with both practical and theoretical
More informationReal Analysis Chapter 3 Solutions Jonathan Conder. ν(f n ) = lim
. Suppose ( n ) n is an increasing sequence in M. For each n N define F n : n \ n (with 0 : ). Clearly ν( n n ) ν( nf n ) ν(f n ) lim n If ( n ) n is a decreasing sequence in M and ν( )
More informationDaniel Akech Thiong Math 501: Real Analysis Homework Problems with Solutions Fall Problem. 1 Give an example of a mapping f : X Y such that
Daniel Akech Thiong Math 501: Real Analysis Homework Problems with Solutions Fall 2014 Problem. 1 Give an example of a mapping f : X Y such that 1. f(a B) = f(a) f(b) for some A, B X. 2. f(f 1 (A)) A for
More informationContinuity. Matt Rosenzweig
Continuity Matt Rosenzweig Contents 1 Continuity 1 1.1 Rudin Chapter 4 Exercises........................................ 1 1.1.1 Exercise 1............................................. 1 1.1.2 Exercise
More information7 Complete metric spaces and function spaces
7 Complete metric spaces and function spaces 7.1 Completeness Let (X, d) be a metric space. Definition 7.1. A sequence (x n ) n N in X is a Cauchy sequence if for any ɛ > 0, there is N N such that n, m
More informationFunctional Analysis. Franck Sueur Metric spaces Definitions Completeness Compactness Separability...
Functional Analysis Franck Sueur 2018-2019 Contents 1 Metric spaces 1 1.1 Definitions........................................ 1 1.2 Completeness...................................... 3 1.3 Compactness......................................
More informationLecture 5 - Hausdorff and Gromov-Hausdorff Distance
Lecture 5 - Hausdorff and Gromov-Hausdorff Distance August 1, 2011 1 Definition and Basic Properties Given a metric space X, the set of closed sets of X supports a metric, the Hausdorff metric. If A is
More informationImmerse Metric Space Homework
Immerse Metric Space Homework (Exercises -2). In R n, define d(x, y) = x y +... + x n y n. Show that d is a metric that induces the usual topology. Sketch the basis elements when n = 2. Solution: Steps
More informationSection 21. The Metric Topology (Continued)
21. The Metric Topology (cont.) 1 Section 21. The Metric Topology (Continued) Note. In this section we give a number of results for metric spaces which are familar from calculus and real analysis. We also
More informationStochastic Processes. Winter Term Paolo Di Tella Technische Universität Dresden Institut für Stochastik
Stochastic Processes Winter Term 2016-2017 Paolo Di Tella Technische Universität Dresden Institut für Stochastik Contents 1 Preliminaries 5 1.1 Uniform integrability.............................. 5 1.2
More informationMathematics for Economists
Mathematics for Economists Victor Filipe Sao Paulo School of Economics FGV Metric Spaces: Basic Definitions Victor Filipe (EESP/FGV) Mathematics for Economists Jan.-Feb. 2017 1 / 34 Definitions and Examples
More informationCharacterisation of Accumulation Points. Convergence in Metric Spaces. Characterisation of Closed Sets. Characterisation of Closed Sets
Convergence in Metric Spaces Functional Analysis Lecture 3: Convergence and Continuity in Metric Spaces Bengt Ove Turesson September 4, 2016 Suppose that (X, d) is a metric space. A sequence (x n ) X is
More informationEconomics 204 Fall 2011 Problem Set 1 Suggested Solutions
Economics 204 Fall 2011 Problem Set 1 Suggested Solutions 1. Suppose k is a positive integer. Use induction to prove the following two statements. (a) For all n N 0, the inequality (k 2 + n)! k 2n holds.
More informationAbstract Measure Theory
2 Abstract Measure Theory Lebesgue measure is one of the premier examples of a measure on R d, but it is not the only measure and certainly not the only important measure on R d. Further, R d is not the
More informationMath General Topology Fall 2012 Homework 6 Solutions
Math 535 - General Topology Fall 202 Homework 6 Solutions Problem. Let F be the field R or C of real or complex numbers. Let n and denote by F[x, x 2,..., x n ] the set of all polynomials in n variables
More informationNotions such as convergent sequence and Cauchy sequence make sense for any metric space. Convergent Sequences are Cauchy
Banach Spaces These notes provide an introduction to Banach spaces, which are complete normed vector spaces. For the purposes of these notes, all vector spaces are assumed to be over the real numbers.
More informationRandom experiments may consist of stages that are performed. Example: Roll a die two times. Consider the events E 1 = 1 or 2 on first roll
Econ 514: Probability and Statistics Lecture 4: Independence Stochastic independence Random experiments may consist of stages that are performed independently. Example: Roll a die two times. Consider the
More informationWeak convergence and Brownian Motion. (telegram style notes) P.J.C. Spreij
Weak convergence and Brownian Motion (telegram style notes) P.J.C. Spreij this version: December 8, 2006 1 The space C[0, ) In this section we summarize some facts concerning the space C[0, ) of real
More informationThe Lebesgue Integral
The Lebesgue Integral Brent Nelson In these notes we give an introduction to the Lebesgue integral, assuming only a knowledge of metric spaces and the iemann integral. For more details see [1, Chapters
More informationRiesz Representation Theorems
Chapter 6 Riesz Representation Theorems 6.1 Dual Spaces Definition 6.1.1. Let V and W be vector spaces over R. We let L(V, W ) = {T : V W T is linear}. The space L(V, R) is denoted by V and elements of
More informationEconomics 204 Fall 2011 Problem Set 2 Suggested Solutions
Economics 24 Fall 211 Problem Set 2 Suggested Solutions 1. Determine whether the following sets are open, closed, both or neither under the topology induced by the usual metric. (Hint: think about limit
More informationMetric spaces and metrizability
1 Motivation Metric spaces and metrizability By this point in the course, this section should not need much in the way of motivation. From the very beginning, we have talked about R n usual and how relatively
More information2. Metric Spaces. 2.1 Definitions etc.
2. Metric Spaces 2.1 Definitions etc. The procedure in Section for regarding R as a topological space may be generalized to many other sets in which there is some kind of distance (formally, sets with
More informationLebesgue Measure. Dung Le 1
Lebesgue Measure Dung Le 1 1 Introduction How do we measure the size of a set in IR? Let s start with the simplest ones: intervals. Obviously, the natural candidate for a measure of an interval is its
More informationMAT1000 ASSIGNMENT 1. a k 3 k. x =
MAT1000 ASSIGNMENT 1 VITALY KUZNETSOV Question 1 (Exercise 2 on page 37). Tne Cantor set C can also be described in terms of ternary expansions. (a) Every number in [0, 1] has a ternary expansion x = a
More informationLEBESGUE MEASURE AND L2 SPACE. Contents 1. Measure Spaces 1 2. Lebesgue Integration 2 3. L 2 Space 4 Acknowledgments 9 References 9
LBSGU MASUR AND L2 SPAC. ANNI WANG Abstract. This paper begins with an introduction to measure spaces and the Lebesgue theory of measure and integration. Several important theorems regarding the Lebesgue
More informationA LITTLE REAL ANALYSIS AND TOPOLOGY
A LITTLE REAL ANALYSIS AND TOPOLOGY 1. NOTATION Before we begin some notational definitions are useful. (1) Z = {, 3, 2, 1, 0, 1, 2, 3, }is the set of integers. (2) Q = { a b : aεz, bεz {0}} is the set
More informationReal Analysis. Jesse Peterson
Real Analysis Jesse Peterson February 1, 2017 2 Contents 1 Preliminaries 7 1.1 Sets.................................. 7 1.1.1 Countability......................... 8 1.1.2 Transfinite induction.....................
More informationCHAPTER I THE RIESZ REPRESENTATION THEOREM
CHAPTER I THE RIESZ REPRESENTATION THEOREM We begin our study by identifying certain special kinds of linear functionals on certain special vector spaces of functions. We describe these linear functionals
More informationMA651 Topology. Lecture 10. Metric Spaces.
MA65 Topology. Lecture 0. Metric Spaces. This text is based on the following books: Topology by James Dugundgji Fundamental concepts of topology by Peter O Neil Linear Algebra and Analysis by Marc Zamansky
More informationFunctional Analysis Exercise Class
Functional Analysis Exercise Class Wee November 30 Dec 4: Deadline to hand in the homewor: your exercise class on wee December 7 11 Exercises with solutions Recall that every normed space X can be isometrically
More information