Compendium and Solutions to exercises TMA4225 Foundation of analysis

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1 Compendium and Solutions to exercises TMA4225 Foundation of analysis Ruben Spaans December 6, Introduction This compendium contains a lexicon over definitions and exercises with solutions. Throughout I will refer to the book, which is A course in real analysis by John N. McDonald and Neil A. Weiss, from Also, the exercise section is incomplete, I did not have enough time to finish this sectiom before the exam. 2 Lexicon over definitions Absolutely continuous function, suppose that f is defined on [a, b] or R, f exists almost everywhere and is Lebesgue integrable on the interval specified, and f(x) = f(a) + x a f (t) dt, x our interval. Then f is said to be absolutely continuous on [a, b] or R. xample 1 See book, example 6.6 on p.344. Accumulation point, x is an accumulation point for a set if for every ɛ > 0, there exists y such that 0 < y x < ɛ. Algebra (set), given a set Ω, a nonempty collection A 0 of subsets of Ω is called an algebra if the following two conditions are satisfied: a) A A 0 implies A C A 0. b) A, B A 0 implies A B A 0. 1

2 Almost everywhere, A property is said to hold µ almost everywhere, or µ-ae, if it holds except on a set of µ-measure zero. That is, except on a set N with µ(n) = 0. Archimedean principle, for each x R, there exists n N such that n > x. Borel measurable functions, the set Ĉ is the smallest set of real-valued functions on R that contains the collection of continuous functions and is closed under pointwise limits. The members of this set are called Borel measurable functions. Borel set, a set B R is called a Borel set if its characteristic function, χ B, is Borel measurable. The collection of all Borel sets is denoted B = {B R χ B Ĉ}. Bounded, a subset K of R is bounded if there exists an M such that for all x K, x M. Bounded variation and total variation, let f be a complex-valued function on [a, b]. The total variation over f over [a, b], denoted by Va b f, is defined by { n } Va b f = sup f(x k ) f(x k 1) a = x 0 < x 1 < < x n = b. k=1 If V b a f <, then f is said to be of bounded variation on [a, b]. xample 1 See book, example 6.4 on p.331. Tip Typical way of proving/disproving V b a f < : Make a partition of [a, b] into n subintervals, let n go to infinity and compare the f(x k ) f(x k 1 ) terms to the terms of the harmonic series { 1 k }n k=1. Cantor set/function, see book ch.2.5. Carathéodory criteron, a set R is said to satisfy the Carathéodory criterion if λ (W ) = λ (W ) + λ (W C ) for all subsets W R. The collection of all such subsets of R which satisfies this criterion is denoted M. Characteristic function, for a subset A of a set Ω, the characteristic function χ A is the real-valued function on Ω defined by { 1 if x A; χ A = 0 if x / A. 2

3 Cluster point, let {x n } n=1 be a sequence of values from R. x R is a cluster point if for every ɛ > 0 and N N, there exists n N such that x x n < ɛ. Compact, a subset K of R is compact if it is closed and bounded. Factoid A compact set contains its inf! Complete measure space, a measure space (Ω, M, µ) is said to be complete if all subsets of A-measurable sets of µ-measurable sets are also A- measurable. That is, if a A and µ(a) = 0, then B A for all B A. xample 1 (R, M, λ) is a complete measure space, while (R, B, λ B ) is not (λ S is the usual λ measure, restricted to S M).) Complex-valued measurable function, let (Ω, A) be a measurable space. A complex-valued function on Ω is said to be an A-measurable function if the inverse image of each open subset of C under f is an A-measurable set. That is, if f 1 (O) A for all open sets O C. Continuous function, let D R, f : D R. f is continuous at x 0 if for every ɛ > 0, there is a δ > 0 such that f(x) f(x 0 ) < ɛ whenever x D and x x 0 < δ. f is continuous on D if it is continuous at every point of D. Countable, finite etc., let A be a set. a) A is finite if it is either empty, or equivalent (has a 1-1 correspondence to another set) to the N first positive numbers for some N N. b) A is infinite if it is not finite. c) A is countably infinite if it is equivalent to N. d) A is countable if it is either finite or countably infinite. e) A is uncountable if it is not countable. Counting measure, let Ω be a non-empty set and A = P(Ω). Define µ on A by { N() if is finite µ() = if is infinite where N() denotes the number of elements of. Then µ is a measure on A and it is called counting measure. Dense Derivative of a real-valued function, a real-valued function f defined in 3

4 some open interval about x R is said to be differentiable at x if f(x + h) f(x) lim h 0 h exists and is finite. In that case the limit is called the derivative of f at x and is denoted by f (x). We are allowed to write f (x) = ± if the limit is ±, but then f is not said to be differentiable at x. xample 1 We know from calculus that if f is differentiable at a point x, then it is continuous at x. The converse is not true, for instance f(x) = x at x = 0. More remarkable is that there exists a function which is continuous at R but differentiable at no points of R. See book, example 6.3 on p Dini-derivates, let f be a real-valued function defined in an open interval about the point x. Set D + f(x + h) f(x) f(x) = lim sup h 0 + h f(x + h) f(x) D + f(x) = lim inf h 0 + h D f(x + h) f(x) f(x) = lim sup h 0 h D f(x) = lim inf h 0 f(x + h) f(x) h These four extended real numbers are called the Dini-derivates of f at x. They are, respectively, the upper right, lower right, upper left and lower left derivates. It follows that f is differentiable at x if and only if all four of the Dini-derivates are equal and finite. xample 1 See book, example 6.2 on p.318. Distance between sets, if x R and R, then the distance from x to is defined to be d(x, ) = inf{ y x y }. If and F are subsets of R, then the distance from to F is defined to be d(, F ) = inf{ y x y, x F }. xtended real numbers, R = R {, }. xtended real-valued measurable function, let (Ω, A) be a measurable space. A function f : Ω R is said to be an A-measurable function if the inverse image of each open subset of R under f is an A-measurable set, that is, if f 1 (O) A for all open sets O R. 4

5 Infimum, let {x n } n=1 be a sequence of elements from R. inf greatest lower bound of the sequence. n x n is the Integral of a nonnegative simple function, let s be a nonnegative simple function with canonical representation s = n k=1 a kχ Ak. Then the Lebesgue integral of s over R is defined by R s(x) dλ(x) = n a k λ(a k ). k=1 If M, then the Lebesgue integral of s over is defined by s(x) dλ(x) = χ (x)s(x) dλ(x). R Integral of a complex-valued function, see book p It s quite similar to the definitions of other Lebesgue integrals. xample 1 See book, example 4.9 on p L 1, L 1 (S) is the set of all functions f on S R such that S f(x) <. L p (S) is the set of all functions f on S R such that S f(x) p < Lebesgue almost everywhere, a property is said to hold Lebesgue almost everywhere or λ-almost everywhere, or λ-ae, if it holds except on a set of Lebesgue measure zero, that is, except on a set N with λ(n) = 0. xample 1 See book, example 3.8 on p.162. Lebesgue integral, let f be a Lebesgue masurable function and M. Then the Lebesgue integral of f over is defined by f(x) dλ(x) = f + (x) dλ(x) f (x) dλ(x) provided that the right-hand side isn t. We say that f is Lebesgue integrable over if f(x) dλ(x) = f + (x) dλ(x) + f (x) dλ(x) <. If f is Lebesgue integrable over R, we say that f is Lebesgue integrable. xample 1 See book, example 3.5 on p Lebesgue integral of a function defined almost everywhere, suppose F is a function defined Lebesgue almost everywhere. Let D be the domaind of f. Then λ(d c ) = 0. Further suppose that there is a Lebesgue measurable 5

6 function g such that f(x) = g(x) for x D. Then, for M, we define the Lebesgue integral of f over by f dλ = g dλ provided that the integrals of the positive and negative parts of g over are not both infinite. Lebesgue integral of a nonnegative function, let f be a nonnegative M-measurable function. Then the lebesgue integral of f over R is defined by f(x) dλ(x) = sup s(x) dλ(x), R s R where the supremum is taken over all nonnegative simple functions that are dominated by f. If M, then the Lebesgue integral of f over is defined by f(x) dλ(x) = χ (x)f(x) dλ(x). R Lebesgue measurable functions, the members of L, the smallest algebra of real-valued functions on R that contains all functions of the form χ where is an element of M (the Lebesgue measurable sets). Alternative definition: A real-valued function f on R is said to be a Lebesgue measurable function if the inverse image of each open set under f is a Lebesgue measurable set; that is, if f 1 (O) M for all open sets O. Functions f M is often said to be M-measurable. Lebesgue measurable sets, the members of M, the collection of all sets satisfying the carathéodory criterion, are called the Lebesgue measurable sets. Lebesgue measure, the restriction of Lebesgue outer measure to the set M, the collection of all sets satisfying the carathéodory criterion is denoted λ and is called the Lebesgue measure. This is a special case of the general measure space (Ω, A, µ) with Ω = R, A = M and µ = λ. Lebesgue outer measure, for each subset A R, the Lebesgue outer measure of A, denoted by λ (A), is defined by { λ (A) = inf l(i n ) {I n } n open intervals, } I n A. n n Limit inferior (sequence), let {x n } n=1 be an infinite sequence of elements from R. The limit inferior of the sequence is lim inf A n = sup inf x k, n k n 6 n

7 the largest inf of the sequence. The limit inferior can be seen as a lower bound of the values of the sequence when n goes to infinity. It is also the smallest cluster point from the sequence. Limit inferior (set), let {A n } n=1 be an infinite sequence of subsets of Ω. The limit inferior of the sequence is ( ) lim inf A n = A k. n n=1 Limit superior (sequence), let {x n } n=1 be an infinite sequence of elements from R. The limit superior of the sequence is k=n lim sup A n = inf sup x k, n n the smallest sup of the sequence. The limit superior can be seen as an upper bound of the values of the sequence when n goes to infinity. It is also the largest cluster point of the sequence. Limit superior (set), consider an infinite sequence of subsets as described above. The limit superior of the sequence is ( ) lim sup A n = A k. n n=1 Lower and upper limits, let g be a real-valued function defined in a deleted interval about the points x, a set of the form (c, d)\{x}, where c < x < d. Then we define lim sup g(y) = inf y x + δ>0 lim inf g(y) = sup y x + lim sup g(y) = inf y x δ>0 lim inf g(y) = sup y x k n k=n sup 0<y x<δ g(y) inf g(y) δ>0 0<y x<δ sup 0<x y<δ g(y) inf g(y) δ>0 0<x y<δ These extended real numbers are called, respectively, the upper right, lower right, upper left and lower left limits of g at x. xample 1 See book, example 6.1 at p.317. Measurable rectangle Let (Γ, S) and (Λ, T ) be measurable spaces. A subset of Γ Λ of the form S T, where S S, T T is called a measurable rectangle. The collection of all measurable rectangles is denoted by U. 7

8 Measurable set, a set Ω is said to be µ -measurable if µ (W ) = µ (W ) + µ (W C ) for all subsets W of Ω. The collection of all µ -measurable sets is denoted by A. xample 1 See book, example 4.14 on p Measurable space, a pair (Ω, A), where Ω is a set and A is a σ-algebra of subsets of Ω, is called a measurable space. Measure, let Ω be a set and A a σ-algebra of subsets of Ω. A measure µ on A is an extended real-valued function satisfying the following conditions: a) µ(a) 0 for all A A. b) µ( ) = 0. c) if A 1, A 2,... are in A, with A i A j = for i j, then ( ) µ A n = µ(a n ). n n (In particular, λ(a 1 A 2 ) = λ(a 1 )+λ(a 2 ) if A 1 A 2 = and A 1, A 2 A.) Measure space, a triple (Ω, A, µ), where Ω is a set, A is a σ-algebra of subsets of Ω and µ is a measure on A, is called a measure space. xamples See book, example 4.1 on p for (non-)examples of measures, measure spaces and measurable spaces. Measure zero, a subset R is said to have measure zero if for each ɛ > 0, there exists a sequence {I n } n of open intervals such that n I n and n l(i n) < ɛ xample 1 Any finite and countable subset of R has measure zero. example, {3}, Z, Q. xample 2 The Cantor set (despite being uncountable) has measure zero. Monotone function, a function f is monotone if it is either nondecreasing or nonincreasing. Open subsets of the extended real numbers, a subset of R is said to be open if it can be expressed as an union of itervals of the form (a, b), [, b), (a, ), where a, br. Outer measure, let Ω be a set, C a semialgebra of subsets of Ω and ι is a nonnegative extended real-valued set function on C satisfying: 1. If C, then ι( ) = 0. 8 For

9 2. If {C k } n k=1 is a finite sequence of pairwise disjoint members of C whose union is in C, then ( n ) n ι C k = ι(c k ). k=1 k=1 3. If C, C 1, C 2,... are in C and C n C n, then ι(c) n ι(c n ). Then the set function µ defined on P(Ω) by µ ( ) = 0 and { µ (A) = inf ι(c n ) {C n } n C, } C n A, n n for A is called the outer measure induced by ι and C. xample 1 See book, example 4.17 on p.210. Pointwise convergence, given a sequence of functions {f n } n=1 (where each f n : Ω R). {f n } n=1 converges pointwise on Ω if for each x Ω, the sequence {f n (x)} n=1 of real numbers converges in R. Positive and negative parts of a function, suppose that f is a realvalued function. Then the positive and negative parts of f, denoted f + and f, are defined by f + = f 0 = max{f, 0}, f = (f 0) = min{f, 0}. Please notice that f is a nonnegative function. Then we have f = f + f. Product measure space, suppose that (Γ, S, µ) and (Λ, T, ν) are σ-finite measure spaces and let U and ι be as in theorem The σ-algebra generated by U, the smallest σ-algebra containing all measurable rectangles, is called the product σ-algebra of S with T and is denoted by S T. The unique extension of ι to a measure on S T is called the product measure of µ with ν and it denoted by µ ν. The measure space (Γ Λ, S T, µ ν) is called the product measure space of (Γ, S, µ) with (Λ, T, ν). Please notice that S T is a notation for the σ-algebra generated by U and is not the Cartesian product of the sets S and T. xample 1 See book, example 4.20 on p.236. Real-valued measurable function, let (Ω, A) be a measurable space. A function f : Ω R is said to be an A-measurable function if the inverse 9

10 image of each open subset of R under f is an A-measurable set, that is, if f 1 (O) A for all open sets O R. xample 1 See book, example 4.2 on p.175. Riemann integral. Let l(i) be the length of the interval I. Then the integral (not Riemann) of a step function is defined to be b a h(x)dx = n a k l(i k ). Let f : [a, b] R be bounded. The upper Riemann integral of f over [a, b] is defined by b a a a k=1 { b } f(x)dx = inf h(x)dx h is a step function and h f. Similarly, the lower Riemann integral of f over [a, b] is defined by b { b } f(x)dx = inf h(x)dx h is a step function and h f. a f is said to be Riemann integrable if the lower and upper Riemann integrals of f are equal. The common value of the lower and upper Riemann integrals is called the Riemann integral of f. TODO make the overline and underline shorter Sections of a function on a product space, suppose that f is a function on Γ Λ. Then the Γ-sections on f and the Λ-sections on f are defined, respectively, by f x (y) = f(x, y), x Γ, f y (x) = f(x, y), y Λ. xample 1 See book, example 4.22 on p.239. Sections of a set in a product space, suppose that A Γ Λ. Then the Γ-sections of A and Λ-sections of A are defined, respectively, by: A x = {y Λ (x, y) A}, x Γ, A y = {x Γ (x, y) A}, y Λ. xample 1 See book, example 4.21 on p.238. Semialgebra of subsets, let Ω be a set. A nonempty collection C of subsets of Ω is called a semialgebra if the following conditions hold: a) If A, B C, then A B C. 10

11 b) If C C, then there is a pairwise disjoint finite (possibly empty) sequence of members of C whose union is C c. C is a semialgebra if it is closed under intersection and the complement of each member of C is a finite (possibly empty) disjoint union of members of C. xample 1 See book, example 4.12 on p.209. σ-algebra (set), given a set Ω, a nonempty collection A of subsets of Ω is called a σ-algebra if the following two conditions are satisfied: a) A A implies A C A. b) {A n } n A implies n A n A. σ-finite measure space, a measure space (Ω, A, µ) is called a σ-finite measure space if there is a sequence {A n } n of A-measurable sets such that n A n = Ω and µ(a n ) < for each n. xamples See book, example 4.19 on p Simple function (canonical representation), an M-measurable function s is said to be a simple function if it takes on only finitely many values, that is, if its range is a finite set. Let a 1, a 2,..., a n denote the distinct nonzero values of s and, for 1 k n, set A k = {x s(x) = a k }. Then we can write s = n a k χ Ak. k=1 This is called the canonical representation of s. xample 1 See book, examples on p Step function, a step function on an interval [a, b] is a function of the form h = n a k χ Ik k=1 where n N, {a k } n k=1 is a sequence of real numbers, and {I k} n k=1 is a finite sequence of pairwise disjoint intervals whose union is [a, b]. Supremum, let {x n } n=1 be a sequence of elements from R. sup n x n is the least upper bound for the sequence. Uniform convergence, given a sequence {f n } n=1 of functions f n : Ω R, {f n } n=1 converges uniformly to the real-valued function f on Ω, if for each ɛ > 0, there is an N N such that n N implies f n (x) f(x) < ɛ for all x Ω. It s usually denoted f n f uniformly to indicate such convergence. Vitali cover, let R. A family V of nondegenerate closed intervals is said to be a vitali cover of if for each x and each δ > 0 there is an 11

12 I V such that x I and l(i) < δ. That is, every point of is contained in arbitrarily small intervals from V. 3 Theorems Proposition 1.9 The image of a countable set (given A, the image f(a)) is countable. Proof See book p.23. Proposition 1.10 A countable union of countable sets is countable. Proof See book p Proposition 1.11 The Cartesian product of two countable sets is countable. Proof See book p.24. Proposition 1.12 The set Q of rational numbers is countable. Proof See book p.24. Proposition The sets Z2, Z 3,..., Z n,... for any finite n is countable. Proposition 2.1 A nonempty subset of real numbers that is bounded below has a greatest lower bound. Proof See book p.38. Proposition 2.2 Archimedean principle. For each x R there is an n N such that n > x. Proof See book p Proposition 2.3 Between any two real numbers there is an irrational number. Proof See book p.39. Proposition 2.4 Between any two real numbers there is a rational number. Proof See book p.39, but apparently the book s proof has a gap. See my notes from TODO write proof here. Theorem 2.2 a) R and are open sets. b) If A and B are open sets, then so is A B. c) If {O ι } ι I is a collection of open sets, then ι I O ι is open. 12

13 Warning, b) does not extend to arbitrary (including countable) intersections. Consider O n = ( 1/n, 1/n). Then each O n is open, but n=1 O n = {0}, which is not open. Proof See book p.58. Proposition 2.13 ach open set O R is a countable union of disjoint open intervals. The representation is unique. Proof See book p Theorem 2.5 Let D R and f : D R. Then f is continuous on D if and only if f 1 (O) is open in D for each open set O in R. Proof See book p.66. Corollary 2.1 A function f : R R is continuous if and only if f 1 (O) is open in R for each open set O in R. Proposition 2.21 Let { n } n be a sequence of subsets of R each having measure zero. Then n n has measure zero. Proof See exercise Theorem 2.7 A bounded function on [a, b] is Riemann integrable if and only if the set of points of discontinuity of the function has measure zero. Proof See Richard Goldberg s Methods of real analysis, section 7.3. Theorem 3.2 The collection of Borel sets, B = {B R χ B Ĉ}, is a σ-algebra of subsets of R. Proof See book p.97. Lemma 3.4 Let F = {f f 1 (O) B for all open sets O}. Then F contains the continuous functions. Proof See book p.98. Lemma 3.5 f F if either of the following conditions hold: a) For each a R, f 1 ((, a)) B. b) For each a R, f 1 ((a, )) B. Proof See book p.98. Lemma 3.6 F is closed under pointwise limits. Proof See book p.98. Corollary 3.1 F contains the Borel measurable functions. Proof See book p

14 Lemma 3.7 Let f F. Then there is a sequence, {f n } n=1, of Borel measurable functions such that f n f pointwise. Proof See book p.99. Theorem 3.3 A function f is Borel measurable if and only if the inverse image of each open set under f is a Borel set; that is, if and only if f 1 (O) B for all open sets O. Proof See book p.99. Theorem 3.4 The collection of Borel sets, B, is the smallest σ-algebra of subsets of R that contains all the open sets. Proof See book p.100. Proposition 3.1 Lebesgue outer measure, λ, has the following properties: a) Nonnegativity: λ (A) 0, for all A R. b) λ ( ) = 0. c) Monotonicity: A B λ (A) λ (B). d) Translation invariance: λ (x + A) = λ(a) for x R, A R, where x + A = {x + y y A}. e) Countable subadditivity: If {A n } n is a sequence of subsets of R, then ( ) λ A n λ (A n ). n n In particular if A, B R, then λ (A B) λ (A) + λ (B). Proof See book p.107. Proposition 3.2 The outer measure of an interval is its length. That is, λ (I) = l(i) for every interval I. Proof See book p Proposition Let d(x, ) and d(, F ) be the distance between an element and a set, and a set and a set respectively. Then the following hold: (1) For a fixed R, d(x, ) is continuous. (2) d(, F ) = inf{d(y, F ) y }. (3) If A and B F, then d(, F ) d(a, B). Proof This proof is, rather cruelly, left as an exercise for the reader. Lemma 3.8 Suppose that I is a finite open interval and ɛ, δ > 0 are given. Then there are a finite number of open intervals J 1, J 2,..., J n such that 14

15 l(j k ) < δ for 1 k n and I n k=1 J k and n k=1 l(j k) < l(i) + ɛ. That is, I can be covered by many small intervals of width < δ and the sum of interval widths are less than λ (I) + ɛ. Proof See book p.111. Lemma 3.9 Suppose that A is a subset of R with λ (A) <. Then for each ɛ, δ > 0, there is a sequence {I n } of open intervals such that l(i n ) < δ for all n, I n A and l(i n ) < λ (A) + ɛ. Proof See book p.111. Theorem 3.8 Suppose that A, B are subsets of R that are a positive distance apart, d(a, B) > 0. Then, λ (A B) = λ (A) + λ (B). Proof See book p.112. Theorem 3.9 Suppose that A, B are subsets of R with the property that there is an open set O with A O and B O C. Then λ (A B) = λ (A) + λ (B). Proof See book p.115. Theorem 3.10 Lebesgue outer measure, λ, is not finitely additive. Proof See book p.117. Proposition 3.3 Let O be an open set. Then λ (W ) = λ (W O) + λ (W O C ). Proof See book p.119. Theorem 3.11 M, the collection of all subsets of R satisfying the Carathéodory criterion, is a σ-algebra and M B. Proof See book p Theorem 3.12 If A 1, A 2,... are in M, with A i A j = for i j, then ( ) λ A n = λ(a n ). n n Proof See book p.122. Proposition 3.4 A subset of R with Lebesgue outer measure zero is a Lebesgue measurable set; that is, λ () = 0 M. 15

16 Proof See book p.123. Proposition 3.5 very countable subset of R has Lebesgue measure zero. Proof See book p.124. Proposition 3.6 The Cantor set, P, has Lebesgue measure zero. Proof See book p.124. Theorem 3.13 Suppose that { n } n=1 is a sequence of Lebesgue measurable sets with 1 2. Then ( ) λ n = lim λ( n). n Proof See book p n=1 Theorem 3.14 The σ-algebra of Borel sets, B, is a proper subcollection of the σ-algebra of Lebesgue measurable sets, M. Proof See exercise Proposition 3.7 Let f be a real-valued function on R. Then the following statements are equivalent: a) f is M-measurable. b) For each a R, f 1 ((, a)) M. c) For each a R, f 1 ((a, )) M. d) For each a R, f 1 ((, a]) M. e) For each a R, f 1 ([a, )) M. Theorem 3.15 The collection of Lebesgue measurable functions forms an algebra. If f and g are M-measurable and α R, then: a) f + g is M-measurable. b) f g is M-measurable. c) αf is M-measurable. Theorem 3.16 Suppose that f and g are M-measurable functions and that {f n } n=1 is a sequence of M-measurable functions that converges pointwise to a real-valued function. Then: a) f g is M-measurable. b) f g is M-measurable. c) lim n f n is M-measurable. 16

17 Lemma 3.14 The integral of a nonnegative linear combination of simple function is linear. Proof See book p Proposition 3.9 a) Let f be a nonnegative M-measurable function. Then there is a nondecreasing sequence of nonegative simple functions that converges pointwise to f. There is a sequence {s n } n=1 of nonnegative simple functions with s 1 (x) s 2 (x) for all x R and lim n s n (x) = f(x). b) If {s n } n=1 is a sequence of nonnegative simple functions that converges pointwise to a real-valued function f, then f is a nonnegative M-measurable function. Proof see book p.135. Proposition 3.10 Let f, g be nonnegative Lebesgue measurable functions, α 0 and M. Then: a) f g f dλ g dλ. b) A and A M A f dλ f dλ. c) f(x) = 0 for all x f dλ = 0. d) λ() = 0 f dλ = 0. e) αf dλ = α f dλ. Proof See book p Theorem 3.17 Monotone convergence theorem (MCT). Suppose that {f n } n=1 is a monotone nondecreasing sequence of nonnegative Lebesgue measurable functions that converges pointwise to a real-valued function; in other words, for each x R, 0 f 1 (x) f 2 (x) f n (x) and lim n f n (x). Then: lim f n dλ = lim n n f n dλ for each R. (Actually, the limit can be equal to and the theorem will still hold.) Proof See book p Proposition 3.11 If f, g are nonnegative Lebesgue measurable functions, then (f + g) dλ = f dλ + g dλ 17

18 for each M. Proof See book p.143. Theorem 3.20 Fatou s lemma. Suppose that {f n } n=1 is a sequence of nonnegative Lebesgue measurable functions that converges pointwise to a real-valued function. Then, for each M, lim f n dλ lim inf f n dλ. n n Proof See book p.146. xample 1 See book, example 3.4 on p.147. Proposition 3.12 Suppose that f is a real-valued function on R. Then: a) f = f + f. b) f = f + + f. c) If f is Lebesgue measurable, then so are f + and f. Lemma 3.16 If f is a Lebesgue measurable function and M and f = f 1 f 2 such that f 1, f 2 are nonnegative and Lebesgue integrable over, then f is Lebesgue integrable over and f dλ = f 1 dλ f 2 dλ. Theorem 3.21 If f, g are Lebesgue integrable over M and α R, then: a) f + g is Lebesgue interable over and (f + g) dλ = f dλ + b) αf is Lebesgue integrable over and αf dλ = α c) f g f dλ g dλ. d) f dλ f dλ. f dλ. g dλ. e) If A, B are measurable subsets of with A B =, then f dλ = f dλ + f dλ. A B A B 18

19 Proof See book p Theorem 3.22 Dominated convergence theorem (DCT). Suppose that {f n } n=1 is a sequence of Lebesgue measurable functions that converges to a realvalued function. Further suppose that there is a nonnegative Lebesgue integrable function g such that f n g for all n N. Then, lim f n dλ = lim f n dλ n n for each M. Proof See book p xample 1 See book, example 3.6 on p.155. Theorem 3.23 Suppose that f is Riemann integrable on [a, b]. Then f is Lebesgue integrable on [a, b] and [a,b] Proof See book p f(x) dλ(x) = b xample 1 See book, example 3.7 on p a f(x) dx. Proposition 3.13 Suppose that f is a Lebesgue measurable function and that g = fλ-ae. Then g is Lebesgue measurable. Proof See book p Proposition 3.14 Suppose that {f n } n=1 is a sequence of Lebesgue measurable functions and that f n fλ ae. Then f is a Lebesgue measurable function. Proof See book p.163. Proposition 3.15 Let f, g be Lebesgue measurable functions with f = gλae. if f is Lebesgue integrable, then so is g and for each M, g dλ = f dλ. Proof See book p Theorem 4.1 Suppose that (Ω, A, µ) is a measure space and that A, B A, that is, they are A-measurable sets. Then the following hold: a) If µ(a) < and A B, then µ(b\a) = µ(b) µ(a). b) Monotonicity: A B λ(a) λ(b). 19

20 c) If { n } n=1 A with 1 2 and µ( 1 ) <, then ( ) µ n = lim µ( n). n n=1 d) If { n } n=1 A with 1 2, then ( ) µ n = lim µ( n). n n=1 e) Countable subadditivity: If { n } n A, then ( ) µ n µ( n ). n n Theorem 4.2 (completion of non-complete measure space) Let (Ω, A, µ) be a measure space. Denote by A the collection of all sets of the form B A where B A and A C for some C A with µ(c) = 0. For such sets, define µ(b A) = µ(b). Then A is a σ-algebra, µ is a measure on A, and (Ω, A, µ) is a complete measure space. Furthermore, A A and µ A = µ. (Ω, A, µ) is called the completion of A. xample 1 The measure space (R, M, λ) is the completion of the measure space (R, B, λ B ). Proof: It can be shown. Proposition 4.1 Let f be a real-valued function on Ω. Then the following statements are equivalent: a) f is A-measurable. b) For each a R, f 1 ((, a)) A. c) For each a R, f 1 ((a, )) A. d) For each a R, f 1 ((, a]) A. e) For each a R, f 1 ([a, )) A. Theorem 4.3 Analogous to Theorem 3.15, but on an arbitrary measurable space (Ω, A) instead of (R, M). Proof See book p Theorem 4.4 A complex-valued function f on C is A-measurable if and only if both its real part, Rf, and its imaginary part, If, are (real-valued) A-measurable functions. 20

21 Proof Unfortunately, left as an exercise for the reader (not by me, but by the book). Proposition 4.3 Similar to proposition 4.1, but for extended real-valued functions. Theorem 4.5 Suppose that f, g are extended real-valued A-measurable functions and that {f n } n=1 is a sequence of extended real-valued A-measurable functions. Then: a) f g and f g are A-measurable. b) sup n f n and inf n f n are A-measurable. c) lim sup n f n and lim inf n f n are A-measurable. d) If {f n } n=1 converges pointwise, then lim n f n is A-measurable. Proof See book p.180. xample See book, example 4.5 on p.181. Proposition 4.4 Analogous to Proposition 3.14, but for arbitrary complete measure spaces (Ω, A, µ) and for complex-valued or extended realvalued functions on Ω. Chapter 4.3 is basically identical to stuff in chapter 3, only in arbitrary measure spaces. Proposition 4.8 Analogous to Proposition 3.10, but for arbitrary A- measurable functions, and that the precondition in a) is f gµ-ae with the same implication as before. Theorem 4.6 Monotone convergence theorem (MCT) Same as theorem 3.17, but in arbitrary measure spaces. Corollary 4.1 Some semi-obvious stuff that follows from theorem 4.6, see book p.188. Theorem 4.7 Fatou s lemma Suppose that {f n } n=1 is a sequence of nonnegative extended real-valued A-measurable functions. Then, for each A, lim inf f n dµ lim inf f n dµ. n n xample 1 See book, example 4.7 on p Theorem 4.8 Suppose that f, g L 1 (Ω, A, µ), and that α C. Then: a) f + g L 1 (µ) and Ω (f + g) dµ = 21 Ω f dµ + g dµ. Ω

22 b) αf L 1 (µ) and αf dµ = α Ω Ω f dµ. c) if f, g are real-valued and f g on Ω, then Ω f dµ Ω g dµ. d) Ω f dµ Ω f dµ. e) µ() = 0 f dµ = 0. f) If A, B are disjoint A-measurable sets, then f dµ = f dµ + A B Also, the abstract Lebesgue integral is linear. A B f dµ. Theorem 4.9 Dominated convergence theorem (DCT). Let (Ω, A, µ) be a measure space. Suppose that {f n } n=1 is a sequence of complex-valued A- measurable function that converges µ-ae. Further, suppose that there is a nonnegative Lebesgue integrable function g such that f n gµ-ae for each n N. Then lim f n dµ = lim f n dµ n n for each A. Proof See book p xample 1 See book, example 4.10 on p Corollary 4.4 Bounded convergence theorem. Let (Ω, A, µ) be a finite measure space. Suppose that {f n } n=1 is a sequence of uniformly bounded, complex-valued, A-measurable functions that converges µ-ae. Then lim f n dµ = lim f n dµ n n for each A. Proof See book p.200. Proposition 4.13 The outer measure, µ, induced by ι and C safisfies: a) µ C = ι, that is, µ (C) = ι(c) for C C. b) Nonnegativity: µ (A) 0 for all A Ω. c) Monotonicity: A B µ (A) µ (B). d) Countable subadditivity: µ ( n A n) n µ (A n ). 22

23 Proof See book, p Proposition 4.14 very set C C is µ -measurable, That is, A C. Proof See book, p Proposition 4.15 A is a σ-algebra of subsets of Ω. Proof See book, p.213. Proposition 4.16 Let µ = µ A. Then µ is a measure on A. Proof See book, p.213. Theorem 4.11 xtension theorem. Suppose Ω is a set, C is a semialgebra of subsets of Ω, and ι is a nonnegative extended real-valued function on C satisfying 1. If C, then ι( ) = If {C k } n k=1 is a finite sequence of pairwise disjoint members of C whose union is in C, then ( n ) n ι C k = ι(c k ). k=1 k=1 3. If C, C 1, C 2,... are in C and C n C n, then ι(c) n ι(c n ). Let µ denote the outer measure induced by ι and C, A the collection of µ - measurable sets, and µ = µ A. Then A is a σ-algebra, A C, µ is a measure on A and µ C = ι. Moreover, the measure space (Ω, A, µ) is complete. xample 1 Let Ω = R n, C = I n and ι = l n =volume. Then µ = λ n and A = M n. The restriction of λ n to M n is denoted by λ n and is called the n-dimensional Lebesgue measure. Theorem 4.12 An extension to a measure is unique. See book p Proposition 4.19 (For the setting, see the definition of measurable rectangles.) The collection U of all measurable rectangles is a semialgebra of subsets of Γ Λ. Proof See book p.232. Theorem 4.14 Suppose that (Γ, S, µ) and (Λ, T, ν) are measure spaces. Let U = {S T S S and T T } and define ι on U by ι(s T ) = µ(s)µ(t ). Then there exists an extension of ι to a measure on a σ-algebra containing U. 23

24 Theorem 4.15 Suppose that (Γ, S, µ) and (Λ, T, ν) are σ-finite measure spaces. Let U = {S T S S and T T } and define ι on U by ι(s T ) = µ(s)µ(t ). Then there exists a unique extension of ι to a measure on the σ-algebra generated by U. Theorem 4.16 Tonelli s theorem. Suppose that {Γ, S, µ} and {Λ, T, ν} are σ-finite measure spaces. Let f be a nonnegative extended real-valued S T -measurable function on Γ Λ. Then: a) f x is T -measurable for all x Γ. b) f y is S-measurable for all y Λ. c) g(x) = Λ f(x, y) dν(y) is S-measurable. d) h(y) = Γ f(x, y) dµ(x) is T -measurable. e) The equalities f(x, y) d(µ ν)(x, y) = Γ Λ = hold. Proof See book p Γ Λ [ ] f(x, y) dν(y) Λ [ ] f(x, y) dµ(x) Γ dµ(x) dν(y) Theorem 4.17 Fubini s theorem. Suppose that {Γ, S, µ} and {Λ, T, ν} are σ-finite measure spaces. Let f be a complex-valued S T -measurable function on Γ Λ such that at least one of the quantities (i) Γ Λ f(x, y) d(µ ν)(x, y), (ii) [ Γ Λ f(x, y) dν(y)] dµ(x), (iii) [ Λ Γ f(x, y) dµ(x)] dν(y) is finite. Then: a) f x L 1 (ν) for µ-almost all x Γ. b) f y L 1 (µ) for ν-almost all y Λ. c) g(x) = Λ f(x, y) dν(y) is defined µ-almost-everywhere and is in L1 (µ). d) h(y) = Γ f(x, y) dµ(x) is defined ν-almost-everywhere and is in L1 (ν). 24

25 e) The equalities f(x, y) d(µ ν)(x, y) = Γ Λ = hold. Proof See book p Γ Λ [ ] f(x, y) dν(y) Λ [ ] f(x, y) dµ(x) Γ dµ(x) dν(y) xample 1 See book, example 4.23 on p (for both Tonelli and Fubini). Theorem 4.18 Suppose that f is Riemann integrable on [a, b] [c, d]. Then f is Lebesgue integrable on [a, b] [c, d] with respect to λ λ and [a,b] [c,d] f(x, y) dλ λ(x, y) = b d a c f(x, y) dx dy. Theorem 4.19 Version of Tonelli s theorem for the completion of a product measure space. See book p Theorem Version of Fubini s theorem for the completion of a product measure space. It isn t even stated in the book, but left to the reader as an exercise... Theorem 4.20 Stuff regarding the product of finitely many measure spaces. See book p.254. xample 1 See book, example 4.24 p.254. Theorem 6.1 Vitali covering theorem. Let R with λ () < and suppose V is a Vitali cover of. Then for each ɛ > 0 there is a finite disjoint collection {I k } n k=1 V such that Proof See book p λ (\ ) n I k < ɛ. k=1 Theorem 6.2 Let f be a monotone function on [a, b]. Then f is differentiable almost everywhere on [a, b]. Proof See book p Proposition 6.1 Let f be a complex-valued function defined in an open interval about x. Let u = Rf and v = If. Then f is differentiable at x if and only if u and v are differentiable at x and, in that case, f (x) = u (x)+iv (x). 25

26 Proof Left as an exercise for the reader by the book. Theorem 6.3 Let f be a real-valued function of bounded variation on [a, b]. Then Va x f + Vx y = Va y f, a x < y b. Moreover, f can be written as the difference of two nondecreasing functions on [a, b]. Proof See book p Theorem 6.4 Any function of bounded variation on [a, b] is differentiable almost anywhere on [a, b]. Proposition 6.2 Suppose f L 1 ([a, b]) and set F (x) = x a f(t) dt, a x b. Then F is continuous and of bounded variation on [a, b]. Moreover, Proof See book p V b a f = b a f(x) dx. Proposition 6.3 Suppose f L 1 (R) and set F (x) = x f(t) dt, < x <. Then F is continuous on R and is of bounded variation on every finite closed interval. Moreover, where, by definition, V F = f(x) dx, V = lim n V n nf. xample 1 See book, example 6.5 on p Theorem 6.5 First fundamental theorem of calculus. Suppose f L 1 ([a, b]) and set F (x) = b a f(t) dt, a < leqx b. Then F is differentiable almost everywhere on [a, b] and, in fact, for almost all x [a, b]. F (x) = f(x) 26

27 Proof See book p Propositions and Lemma Stuff regarding absolutely continuous functions. See book p Theorem 6.6 Second fundamental theorem of calculus. Suppose f is defined on [a, b]. A necessary and sufficient condition for f to exist almost everywhere and be Lebesgue integrable on [a, b], and for x a f (t) dt = f(x) f(a), a x b, is that for each ɛ > 0, there is a δ > 0 such that n f(b k ) f(a k ) < ɛ k=1 whenever {(a k, b k )} n k=1 is a finite sequence of pairwise disjoint subintervals of [a, b] with n k=1 (b k a k ) < δ. Theorem 6.7 A function f is absolutely continuous on R if and only if it is absolutely continuous on every finite closed interval, V < and lim x f(x) = 0. 4 Problem set 1 xercise 1.8 Let {A n } n=1 be an infinite sequence of subsets of Ω. a) Prove that ( ) A k n=1 k=n ( ) A k. The set on the left is called the limit inferior of {A n } n=1 and is denoted by lim inf n A n ; the set on the right is called the limit superior of {A n } n=1 and is denoted by lim sup n A n. b) Describe in words the limit inferior and limit superior of {A n } n=1, and use that description to interpret the relation in part (a). n=1 k=n c) Let Ω = R, and define { [0, 1 + 1/n], if n is an even positive integer; A n = [ 1 1/n, 0], if n is an odd positive integer. Determine lim inf n A n and lim sup n A n. 27

28 Solution N/A xercise 1.19 xercise 1.21 xercise 1.23 Let Ω be a nonempty set. Construct a one-to-one function from P (Ω) onto {0, 1} Ω. Solution For an arbitrary A Ω, define the characteristic function { 1 if x A X A = 0 if x / A Show that for A, B Ω, X A = X B A = B. Clearly A = B X A = X B. Assume A B. Pick x A B. Then X A (x) = 1 and X B (x) = 0, hence X A X B. (For propositions A, B, showing A B is the same as showing B A.) Hence A = B X A = X B. Hence it s shown. xercise 2.9 Let x R and set A = {z Z z x}. a) Prove that A is nonempty. b) xplain why A has a least upper bound. c) Prove that sup A A and, hence, that sup A is an integer. d) The integer sup {z Z z x} is called the greatest integer in x and is denoted by [x]. Prove that [x] x [x] + 1 or, equivalently, that x 1 < [x] x. e) The function f : R Z defined by f(x) = x is called the greatest integer function. Prove that for each z Z, f(z + x) = z + f(x). Solution a) Archimedean principle. (x, ) N for all x R if and only if (, x) ( N) for all x R. Hence for all x R, there exists z Z such that z < x. We can negate, because it s a simple function f : R R, f(x) = x or f(a) = { x x A} = f( A). b) A is bounded from above by x. c) First show M = sup A Z. If sup A / Z, then supa has a positive distance δ to the nearest integer. But (M δ, M + δ) contains no integer. But this contradicts the fact that there is z Z such that M δ < z M. Contradiction! Hence M = sup A must be an integer, Since M is an integer, it is the only integer on the interval (M 1 2, M ). But A (M 1 2, M ) (since M = sup A), and A consists of integers, hence that integer must be M. Hence M = sup A. d) For any x R, define x = sup {z Z z x}. This exists and is an integer. x = n for n x < n + 1 or x x x + 1 for x R. 28

29 e) In order words, show that f(x+z) = z +f(x) for all x R for f(x) = x. Let A R, b R. Then In general: holds for all A R, b R. A + b = {a + b a A}. sup(a + b) = (sup A) + b 1) sup A = if and only if sup(a + b) =. 2) sup A < if and only if sup(a + b) <. for all x A, hence sup A + b x + b sup A + b sup(a + b). Suppose sup(a + b) < sup A + b. Then, sup A > sup(a + b) b, hence there exists x A such that x > sup(a + b) b, hence there exists an x A such that x + b > sup (A + b). Contradiction! (x + b A + b.) Hence sup(a + b) = sup A + b. 5 Problem set 2 xercise 2.19 xercise 2.25 Prove that inf k n x k sup k m x k for all n, m N, where {x n } n=1 is any sequence of real numbers. Solution inf k n x k sup k n x k follows from the definition of inf and sup. We realise that inf x k is nondecreasing, and sup x k is nonincreasing. When assuming m n (without loss of generality), we get inf x k inf k m and hence we have shown the claim. xercise 2.26 k n x k sup k n x k sup x k k m xercise 2.27 Let {x n } n=1 and {y n} n=1 be sequences of real numbers. Verify that each of the following holds, provided that the right-hand side makes sense. 29

30 a) lim sup (x n + y n ) lim sup x n + lim sup y n. b) lim sup (x n + y n ) lim sup x n + lim inf y n. c) lim inf (x n + y n ) lim inf x n + lim inf y n. d) lim inf (x n + y n ) lim sup x n + lim inf y n. Solution The right-hand side makes sense if it isn t or +. a) Choose n N. When k n, we have x k + y k sup k n x k + sup x n y k, which implies sup k n (x y + y k ) sup k n x k + sup x n y k (because we have a n b n lim a n lim b n ). Then, lim sup n k n(x k + y k ) lim (sup n k nx k + sup k n y k ) = lim sup k n x n + lim sup k n y n b) TODO c) TODO d) TODO xercise 2.28 TODO Solution TODO xercise 2.29 Let {x n } n=1 and {y n} n=1 be sequences of real numbers. Suppose x n y n for n sufficiently large; that is, there is an N N such that x n yy n for n N. a) Prove that lim sup x n < lim sup y n and lim inf x n < lim inf y n. b) Suppose a and b are extended real numbers such that for n sufficiently large, a x b. Show that a lim inf x n lim sup x n b. Solution TODO xercise 2.33 Prove that an extended real number is a cluster point of a sequence if and only if the sequence has a subsequence converging to that number. Conclude that the limit superior of a sequence is the limit of a subsequence of the sequence and likewise for the limit inferior. Solution TODO xercise 2.47 Let A and B be subsets of R. We recall that A is the closure of A. stablish each of the following facts: a) A B = A B. b) A B A B. Provide an example to show that the reverse inclusion does not hold in general. 30

31 c) A is closed. d) If A and B are closed, then so is A B. e) If {F ι } ι I is a collection of closed sets, then ι I F ι is closed. Solution a) : We have A A B A A B and B A B B A B which together give A B A B. : Assume that x A B. Then there exists a sequence {x n } A B such that limx n = x. At least one of A, B must contain an infinite number of elements from {x n }. Let {x nk } k=1 be those elements, which reside in A. Then we have lim k x nk = x, hence x A and hence A B A B. b) N/A c) N/A d) N/A xercise 2.49 xercise 2.55 xercise 2.56 xercise 2.60 xercise Problem set 3 xercise 2.59 Suppose that f : (a, b) R is nondecreasing. For x (a, b), let L x = {f(t) a < t < x} and R x = {f(t) x < t < b}, and define f(x ) = sup L x and f(x+) = inf R x. a) Show that f(x ) f(x) f(x+) for all x (a, b). b) Prove that f is discontinuous at x if and only if f(x ) = f(x+). c) Prove that f has countably many discontinuities; that is, the set of points at which f is discontinuous is countable. 31

32 d) Deduce that a nonincreasing function on (a, b) has countably many discontinuities. Solution TODO xercise 2.67 Prove Proposition 2.16 on page 73 (restate it here ploks) by using the following steps. We can assume that x (0, 1). Why? a) TODO b) TODO c) TODO d) TODO Solution TODO xercise 2.69 TODO xercise 2.71 TODO xercise 2.72 TODO xercise 2.73 xercise Problem set 4 xercise 2.76 xercise 2.77 TODO xercise 2.78 TODO xercise 2.79 TODO xercise 2.80 TODO xercise 2.84 TODO xercise 2.85 Show that a subset of a set with measure zero also has measure zero. Solution A B and λ (B) = 0. This is trivially trivial because of the monotonicity of λ, so λ (A) λ (B) = 0 combined with λ () 0 for all R (proposition 3.1 a)) gives λ (A) = 0. xercise 2.88 TODO xercise 2.91 TODO 32

33 xercise 2.92 TODO 8 Problem set 5 xercise 1.40 TODO xercise 1.41 xercise 1.42 TODO xercise 1.47 TODO xercise 3.2 TODO xercise 3.3 TODO xercise 3.5 TODO xercise 3.6 TODO xercise 3.7 done in lecture? xercise 3.8 done in lecture? xercise 3.11 xercise 3.12 TODO xercise Problem set 6 xercise 3.15 TODO xercise 3.16 TODO xercise 3.17 xercise 3.18 TODO xercise 3.19 TODO xercise 3.20 Suppose that f : R R is differentiable at each point of R. a) If f (x) 1 for each x R, prove that for each A R, λ (f(a)) λ (A). b) Provide an example to show that the previous inequality may fail to hold if f (x) > 1 for some x R. 33

34 Solution a) Since f is differentiable at each point, f is continuous at every x R and hence f(i) for any interval I is an interval, by the mean value theorem. Let {I n } n=1 cover A. Then f({i n} n=1 ) covers f(a) (but isn t necessarily open). Let I be an interval. We have λ (f(i)) = l(f(i)) = sup x,y I f(y) f(x) (because f can wiggle). Also, sup x,y I f(y) f(x) sup x,y I f (z) y x sup x,y I y x = l(i) = λ (I) for some z I. Let ɛ > 0. Take {I n } n=1, n=1 I n A, and n=1 l(i n) < λ (A) + ɛ 2. Let J n be an open interval, J n f(i n ) and l(j n ) < l(f(i n )) + then have λ (f(a)) = l(j n ) n=1 n=1 ( l(f(i n )) + l(f(i n )) + n=1 λ (A) + ɛ 2 + ɛ 2 = λ (A) + ɛ for all ɛ > 0. This shows that λ (f(a)) λ (A). ɛ ) 2 n+1 ɛ 2 n+1 n=1 } {{ } ɛ/2 ɛ 2 n+1. We b) For instance f(x) = 2x gives f([0, 1]) = [0, 2] and 2 = λ ([0, 2]) λ ([0, 1]) = 1. xercise 3.21 TODO xercise 3.33 xercise 3.34 TODO xercise 3.35 TODO xercise 3.36 TODO xercise 3.37 TODO xercise 3.39 TODO xercise 3.51 TODO 34

35 10 Problem set 7 xercise 3.40 TODO xercise 3.41 TODO xercise 3.50 xercise 3.53 TODO xercise 3.54 TODO xercise 3.56 TODO 11 Problem set 8 xercise 3.67 TODO xercise 3.68 TODO (part of 2009 exam!) xercise 3.69 TODO xercise 3.70 TODO xercise 3.71 TODO xercise 3.75 TODO xercise Problem set 9 xercise 3.95 TODO (foreleser var borte) xercise 3.96 TODO (foreleser var borte) xercise 3.97 TODO xercise 3.99 xercise TODO xercise TODO xercise TODO xercise 4.3 TODO xercise 4.8 xercise

36 xercise 4.12 xercise Problem set 10 xercise 3.44 TODO xercise 4.16 TODO xercise 4.21 xercise 4.22 TODO xercise 4.53 TODO xercise 4.61 TODO xercise 4.63 TODO xercise 4.67 TODO xercise 4.74 TODO xercise 4.81 TODO xercise 4.82 TODO 14 Problem set 11 (xercise 4.134) TODO apparently this exercise was done in class. xercise xercise xercise xercise xercise xercise TODO xercise xercise TODO xercise xercise

37 xercise Problem set 12 xercise 6.20 Using only definition 6.6, prove that if f is of bounded variation on [a, b], then it is bounded thereon. Solution f(x) = f(x) f(a) + f(a) f(x) f(a) + f(a) V b a f + f(a) <. xercise 6.21 TODO xercise 6.22 TODO xercise 6.34 TODO xercise 6.36 TODO xercise 6.37 TODO xercise 6.38 TODO (sketch) xercise 6.39 xercise 6.42 TODO xercise 6.48 TODO xercise 6.53 TODO xercise 6.55 TODO 37