Daniel Akech Thiong Math 501: Real Analysis Homework Problems with Solutions Fall Problem. 1 Give an example of a mapping f : X Y such that

Transcription

1 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 some A Y and f 1 (f(a)) A for some A X. 1. Consider f : [ 1, 1] [ 1, 1] : f(x) = x 2. Then let A = [ 1, 0] and B = [0, 1]. Then we have f(a B) = {0} = B = f(a) f(b). The reason for this failure is due to the fact that f is not injective. 2. Choose A [ 1, 1], then y f(f 1 (A)) implies that there exists some x f 1 (A) such that f(x) = y but x f 1 (A) implies that f(x) A and hence y A. Hence f(f 1 (A)) A. To show that the inclusion can be strict, we let A = [ 1, 1]. Then f 1 (A) = A and f(a) = [0, 1]. Therefore, we have f(f 1 (A)) = f(a) = [0, 1] = [ 1, 1] = A. Choose a A. Then f(a) f(a) so that a f 1 (f(a)). Now let A = [0, 1]. Then f(a) = A and f 1 (f(a)) = f 1 (A) = [ 1, 1] = [0, 1] = A. Problem. 2 If f : X Y, g : Y Z and both f and g are one-to-one, show that g f : X Z is one to one mapping and (g f) 1 = f 1 g 1. Choose x, y X. Then (g f)(x) = (g f)(y) g(f(x)) = g(f(y)) f(x) = f(y) x = y. The first double implication is by definition, the second double implication is because g is one-to-one and the last double implication is because f is one-to-one. Therefore, g f is one -to- one. Set h = g f. We must show that h 1 (x) = f 1 (g 1 (x)) for all x X. Write h 1 (x) = y so that h(y) = x. Then f 1 (g 1 (x)) = f 1 (g 1 (h(y))) = f 1 (g 1 (g(f(y)))) = f 1 (f(y)) = y = h 1 (x). Therefore, (g f) 1 = f 1 g 1. Problem. 3 A complex number is said to be algebraic if it is a zero of some polynomial with rational coefficients. Show that the set of all algebraic numbers is countable. Choose a = (a 0,..., a n 1 ) Q n. Let R a be the roots of the polynomial x n + a n 1 x n a 1 x + a 0. Then card(r a ) = n by the Fundamental Theorem of Algebra. Let A denote the set of algebraic numbers. Then A = n N a Q n R α, which is a countable union of countable unions of finite sets and it is therefore countable. 1

2 Problem. 4 If f is an increasing real-valued function of real variable, show that f has at most countable number of discontinuities. Let [a, b] be an arbitrary interval of R. For c [a, b], define j(c) := lim x c + f(x) lim x c f(x) to be the number of jump discontinuities in [a, b]. First we will show that j(c) is welldefined. Set A = {y = f(x) : a < x < c}. Since f(x) is increasing, f(x) < f(c) for all x so that f(c) is an upper bound of A. By the Axiom of Completeness, the supremum of A exists, call it L. We wish to show that lim x c f(x) = L. Note that lim x c f(x) = L for all ε > 0, there exists δ > 0 such that : L ε < f(x) < L + ε whenever c δ < x < c. Choose ε > 0. Then L ε is not an upper bound for A so that there exists y ε A such that L ε < y ε < L + ε. Since y ε A, there exists x ε < c such that f(x ε ) = y ε so that L ε < f(x ε ) < L + ε. Suppose that {x n } is a sequence such that x n < c such that it converges to c as n grows large. Then for every γ > 0, there exists N N such that c γ < x n < c + γ for all n > N. We have x ε γ < c γ < x n < c for all n > N = x ε < x n < c for all n > N and then by the monotonicity of f, we get f(x ε ) < f(x n ) < f(c) for all n > N. We have L ε < f(x n ) < L + ε for all n > N. It follows that lim x c f(x) = L. A similar argument shows that lim x c + f(x) = inf{y = f(x) : c < x < b}. It follows from the above proof that lim x c f(x) < lim x c + f(x) and that whenever c < d we have lim x c + f(x) < lim x d f(x). Now, by the density of rational numbers, for each c at which f is discontinuous we associate a rational number r(c) with it such that lim x c f(x) < r(c) < lim x c + f(x). Since lim x c + f(x) < lim x d f(x), r(c) = r(d) whenever c = d. Therefore, r is a one to one map from the set at which f is discontinuous to Q and by the countability of Q, the number of discontinuities is countable. Problem. 5 Show that if E n, n = 1, 2, 3,.. are countable sets then the sets n 1 E n and m n=1 E n, m 1 are also countable. We will use the fact that N N is countable, which is proved in the book as Corollary 4. For each i = 1, 2,..., we enumerate the elements of E i by E i = {x ij } j N. Then the map φ : N N i=1 E i : φ(i, j) = x ij is a surjective map. Therefore, φ(n N) = i=1 E i so that the latter is the image of a function whose domain is a countable set and it is therefore countable by Theorem 5 in the book. There is a bijection between m n=1 E n = m n=1 E n and N m. So it suffices to show that N m is countable. We do this by embedding N m into a countable set, namely, N. Consider φ : N m N : φ(n 1, n 2,..., n m ) = m i=1 pn i i, where p i are distinct primes. Then φ(n 1, n 2,..., n m ) = φ(k 1, k 2,..., k m ) m i=1 pn i i = m i=1 pk i i. If n j k j > 0 for some 2

3 1 j m, then p j p n j k j j and m i=1 pn i i = m i=1 pk i i imply p j m i =j pk i i, a contradiction since p i s are distinct. So n i = k i for all i and φ is injective. Problem. 6 Show that each of the relations below is an equivalence relation and describe equivalence classes: 1. (a, b) R in R if a b is a rational number; 2. (a, b) R in Z if m (b a), where m is a given number. 1. Note that (a, a) R in R since a a = 0 is a rational number. (a, b) R in R implies (b, a) R in R because (a b) = b a is a rational number since the additive inverse of (a b) is a rational number. If (a, b) R in R and (b, c) R in R, then (a b) and (b c) is a rational number and hence a c = (a b) + (b c) is a rational number since rational numbers are closed under addition as an additive group and we therefore get (a, c) R in R. Therefore we have an equivalent relation. To find the equivalence class, note that the equivalence class of 0 is Q and since any rational number is equivalent to 0 its equivalence class is also Q. The equivalence class of α Q c is the set {r + α r Q}. These are the two conditions we have. 2. Note that (a, a) R in Z since m 0 = a a. If (a, b) R in Z, then (b, a) R in Z since m (b a) = m (a b) = (b a). If (a, b) R in Z and (b, c) R in Z. Then (a, c) R in Z since m (a b) + (b c) = (a c) since a number dividing two numbers divides their sum. By Division algorithm, there exist q and r such that a = mq + r, where 0 r m 1. Then consider this equation modulo m we get [a] = [r]. Hence, the equivalence class is R/R = Z/mZ. Problem. 7 Show that the following sets have the power of continuum: 1. the unit interval [0, 1] and the set of all real numbers R; 2. the unit cube in R n, n 2; 3. the set of all sequences of real numbers; 4. the set of all continuous functions on the real line. 3

4 1. Note that the inclusion map i : (0, 1) [0, 1] : i(x) = x is an injection map and the map f : [0, 1] (0, 1) : f(x) = 2x+1 4 is also an injective map. Hence by Schroeder - Bernstein Theorem card((0, 1)) = card([0, 1]). Now g : (0, 1) R : g(x) = tan(π(x 2 1)) is a bijection with g 1 (x) = tan 1 (x) π. Hence, card([0, 1]) = card((0, 1)) = card(r) and the latter has the power of continuum by the usual Cantor s diagonal argument. 2. It is enough to show this for n = 2 and by what we have shown in the first part, it suffices to only consider (0, 1) n. Let S = (0, 1) (0, 1). Choose (x, y) (0, 1) (0, 1). Then we can write x = 0.a 1 a 2 a 3... and y = 0.b 1 b 2 b 3... Then we define z = 0.c 1 c 2 c 3..., where c i = a i b i and this process is reversible in the sense that if we are given z, we can construct (x, y). Then the map f : S (0, 1) is well-defined bijection. Hence card(s) = card((0, 1)). Now, consider g : (0, 1) R : g(x) = tan(π(x 2 1 )). Then define F : S R 2 : F(x, y) = (g(x), g(y)), which is a bijection. Hence, we have card(s) = card(r 2 ). The exact same argument extends to n. 3. Let S = {a n R : a n {1, 1}}. Then S is the same as the set of functions f : N {1, 1}, which is the same as the power set of the natural 2 N, which is uncountable so that S is uncountable. Since S is contained in the set of all real sequences of R, the latter cannot be countable. 4. The cardinality is at least that of the continuum because α R corresponds to a constant function, which is continuous. If f : R R is continuous then for every x R, there exists a sequence of rational numbers q n converging to x by the density of rationals and then by continuity we have lim n f(q n ) = f(x). Thus the values of f at rational numbers already determined f. The cardinality is at most that of the continuum because the set of real continuous functions inject into the sequence space as shown above. Hence, by the Schroeder- Bernstein Theorem, the set of real continuous functions as the power of continuum. Problem. 8 Show that the Cantor set contains no open interval. Note that the Cantor set is given as follows: C = k N A k, where A k is the union of 2k closed intervals of [0, 1] each of length 3 k. Suppose by contradiction that (a, b) [0, 1] is an open interval that is contained in C. We will show that there exists A m such that (a, b) A m and we do this by showing that b a > 3 m. Note that b a < 1 = log 3 (b a) > 0. Then by the Archimedian property, there exists m > 0 such that m 1 > log 3 (b a). Then m < log 3 (b a) = 3 m < 3 log 3 (b a) = b a = b a. Therefore, (a, b) A m and it follows that (a, b) C. 4

5 Problem. 9 Let (X, d) be a metric space. Show that: 1. d is a continuous real-valued function the metric space X X (endowed with the metric of the product space); 2. A subset A X is bounded if the function d is bounded on A A. 1. Choose ε > 0. Let δ = ε. Suppose d X X (x, y) < δ, where x = (x 1, x 2 ), y = (y 1, y 2 ). Then d(x) d(y) 2 = ( d(x 1, y 1 ) ) 2 + ( d(x2, y 2 ) ) 2 2 d(x1, y 1 ) d(x 2, y 2 ) ( d(x 1, y 1 ) ) 2 + ( d(x2, y 2 ) ) 2 = ( dx X (x, y) ) 2 = d(x) d(y) dx X (x, y) < ε. 2. The diameter of a non-empty subset A X is defined as diam(a) = sup{d(x, y) x, y A}. Then A is bounded if and only if its diameter is finite, but this is true if and only if the function d is bounded on A A as follows: d A A (x, y) = (da (x 1, y 1 ) 2 + ( da (x 2, y 2 ) 2 2diam(A). Problem. 10 Let A X be a subset and x X a point in a metric space (X, d). Define d(x, A) = inf y A d(x, y). Show that: 1. for a fixed A the function f(x) = d(x, A) is continuous; 2. {x : d(x, A) = 0} = A; 3. a set A is closed if d(x, A) > 0 for any x A c. 1. Let {x n } be a sequence in X converging to x X. Let ε > 0 be given. Then there exists n ε such that d(x n, x) < ε for all n > n ε. Note that by triangle inequality, we have, for all y A, ± ( d(x, y) d(x n, y) ) d(x n, x). It follows that ± (d(x n, A) d(x, A)) d(x n, x). Therefore, f(x n ) f(x) = d(x n, A) d(x, A) d(x n, x) < ε for all n > n ε. Thus, f is continuous. 2. Let S = {x A : d(x, A) = 0}. Note that for all y, z A, we have d(y, z) 0 so that f(y) = d(y, A) = 0 whenever y A. Choose t A. Let ε > 0 be given. Then B(t, ε) A =. Hence, there exists y A such that d(t, y) < ε. Then by continuity of f, proven in part (1), we have d(t, A) = f(t) = f(t) = f(t) f(y) < ε = d(t, A) = 0 = t S. Let r > 0 be any positive number. Then x S = d(x, A) < r = B(x, r) A = = x A. 3. Let {x n } A be a sequence such that x n x. The sequence {x n } is Cauchy. If x A, then A is closed. If x A c, then we have 0 < d(x, A) d(x n, x) and since x n x, we have 0 lim n d(x n, x) 0 so that x n = x for n sufficiently large and therefore, x A so that A is closed. 5

6 Problem. 11 Prove that the Paris metric is not equivalent to the standard metric in R 2. Note that d p (x, y) = d(x, p) + d(p, y) d(x, p) + d(p, y) + d(z, p) + d(p, z) = d p (x, z) + d p (y, z) so that d p is indeed a matrix as the other two properties follow trivially. ( ) ) We wish you to show that the identity map id : R 2, d (R 2, d p is not an isometry so that the two spaces cannot be homeomorphic. We simply compare the balls in both spaces. Fix x 0 R 2. Let ε = 1 2. Then for each δ > 0, there exists x = x 0 such that x B d (x 0, δ). Then x B dp (x 0, ε) if and only if d p (x, x 0 ) < ε x B d (x 0, ε d(p, x 0 ) provided p B d (x 0, ε) but this fails when the city x 0 sits more than half distance away from Paris, p. Problem. 12 Show that a sequence {x n } in a metric space converges to a point x if and only if every its subsequence has x as an accumulation point. Suppose that x n x. Then {x n } is Cauchy. Let ε > 0 be given. Then x n x implies that there exists n ε such that d(x n, x) < ε 2 for all n > n ε and Cauchyness of {x n } implies that there exists m ε such that d(x m, x n ) < ε 2. Let N = max(n ε, m ε ). Then d(x nk, x) d(x nk, x n ) + d(x n, x) < ε for all k > N. Hence x is an accumulation point for the set of subsequences. Suppose that x n does not converge to x. Then there exists ε 0 > 0 such that for all n, there exists k n > n such that x kn / B(x, ε 0 ). Since we can do this for all n = 1, 2, 3,...,, we construct a set of subsequences {x kn } such that x is not its accumulation point. Problem. 13 Let X be a totally bounded metric space and f : X Y a uniformly continuous map onto Y. Prove that Y is totally bounded. Let r > 0 be given. Since f is uniformly continuous on X, there exists a δ r > 0 such that d(u, v) < δ r = d(f(u), f(v)) < r. Since X is totally bounded, there exists a finite set {x 1,..., x n } such that X n i=1 B(x i, δ r ). Choose y Y. Since f is onto, Y = f(x) so that there exists x X such that f(x) = y. Then there exists k such that x B(x k, δ r ) = d(x, x k ) < δ r. Then by the uniform continuity of f, d(f(x), f(x k )) < ε = y = f(x) B(f(x k ), ε). Since y was arbitrarily chosen, we have, Y = f(x) n i=1 B(f(x i), ε). Therefore, Y is totally bounded. Problem. 14 A function f : [a, b] R is called Holder continuous if there are C > 0 and 0 < α 1 such that for every x, y [a, b], f(x) f(y) C x y α. A function f : [a, b] R is called Lipschitz continuous if it is Holder continuous with Holder exponent α = 1. Find a differentiable function on [a, b], which is: 1. Holder continuous but not Lipschitz continuous; 2. continuous but not Holder continuous. 6

7 1. Consider f(x) = x α. Then f is holder continuous near zero, but since f (x) = α f(x) x is unbounded near zero, f is not lipschitz continuous on a closed set containing zero. 2. The follow function f is continuous, but can t be Hlder continuous at 0 for any α because we know that C x α log x 0 as x 0 for any positive C and α: 0, if x = 0 1 f(x) = log x, if x (0, 1 2 ], log 1 2, if x [1 2, 1] Problem. 15 Show that the coding map h from ( Σ + 2, d a) (a 2) where C is the Cantor set (endowed with the standard metric d(x, y) = x y is Holder continuous. Two points x, y Σ + 2 are close if x x j y j = y j for all j n and we have d a (x, y) = j j>n a j 1 a 1 a n+1 j 0 a j = a 1 n. Conversely, if x j = y j for at least one j n, then d a (x, y) a 1 n. Note that h(x) C h(x) = k=1 a k 3 k, where a k {0, 2}. We have h(x) h(y) = ( ) k=1 a k b k (n 1) ( ) loga 3 13 = 3 1 ( 3 k a n 3 da (x, y) ) log a 3. Therefore h is Holder continuous. Note that for a = 3, h(x) is not Lipschitz continuous. Problem. 16 Let {f n } be an equicontinuous sequence of functions from a metric space X to a metric space Y. Show that every subsequence {f nk } is equicontinuous. Is the opposite statement If every subsequence {f nk } of the sequence {f n } is equicontinuous then so is the sequence {f n } true? If the sequence of functions is equicontinuous then for any x X and ε > 0, there are N > 0 and δ > 0 such that ( ) d(f n (y), f n (x)) < ε for all n > N. For a subsequence {f nk } there exists K > 0 such that n k > N for all k > K. Thus for any k > K and y B(x, δ) the inequality above ( ) holds and hence {f nk } is equicontinuous. Assume any subsequence {f nk } is equicontinuous. Then since the whole sequence is a subsequence of itself the equicontinuity holds. Problem. 17 Let F be a set of functions f on [a, b] that is uniformly bounded and satisfies the Holder condition of order α, 0 < α 1, with a fixed constant M. Show that F is relatively compact in C ([a, b]). Show also that the unit ball in the Holder space C α ([a, b]) (the space of all Holder continuous functions with a given Holder exponent α ) is compact in C ([a, b]). 7

8 1. Note that F is relatively compact in C ([a, b]) if and only if F is equicontinuous and uniformly bounded. Since F is uniformly bounded, it suffices to show that F is equi-continuous. Let ε > 0 be given. Let δ = ( ε M ) 1 α. Then for all x, y [a, b] with x y < δ, we have f(x) f(y) M x y α < Mδ α = ε for all f F. Hence, F is equicontinuous. 2. Since the unit ball is closed and every closed subset of a compact set is compact, it suffices to find a compact set containing the unit ball. What about its closure? The unit ball in C 0 ([a, b]) has compact closure if and only if it is bounded in C 0 ([a, b]) and equicontinuous. The unit ball in C 0,α ([a, b]) is certainly bounded in C 0 ([a, b]). If f C 0,α 1 then for any ε > 0, take δ = ε, f(x) f(y) x y < ε whenever x y < δ for all x, y [a, b] and therefore we have equicontinuous and the result follows. Problem. 18 Let {f n } be a uniformly bounded sequence of real-valued differentiable functions on [a, b] such that the derivatives {f n} are uniformly bounded. Show that there exists a subsequence {f nk } that converges uniformly on [a, b]. Since {f n } are uniformly bounded, if we can show that {f n } is equicontinuous then the result will follow from the Arzela-Ascoli Theorem. Since f n are uniformly bounded, there exists a uniform bound, M such that f n(z) M for all z. For a given ε > 0, let δ = ε 2M. Then for each x, y [a, b], there exists some z between x and y, by the Mean Value Theorem, such that f n (x) f n (y) = f n (z) x y ε whenever x y < δ. Now, by the Arzela-Ascoli Theorem there exists a subsequence of {f n } that converges uniformly on [a, b]. Problem. 19 Show that on [0, 1] there is a nowhere dense closed set E whose length is at least 1 n 1 (i.e., Ec is a countable union of open intervals of total length at most n 1 ). Let (a m, b m ) be an open interval in [0, 1] with rational endpoints. For each m N, let (c m, d m ) be an open interval in (a m, b m ) of length at most 2 m 1 n. Set E = [0, 1]\ m N (c m, d m ). Then E is closed because its complement is open and and it is nowhere dense because all the rational numbers are deleted and finally, the length of E is at least 1 1 n m=1 (1 2 )m = 1 1 n. Ec = m N (c m, d m ) is a countable union of open intervals of total length at most 1 n. Problem. 20 Construct a set E [0, 1] of first category that has length 1 (i.e., E c has length zero). 8

9 Let x (0, 1) and let E 0 = [0, 1]. Let {a n } be a sequence of positive numbers such that n=1 a n < 1. Remove from the center of the unit interval an open interval of length a 1 ; then from the center of each remaining interval remove an open interval of length 2 1a 2; and so on. Suppose that we have constructed E k, 0 k < n and each E k consists of 2 k disjoint uniformly distributed closed intervals of lengths 2 k (1 x(1 2 k )) > x2 2k. Each of the closed intervals making up E n 1 has length greater than x2 2(1 n) and therefore we can remove from each the middle open interval of length x2 (1 2n) < x2 2(1 n). This construction yields E n. Set E(x) = n=1 E n. The longest interval in E n has length 2 n (1 x(1 2 n )) < 2 n. Hence, E(t) contains no nonempty intervals and therefore E(x) =. Since E n E(x) and E n has length 2 n 2 n (1 x(1 2 n )) we see that E(x) has length 1 x. Since E(x) is closed we have E(x) is a closed nowhere dense subset of [0, 1] with length 1 x. Setting E n = E( n 1 ) and letting E = n 1E( n 1 ). Then E is of first category subset of [0, 1] and E has length 1. Problem. 21 Let X be a metric space and {f n } a sequence of continuous functions from X to a metric space Y. Assume that f n converge to a function f uniformly on each compact subset K X. Show that f is continuous. Let {x n } be a sequence in X such that x n x. Set K = {x n } {x}. We will show that K is compact. Consider a subsequence {y m : m 1} K. If {y m : m 1} is finite, then {y m } is eventually a constant and hence it has a convergence subsequence. If {y m : m 1} is infinite, we construct a subsequence of (y m ) as follows: If y m = x, then y m = x n for some n 1. Let m 1 be the smallest index such that y m1 = x, then y m1 = x n1 for some n 1 1. Let m 2 > m 1 be the smallest index such that y m2 = x and y m2 = x n2 with n 2 > n 1 (note that we can always find the next index because the range is infinite). By induction we obtain a subsequence {y mk } k 1, which is also a subsequence of x n and since x n x, y mk x. Therefore we have constructed a convergence subsequence of an arbitrary chosen sequence {y m : m 1} and therefore K must be compact. By the hypothesis, f n converges uniformly to f on K. Let ε > 0 be given, then there exists N ε such that f n (x) f(x) < 3 ε for all n N ε and for all x K. Keep the same ε and Fix n, then by the continuity, there exists a δ > 0 such that f n (x) f n (y) < 3 ε whenever x y < δ. Now, suppose x k x, then f(x k ) f(x) f(x k ) f n (x k ) + f n (x k ) f n (x) + f n (x) f(x) < ε. Problem. 22 Let (X, A) be a topological space. Prove that the sum and product of two real-valued continuous functions on X are themselves continuous. Let f(x) and g(x) be continuous real-valued functions on X. By exercise 8 below, it suffices to show that for each b R, the sets {x X : f(x) + g(x) < b} and {x X : f(x) + g(x) > b} are open and the sets {x X : f(x)g(x) < b} and {x X : f(x)g(x) > b} are open. Since 9

10 f and g are real-valued, f + g, fg are real-valued. It is easier to just work with an arbitrary interval of the forms: (a, b). Since f(x) <, subtractions in the following manipulations make sense. We have: (f + g) 1 (a, b) = {x X : a f(x) < g(x) < b f(x)} = g 1 (a f(x), b f(x)) = g 1 ( {c X : a c < f(x) < b c} ) ( ) = g 1 f 1 (a c, b c) and by the continuity of f, O = f 1 (a c, b c) is open by the continuity of g, g 1 (O) is open in X and therefore (f + g) 1 (a, b) is open in X. Next, consider A = {x X : f(x) = 0}. Then h(x) = f(x)g(x) is continuous on A because as it is identically zero we have h 1 (a, b) = if 0 / (a, b) and h 1 (a, b) = (a, b) if 0 (a, b). For x X\A, Let (a, b) be an arbitrary open interval of the real numbers. Then: h 1 (a, b) = {x X : a < f(x)g(x) < b} = {x X : a f(x) < g(x) < f(x) b } ( ) ( ( )) ( ( )) = g 1 a f(x), b f(x) = g 1 f 1 a c, b c g 1 f 1 b c, a c whenever c = 0 If g(x) = 0, then h 1 (a, b) = {x X : a < f(x)g(x) < b} = {x X : a ( ) f(x) < g(x) < a f(x), b f(x). In either case we see that h 1 (a, b) is open and hence h(x) is continuous. ( ) a f(x), b f(x). b f(x) } = 10

11 Problem. 23 Let (X, A) be a topological space. Prove that a real-valued function f on X is continuous if and only if for every real number a the sets {x X : f(x) < a} and {x X : f(x) > a} are open. Since f is real-valued, f(x) <. Let a R. Then f 1 (, a) = {x X : < f(x) < a} = {x X : f(x) < a} and f 1 (a, ) = {x X : a < f(x) < } = {x X : f(x) > a}. If f is continuous then f 1 (, a) is open in X and f 1 (a, ) is open in X so that the two sets in the problem are both open. Next, suppose that f 1 (, a) is open in X and f 1 (a, ) is open in X for each a. We wish to show that f is continuous. Choose U to be an open subset of R. Choose x f 1 (U), then f(x) U. Since U is open, there exists ε such that (f(x) ε, f(x) + ε) U. We have f 1 (U) = x f 1 (U) f 1 ((f(x) ε, f(x) + ε)) = ( ) x f 1 f 1 (f(x) ε, ) f 1 (, f(x) + ε), and the latter is open. (U) Problem. 24 Let E be a collection of subsets in a space X. Show that 1. if A = A(E) is the ring that is generated by E, then for every set A A there is a finite collection of subsets {E i } n i=1 E such that A n i=1 E i. 2. if A = A(E) is the σ- algebra that is generated by E, then for every set A A there is a countable collection of subsets {E i } i=1 E such that A i=1 E i. 1. Let A be a non-empty collection of all finite unions of sets in the semi-ring S generated bye. Note that A A by definition. It remains to show that A A, but since A is the smallest ring containing E by definition, it suffices to show that A is a ring. A is closed under finite union. With A = n i=1 A i and B = n i=1 B i. Then A B = i n,j k A i B j and A\B = n i=1 A is a ring. ( A i \ k j=1 B j ) ) = n i=1 k j=1 (A i \B j = n i=1 k j=1 ( ) A i \(A i B j ). Therefore 2. Let A be the collection of all countable unions of sets in the semi-ring S generated by E. Note that A A by definition. It remains to show that A A, but since A is the smallest σ- ring containing E by definition, it suffices to show that A is a σ- ring, but the argument in the first part goes through just fine. Problem. 25 Let E be a countable collection of sets in a space X. Show that the generated algebra A(E) is a countable collection of sets. Does this state remain true if the generated algebra is replaced with the generated σ - algebra? 11

12 Write E = {E 1, E 2,..., }. Then each A A(E) can be written as an expression using the E i and the operations of intersection, union and complement, and thus encoded as a finite string over a finite alphabet. But there are only countably many such strings and so A A(E) is countable. Does this state remain true if the generated algebra is replaced with the generated σ - algebra? The answer is No. The counterexample follows: Take X = R. Suppose {A n } is a countable collection of countable sets, let {B n } be an enumeration of the countable collection of open intervals with rational end points. Now consider the family E = {A n B m n, m ω}. It is a countable family of countable sets, and every open interval B k is the countable union n(a n B k ). Therefore the σ-algebra generated by this family is 2 R, which is uncountable. Problem. 26 Let μ be a finite measure on X. Show that if E 1, E 2 X are measurable sets then μ(e 1 E 2 ) + μ(e 1 E 2 ) = μ(e 1 ) + μ(e 2 ). We have E 1 E 2 = E 1 \(E 1 E 2 ) E 2 \(E 1 E 2 ) E 1 E 2 E 2 )) + μ(e 2 \(E 1 E 2 )) + μ(e 1 E 2 ). = μ(e 1 E 2 ) = μ(e 1 \(E 1 We have E 1 = E 1 \(E 1 E 2 ) E 1 E 2 = μ(e 1 ) = μ(e 1 \(E 1 E 2 )) + μ(e 1 E 2 ). We also have E 2 = E 2 \(E 1 E 2 ) E 1 E 2 = μ(e 2 ) = μ(e 2 \(E 1 E 2 )) + μ(e 1 E 2 ). Therefore: μ(e 1 E 2 ) + μ(e 1 E 2 ) = μ(e 1 \(E 1 E 2 )) + μ(e 2 \(E 1 E 2 )) + μ(e 1 E 2 ) + μ(e 1 E 2 ) = ( μ(e 1 \(E 1 E 2 )) + μ(e 1 E 2 ) ) + ( μ(e 2 \(E 1 E 2 )) + μ(e 1 E 2 ) ) = μ(e 1 ) + μ(e 2 ) Problem. 27 Let μ be an outer measure on a set X and A X a subset with μ (A) = 0. Prove that μ (A B) = μ (B) for any subset B X. Since A B A, 0 μ (A B) μ (A) = 0 = μ (A B) = 0 and μ (A\(A B)) = 0 by the same reason. We have μ (B\(A B)) μ (B). Therefore, using monotonicity and finite sub-additivity, we have: μ (B) μ (A B) μ (A B) + μ (A\(A B)) + μ (B\(A B)) μ (B) and the result follows. Problem. 28 Let X, A, μ) be a measure space and let {E n A} n 1 be a sequence of subsets. Show that μ(lim infe n ) lim n infμ(e n ) and μ(lim n sup E n ) lim n sup μ(e n ) provided that μ( i=n E i) < for some n 1. 12

13 Set L = lim inf n E n = n=1 m=n E m. Then L = n=1 B n, B n = m=ne m L. We have L = B 1 n=1 (B n+1\b n ), a disjoint union. We get μ(l) = lim n μ(b n ) = sup n N μ(b n ) sup n N inf k n μ(e k ) = lim n infμ(e n ). Set L = lim sup n E n. Then L = n=1 B n, where B n = m=ne m L. Write B 1 = n=1 (B k \B k+1 ) L, which is a disjoint union, and we have: μ(b 1 ) = lim n μ(b n ) + μ(b 1 ) + μ(l) and since μ(b 1 ) <, we have μ(l) = lim n μ(b n ) = inf n N μ(b n ) inf n N sup m n μ(e m ) = lim sup n μ(e n ). Problem. 29 Let B be the σ- algebra of Borel sets in X = [0, 1]. Show that 1. B = B(E) where E is the collection of all open intervals (a, b) X and that B = B(F) where F is the collection of all closed intervals [a, b] X. 2. Every half -open interval [a, b) X is an F σ set as well as a G δ set. 3. Let E be a collection of the subsets of a set X and let A X be a subset. Show that, A(E) A = A(E A) where A(E) and A(E A) are the σ- algebras generated by E and E A respectively. 1. Since every open set is a union of a sequence of open intervals, we have the first part of (1). Since [a, b] = n=1 (a n 1, b] and (a n 1, b] = m=1 (a n 1, b + m 1 ), the second result follows as well. 2. We just note that [a, b) = (, a) c (, b) 3. Since E A(E), we have E A A(E) A. Then from the fact that A(E) is a σ- algebra of subsets of X and A X it follows that A(E) A is a σ- algebra of subsets of A. Thus A(E A) A(E) A. It remains to show that A(E) A A(E A). Problem. 30 Given a measurable set E R n and ε > 0, there is an open set U and a closed set F such that F E U and m(u\f) < ε. Suppose that E R n is measurable. Then m(e) = m (E). We can cover E with collections of open parallelepipeds. Since m(e) is the infimum, for ε > 0, we can find a collection of open parallelepipeds P 1 that covers E such that m( A P1 A\E) < 2 ε. Similarly, for the measurable sets E c we can find a collection of open parallelepipeds P 2 that covers E c such that m( B P2 B\E c ) < 2 ε. Set F = ( B P 2 B) c, which is closed as its complement is a union of open sets. Set U = A P1 A is a union of open sets and so it is open. We have m(f c \E c ) = m(f c E) < 2 ε. We have U\F = (U\E) (E\F) = m(u\f) m((u\e)) + m((e\f)) < ε. 13

14 Problem. 31 Given an example of a sequence {E n } of disjoint subsets of R such that m ( i E i ) < i m (E i ). Let R be the equivalence relation: (a, b) R in [0, 1) if a b is a rational number, which we showed in homework number 1 problem number 6 to be an equivalence relation. Set E = [0, 1). Denote the set of all equivalence classes that have representatives in E as R E. We can choose a representative from each element in R E that lies in E, by Axiom choice. Let F be the set of such representatives. For any y E there are x F and q Q for which y = x + q. Hence, E q Q ((F + q) [0, 1)). Set E n = (F + q n ) [0, 1), which is countable since Q is. m (E n ) = m (F) by translation invariant of measure. m (F) = 0 leads to a contradiction 1 0 as seen below. Thus m (F) is strictly positive and the sum is infinite. We get 1 = m (E) < m ( q Q ((F + q) [0, 1))) = n=1 m (E n ) =. 14

15 Problem. 32 Let (X, A) be a measurable space. Let also μ : A [0, ) be a finitely additive set function. Show that μ is a measure if and only if whenever {A k A} k=1 is an increasing sequence of sets, then μ( k=1 A k ) = lim k μ(a k ). Suppose that {A k A} k=1 is an increasing sequence of sets. Write B n = A n \A n 1, where A 0 = (note that for each A A, we have μ(a) [0, ) so that μ(a ) = μ(a) + μ( ) = μ( ) = 0). Then B n are pairwise disjoint and n=1 B n = n=1 A n. Suppose that μ is a measure. Then μ( n=1 A n) = μ( n=1 B n) = n=1 μ(b n) = lim N N [ n=1 μ(an ) μ(a n 1 ) ] = lim n μ(a n ). Suppose that {A k A} k=1 is a sequence of sets that are disjoint. Then set B n = n k=1 A k. Then B n is a sequence of increasing sets. Then by the hypothesis and using the fact that μ is finitely additive in the second to the final equation: μ( k=1 A k ) = μ( n=1 B n) = lim n μ(b n ) = lim n μ( n k=1 A k ) = lim n n k=1 μ(a k ) = lim n μ(a n ). Thus, μ is a measure. Problem. 33 Let (X, A) be a measurable space. Let also μ and ν be measures on A. Show that the set function λ on A defined by λ(e) = μ(e) + ν(e) is a measure. We have λ( ) = μ( ) + ν( ) = = 0. Suppose {A n A} n=1 is a disjoint sequence sets. Then λ( n=1 A n) = μ( n=1 A n) + ν( n=1 A n) = n=1 μ(a n) + n=1 ν(a n) = n=1 [ μ(an ) + μ(a n ) ] = n=1 λ(a n). Problem. 34 Let (X, A) be a measurable space. Show that if E 1, E 2 A and μ(e 1 ΔE 2 ) = 0, then μ(e 1 ) = μ(e 2 ). Since E 2 \E 1 E 1 ΔE 2, then by monotonicity, we have 0 μ(e 2 \E 1 ) μ(e 1 ΔE 2 ) = 0 = μ(e 2 \E 1 ) = 0. We have E 2 = E 2 \E 1 E 1 is a disjoint union and hence we have: μ(e 2 ) = μ(e 2 \E 1 ) + μ(e 1 ) = 0 + μ(e 1 ) = μ(e 1 ). 15

16 Problem. 35 On the collection E = 2 R of all subsets of R define the set function μ : E R by setting μ(a) to be the number of integers in the set A. Determine the outer measure μ induced by μ and the σ - algebra of measurable sets. μ (A) = inf{ k=1 μ(a k ) : A } k=1 A k, A k E. On taking A 1 = A and A k = for all k > 1, we get μ(a) = μ(a). Problem. 36 Let X be a set. If we start with an outer measure μ on 2 X and form the induced measure μ on the σ-algebra of μ -measurable subsets, we can view μ as a set function and denote by μ + the outer measure induced by μ. Show that μ + (E) μ (E) for any E X. We have μ (E) μ(e) and since μ + (E) is the greatest lowest bound for μ(e), we must have μ + (E) μ (E). Problem. 37 Let E be the class of all open sub-intervals (a, b) (0, 2 1 ) and let μ be a set function. Determine the outer measure μ, the σ - algebra of measurable sets and the induced measure μ assuming that μ is given by the formula: 1. μ((a, b)) = (b a) 2 ; 2. μ((a, b)) = (b a); 1. In order for μ to be an outer measure, it must be monotone and countable subadditive, and μ ( ) = 0. The first two properties do not rely on the particular form of μ and hence they continue to hold. However, μ ( ) = 0 depends on the particular form of μ. In order for μ ( ) = 0 to hold we must be able to find sets (c, d) for which μ(c, d) is arbitrarily small. Since = (a, a) E, we must have μ(a, a) = 0, but this is true by our definition. Hence, the set function μ((a, b)) = (b a) 2 defines an outer measure m, which unfortunately does not agree with μ on the sets of E: Suppose that b = a. Note that (b a) 2 = [(c a) + (b c)] 2 = (c a) 2 + 2(c a)(b c) + (b c) 2 and if m (a, b) = m (a, c) + m (c, b) and m (a, b) = μ(a, b), then 2(c a)(b c) = 0 so that we have (b c) 2 = 0 and b = c giving us m (a, b) = 0, a contradiction. 2. In order for μ to be an outer measure, it must be monotone and countable subadditive, and μ ( ) = 0. The first two properties do not rely on the particular form of μ and hence they continue to hold. However, μ ( ) = 0 depends on the particular form of μ. In order for μ ( ) = 0 to hold we must be able to find sets (c, d) for which μ(c, d) is arbitrarily small. Since = (a, a) E, we must have μ(a, a) = 0, but this is true by our definition. Hence, the set function μ((a, b)) = (b a) defines an outer 16

17 measure m. But the intervals are not measurable with respect to this outer measure since μ ((a, b)) = (b a) = b c) + (c a) = μ ((a, c)) + μ ((c, b)). Problem. 38 Let C be a Cantor-like set, that is the result of a Cantor-like construction, which starts with k disjoint open intervals Δ i, i = 1, 2,..., k of the unit interval. Given n - tuple (i 1,..., i n ), where i j = 1,..., k, let Δ i1...i n be the corresponding interval on the step n of the construction and denote by C i1,...,i n = Δ i1...i n C. Let E be the collection of all sets C i1,...,i n over all n and all n-tuples. Given a probability vector P = {p 1,..., p k }, where p i 0 and k j=1 p j = 1, define a set function μ on E by the formula μ(c i 1,...,i n ) = k j=1 p j. If μ is the outer measure generated by μ and μ is the induced measure, show that every set C i1,...,i n is measurable and its μ-measure is equal to its μ-measure. We show first that μ is finitely additive on the class of cylinders. We have k i=1 μ(c ω 1,...,ω n i) = k i=1 p ω 1 p ω2...p ωn p i = p ω1 p ω2...p ωn k i=1 p i = μ(c ω 1,...,ω n ). We get: (ω1,...ω n ) μ(c ω 1,...,ω n ) = (ω1,...ω n 1 ) μ(c ω 1,...,ω n 1 1) + μ(c ω1,...,ω n 1 2) μ(c ω1,...,ω n 1 k ) = (ω1,...ω n 1 ) μ(c ω 1,...,ω n 1 ) k i=1 p i = ω 1 μ(c ω1 ). We therefore have additivity. One way of defining topology on the space of two-sided sequences of N symbols is to declare that all cylinders are open sets. Then every cylinder is also closed because the complement to a cylinder is a finite union of cylinders. Therefore the collection E is closed under differences and therefore, we we have the result by a theorem proven in class. Problem. 39 Show that for given a measurable set A (0, 1) of positive Lebesgue measure (μ(a) > 0), there is a non-measurable subset E A. Define the relation x y on R to mean that x y Q. This relation defines an equivalence relation on R and the equivalence classes are of the form {x + r : r Q}. Each equivalence class contains a point in the interval A [0, 1]. Let E A be the set that consists of exactly one point from each equivalence class (the choice here is possible by the Axiom of Choice). Let {r i : i = 1, 2, 3,...} be an enumeration of the countable set of all rational numbers in the closed interval [ 1, 1], and let E i = E + r i. Clearly, E i = E + r i [ 1, 2]. Choose x [0, 1], then since each equivalence class contains a point in the interval [0, 1], there exists a point y E such x y Q. So x = y + r j E j so that [0, 1] i=1 E i [ 1, 2]. Note also that E i E j = whenever i = j. For otherwise, there would exist y and z in E such that y + r i = z + r j, which would imply that y z, a contradiction. Now, suppose E is measurable then each E i is measurable and by translation - invariance we have μ(e) = μ(e i ) so we have [0, 1] i=1 E i [ 1, 2] = μ([0, 1]) i=1 μ(e i) μ([ 1, 2]) = 1 i=1 μ(e) 3, a contradiction. Hence, E is not measurable. 17

18 Problem. 40 Let X = [0, 1] and x 1,..., x n [0, 1] be a finite collection of points. Define a set function μ : X [0, ] such that μ (E), E X, is the number of points in the collection that lie in E. Is μ an outer measure? We have to show that: 1. We must show that μ is non-negative. We have this by definition. 2. We must show that μ ( ) = 0. Since μ ( ) counts the the number of points in the collection that lie in and there is no collection of points in, we have μ ( ) = Suppose that A B. Then every collection of points {x i } that lies in A lies also in B = A B\A. Hence, we must have μ (A) μ (B). Let E k be a countable collection of subsets of X. Consider E = k=1 E k. Let {x k } be a collection of points lying in E. Then there exists n such that {x k } belongs to E n. If the number of points in {x k } is infinite, then we have infinity on both sides of mu (E) k=1 μ (E k ).( ) The only time we have the left hand side finite is when the collection {E n : n N}contains only finitely many non-empty sets and each of the non-empty set is finite, but then the ( ) holds because the cardinality of A B is always less than or equal the sum of the cardinalities of A and B with equality when both sets are disjoint. Problem. 41 Let X = N and a 0 = 0, a 1 = 2 1 < a 2 <... <.. with lim n a n = 1. Set μ (E) = a n, if E is finite and has n-elements, and μ (E) = 1 if E is infinite. Show that μ is an outer measure. Note that μ (E) 0 for all E by definition. Since the cardinality of is 0, we have μ (E) = a 0 = 0. Suppose that A B, then if A is infinite, then B is infinite and we have μ (A) = 1 = μ (B). If A is finite then the cardinality of A, say n is less than or equal the cardinality of B, say m or infinity. Since a n is an increasing sequence converging to sup a n = 1, we have μ (A) μ (B). Take a collection E i, then E = i=1 E i. Then if the collection contains infinitely many non-empty sets, E is infinite and then μ (E) = 1 and also i=1 μ (E i ) =. E is only finite if there are only finitely many non-empty sets E i and all of whose are finite, but then the cardinality of E is less than the sum of the cardinalities of the sets E 1,...E N making up E. Since a n < a m whenever n < m, we must have μ (E) N i=1 μ (E i ). Problem. 42 Show that the set of those points in the closed unit interval whose binary expansion has 0 in the even places is Lebesgue measurable of measure zero. 18

19 Let E be the collection of all such points in [0, 1] with binary expansion has 0 in the even places. We will construct a cantor-like set F whose Lebesgue measure is zero and which contains E and then by the completeness of the Lebesgue measure we will have μ(e) = 0 and every set of measure zero is measurable by a simple Caratheodory Criterion examination. Our main task is to construct F by deleting intervals that do not contain the digits of the binary expansion type in question. } Divide I 2 0 = [0, 1] into four equal parts: {[0, 1 4 ], [1 4, 2 4 ], [2 4, 3 4 ], [3 4, 4 4 ]. Set I 2 = [0, 4 1] [2 4, 3 4 ] and do likewise to get I 2 2 = [0, ] [ 2 4 2, ] [ 8 4 2, ] [10 4 2, ] and so on and so forth. Let J k be the union of k open intervals deleted in this construction, that is, J k = 2k 1 l=1 I l.k, where I l,k are open disjoint open intervals of lengths 4 k. Now let F = [0, 1]\ k N J k. We have μ([0, 1]\F) = k μ(j k ) = 1 so that μ(f) = 0 and since E F, the result follows as noted in the first paragraph above. Problem. 43 For any set A N, let μ(a) = n A 2 n if the set A is finite and μ(e) = otherwise. Show that μ is an additive set function but not a measure on 2 N. μ( ) is not defined and so μ is not a measure. If A B = and neither A nor B is infinite, then μ(a B) = n A B 2 n = k A,k/ B 2 k + m A,m/ B 2 m = μ(a) + μ(b). If A or B is infinite, then μ(a B) = and μ(a) = or μ(b) = and we have the result. Problem. 44 Give an example of a nowhere dense compact subset of [0, 1] which has positive Lebesgue measure. Let ε > 0 be given. Let (a m, b m ) be an open interval in [0, 1] with rational endpoints. For each m N, let (c m, d m ) be an open interval in (a m, b m ) of length at most 2 m ε. Set E = [0, 1]\ m N (c m, d m ). Then E is closed because its complement is open (and since a closed subset of a compact set is compact E is also compact) and and it is nowhere dense because all the rational numbers are deleted and finally, the length of E is at least 1 ε m=1 (1 2 )m = 1 ε, which is positive and since the Lebesgue measure is defined to be the length, we are done. Problem. 45 Let D be a dense subset of real numbers and f a real-valued function on R such that {x : f(x) > a} is measurable for every a D. Show that f is measurable. Choose c R. Then by the density of D in R, there exists a sequence {a n } D such that lim n a n = c, which means that for each k N, there exists N k such that c + 1 k > a N k. 19

20 We have {x : f(x) > c} = k N {x : f(x) c + 1 k } = k N {x : f(x) > a N k } and the righthand side is measurable by the hypothesis and so the left hand side is also measurable and hence f is measurable. Problem. 46 Let D, E R n be measurable sets and f a real-valued function on D E. Show that f is measurable if and only if the restrictions f D and f E are measurable functions. If f is measurable on D E, then choose c R and {x D E : f(x) < c} is measurable and since E and D are measurable we have {x D E : f(x) < c} D = {x D : f(x) < c} and {x D E : f(x) < c} E = {x E : f(x) < c} are measurable and so f E and f D are measurable. Suppose that f E and f D are measurable, then for each c R, the two sets {x E : f(x) < c} and {x D : f(x) < c} are measurable and so is their union and hence {x D E : f(x) < c} = {x D : f(x) < c} {x E : f(x) < c} and it follows that f is measurable on D E. Problem Show that if f is measurable real-valued function and g a continuous function on R then the function g f is measurable. 2. Give an example of a continuous function g and a measurable function f such that f g is not measurable. 1. Let O be an open subset of R, then g 1 (O) = U is open set by the continuity of g and by the measurability of f, f 1 (U) is measurable. Hence h = g f is measurable by Proposition 2 of chapter 3 of the textbook since h 1 (O) = f 1 (g 1 (O)) = f 1 (U) is measurable. 2. Let f : [0, 1] [0, 1] be the Cantor-Lebesgue function. This is a monotonic and continuous function, and the image f(c) of the Cantor set C is all of [0, 1]. Define g(x) = x + f(x) as done in the class. Then g : [0, 1] [0, 2] is a strictly monotonic and continuous map, so its inverse h = g 1 is continuous, too. Observe that g(c) measure one in [0, 2]: this is because f is constant on every interval in the complement of C, so g maps such an interval to an interval of the same length. It follows that there is a non-lebesgue measurable subset A of g(c) (Vitali s theorem: a subset of R is a Lebesgue null set if and only if all its subsets are Lebesgue measurable). Put B = g 1 (A) C. Then B is a Lebesgue measurable set as a subset of the Lebesgue null set C, so the characteristic function 1 B of B is Lebesgue measurable. 20

21 The function k = 1 B h is the composition of the Lebesgue measurable function 1 B and and the continuous function h, but k is not Lebesgue measurable, since k 1 (1) = (1 B h) 1 (1) = h 1 (B) = g(b) = A. Problem. 48 If f and g are real-valued (Borel) measurable functions, show that the set {x : f(x) > g(x)} is (Borel) measurable. By the density of Q, there exists r n Q such that f(x) > r n > g(x) for some n Q. Hence, {x : f(x) > g(x)} = n N {x : f(x) > r n } {x : g(x) < r n } and the measurability of the set in question follows from that of f and g since the right hand side is measurable as it is a union of measurable sets. Problem. 49 measurable. If f 0, show that f is measurable if for some p > 0 the function f p is Since f 0 and p > 0, we can legally write h(x) 1 p = f(x). Choose c R, then {x : f(x) > c} = {x : h(x) 1 p > c} = {x : h(x) > c p } and the latter is measurable so that f is measurable. Problem. 50 Let f be a non-negative measurable function. Show that there is an increasing sequence φ n of non-negative simple functions each of which vanishes outside a set of finite measure such that f = lim n φ n. Note that the {[ range of ) f is a subset of [0, ). Partition the range so that we have J n = [n, ) and I n,k = k 1 2 n, 2 k n : k = 1, 2,..., n2 n}. Let E n,k = f 1 (I n,k ) and F n = f 1 (J n ). Define φ n (x) = n2n k=1 k 1 2 n χ En,k (x) + n χ Fn (x). From the definition we have 0 φ n (x) f(x) and φ n (x) are measurable simple functions. Also, since the dissection of the range of f giving φ n+1 (x) is basically a refinement of that giving φ n (x) we have φ n+1 (x) φ n (x) for all x and for all n. If f is finite, then x F c n for all n large enough and then f(x) φ n 2 n. So φ n (x) f(x). If f(x) =, then x n=1 F n, so φ n (x) = n for all n and again φ n (x) f(x). Problem. 51 Let f be a measurable function on a measure space (X, A, μ). Show that 1. if f 0, then f = 0 if and only if f = 0 a.e; 2. if E A and μ(e) = 0, then E f = 0; 3. if f > 0 a. e, E A and E f = 0, then μ(e) = 0; 4. if E f = 0 for any E A, then f = 0. a. e; 21

22 1. Set E n = {x : f(x) 1 n }, then we have E = {x : f(x) > 0} = n NE n. Since E n E, we have μ(e) = lim n μ(e n ). We have for all n, 0 = E n f 1 n μ(e n) = μ(e n ) = 0 = 0 = lim n μ(e n ) = μ(e) = f = 0 a. e. Write E = {x : f(x) > 0}. Then assuming that f = 0 a.e, then μ(e) = 0. Since f is non-negative, then there exists an increasing sequence of measurable simple functions φ n (x) f(x) on E. Then writing E = m i=1 E i E, then by the regularity of the Lebesgue measure we have μ(e i ) μ(e) = 0 and we get E fdμ = lim n E φ ndμ = lim n m i=1 a i μ(e i ) = If f is non-negative, then there exists an increasing sequence of measurable simple functions φ n (x) f(x) on E. Then writing E = m i=1 E i E, then by the regularity of the Lebesgue measure we have μ(e i ) μ(e) = 0 and we get E fdμ = lim n E φ ndμ = lim n m i=1 a i μ(e i ) = 0. For the general f, we have apply our result to f + and f and we have the result. 3. Write E 0 = {x : f(x) = 0}. Then μ(e 0 ) > 0 contradicts f > 0 a. e. Define E n = {x : f(x) 1 n }. Then E 0 E n E so that lim n μ(e 0 E n ) = μ(e). We get 0 = E f E n f 1 n μ(e n) = μ(e n ) = 0 for all n and we therefore have 0 = lim n μ(e n ) = μ(e). 4. The sets E n = {x : f(x) 1 n } are increasing. We have E 0 = {x : f(x) > 0} = n=1 E n. Thus if 0 < μ(e 0 ) = lim n μ(e n ), then there exists n 0 such that μ(e n0 ) > 0, Then 0 = E n0 fdμ 1 n 0 μ(e n0 ) > 0, a contradiction. Similar if we consider F 0 = {x : f(x) < 0} and assume the μ(f 0 ) > 0, then we get a contradiction. Problem. 52 Let f be a non-negative measurable function. Show that F(x) = x fdμ is continuous in x. Fix x R. Let {x n } be an increasing sequence so that x n x from the left as n. Write f n (t) = χ (,xn ) (t)f(t). Then f n(t) is a non-negative measurable increasing sequence (because f is non-negative) converging to χ (,x) (t)f(t). Hence by the Monotone Convergence Theorem, we can swap the integral sign with the limit sign: lim n F(x n ) = lim n fn (t)dμ = χ (,x) fdμ = F(x). Similarly, the same argument with a decreasing sequence approaching x gives the other direction. Problem. 53 Construct an example of a sequence of measurable functions f n, which converge to a function f a.e on a set E, and for which E f lim n i.e., there is strong inequality in Fatou s Lemma. 22 E f n(x)

23 Let f n (x) = nχ (0, 1 n ) (x). Then E fdμ = 1 and lim n f n(x) = 0 Problem. 54 Let f be an integrable function on (, ). Show that for any t R, f(x)dx = f(x + t)dx. Write f = f + f. Then since both f + and f are non-negative measurable functions they can be approximated by increasing sequences of simple functions vanishing outside sets of non-zero measures as it was proved in homework number 7. Hence, it suffices to show the result for simple functions. Write φ(x) = n i=1 a iχ Ai (x). Then φ(x)dx = n i=1 a i χai (x) = n i=1 a iμ(a i ) = n i=1 a iμ(a i t) = n i=1 a iχ Ai t(x)dx = n i=1 a iχ Ai (x + t)dx = φ(x + t)dx. 23

24 Problem. 55 Show that if f n is a sequence of measurable functions that converges to f in measure then each subsequence f nk converges to f in measure. Suppose that f n converges to f in measure on X. Then for each ε > 0 and λ > 0, there exists N ε,λ N such that ( ) μ({x X : f n (x) f(x) ε}) λ for all n N ε,λ. For each subsequence {f nk }, there exists K > 0 such that n k > N ε,λ for all k > K. Thus for any k > K, the inequality ( ) holds. Problem. { 56 Let {f n } be a sequence of real-valued functions on [0, 1] given by 1, if x [ k f n (x) = 2 m, (k+1) 2 m ] 0, if x / [ 2 k m, (k+1) 2 m ] where the integers k and m are such that n = k + 2 m and 0 k < 2 m (note that for every positive integer n the integers k and m are uniquely defined). Show that the sequence {f n } converges to zero in measure but for every x [0, 1] the sequence {f n (x)} does not converge. For each ε (0, 1] and f = 0, we have lim n μ L({x [0, 1] : f n f ε}) = lim n μ L([ k (k + 1) 1 2m, 2m ]) = lim n 2 m = lim n n k = 0. This shows that f n converges to f = 0 in measure. However, the sequence does converge to two distinct functions f(x) = 0 and g(x) = 0 and so it does not converge as this contradicts the uniqueness of the limit. Problem. 57 Let μ be a signed measure and let f be a real-valued function which is integrable with respect to the total variation μ. Show that f is integrable with respect to the measures μ + and μ and that fdμ = fdμ + fdμ. Note that if f(x) = χ E (x), then we have fd μ = μ (E) = μ + (E) + μ (E) = fdμ + + fdμ so that the integrability of f with respect to the total variation implies the integrability of f with respect to the measures μ + and μ. By the linearity of the integral, it follows that the equality holds when f is nonnegative simple function. If f is a nonnegative real-valued then we can approximate it by a sequence of increasing simple functions according to problem number 6 of homework number 7. In generality, we can write f = f + f and each of the terms in the sum is a non-negative real-valued function and so we have the result. We have: fdμ = fd( u + ( 2μ )) = fd μ + d( 2u ) = fd μ 2 fdμ = fdμ + fdμ. 24