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Problem Set 6 Richard Bamler Jeremy Leach office 382-N, phone: 650-723-2975 office 381-J [email protected] [email protected] office hours: office hours: Mon 1:00pm-3pm, Thu 1:00pm-2:00pm Thu 3:45pm-6:45pm MATH 172: Lebesgue Integration and Fourier Analysis (winter 2012) Problem set 6 due Wed, 2/22 in class (1) (20 points) Let X be an arbitrary space, M a σ-algebra on X and λ a measure on M. Consider an M-measurable function ρ : X ! [0; 1] which only takes non-negative values. (a) Show that the function, Z µ : M! [0; 1];A 7! (ρχA)dλ is a measure on M. We will also denote this measure by µ = (ρλ). (b) Show that any M-measurable function f : X ! R is integrable with respect to µ if and only if fρ is integrable with respect to λ. In this case we have Z Z Z fdµ = fd(ρλ) = fρdλ. (Hint: Show first that this is true for simple functions, then for non-negative functions, then for general functions). (c) Show that if A 2 M is a nullset with respect to λ, then it is also a nullset with respect to µ = ρλ. (Remark: A measure µ with this property is called absolutely continuous with respect to λ in symbols µ λ.) (d) Assume that X = Rn, M = L and λ is the Lebesgue measure. Give an example for a measure µ0 on M such that there is no non-negative, M- measurable function ρ : X ! [0; 1] with µ0 = ρλ. (2) (20 points) This is a continuation from problem 3 on problem set 5. (a) Let f :[a; b] ! R be a Riemann integrable function on the interval [a; b]. Show that f is integrable in the Lebesgue sense and that Z b Z f(x)dx = fdλ. a [a;b] (b) Let I ⊂ R be an interval (not necessarily closed or bounded) and f : I ! R a function such that for any compact subinterval [a; b] ⊂ I, the restriction fj[a;b] is Riemann integrable. Assume moreover, that the function f itself is 1 2 integrable with respect to the Lebesgue measure. Show that the improper Riemann integral of f over I exists and is equal to the Lebesgue integral of f, i.e. Z Z bk Z f(x)dx = lim f(x)dx = fdλ k!1 I ak I for any sequence [a1; b1] ⊂ [a2; b2] ⊂ ::: of closed intervals with I = S1 k=1[ak; bk]. (c) Let I ⊂ R be again an arbitrary interval and let f : I ! R be a function for which the improper Riemann integral exists and is finite but for which R R R one of the quantities I jfjdλ, I f+dλ, I f−dλ is infinite. Show that then all the other quantities are infinite. (d) Give an example for a continuous function f : (0; 1) ! R for which the R 1 improper Riemann integral 0 f(x)dx exists and is finite, but which is not integrable with respect to the Lebesgue measure. (3) (10 points) (a) Give an example for a sequence of nonnegative continuous functions fk : R 1 [0; 1] ! [0; 1) such that limk!1 fk(x) = 0 for all x 2 [0; 1], but 0 fk(x)dx goes to 1 as k ! 1. (k) (b) Use Fatou's Lemma to show the following statement: Let (ai )i2N be a sequence of sequences of non-negative real numbers, i.e. for every k 2 N (k) (k) there is a sequence a1 ; a2 ;:::. Then 1 1 X (k) X (k) lim inf ai ≤ lim inf ai : k!1 k!1 i=1 i=1 Give an example for which the inequality is strict. (c) Use Lebesgue's Dominated Convergence Theorem to show the following (k) statement: Let (ai )i2N be a sequence of sequences of (general) real numbers and assume that there is a sequence (bi)i2N of non-negative real numbers (k) P1 such that jai j ≤ bi for all i; k 2 N and i=1 bi < 1. Moreover, assume (k) that limk!1 ai exists of every i 2 N. Then 1 1 X (k) X (k) lim ai = lim ai : k!1 k!1 i=1 i=1 (4) (30 points) P1 (a) Let a1; a2;::: 2 R be a sequence of real numbers such that k=1(ak)− = P1 k=1(ak)+ = 1, but jakj ! 0 as k ! 1. Show that for any z 2 [−∞; 1], 0 0 we can create a sequence a1; a2;::: 2 R by rearranging a1; a2;::: such that 1 X 0 ak = z: k=1 (Remark: By rearranging, we mean that there is a bijective map f : N ! N 0 such that ak = af(k).) sin x (b) Consider the function f : R ! R, f(x) = x . Prove that for every number z 2 [−∞; 1] there is a sequence of bounded Lebesgue sets Ak ⊂ R which 3 S1 is increasing (i.e. A1 ⊂ A2 ⊂ :::) and which satisfies k=1 Ak = R such that fχAk is integrable for all k and we have Z lim fχA = z: k!1 k (c) Let f : R ! [0; 1) be an L-measurable function which takes only non- negative and finite values. Show that for any b > 0 there is a Lebesgue set B ⊂ R such that f is bounded on B and λ(B) > b. (d) Prove that the conclusion of (b) holds for any non-integrable, L-measurable R R function f : R ! R for which f+dλ = f−dλ = 1. (Hint: Use (a) and (c)). (e) Given any z 2 R, show that there is a sequence fk : R ! R of continuous sin x functions such that limk!1 fk(x) = x for all x 2 R and such that Z 1 lim fk(x)dx = z: k!1 1 Maximum total points: 80.
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