Chapter 4. The dominated convergence theorem and applications
|
|
- Rafe Jones
- 6 years ago
- Views:
Transcription
1 Chapter 4. The dominated convergence theorem and applications The Monotone Covergence theorem is one of a number of key theorems alllowing one to exchange limits and [Lebesgue] integrals (or derivatives and integrals, as derivatives are also a sort of limit). Fatou s lemma and the dominated convergence theorem are other theorems in this vein, where monotonicity is not required but something else is needed in its place. In Fatou s lemma we get only an inequality for lim inf s and non-negative integrands, while in the dominated convergence theorem we can manage integrands that change sign but we need a dominating integrable function as well as existence of pointwise limits of the sequence of inetgrands. Contents 4.1 Fatou s Lemma Almost everywhere Dominated convergence theorem Applications of the dominated convergence theorem What s missing? Fatou s Lemma This deals with non-negative functions only but we get away from monotone sequences. Theorem (Fatou s Lemma). Let f n : [0, ] be (nonnegative) Lebesgue measurable functions. Then lim inf f n dµ lim inf f n dµ Proof. Let g n (x) inf k n f k (x) so that what we mean by lim inf f n is the function with value at x given by ( ) ( ) lim inf f n (x) lim inf f n(x) lim inf f k(x) lim g n (x) k n Notice that g n (x) inf k n f k (x) inf k n+1 f k (x) g n+1 (x) so that the sequence (g n (x)) n1 is monotone increasing for each x and so the Monotone convergence theorem says that lim g n dx lim g n dµ lim inf f n dµ But also g n (x) f k (x) for each k n and so g n dµ f k dµ (k n)
2 Mathematics MA2224 or Hence lim inf g n dµ inf f k dµ k n ( ) f n dµ lim inf f k dµ lim g n dµ lim inf k n f n dµ Example Fatou s lemma is not true with equals. For instance take, f n χ [n,2n] and notice that f n dµ n as n but for each x, lim f n (x) 0. So lim inf f n dµ > lim inf f n dµ 0 dµ 0 This also shows that the Monotone Convergence Theorem is not true without Monotone. 4.2 Almost everywhere Definition We say that a property about real numbers x holds almost everywhere (with respect to Lebesgue measure µ) if the set of x where it fails to be true has µ measure 0. Proposition If f : [, ] is integrable, then f(x) holds almost everywhere (or, equivalently, f(x) < almost everywhere). Proof. Let E {x : f(x) }. What we want to do is show that µ(e) 0. We know f dµ <. So, for any n N, the simple function nχ E satisfies nχ E (x) f(x) always, and so has nχ E dµ nµ(e) f dµ <. But this can t be true for all n N unless µ(e) 0. Proposition If f : [, ] is measurable, then f satisfies f dµ 0 if and only if f(x) 0 almost everywhere. Proof. Suppose f dµ 0 first. Let E n {x : f(x) 1/n}. Then 1 n χ E n f
3 Lebesgue integral 3 and so 1 n χ E n dµ 1 n µ(e n) f dµ 0. Thus µ(e n ) 0 for each n. But E 1 E 2 and So E n {x : f(x) 0}. n1 µ({x : f(x) 0}) µ ( n1 E n ) lim µ(e n ) 0. Conversely, suppose now that µ({x : f(x) 0}) 0. We know f is a non-negative measurable function and so there is a monotone increasing sequence (f n ) n1 of measurable simple functions that converges pointwise to f. From 0 f n (x) f(x) we can see that {x : f n (x) 0} {x : f(x) 0} and so µ({x : f n (x) 0}) µ({x : f(x) 0}) 0. Being a simple function f n has a largest value y n (which is finite) and so if we put E n {x : f n (x) 0} we have f n y n χ En f n dµ y n χ En dµ y n χ En dµ y n µ(e n ) 0. From the Monotone Convergence Theorem ( ) f dµ lim f n dµ lim f n dµ lim 0 0. The above result is one way of saying that integration ignores what happens to the integrand on any chosen set of measure 0. Here is a result that says that in way that is often used. Proposition Let f : [, ] be an integrable function and g : [, ] a Lebesgue measurable function with f(x) g(x) almost everywhere. Then g must also be integrable and g dµ f dµ. Proof. Let E {x : f(x) g(x)} (think of E as standing for exceptional ) and note that f(x) g(x) almost everywhere means µ(e) 0. Write f (1 χ E )f + χ E f. Note that both (1 χ E )f and χ E f are integrable because they are measurable and satisfy (1 χ E )f f and χ E f f. Also χ E f dµ χ E f dµ 0 as χ E f 0 almost everywhere. Similarly χ Eg dµ 0.
4 Mathematics MA2224 So f dµ ((1 χ E )f + χ E f) dµ (1 χ E )f dµ + χ E f dµ (1 χ E )f dµ + 0 (1 χ E )g dµ The same calculation (with f in place of f) shows (1 χ E) g dµ f dµ <, so that (1 χ E )g must be integrable. Thus g (1 χ E )g + χ E g is also integrable (because χ Eg dµ 0 and so χ E g is integrable and g is then the sum of two integrable functions). Thus we have g dµ (1 χ E)g dµ + χ Eg dµ f dµ + 0. emark It follows that we should be able to manage without allowing integrable functions to have the values ±. The idea is that, if f is integrable, it must be almost everywhere finite. If we change all the values where f(x) to 0 (say) we are only changing f on a set of x s of measure zero. This is exactly changing f to the (1 χ E )f in the above proof. The changed function will be almost everywhere the same as the original f, but have finite values everywhere. So from the point of view of the integral of f, this change is not significant. However, it can be awkward to have to do this all the time, and it is better to allow f(x) ±. 4.3 Dominated convergence theorem Theorem (Lebesgue dominated convergence theorem). Suppose f n : [, ] are (Lebesgue) measurable functions such that the pointwise limit f(x) lim f n (x) exists. Assume there is an integrable g : [0, ] with f n (x) g(x) for each x. Then f is integrable as is f n for each n, and lim f n dµ lim f n dµ f dµ Proof. Since f n (x) g(x) and g is integrable, f n dµ g dµ <. So f n is integrable. We know f is measurable (as a pointwise limit of measurable functions) and then, similarly, f(x) lim f n (x) g(x) implies that f is integrable too. The proof does not work properly if g(x) for some x. We know that g(x) < almost everywhere. So we can take E {x : g(x) } and multiply g and each of the functions f n and f by 1 χ E to make sure all the functions have finite values. As we are changing them all only on the set E of measure 0, this change does not affect the integrals or the conclusions. We assume then all have finite values.
5 Lebesgue integral 5 Let h n g f n, so that h n 0. By Fatou s lemma lim inf (g f n ) dµ lim inf (g f n) dµ (g f) dµ and that gives ( lim inf g dµ or ) f n dµ g dµ lim sup f n dµ g dµ f dµ lim sup f n dµ f dµ (1) epeat this Fatou s lemma argument with g + f n rather than g f n. We get lim inf (g + f n ) dµ lim inf (g + f n) dµ (g + f) dµ and that gives ( lim inf g dµ + or Combining (1) and (2) we get f dµ lim inf which forces ) f n dµ g dµ + lim inf f n dµ g dµ + f dµ lim inf f n dµ f dµ (2) f dµ lim inf f n dµ lim sup f n dµ f dµ f n dµ lim sup f n dµ and that gives the result because if lim sup a n lim inf a n (for a sequence (a n ) n1), it implies that lim a n exists and lim a n lim sup a n lim inf a n. emark The example following Fatou s lemma also shows that the assumption about the existence of the dominating function g can t be dispensed with. 4.4 Applications of the dominated convergence theorem Theorem (Continuity of integrals). Assume f : is such that x f [t] (x) f(x, t) is measurable for each t and t f(x, t) is continuous for each x. Assume also that there is an integrable g : with f(x, t) g(x) for each x, t. Then the function f [t] is integrable for each t and the function F : defined by F (t) f [t] dµ f(x, t) dµ(x) is continuous.
6 Mathematics MA2224 Proof. Since f [t] is measurable and f [t] g we have f [t] dµ g dµ < and so f [t] is integrable (for each t ). This F (t) makes sense. To show that F is continuous at t 0 it is enough to show that for each sequence (t n ) n1 with lim t n t 0 we have lim F (t n ) F (t 0 ). But that follows from the dominated convergence theorem applied to f n (t) f(x, t n ), since we have lim f n(t) lim f(x, t n ) f(x, t 0 ) by continuity of t f(x, t). We also have f n (t) f(x, t n ) g(x) for each n and each x. Example Show that is continuous. F (t) [0, ) e x cos(πt) dµ(x) Proof. The idea is to apply the theorem with dominating function g(x) given by { g(x) χ [0, ) (x)e x e x for x 0 0 for x < 0 We need to know that g dµ < (and that g is measurable and that x χ [0, )(x)e x cos(πt) is measurable for each t but we do know that these are measurable because e x is continuous and χ [0, ) is measurable). By the Monotone Convergence Theorem, n g dµ lim χ [ n,n] g dµ lim χ [0,n] e x dµ(x) lim e x dµ(x) 0 You can work this out easily using ordinary iemann integral ideas and the limit is 1. So g dµ <. Now the theorem applies because χ [0, ) (x)e x cos(πt) g(x) for each (x, t) 2 (and certainly t χ [0, ) (x)e x cos(πt) is continuous for each x). Theorem (Differentiating under the integral sign). Assume f : is such that x f [t] (x) f(x, t) is measurable for each t, that f [t0] (x) f(x, t 0 ) is integrable for some t 0 and f(x,t) exists for each (x, t). Assume also that there is an integrable g : t with f t (x,t) g(x) for each x, t. Then the function x f(x, t) is integrable for each t and the function F : defined by F (t) f [t] dµ f(x, t) dµ(x) is differentiable with derivative F (t) d dt f(x, t) dµ(x) f(x, t) dµ(x). t
7 Lebesgue integral 7 Proof. Applying the Mean Value theorem to the function t f(x, t), for each t t 0 we have to have some c between t 0 and t so that f(x, t) f(x, t 0 ) f t (x,c)(t t 0 ) It follows that and so f(x, t) f(x, t 0 ) g(x) t t 0 f(x, t) f(x, t 0 ) + g(x) t t 0. Thus f(x, t) dµ(x) ( f(x, t 0 ) + g(x) t t 0 ) dµ(x) f(x, t 0 ) dµ(x) + t t 0 g dµ <, which establishes that the function x f(x, t) is integrable for each t. To prove the formula for F (t) consider any sequence (t n ) n1 so that lim t n t but t n t for each t. We claim that We have where F (t n ) F (t) t n t Notice that, for each x we know F (t n ) F (t) lim t n t f(x, t n ) f(x, t) t n t f(x, t) dµ(x). (3) t dµ(x) f n (x) f(x, t n) f(x, t). t n t lim f n(x) f t (x,t) f n (x) dµ(x) and so (3) will follow from the dominated convergence theorem once we show that f n (x) g(x) for each x. That follows from the Mean Value theorem again because there is c between t and t n (with c depending on x) so that f n (x) f(x, t n) f(x, t) t n t f t (x,c). So f n (x) g(x) for each x.
8 Mathematics MA What s missing? There are quite a few topics that are very useful and that we have not covered at all. Some of the things we have covered are simplified from the way they are often stated and used. An example in the latter category is that the Monotone Convergence Theorem and the Dominated Convergence Theorem are true if we only assume the hypotheses are valid almost everywhere. The Monotone Convergence Theorem is still true if we assume that the sequence (f n ) n1 of measurable functions satisfies f n 0 almost everywhere and f n f n+1 almost everywhere (for each n). Then the pointwise limit f(x) lim f n (x) may exist only almost everywhere. Something similar for the Dominated Convergence Theorem. We stuck to integrals of functions f(x) defined for x (or for x X L which is more or less the same because we can extend them to be zero on \ X) and we used only Lebesgue measure µ on the Lebesgue σ-algebra L. What we need abstractly is just a measure space (X, Σ, λ) and Σ-measurable integrands f : X [, ]. By definition f is Σ- measurable if f 1 ([, a]) {x X : f(x) a} Σ ( a ), and it follows from that condition that f 1 (B) Σ for all Borel subsets B. We can then talk about simple functions on X (f : X with finite range f(x) {y!, y 2,..., y n }), their standard form (f n j1 y jχ Fj where F j f 1 ({y j })), integrals of non-negative Σ- measurable simple functions ( f dλ n X j1 y jλ(f j )), integrals of non-negative measurable f : X [0, ], (generalised) Monotone convergence theorem, λ-integrable Σ-measurable f : X [, ], (generalised) Dominated Convergence Theorem. To make this applicable, we would need some more examples of measures, other that just Lebesgue measure µ: L [0, ]. We did touch on some examples of measures that don t correspond so obviously to total length as µ does. For instance if f : [0, ] is a nonnegative (Lebesgue) measurable function, there is an associated measure λ f : L [0, ] given by λ f (E) f dµ. (We discussed λ E f when f was simple but the fact that λ f is a measure even when f is not simple follows easily from the Monotone Convergence Theorem.) If we choose f so that f dµ 1, then we get a probability measure from λ f (and f is called a probability density function). The standard normal distribution is the name given to λ f when f(x) (1/ 2π)e x2 /2. That s just one example. The adon Nikodym theorem gives a way to recognise measures λ: L [0, ] that are of the form λ λ f for some non-negative (Lebesgue) measurable function f. The key thing is that µ(e) 0 λ(e) 0. The full version of the adon Nikodym theorem applies not just to measures on (, L ) with respect to µ, but to many more general measure spaces. And then we could have explained area measure on 2, volume measure on 3, n-dimensional volume measure on n, Fubini s theorem relating integrals on product spaces to iterated integrals, the change of variables formula for integrals on n, and quite a few more topics. One place where measure theory comes up is in defining so-called Lebesgue spaces, which are Banach spaces defined using (Lebesgue) integration. For example L 1 () is the space of integrable functions f : with norm f 1 f dµ. Or to be more precise it is the space of almost everywhere equivalence classes of such functions. (That is so that f 1 0 only for the zero element of the space, a property that norms should have.) To make L 1 () complete we need
9 Lebesgue integral 9 the Lebesgue integral. Hilbert spaces like L 2 () come into Fourier analysis, for instance. By definition L 2 () is the space of almost everywhere equivalence classes of measurable f : that satisfy f 2 dµ < with norm given by f 2 ( f 2 dµ ) 1/2. In short then, there is quite a range of things that the Lebesgue theory is used for (probability theory, Fourier analysis, differential equations and partial differential equations, functional analysis, stochastic processes,... ). My aim was to lay the basis for studying these topics later.. Timoney March 28, 2018
Chapter 6. Integration. 1. Integrals of Nonnegative Functions. a j µ(e j ) (ca j )µ(e j ) = c X. and ψ =
Chapter 6. Integration 1. Integrals of Nonnegative Functions Let (, S, µ) be a measure space. We denote by L + the set of all measurable functions from to [0, ]. Let φ be a simple function in L +. Suppose
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 informationHomework 11. Solutions
Homework 11. Solutions Problem 2.3.2. Let f n : R R be 1/n times the characteristic function of the interval (0, n). Show that f n 0 uniformly and f n µ L = 1. Why isn t it a counterexample to the Lebesgue
More informationSolutions: Problem Set 4 Math 201B, Winter 2007
Solutions: Problem Set 4 Math 2B, Winter 27 Problem. (a Define f : by { x /2 if < x
More information6.2 Fubini s Theorem. (µ ν)(c) = f C (x) dµ(x). (6.2) Proof. Note that (X Y, A B, µ ν) must be σ-finite as well, so that.
6.2 Fubini s Theorem Theorem 6.2.1. (Fubini s theorem - first form) Let (, A, µ) and (, B, ν) be complete σ-finite measure spaces. Let C = A B. Then for each µ ν- measurable set C C the section x C is
More informationMATH MEASURE THEORY AND FOURIER ANALYSIS. Contents
MATH 3969 - MEASURE THEORY AND FOURIER ANALYSIS ANDREW TULLOCH Contents 1. Measure Theory 2 1.1. Properties of Measures 3 1.2. Constructing σ-algebras and measures 3 1.3. Properties of the Lebesgue measure
More information3 (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 informationChapter 5. Measurable Functions
Chapter 5. Measurable Functions 1. Measurable Functions Let X be a nonempty set, and let S be a σ-algebra of subsets of X. Then (X, S) is a measurable space. A subset E of X is said to be measurable if
More informationNotes on the Lebesgue Integral by Francis J. Narcowich November, 2013
Notes on the Lebesgue Integral by Francis J. Narcowich November, 203 Introduction In the definition of the Riemann integral of a function f(x), the x-axis is partitioned and the integral is defined in
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 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 informationIntegration on Measure Spaces
Chapter 3 Integration on Measure Spaces In this chapter we introduce the general notion of a measure on a space X, define the class of measurable functions, and define the integral, first on a class of
More informationSection Integration of Nonnegative Measurable Functions
18.2. Integration of Nonnegative Measurable Functions 1 Section 18.2. Integration of Nonnegative Measurable Functions Note. We now define integrals of measurable functions on measure spaces. Though similar
More information2 Measure Theory. 2.1 Measures
2 Measure Theory 2.1 Measures A lot of this exposition is motivated by Folland s wonderful text, Real Analysis: Modern Techniques and Their Applications. Perhaps the most ubiquitous measure in our lives
More information(U) =, if 0 U, 1 U, (U) = X, if 0 U, and 1 U. (U) = E, if 0 U, but 1 U. (U) = X \ E if 0 U, but 1 U. n=1 A n, then A M.
1. Abstract Integration The main reference for this section is Rudin s Real and Complex Analysis. The purpose of developing an abstract theory of integration is to emphasize the difference between the
More informationAnalysis Comprehensive Exam Questions Fall 2008
Analysis Comprehensive xam Questions Fall 28. (a) Let R be measurable with finite Lebesgue measure. Suppose that {f n } n N is a bounded sequence in L 2 () and there exists a function f such that f n (x)
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 information02. Measure and integral. 1. Borel-measurable functions and pointwise limits
(October 3, 2017) 02. Measure and integral Paul Garrett garrett@math.umn.edu http://www.math.umn.edu/ garrett/ [This document is http://www.math.umn.edu/ garrett/m/real/notes 2017-18/02 measure and integral.pdf]
More informationNotes on the Lebesgue Integral by Francis J. Narcowich Septemmber, 2014
1 Introduction Notes on the Lebesgue Integral by Francis J. Narcowich Septemmber, 2014 In the definition of the Riemann integral of a function f(x), the x-axis is partitioned and the integral is defined
More informationExercise 1. Let f be a nonnegative measurable function. Show that. where ϕ is taken over all simple functions with ϕ f. k 1.
Real Variables, Fall 2014 Problem set 3 Solution suggestions xercise 1. Let f be a nonnegative measurable function. Show that f = sup ϕ, where ϕ is taken over all simple functions with ϕ f. For each n
More information2 Lebesgue integration
2 Lebesgue integration 1. Let (, A, µ) be a measure space. We will always assume that µ is complete, otherwise we first take its completion. The example to have in mind is the Lebesgue measure on R n,
More informationFACULTY OF SCIENCE FINAL EXAMINATION MATHEMATICS MATH 355. Analysis 4. Examiner: Professor S. W. Drury Date: Tuesday, April 18, 2006 INSTRUCTIONS
FACULTY OF SCIENCE FINAL EXAMINATION MATHEMATICS MATH 355 Analysis 4 Examiner: Professor S. W. Drury Date: Tuesday, April 18, 26 Associate Examiner: Professor K. N. GowriSankaran Time: 2: pm. 5: pm. INSTRUCTIONS
More informationThree hours THE UNIVERSITY OF MANCHESTER. 24th January
Three hours MATH41011 THE UNIVERSITY OF MANCHESTER FOURIER ANALYSIS AND LEBESGUE INTEGRATION 24th January 2013 9.45 12.45 Answer ALL SIX questions in Section A (25 marks in total). Answer THREE of the
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 informationL p Spaces and Convexity
L p Spaces and Convexity These notes largely follow the treatments in Royden, Real Analysis, and Rudin, Real & Complex Analysis. 1. Convex functions Let I R be an interval. For I open, we say a function
More informationChapter 8. General Countably Additive Set Functions. 8.1 Hahn Decomposition Theorem
Chapter 8 General Countably dditive Set Functions In Theorem 5.2.2 the reader saw that if f : X R is integrable on the measure space (X,, µ) then we can define a countably additive set function ν on by
More informationLEBESGUE INTEGRATION. Introduction
LEBESGUE INTEGATION EYE SJAMAA Supplementary notes Math 414, Spring 25 Introduction The following heuristic argument is at the basis of the denition of the Lebesgue integral. This argument will be imprecise,
More information(2) E M = E C = X\E M
10 RICHARD B. MELROSE 2. Measures and σ-algebras An outer measure such as µ is a rather crude object since, even if the A i are disjoint, there is generally strict inequality in (1.14). It turns out to
More informationANALYSIS QUALIFYING EXAM FALL 2017: SOLUTIONS. 1 cos(nx) lim. n 2 x 2. g n (x) = 1 cos(nx) n 2 x 2. x 2.
ANALYSIS QUALIFYING EXAM FALL 27: SOLUTIONS Problem. Determine, with justification, the it cos(nx) n 2 x 2 dx. Solution. For an integer n >, define g n : (, ) R by Also define g : (, ) R by g(x) = g n
More informationSolutions to Tutorial 11 (Week 12)
THE UIVERSITY OF SYDEY SCHOOL OF MATHEMATICS AD STATISTICS Solutions to Tutorial 11 (Week 12) MATH3969: Measure Theory and Fourier Analysis (Advanced) Semester 2, 2017 Web Page: http://sydney.edu.au/science/maths/u/ug/sm/math3969/
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 informationExercise 1. Show that the Radon-Nikodym theorem for a finite measure implies the theorem for a σ-finite measure.
Real Variables, Fall 2014 Problem set 8 Solution suggestions xercise 1. Show that the Radon-Nikodym theorem for a finite measure implies the theorem for a σ-finite measure. nswer: ssume that the Radon-Nikodym
More informationLecture 6 Basic Probability
Lecture 6: Basic Probability 1 of 17 Course: Theory of Probability I Term: Fall 2013 Instructor: Gordan Zitkovic Lecture 6 Basic Probability Probability spaces A mathematical setup behind a probabilistic
More informationLebesgue Integration: A non-rigorous introduction. What is wrong with Riemann integration?
Lebesgue Integration: A non-rigorous introduction What is wrong with Riemann integration? xample. Let f(x) = { 0 for x Q 1 for x / Q. The upper integral is 1, while the lower integral is 0. Yet, the function
More informationProduct measures, Tonelli s and Fubini s theorems For use in MAT4410, autumn 2017 Nadia S. Larsen. 17 November 2017.
Product measures, Tonelli s and Fubini s theorems For use in MAT4410, autumn 017 Nadia S. Larsen 17 November 017. 1. Construction of the product measure The purpose of these notes is to prove the main
More informationReal Analysis Qualifying Exam May 14th 2016
Real Analysis Qualifying Exam May 4th 26 Solve 8 out of 2 problems. () Prove the Banach contraction principle: Let T be a mapping from a complete metric space X into itself such that d(tx,ty) apple qd(x,
More information3 Integration and Expectation
3 Integration and Expectation 3.1 Construction of the Lebesgue Integral Let (, F, µ) be a measure space (not necessarily a probability space). Our objective will be to define the Lebesgue integral R fdµ
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 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 informationFinal Exam Practice Problems Math 428, Spring 2017
Final xam Practice Problems Math 428, Spring 2017 Name: Directions: Throughout, (X,M,µ) is a measure space, unless stated otherwise. Since this is not to be turned in, I highly recommend that you work
More informationThe Lebesgue Integral
The Lebesgue Integral Having completed our study of Lebesgue measure, we are now ready to consider the Lebesgue integral. Before diving into the details of its construction, though, we would like to give
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 informationReview of measure theory
209: Honors nalysis in R n Review of measure theory 1 Outer measure, measure, measurable sets Definition 1 Let X be a set. nonempty family R of subsets of X is a ring if, B R B R and, B R B R hold. bove,
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 informationNOTES ON THE REGULARITY OF QUASICONFORMAL HOMEOMORPHISMS
NOTES ON THE REGULARITY OF QUASICONFORMAL HOMEOMORPHISMS CLARK BUTLER. Introduction The purpose of these notes is to give a self-contained proof of the following theorem, Theorem.. Let f : S n S n be a
More informationClass Notes for Math 921/922: Real Analysis, Instructor Mikil Foss
University of Nebraska - Lincoln DigitalCommons@University of Nebraska - Lincoln Math Department: Class Notes and Learning Materials Mathematics, Department of 200 Class Notes for Math 92/922: Real Analysis,
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 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 informationA List of Problems in Real Analysis
A List of Problems in Real Analysis W.Yessen & T.Ma December 3, 218 This document was first created by Will Yessen, who was a graduate student at UCI. Timmy Ma, who was also a graduate student at UCI,
More information2) Let X be a compact space. Prove that the space C(X) of continuous real-valued functions is a complete metric space.
University of Bergen General Functional Analysis Problems with solutions 6 ) Prove that is unique in any normed space. Solution of ) Let us suppose that there are 2 zeros and 2. Then = + 2 = 2 + = 2. 2)
More informationFor example, the real line is σ-finite with respect to Lebesgue measure, since
More Measure Theory In this set of notes we sketch some results in measure theory that we don t have time to cover in full. Most of the results can be found in udin s eal & Complex Analysis. Some of the
More informationMath 5051 Measure Theory and Functional Analysis I Homework Assignment 3
Math 551 Measure Theory and Functional Analysis I Homework Assignment 3 Prof. Wickerhauser Due Monday, October 12th, 215 Please do Exercises 3*, 4, 5, 6, 8*, 11*, 17, 2, 21, 22, 27*. Exercises marked with
More informationProbability and Measure
Part II Year 2018 2017 2016 2015 2014 2013 2012 2011 2010 2009 2008 2007 2006 2005 2018 84 Paper 4, Section II 26J Let (X, A) be a measurable space. Let T : X X be a measurable map, and µ a probability
More informationProblem 3. Give an example of a sequence of continuous functions on a compact domain converging pointwise but not uniformly to a continuous function
Problem 3. Give an example of a sequence of continuous functions on a compact domain converging pointwise but not uniformly to a continuous function Solution. If we does not need the pointwise limit of
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 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 information1.3.1 Definition and Basic Properties of Convolution
1.3 Convolution 15 1.3 Convolution Since L 1 (R) is a Banach space, we know that it has many useful properties. In particular the operations of addition and scalar multiplication are continuous. However,
More informationProblem Set. Problem Set #1. Math 5322, Fall March 4, 2002 ANSWERS
Problem Set Problem Set #1 Math 5322, Fall 2001 March 4, 2002 ANSWRS i All of the problems are from Chapter 3 of the text. Problem 1. [Problem 2, page 88] If ν is a signed measure, is ν-null iff ν () 0.
More informationf(x)dx = lim f n (x)dx.
Chapter 3 Lebesgue Integration 3.1 Introduction The concept of integration as a technique that both acts as an inverse to the operation of differentiation and also computes areas under curves goes back
More informationProbability and Random Processes
Probability and Random Processes Lecture 4 General integration theory Mikael Skoglund, Probability and random processes 1/15 Measurable xtended Real-valued Functions R = the extended real numbers; a subset
More informationINTRODUCTION TO MEASURE THEORY AND LEBESGUE INTEGRATION
1 INTRODUCTION TO MEASURE THEORY AND LEBESGUE INTEGRATION Eduard EMELYANOV Ankara TURKEY 2007 2 FOREWORD This book grew out of a one-semester course for graduate students that the author have taught at
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 information13. Examples of measure-preserving tranformations: rotations of a torus, the doubling map
3. Examples of measure-preserving tranformations: rotations of a torus, the doubling map 3. Rotations of a torus, the doubling map In this lecture we give two methods by which one can show that a given
More informationIt follows from the above inequalities that for c C 1
3 Spaces L p 1. In this part we fix a measure space (, A, µ) (we may assume that µ is complete), and consider the A -measurable functions on it. 2. For f L 1 (, µ) set f 1 = f L 1 = f L 1 (,µ) = f dµ.
More informationNAME: MATH 172: Lebesgue Integration and Fourier Analysis (winter 2012) Final exam. Wednesday, March 21, time: 2.5h
NAME: SOLUTION problem # 1 2 3 4 5 6 7 8 9 points max 15 20 10 15 10 10 10 10 10 110 MATH 172: Lebesgue Integration and Fourier Analysis (winter 2012 Final exam Wednesday, March 21, 2012 time: 2.5h Please
More informationGeneral Instructions
Math 240: Real Analysis Qualifying Exam May 28, 2015 Name: Student ID#: Problems/Page Numbers Total Points Your Score Problem 1 / Page 2-3 40 Points Problem 2 / Page 4 Problem 3 / Page 5 Problem 4 / Page
More informationMTH 404: Measure and Integration
MTH 404: Measure and Integration Semester 2, 2012-2013 Dr. Prahlad Vaidyanathan Contents I. Introduction....................................... 3 1. Motivation................................... 3 2. The
More informationIt follows from the above inequalities that for c C 1
3 Spaces L p 1. Appearance of normed spaces. In this part we fix a measure space (, A, µ) (we may assume that µ is complete), and consider the A - measurable functions on it. 2. For f L 1 (, µ) set f 1
More informationMeasure and Integration: Solutions of CW2
Measure and Integration: s of CW2 Fall 206 [G. Holzegel] December 9, 206 Problem of Sheet 5 a) Left (f n ) and (g n ) be sequences of integrable functions with f n (x) f (x) and g n (x) g (x) for almost
More informationFACULTY OF SCIENCE FINAL EXAMINATION MATHEMATICS MATH 355. Analysis 4. Examiner: Professor S. W. Drury Date: Wednesday, April 18, 2007 INSTRUCTIONS
FACULTY OF SCIENCE FINAL EXAMINATION MATHEMATICS MATH 355 Analysis 4 Examiner: Professor S. W. Drury Date: Wednesday, April 18, 27 Associate Examiner: Professor K. N. GowriSankaran Time: 2: pm. 5: pm.
More informationMeasurable functions are approximately nice, even if look terrible.
Tel Aviv University, 2015 Functions of real variables 74 7 Approximation 7a A terrible integrable function........... 74 7b Approximation of sets................ 76 7c Approximation of functions............
More informationPart II Probability and Measure
Part II Probability and Measure Theorems Based on lectures by J. Miller Notes taken by Dexter Chua Michaelmas 2016 These notes are not endorsed by the lecturers, and I have modified them (often significantly)
More informationLECTURE NOTES FOR , FALL Contents. Introduction. 1. Continuous functions
LECTURE NOTES FOR 18.155, FALL 2002 RICHARD B. MELROSE Contents Introduction 1 1. Continuous functions 1 2. Measures and σ-algebras 9 3. Integration 16 4. Hilbert space 27 5. Test functions 30 6. Tempered
More informationx 0 + f(x), exist as extended real numbers. Show that f is upper semicontinuous This shows ( ɛ, ɛ) B α. Thus
Homework 3 Solutions, Real Analysis I, Fall, 2010. (9) Let f : (, ) [, ] be a function whose restriction to (, 0) (0, ) is continuous. Assume the one-sided limits p = lim x 0 f(x), q = lim x 0 + f(x) exist
More informationLebesgue s Differentiation Theorem via Maximal Functions
Lebesgue s Differentiation Theorem via Maximal Functions Parth Soneji LMU München Hütteseminar, December 2013 Parth Soneji Lebesgue s Differentiation Theorem via Maximal Functions 1/12 Philosophy behind
More informationAdvanced Analysis Qualifying Examination Department of Mathematics and Statistics University of Massachusetts. Tuesday, January 16th, 2018
NAME: Advanced Analysis Qualifying Examination Department of Mathematics and Statistics University of Massachusetts Tuesday, January 16th, 2018 Instructions 1. This exam consists of eight (8) problems
More informationPartial Solutions to Folland s Real Analysis: Part I
Partial Solutions to Folland s Real Analysis: Part I (Assigned Problems from MAT1000: Real Analysis I) Jonathan Mostovoy - 1002142665 University of Toronto January 20, 2018 Contents 1 Chapter 1 3 1.1 Folland
More informationFunctional Analysis, Stein-Shakarchi Chapter 1
Functional Analysis, Stein-Shakarchi Chapter 1 L p spaces and Banach Spaces Yung-Hsiang Huang 018.05.1 Abstract Many problems are cited to my solution files for Folland [4] and Rudin [6] post here. 1 Exercises
More informationSOME PROBLEMS IN REAL ANALYSIS.
SOME PROBLEMS IN REAL ANALYSIS. Prepared by SULEYMAN ULUSOY PROBLEM 1. (1 points) Suppose f n : X [, ] is measurable for n = 1, 2, 3,...; f 1 f 2 f 3... ; f n (x) f(x) as n, for every x X. a)give a counterexample
More information+ 2x sin x. f(b i ) f(a i ) < ɛ. i=1. i=1
Appendix To understand weak derivatives and distributional derivatives in the simplest context of functions of a single variable, we describe without proof some results from real analysis (see [7] and
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 informationMATH 650. THE RADON-NIKODYM THEOREM
MATH 650. THE RADON-NIKODYM THEOREM This note presents two important theorems in Measure Theory, the Lebesgue Decomposition and Radon-Nikodym Theorem. They are not treated in the textbook. 1. Closed subspaces
More informationReal Analysis, October 1977
Real Analysis, October 1977 1. Prove that the sequence 2, 2 2, 2 2 2,... converges and find its limit. 2. For which real-x does x n lim n 1 + x 2n converge? On which real sets is the convergence uniform?
More informationREAL ANALYSIS I Spring 2016 Product Measures
REAL ANALSIS I Spring 216 Product Measures We assume that (, M, µ), (, N, ν) are σ- finite measure spaces. We want to provide the Cartesian product with a measure space structure in which all sets of the
More informationMeasure Theory & Integration
Measure Theory & Integration Lecture Notes, Math 320/520 Fall, 2004 D.H. Sattinger Department of Mathematics Yale University Contents 1 Preliminaries 1 2 Measures 3 2.1 Area and Measure........................
More informationLecture 21: Expectation of CRVs, Fatou s Lemma and DCT Integration of Continuous Random Variables
EE50: Probability Foundations for Electrical Engineers July-November 205 Lecture 2: Expectation of CRVs, Fatou s Lemma and DCT Lecturer: Krishna Jagannathan Scribe: Jainam Doshi In the present lecture,
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 informationAnother Riesz Representation Theorem
Another Riesz Representation Theorem In these notes we prove (one version of) a theorem known as the Riesz Representation Theorem. Some people also call it the Riesz Markov Theorem. It expresses positive
More informationCONTENTS. 4 Hausdorff Measure Introduction The Cantor Set Rectifiable Curves Cantor Set-Like Objects...
Contents 1 Functional Analysis 1 1.1 Hilbert Spaces................................... 1 1.1.1 Spectral Theorem............................. 4 1.2 Normed Vector Spaces.............................. 7 1.2.1
More informationOverview of normed linear spaces
20 Chapter 2 Overview of normed linear spaces Starting from this chapter, we begin examining linear spaces with at least one extra structure (topology or geometry). We assume linearity; this is a natural
More information4. Product measure spaces and the Lebesgue integral in R n.
4 M. M. PELOSO 4. Product measure spaces and the Lebesgue integral in R n. Our current goal is to define the Lebesgue measure on the higher-dimensional eucledean space R n, and to reduce the computations
More informationMAT 571 REAL ANALYSIS II LECTURE NOTES. Contents. 2. Product measures Iterated integrals Complete products Differentiation 17
MAT 57 REAL ANALSIS II LECTURE NOTES PROFESSOR: JOHN QUIGG SEMESTER: SPRING 205 Contents. Convergence in measure 2. Product measures 3 3. Iterated integrals 4 4. Complete products 9 5. Signed measures
More informationA VERY BRIEF REVIEW OF MEASURE THEORY
A VERY BRIEF REVIEW OF MEASURE THEORY A brief philosophical discussion. Measure theory, as much as any branch of mathematics, is an area where it is important to be acquainted with the basic notions and
More informationMath 240 (Driver) Qual Exam (9/12/2017)
1 Name: I.D. #: Math 240 (Driver) Qual Exam (9/12/2017) Instructions: Clearly explain and justify your answers. You may cite theorems from the text, notes, or class as long as they are not what the problem
More informationLebesgue measure and integration
Chapter 4 Lebesgue measure and integration If you look back at what you have learned in your earlier mathematics courses, you will definitely recall a lot about area and volume from the simple formulas
More informationMath212a1413 The Lebesgue integral.
Math212a1413 The Lebesgue integral. October 28, 2014 Simple functions. In what follows, (X, F, m) is a space with a σ-field of sets, and m a measure on F. The purpose of today s lecture is to develop the
More informationANALYSIS QUALIFYING EXAM FALL 2016: SOLUTIONS. = lim. F n
ANALYSIS QUALIFYING EXAM FALL 206: SOLUTIONS Problem. Let m be Lebesgue measure on R. For a subset E R and r (0, ), define E r = { x R: dist(x, E) < r}. Let E R be compact. Prove that m(e) = lim m(e /n).
More informationAdvanced Analysis Qualifying Examination Department of Mathematics and Statistics University of Massachusetts. Monday, August 26, 2013
NAME: Advanced Analysis Qualifying Examination Department of Mathematics and Statistics University of Massachusetts Monday, August 26, 2013 Instructions 1. This exam consists of eight (8) problems all
More informationLECTURE NOTES FOR , FALL 2004
LECTURE NOTES FOR 18.155, FALL 2004 RICHARD B. MELROSE Contents Introduction 1 1. Continuous functions 2 2. Measures and σ algebras 10 3. Measureability of functions 16 4. Integration 19 5. Hilbert space
More informationFolland: Real Analysis, Chapter 7 Sébastien Picard
Folland: Real Analysis, Chapter 7 Sébastien Picard Problem 7.2 Let µ be a Radon measure on X. a. Let N be the union of all open U X such that µ(u) =. Then N is open and µ(n) =. The complement of N is called
More information