Homework SetsMath 4121, Spring ’06, I. Krishtal WARNING: “Solutions” given here are, in general, not complete. Some of them would be acceptable if you encountered similar problems in another (more advanced) course but for this course YOUR solutions should be more detailed. If you are using these to prepare for the test it would be a good idea to check that you can supply all the tiny details which I did not spell out here. Solutions below is a Guide, not a Bible. 1 M4121 Homework 10, Due 4/22/2006 1 Homework 10 This is the last homework in the course and it contains miscellanious exercises referring to what you should already know as well as to the material I will be covering in the remaining 6 lectures. This homework is harder than most of the ones you had before, hence, the deadline is extended until the end of Saturday, 4/22. I will leave an envelope on my office door where it should be put. Problems for all. 1. Consider f : R → R defined by f (x) = x−1/2 χ(0,1)(x). Let {rn }∞ n=1 be a fixed enumeration of the rationals Q and F (x) = ∞ X 2−n f (x − rn ). n=1 Prove that F is integrable, hence the series defining F converges for almost every x ∈ R. Show that, nevertheless, any F˜ such that F˜ = F a.e. is unbounded in any interval. Solution. By Monotone Convergence Theorem Z F dx = ∞ Z X n=1 2 −n f (x − rn )dx = ∞ Z X n=1 2 −n f (x)dx = ∞ X n=1 2 −n Z 1 0 dx √ = 2. x Let I ⊂ R be an arbitrary open interval and n ∈ N be such that rn ∈ I. For any M > 0 we have µ({x ∈ I : F (x) > M }) ≥ µ({x ∈ I : f (x − rn ) > 2n M }) > 0. 2. Let f, g be integrable functions on Rd . Prove that f (x − y)g(y) is measurable and integrable on R2d. Solution. By 10.H, F (x, y) = f (x − y) is measurable. The function G(x, y) = g(y) is measurable, becuase measurability of E ⊆ R implies measurability of R × E ⊆ R2d. Hence, H(x, y) = f (x − y)g(y) is measurable (see p.13 in Bartle). By Tonelli, Z Z Z |f (x − y)g(y)|dµ(x, y) = |g(y)| |f (x − y)|dx dy = ||f ||1||g||1. 3. Let F (x) = x2 sin(1/x2), x 6= 0, and F (0) = 0. Show that F 0 (x) exists for every x ∈ R but F 0 is not integrable on [−1, 1]. Solution. Direct computation shows that 1 cos( x12 ), x 6= 0; 2x sin( x12 ) − 2x 0 F (x) = 0, x = 0. One way to show that F 0 is not integrable is to prove that F does not have bounded variation. Alternatively, since the measure of the set {x ∈ (0, ) : | cos( x12 )| > 12 } is R1 R 1 bigger than 3 for every > 0 we have that −1 |F 0 (x)|dx ≥ 0 2x | cos( x12 )|dx = +∞. x χ (t)dt > 0 because K is open and C contains no intervals. i. {x ∈ (a. b) : ∃y ∈ (x.M4121 Homework 10. The sets in the curly brackets are open because whenever x is in such a set there will be an interval around it contained in that set.. b] → R such that F 0 (x) = 0 on a set of positive measure.h] t∈(0. Give an example of an absolutely continuous strictly increasing function F : [a.e. Since F is continuous. lim h→0+ F (x + t) − F (x) F (x + t) − F (x) sup = lim . K y 0 F = χK vanishes on a set of positive measure. Let F be continuous on [a. Why can one restrict to countably many h in the lim sup above? Solution. Due 4/22/2006 2 4. if x > y. for any α ∈ R. Let C be a fat Cantor set (see one of the extra problems before) R x and K be the complement of C in [a. 5. Show that D+ (F )(x) = lim sup h→0+ F (x + h) − F (x) F (x + t) − F (x) = lim sup + h t h→0 t∈(0. b]. a countable intersection of open sets. b) : D+ (F )(x) > α} = \ h∈Q+ \ F (x + t) − F (x) > α} = t t∈(0. F is absolutely continuous (as an integral) and. Define F : [a. x + h] such that F (y) − F (x) > α(y − x)} = h∈Q+ \ {x ∈ (a. h∈Q+ This set is Gδ .h] is measurable. You will need a “fat Cantor set” for that. b] → R by F (x) = a χK (t)dt. h∈Q t∈(0. it is measurable. and. then F (x) − F (y) = RClearly.h) sup Hence. b) : sup {x ∈ (a.h] {x ∈ (a. x + h]}. . + t t h→0 . b) : x is invisible from the right for F (·) − α on [x. b]. On the other hand. Solution. hence. if you want). y) = χ{0} (x − y).M4121 Homework 9. y) = x−y). 10. prove the first equality in 10.O After correcting the obvious typo.D. The integrability condition in this case would be the absolute convergence of the double sum. S from Bartle’s book. although. This follows from (A1 × B1 )\(A2 × B2 ) = ((A1 ∩ A2 ) ∪ (A1 \A2)) × ((B1 ∩ B2 ) ∪ (B1 \B2 )) \ n ((A1 ∩ A2 ) ∪ (A2 \A1)) × ((B1 ∩ B2 ) ∪ (B2 \B1 )) = (A1 ∩ A2 ) × (B1 ∩ B2 ) ∪ o (A1 ∩ A2 ) × (B1 \B2 ) ∪ (A1 \A2) × (B1 ∩ B2 ) ∪ (A1 \A2) × (B1 \B2 ) \ n (A1 ∩ A2) × (B1 ∩ B2 ) ∪ (A1 ∩ A2 ) × (B2 \B1 ) ∪ (A2 \A1) × (B1 ∩ B2 ) o ∪ (A2\A1 ) × (B2 \B1 ) = (A1 ∩ A2 ) × (B1 \B2 ) ∪ (A1 \A2) × B1 . Since ν(Dx ) ≡ 1 and µ(Dy ) ≡ 0. This implies the two equalities in 10. Since the Borel σ-algebra is certainly smaller than the σ-algebra in question. As soon as you get this picture all the verifications are trivial. Problems for all. we have that D is measurable. solve 10.2 and solve problems (10. (amn ) form an infinite matrix with +1 on the main diagonal and −1 on the diagonal above. D ∈ B × B. Due 4/13/2006 3 Homework 9 Again. you take ψ = χE and f (x.R (for the first part. 10. I strongly recommend to you to look at ALL the problems after chapter 10. Fill in all the details in the proof of Lemma 10. which obviously fails.H Follows immediately from 2. we have Z Z 1 = ν(Dx)dµ 6= µ(Dy )dν = 0. Therefore.E. 10. 10.)H.S Here you were supposed to figure out that π(E) = 0 if and only if E is contained in a countable union of lines parallel to coordinate axes. while ρ(E) = 0 if and only if E is contained in a countable union of lines parallel either to one of the coordinate axes or the line x + y = 0. I want you to submit only the problems below (+ the extra. Solutions. O. apply De Morgan’s laws in an obvious way and supply the necessary words which would make the exposition self-contained. . Finally. 10.O. K. This follows from (A1 × B1 ) ∪ (A2 × B2 ) = ((A1 ∩ A2 ) ∪ (A1 \A2)) × ((B1 ∩ B2 ) ∪ (B1 \B2 )) ∪ ((A1 ∩ A2 ) ∪ (A2\A1 )) × ((B1 ∩ B2 ) ∪ (B2 \B1 )) = (A1 ∩ A2 ) × (B1 ∩ B2 ) ∪ (A1 ∩ A2 ) × (B1 \B2 ) ∪ (A1\A2 ) × (B1 ∩ B2 ) ∪ (A1\A2 ) × (B1 \B2 ) ∪ (A1 ∩ A2 ) × (B2 \B1 ) ∪ (A2\A1 ) × (B1 ∩ B2 ) ∪ (A2\A1 ) × (B2 \B1 ) . every row in the matrix and every column but first sum to 0.K Since χD (x.2 First. Next. The sum of the elements in the first column is 1. Solution. −1 Hence. Come up with two more. we define f : E = [0.e. The first example should feature a function such that the double integral is infinite but both iterated integrals are finite and equal. y)dy = 0 a. . y) = ∞ X 2n 2 χ(2−n−1 . we define f : E = [−1.2−n+1 ] (y). y)dx dy = 0.2−n ] (y)− n=0 22n−1 χ(2−n−1 . Z Z 1 Z 2π Z 1 | cos φ sin φ| dr |f (x. one computes that Z 1 Z 1 f (x. and −1 Z 1 f (x. but 0 ∞ X 0 1 Z 1 Z 0 0 f (x. (x2 + y 2 )2 in which case Z 1 f (x. For the first case. y) = xy .e. In the second example both iterated integrals should be finite but not equal. 1]2 → R by f (x. 1]2 → R by f (x. r E 0 0 0 r For the second case. 10* In the above exercises you have seen a few instances when Tonelli/Fubini fails. this time on a compact set in R2 with the usual Lebesgue mesure. y)dy dx = 1/4. both iterated integrals are equal to 0. Due 4/13/2006 4 Extra problems. y)dx = 0 a. y)|dA ≥ dφ dr = 2 = +∞. n=1 Drawing a picture.2−n ] (x)χ(2−n−1. However..2−n] (x)χ(2−n.M4121 Homework 9. bn]) ≤ l∗(A) + /2. n ∈ N be such that l∗(A) ≤ ∞ X l((an .C.)C. The m=1 ∞ S m. n=1 and the sum is infinite. Solve problems (9.K. 108–112 of Bartle’s book. Set G = ∞ S (an .I. m ∈ N. M. Hence. we get (∗∗) l∗(A) = lim m→∞ m X n l∗ (K1/m ). implies the first of the equalities in 9. Due 4/6/2006 5 Homework 8 Again. n + 1] and An = A ∩ In . 9. a + M ] ⊂ ∞ X S (an . l((an. Problems for all. bn]. The only thing left is to find a compact subset of A of arbitrarily big measure. m ∈ N.n=1 n .H. K1/m . K. n=1 Such a sequence exists almost by definition of l∗. In this exercise.M4121 Homework 8. S on pp. So I provide the solution of all three. let Bn = In \An and Kn = In \G . Taking = 2−|n| /m. Taking = 1/m. although.I. bn + 2−n−2 ) to n=1 get l∗ (A) ≤ l∗(G ) ≤ l∗(A) + . n=−m Since a finite union of compact sets is compact we get the second equality in 9. I strongly recommend to you to look at ALL the problems after chapter 9. I really wanted you to also solve the previous two as well. I. I want you to submit only the problems below (+ the extra. For every M > 0. if you want). bn]) ≥ M. The first assertion follows immediately from (*) and by taking B = second assertion follows immediately from (**) and by taking C = ∞ T G1/m . bn). Now.I. 9. Next. First. m→∞ which together with A ⊆ G1/m . Solutions. 9. This is easily done by taking a sequence of sets of finite measure increasing to A and using what we already proved. we get (∗) l∗(A) = lim l∗(G1/m). let l∗(A) < +∞ and En = [an . we have (a. where In = [n. observe that if l∗(A) = +∞ there is almost nothing to prove. m ∈ N. Then Kn ⊆ Bn implies l∗(Kn ) ≤ l∗ (Bn ) and l∗(Bn ) − = 1 − l∗ (An ) − ≤ 1 − l∗ (G) ≤ l∗(Kn ) proves 9. M. . Here is a clever way of proving it. You will have to use right continuity of g in the end. (ii) follows since any compact set in R is contained in a finite union of intervals of finite length.I. Due 4/6/2006 6 9. µ) define λf : R → R by λf (α) = µ{x ∈ X : |f (x)| > α} Show that ||f ||pp = Z +∞ pαp−1 λf (α)dα. We let Eα = {x ∈ X : |f (x)| > α}.M4121 Homework 8. The function λf (α) is very important in analysis because it paves way to the study of weak Lp spaces (see. 9. 0 where we used Tonelli’s theorem.g. the first equality in 9. e. Folland’s book). Follow carefully the proof of lemma 9.. say. (i) follows since any nonempty open set contains an interval. Extra problems.3.S. Z +∞ pα 0 p−1 λf (α)dα = Z +∞ 0 Z pα X p−1 χEα dµdα = Z Z X |f (x)| pαp−1 dαdµ = ||f ||pp. (iii) is obviously valid for intrervals and extends to B via. 0 Solution. 9* For any f ∈ Lp (X. Solutions. Therefore we must have f = lim inf fn a.e.6) Applying the usual Fatou’s lemma we get Z Z Z Z Z f dµ = lim fnkj dµ ≤ lim inf fnkj dµ = lim fnk dµ = lim inf fn dµ. 1+α 1 + |fn (x) − f (x)| This implies that µ(Enα ) = Z dµ ≤ Enα 1+α α Z Enα |fn − f | 1+α dµ ≤ r(fn − f ). N. 7. Therefore. n ∈ N.e.e. α > 0. I want you to submit only the problems below (+ the extra. Notice that for this direction you do not need that µ(X) < +∞. Clearly. Problems for all. it is enough to show that f = lim inf fn and apply the usual Fatou’s lemma. As usually. I strongly recommend to you to look at ALL the problems after chapter 7 and 8. 1 + |fn − f | α which. let Enα = {x ∈ X : |fn (x) − f (x)| ≥ α}. 1+α . yields the first of the required implications.)M. 78 and (8. We still have that fnk converges to f in measure.e. Solve problems (7. but it will do you good to find an error. Let fnkj be a subsequence of the sequence fnk which converges to f a. unique. Let (fn ) be a sequence of functions in M + that converges in measure to f ∈ M + (which means among other things that these R functions areR essentially real valued rather than extended real valued).e. Let fnk be a subsequence of the original sequence such that lim fnk (x) = lim inf fn (x) a. (Theorem 7. there is a subsequence of fnk which converges a.)N.e. Q on p.M4121 Homework 7. although. you are supposed to do the following. Observe that for x ∈ Enα α |fn (x) − f (x)| ≤ ≤ 1. limit is a. V on pp. 94–95 of Bartle’s book. Let And now the proof that is supposed to be correct. The following is a WRONG proof.N. if you want). R R Let fnk be a subsequence of the original sequence such that lim fnk dµ = lim inf fn dµ. to f . Show that f dµ ≤ lim inf fn dµ. The opposite implication follows from the following inequalities: Z Z |fn − f | |fn − f | r(fn − f ) = dµ + dµ ≤ 1 + |fn − f | c Enα 1 + |fn − f | Enα Z dµ + Enα Z c Enα α dµ ≤ µ(Enα ) + αµ(X). But the a. in turn. 7. Due 3/30/2006 7 Homework 7 Again. As I indicated in my e-mail.Q. M4121 Homework 7. In both cases λ is absolutely continuous with respect to µ because µ(E) = 0 implies R P E = ∅. φf dµ for all simple functions φ. G is positive by definition and . 8.M. if we assume λ(E) = f dµ. First of all. G(f ) = 0 iff f = 0 a. By Radon-Nikodym Theorem. R Again in both cases E f dµ = x∈E f (x). G(αf + βg) = αG(f ) + βG(f ) by linearity of the integral. There are a few little details to check here. It remains to apply the monotone convergence theorem. Due 3/30/2006 8 8.V. and. Secondly. This. Next. is a contradiction. hence.N. clearly. we get 0 = λ({x}) = f (x) which yields λ = 0.e. 8. The problem is that µ is not a σ-finite measure. ||G|| > 0. λ(E) = By linearity of an integral we have Z Z χE dλ = φdλ = Z Z χE f dµ. In particular. Z . Z Z 1/2 Z . . 2 . f dλ. ≤ |f |dλ ≤ |f |dν ≤ |f | dν (ν(X))1/2 . . there exists g ∈ L2 (ν) such that Z Z f dλ = f gdν. we get g ≤ 1 ν-a. Z h(1 − g)dλ = Z hgdµ. k = 1. λ(E) = E gdλ + E gdµ and. The rest is in the text. k=1 satisfies n X |f (bk ) − f (ak )| < . λ(E) = E gdν implies g ≥ 0. f ∈ L2 (ν). . E measurable. k=1 Show that every absolutely continuous function is continuous. Provide an example of a uniformly continuous function which is not absolutely continuous. Comment I will probably find time to do this in class. 1] → R is called absolutely continuous if for every > 0 there exists δ > 0 such that any finite collection of mutually disjoint subintervals (ak . Since all simple functions are in L2 we get that the above equality for all h ∈ M + due to Monotone Convergence Theorem. 1].. 2. R RHence. By the “baby” RRT. . Hence. . . Hence. n. taking E = {x : g(x) ≥ 1} or E = {x : g(x) = 1}. for nonnegative h ∈ L2 (ν) we have Z Z Z Z G(h) = hdλ = hgdν = hgdλ + hgdµ and. bk ) ⊂ [0. Extra problems. implies ||G|| < +∞. 8* A function f : [0. consequently.e. forR f = χE . and µ{x : g(x) = 1} = 0. for which n X (bk − ak ) < δ. Then .I.)J.M4121 Homework 6. then the real and imaginary parts.30 am. 5. |=f |}. R Let f dµ = reiθ with r. so are <f and =f . I strongly recommend to you to look at ALL the problems after chapter 5 and 6. θ ∈ R. Assume now that |f | is integrable. If f is integrable. O. Solve problems (5. r ≥ 0. 49–50 and (6. Also |<f | ≤ |f | and |=f | ≤ |f | imply that <f and =f are integrable and.4). No exclusions will be made. p Since |f | = (<f )2 + (=f )2 ≤ 2 max{|<f |. Since f is measurable by assumption. if you want). |f | is measurable (as a composition of a continuous and measurable functions) and integrable by Corollary (5. Due 3/2/2006 9 Homework 6 Again. hence. are also integrable. I want you to submit only the problems below (+ the extra. 62–63 of Bartle’s book. <f and =f . O on pp. The due time for this homework is 11. Solutions. Problems for all. March 2 sharp. although. R on pp. f is.)I. Z . Z Z Z . . −iθ . . r = . f dµ. O. t)dµ(x). n=1 sdµ = lim m Z sm dµ = lim m m Z X |fn |dµ = n=1 ∞ Z X |fn |dµ < ∞. Let F (t) = The final step is exactly the same as in the proof of Corollary 5. t) is integrable on X for every t ∈ [a. t1 ) is integrable on X. this is an instance of the celebrated Fubini’s theorem which we will learn in full generality towards the end of the course. t)| ≤ |f (x. R f (x. Second. 5. s is integrable and can attain infinite value only on a set of measure 0. This m P implies that fn ≤ s converges a. By MCT. As I mentioned in class. t1)| ≤ |f (x. .e.9. Let sm = m P |fn | and s = lim sm = n=1 Z ∞ P |fn |. |f (x. |f (x. (sketch) First. where we used the following simple relations: |f |2 − (<f cos θ + =f sin θ)2 = (<f sin θ − =f cos θ)2 ≥ 0.R. = e f dµ = (<f cos θ + =f sin θ) dµ ≤ |f |dµ. t1)| + g(x)|t1 − t| implies that f (x. 5. b]. and applying LDCT we obtain the desired n=1 equality. t0)| + g(x)|t1 − t0 | implies that f (x. n=1 Hence. M4121 Homework 6. Clearly. χE1 ∞ ∞ ∞ X X 1X + nχEn ≤ (n − 1)χEn ≤ |f | ≤ nχEn .O. Due 3/2/2006 10 6. for En = {x ∈ X. ∞) we get 2−p ∞ X np µ(En ) + µ(E1 ) ≤ Z |f |pdµ ≤ n=2 ∞ X np µ(En ). n=1 6. It is easily seen that g0 is measurable and straightforward computation shows that Z Z q p |g0| dµ = cq |f |(p−1)q dµ = ||f ||−p p ||f ||p = 1 and .J. n − 1 ≤ |f (x)| < n}. 2 n=2 n=1 n=1 Integrating with power p ∈ [1. Z . Z p . . q . f g0 dµ. = c|f |pdµ = ||f ||p− = ||f ||p. p . . Let φ = ciχEi be a step function in standard representation.berkeley. For p = ∞ the proof is obvious. R (b) Let f ∈ Lp (Rn ). WLOG kgk1 = 1. Solution. X. 1 ≤ p ≤ ∞.g. Then for every integrable function f ∈ M + (X. Use Jensen’s inequality to show that ||h||p ≤ ||f ||p||g||1. e. Then. ∞) and we can use Jensen’s inequality with dµ = |g(y)|dy: . http://mathcircle. µ) Z Z Φ( f dµ) ≤ (Φ ◦ f )dµ. Extra problems. X.edu/trig/node2. µ) be a measure space with µ(X) = 1 and Φ : R → R be a convex function. 7* (a) Prove the following Jensen’s inequality: Let (X. g ∈ L1(Rn ).html) we get Z Z X X Φ φdµ = Φ ci µ(Ei ) ≤ Φ(ci)µ(Ei ) = (Φ ◦ φ)dµ. Solution. for 1 ≤ p < ∞ the function φ = xp is convex on [0. it’s enough to restrict our attantion to P step functions. and h(x) = f ∗ g(x) = f (x − y)g(y)dy be the convolution of f and g. Using Jensen’s inequality for sums (see. By monotone Convergence theorem. p p Z . Z Z Z . . . f (x − y)g(y)dy . dx ≤ |f (x − y)g(y)|dy dx . . . ≤ ZZ |f (x − y)|p |g(y)|dydx = Z Z |f (x − y)|pdx · |g(y)|dy = kf kpp · kgk1. For every > 0 and x ∈ R there exists N > 1 such that |fn (x)| < for every n > N . 4. Moreover. I want you to submit only the problems below (+ the extra. g = 0. Due 2/23/2006 11 Homework 5 Again. Fatou’s lemma obviously applies. The MCT still does not apply because the sequence is not monotone increasing and Fatou’s lemma does apply. 2 ] . I strongly recommend to you to look at ALL the problems after chapter 4. Again.n] . this time convergence is not uniform (apply the definition). fn converges to f = 0 uniformly. O. The MCT does not apply because the sequence is not monotone increasing. Then . K.)J. R on pp. for a given > 0 let N ∈ N be such that sup |f (x) − fn (x)| < for all n > N . Problems for all. 4. Solutions. (b) Let gn = nχ[ 1 . Z Z 0 = f dλ 6= lim fn dλ = 1.K. f ∈ M + by Corollary 2. Hence. if you want). 37–39 of Bartle’s book. (a) Let fn = n1 χ[0. L. although. However.M4121 Homework 5. Solve problems (4.10 and the usual properties of the limit. However. n n 0= Z gdλ 6= lim Z gn dλ = 1.J. Z . Z Z . . . f dµ − fn dµ. ≤ dµ = µ(X) . . kn [ [ n∈N j=1 Ej. Let φn be an increasing sequence of real-valued step functions that converges to f kn P pointwise. j=1 µ(Ej. See Prof. Extra problems.n . Wilson’s handout.n be the canonical representation of φn . implies the desired equality. Clearly. Apply Fatou’s lemma to fn + h. Then [ N= {x ∈ X : φn (x) > 0} = n∈N implies that N is σ-finite.R.n ) < ∞ for all j. Let φn = λj. 4.L. 4. n because f is integrable.n χEj.O. 4. M4121 Homework 5. f dµ ≥ α α Eα α Eα 12 . Solution. Follows from Z Z Z 1 1 1 f dµ ≥ αdµ = µ(Eα). Due 2/23/2006 6* Prove the following easy inequality due to Chebyshev: For f ∈ M + and Eα = {x ∈ X : f (x) ≥ α}. µ(Eα) ≤ 1 α Z f dµ. indeed. limit of a sequence of functions is not necessarily unique or measurable.M. i. Below are the steps we need to verify: • µ0 is well-defined. µ) where µ is the counting measure. Problems for all. if E ∪ Z = E 0 ∪ Z 0 then µ(E) = µ(E 0).C. M.e. I want you to submit only the problems below (+ the extra.) from a P −n sequence of existing ones. X ). we can apply Lemma 3. • µ0 is countably additive. This is a standard way of constructing a new measure (metric. Solve problems (3. the counterexample provided in class still works: for (R. Z ⊂ F ∈ Z} be the completion of X. λ(E) ≥ 0 and λ(X) = ∞ · 1 = 1. Let (X. 23–26 of Bartle’s book and the following problem 3. This is a preview of the difference between Borel measurability and Lebesgue measurability.4(b) to obtain µ(lim sup En ) = µ( ∞ \ Fm ) = lim µ(Fm ) = lim µ( m m=1 m ∞ [ n=m En ) ≥ inf sup µ(En ) = lim sup µ(En ). m=1 The interchange of the order of summation is permitted because the double series converges absolutely. J.e.. and En = [− n1 . 0 µ ∞ [ n=1 (En ∪ Zn ) ! = ∞ X n=1 µ0 (En ∪ Zn ). 2R. We need to show that µ0 : X → R 0 0 by µ (E ∪ Z) = µ(E) is a measure on (X. Clearly. 3. if you want). although. and X0 = {E ∪ Z. Due 2/16/2006 13 Homework 4 Again. ∞ ∞ ∞ T S T 3. Z = {F ∈ X : µ(F ) = 0}. the completion of Borel σ-algebra (measure). Since Fm is a decreasing family of m=1 n=m m=1 subsets with µ(F1 ) < ∞.. Show that a µ-a. 3.)C. Recall that lim sup En = En := Fm .. Solutions. 3. X.W.. I strongly recommend to you to look at ALL the problems after chapter 3. norm.J. . Q on pp. m n≥m If µ(Fm ) = ∞ for all m.M4121 Homework 4. µ) be a measure space. E ∈ ¯ + defined X.W.e. n1 ] we have lim sup µ(En ) = ∞ but µ(lim sup En ) = 1. since Lebesgue σ-algebra (measure) is. i. The n=1 2 countable additivity follows from λ( ∞ [ ∞ X Em ) = m=1 2−n µn ( n=1 ∞ [ Em ) = m=1 ∞ ∞ X X ∞ ∞ X X 2−n µn (Em ) = n=1 m=1 2−n µn (Em ) = m=1 n=1 ∞ X λ(Em). 3. This is one of the first steps on the road leading to the decomposition of charges or functions with bounded variation. Since F. n=1 where we used the fact that a countable union of measure-zero sets also has measure zero. F 0 ∈ Z be such that Z ⊆ F and Z 0 ⊆ F 0 . let F.M4121 Homework 4. we clearly have µ(E) = µ(E 0). Then E 0 ⊆ E∪F and E ⊆ E 0 ∪ F 0 . S ∞ N n=1 En = j=1 Aj . theSonly property to be verified is countable additivity of ν. which implies µ(E 0) ≤ µ(E) + µ(F ) and µ(E) ≤ µ(E 0) + µ(F 0 ). F 0 ∈ Z. Due 2/16/2006 14 To verify the first one. I am not sure how much of this road we will be able to cover but it is worth knowing that it exists. The second one is equally easy: ∞ [ µ0 ! (En ∪ Zn ) = µ0 n=1 µ ∞ [ En ! ∪ n=1 ∞ [ En ! = n=1 ∞ X µ(En ) = n=1 ∞ [ Zn !! = n=1 ∞ X µ0 (En ∪ Zn ).Q. Then sub-additivity follows easily from ν ∞ [ En ! = sup n=1 N X j=1 . Clearly. N . X . |µ(Aj )| = sup . µ . j=1 ∞ [ n=1 Let E = !. . . (Aj ∩ En ) . = . . . N . X ∞ ∞ N ∞ . X X X X . . sup µ(A ∩ E ) ≤ sup |µ(A ∩ E )| = ν(Aj ∩ En ). . j n . j n . . for each j ∈ N.j )|. 4* Solve problem 3. taking the sup over all such sequences (aj ) we obtain the desired inequality. hence.html . http://classes.j X |µ(Ai.j )| = sup i. The reverse inequality is trickier. Observe that E = Ai.j )| ≤ ν(E).j implies that i X j aj ≤ X i.yale. for example.U.edu/fractals/Labs/PaperFoldingLab/FatCantorSet.j |µ(Ai.V.W. Follows immediately from an example in 3. By the definition P of ν. Extra problems. absolutely. j=1 n=1 n=1 j=1 n=1 Again. See. 26 of Bartle’s book. Solution. 3. – finite Finally. there S exists a finite partition E = Ai.j such that aj ≤ |µ(Ai.j i. change of the order of summation is permissible because the definition of the charge implies that the series converges unconditionally and. Consider a sequence (aj ) of real numbers such that aj <Sν(Ej ). on p. b ∈ X.M4121 Homework 4. Let X = [0. 1] ∩ Q. and X be the restriction of the Borel σ-algebra to X.1 on p. For a. r∈Q . define µ((a. Sketch.e.1) remains true for finite unions and sums but fails for a countable union.. It’s not too hard to see that µ extends additively but X 1 = µ(X) 6= µ({r}) = 0. a function which satisfies Definition 3.19 except that formula (3. Due 2/16/2006 15 5* Give an example of an additive but not σ-additive measure. b) ∩ X) = b − a. i. Solution. I strongly recommend to you to look at ALL the problems after chapter 2.M4121 Homework 3. This implies that Y is a σ-algebra. 2. Solutions.H. b) = (a. I want you to submit only the 5 problems below (+ the extra. H. on p. Follows immediately from 2. Let A consist of An = [n. K. +∞)c. A = lim inf An = ∞ [ m=1 ∞ \ An and n=m B = lim sup An = ∞ \ m=1 " # ∞ [ An . Due 2/9/2006 16 Homework 3 This set is the first from the Bartle’s book. 18 of Bartle’s book. Solve problems (2. Then A 6= M because ∅ = and M 6= S because Acn ∈ / M for any n ∈ N. 2. Now F is in the smallest n=1 f σ-algebra that contains A. 2. We first solve the preceding exercise. Although. A ≤ α. we have f ( n=1 En ) = S∞ −1 (En ) ∈ X. +∞)). P on pp. The sequence An = 2N if n is even and An = 2N − 1 if n is odd gives A = ∅ and B = N = X. if you want). Extra problems. ∅. Therefore.)B.B. b] = (a. Problems for all. +∞) ∩ (b. n=m S We have (x ∈ A) ⇒ ((∃m ∈ N)((∀n ≥ m)(x ∈ An ))) ⇒ ((∀m ∈ N)(x ∈ ∞ n=m An )). Let Y = {E ⊆ Y : f −1 (E) ∈ X}. 14–17 of Bartle’s book. b − 1 ] and n " # n=1 (a. 3* Solve problem 2. n ∈ N .P. We have α < −A. Hence. if En ∈ Y. we have that both ∅ and Y are in Y. T∞ n=1 An ∈ /A . −1 c −1 −1 f (E ) = f (E) ∈ X. X. −1 Aα = f ((α.W. Since f −1 (∅) = ∅ and f −1 (Y c ) = X. A ⊆ B.K. Finally.O. F ∈ Y and the statement follows. 2. By definition.O. 2. The sequence An = N\{n} is a non-monotonic example of A = B = N. −A ≤ α < A. +∞). S∞if E ∈ Y.B and 2. We need to calculate the inverse images Aα = {x ∈ X : fA (x) > α}. Next. Clearly. these sets are measurable. O. Follows from the fact that ∞ [ (a. Hint: express the sum as a Riemann sum of a certain integral. The sum in the limit is precisely the upper D’Arboux sum for the function f : [0. and. x→0 n→∞ n→∞ x→0 . the celebrated Mean Value Theorem.134 of Rosenlicht’s book. f (xi)) = lim P ∆xi →0 i=1 i=1 j=1 v v N uX N uX X X u n u n 0 0 ∗ 2 t t (f (xi ) + o(1))2∆xi = lim ((fj (xij ))∆xi) = lim j ∆xi →0 i=1 ∆xi →0 j=1 i=1 j=1 v uX Z N X u n 0 t (f (xi ))2 + o(1) ∆xi = lim j ∆xi →0 i=1 j=1 v n bu uX a t (fj0 (x))2dx. the integrals on both sides make sence and d the FTC applies to the function f (x) = u(x)v 0(x) + v(x)u0(x) = dx (u(x)v(x)): Z b f (x)dx = Z a b 0 u(x)v (x)dx + a Z b v(x)u0(x)dx = u(b)v(b) − u(a)v(a). You only need to add a few words to the “proof” you shoud have had in a calculus course. boundedness and the uniform(!) continuity of the integrand to justify our manipulations with o(1). Solution. j=1 where we used the triangle inequality to replace the “sup” with the “lim”. Z 1 n 1k + 2 k + · · · + n k 1X j k 1 lim = lim = xk dx = . Solve problem 1 on p. f (x) = xk associated with the uniform n-element partition of [0.so let us seize one when it presents itself. Solution.133 of Rosenlicht’s book. By definition. This set seems to be easier than the first one but contains many facts and examples that I feel necessary for you to know. Solve problem 21(a) on p. Clearly. 1. 1] → R. Consider fn (x) = |x| n . k+1 n→∞ n→∞ n n n k+1 0 j=1 4. 3. Hence. Solve problem 20 on p.M4121 Homework 2. In the last 3 problems do not forget to prove that the examples you provide work.160 of Rosenlicht’s book. lim lim fn (x) = 1 6= 0 = lim lim fn (x). I’m afraid. finally. 1]. a 2. Solution. the length `(f ) of the curve defined by the vector function f is v N uX N X X u n t (fj (xi ) − fj (xi−1 ))2 = `(f ) = sup d(f (xi−1). Since u0 and v 0 are continuous. This is an easy example.134 of Rosenlicht’s book. Due 2/2/2006 17 Homework 2 Problems for all. we won’t have many oportunities in this course to explore geometric applications of integral calculus . Solve problem 17 on p. 1 Solution. However the pointwise limit is the Dirichlet function which is discontinuous at every point and. 138 of Rosenlicht’s book 1 0 ≤ x ≤ 2n . −nx 2* Let fn = e√x . 6. R∞ nevertheless. not uniform because lim fn (x) = ∞ x→0+ for all n ∈ N. Let f : [0. This. Solution.161 of Rosenlicht’s book.90. Therefore. contrudicts the definition of uniform convergence. where In = 0 fn (x)dx. Clearly. R and let gn = nfn . Consider the sequence of “hat” functions defined on p.161 of Rosenlicht’s book. not (Riemann) integrable. This might be tricky but not too much so. for every N ∈ N there exists x ∈ (0. 1] → R be the Riemann function defined on p. Show that fn → 0 pointwise but not uniformly on (0. Solution. 2n < x ≤ n1 . therefore. gn → 0 pointwise but gn = n. 4n2 x. Sketch. However. obviously. Define the sequence fn = f n . using the change of variable formula. Solve problem 3 on p. 1 0. indeed. +∞). Solve problem 5 on p. √ dy ≤ √ √ dy + n 0 y n y n 0 0 1 . we get Z 1 Z ∞ Z ∞ −y Z ∞ e 1 1 1 1 −y In = fn (x)dx = √ e dy = √ (2+e−1 ) → 0. The convergence fn → 0 is. These functions are all integrable (see the proposition about composite functions that we proved in class). Let’s see if you are up to Bartle’s hope (see p. This one is slightly harder. Due 2/2/2006 18 5. of 1 Rosenlicht’s book (see also the previous homework set).M4121 Homework 2. In → 0. n < x ≤ 1. Extra problems.2). 1 2 fn = 4n − 4n x. Problem 1(d). Show that. ∞) such that fN (x) > 1. Let f be (Riemann) integrable and concave on [a. b] → R by φ(x) = f (a) + Z b a Z b f (x)dx. 1. it has measure 0. do not just quote it. b] = [0.) Solution. Show that f is integrable on any finite interval. To see this directly. we will show that for any fixed > 0 there exists a partition of [0.90. This implies integrability of f . 1] such that U (P. Problem 1(d). φ ≤ f . f (a) + f (b) φ(x)dx = (b − a) ≤ 2 Z b f (x)dx. b]. we can use a specific sequence of Riemann sums to obtain Z b 2n X 1 b−a b−a f (x)dx = lim f a + (k − ) = n→∞ 2 2n 2n a k=1 n 1 b−a b−a X 1 b−a lim f a + (k − ) + f a + (2n − k + ) ≤ n→∞ 2n 2 2n 2 2n k=1 n b−a X lim 2f n→∞ 2n k=1 a+b 2 = (b − a)f a+b 2 . Let f be the Riemann function defined on p. 2. This one is in a sense a preview of one of the main theorems on Riemann integrability that will be covered in class. f ) = 0 for all P . a To prove the second inequality we use a different method (there might be easier ones but I thought it would be useful for you to see the one below). Due 1/26/2006 19 Homework 1 Problems for all. Since the set of rationals is countable. Since L(P. I promise to help. this would imply integrability of f . b].:) Unlike the exam situation. of Rosenlicht’s book. Can you compute the integral? (You can assume that the set Q is countable. From the solution of the cited problem (see last semester’s homework). Let us first prove the inequality (b − a) f (a) + f (b) ≤ 2 Define φ : [a. If the theorem is covered before this homework is due. By concavity. Show that Z b f (a) + f (b) a+b (b − a) f (x)dx ≤ (b − a)f ≤ . The points in the Riemann sum are computed as midpoints of subintervals of length b−a 2n that partition [a. Hence. we know that the set of discontinuities of f is precisely Q. however. Prove the statement below with “bare hands”. Last time the first homework was a warm-up. Since f is concave and integrable. This one is a bucket of cold water.M4121 Homework 1. we can assume [a. 1]). f ) < (WLOG. 2 2 a Solution. . a f (b)−f (a) (x − b−a a). . .b] for all x ∈ [c. Some of you may notice some connection with the Gelfand’s spectral radius formula (don’t worry. Solve problem 13 on p. b]. Then n 1X U (Pn .M4121 Homework 1.b] . 4. For each n ∈ N let us choose a uniform partition Pn = {xj = nj . . it is bounded. yi − 2−i−2 ) and P be a partition containing the end points of these intervals.133 of Rosenlicht’s book. f ) = f (xj ) n n 1X L(Pn . Solution. while the formula on the right hind side is essentially the L∞ norm. 1] and f is increasing. for a given > 0 let [c. Let these points be centers of the intervals Ii = (yi − 2−i−2 . n and j=1 j=1 Therefore.. i = 1. + 2 i=1 3. d] ⊂ [a. 1] we have U (Pn . . n}.133 of Rosenlicht’s book. n n It remains to choose n sufficiently large. Above I mentioned the spectral radius formula. b] = [0. if |f (x)| ≤ M for all x ∈ [0. First of all observe that since f is defined and monotonic on the closed interval [a. f ) ≤ X −i−1 2 < . You may want to look at problem 12 to get some ideas. Solve problem 8 on p. . but a more significant reason to look at this formula is the fact (probably. Again WLOG [a. b] be such that max (f (y)) − f (x) ≤ y∈[a. j = 0. ≥ Z d n (f (x)) dx n1 c 1 ≥ (d − c) n ( max f (y) − ) y∈[a. f ) = f (xn ) − f (x0) 2M ≤ . First observe that Z n1 b n (f (x)) dx ≤ Z b n1 1 max (f (y)) dx = (b − a) n max (f (x)) n a y∈[a.b] implies LHS ≤ RHS. We will learn quite a bit about these spaces later in this course. Solution. Then N U (P. N . Due 1/26/2006 20 From the definition of f we see that there are only finitely many points yi . known to some of you) that the integrals on the left hand side of the formula gives rise to norms in Lp spaces for p = n ∈ N.b] a x∈[a. if you’ve never heard about it before). . for which f (yi ) > 2 . d]. Then Z b n (f (x)) dx n1 a implies RHS ≤ LHS. This may help you to solve the extra problem. . f ) = f (xj−1 ). f ) − L(Pn . On the other hand. M4121 Homework 1. this is the easiest problem in this set.133 of Rosenlicht’s book. To me. Solve problem 16 on p. We need to show that δ > 0 such that whenever . Due 1/26/2006 21 5. Solution. This is another one of my numerous attempts to make you comfortable with abstract material. R for any >R 0 there exists . . b . b]) < δ we have . b ||f − g||C([a. a f (x)dx − a g(x)dx. < . Since ||f − g||C([a.b]) = . R . R . b . b max |f − g|(x) and . a (f (x) − g(x)). Comment. This time. 1* Solve problem 1 without assuming integrability of f . .b] Extra problems. x∈[a. b] can be decomposed on at most two intervals on each of which f is monotonic. ≤ a |f − g|(x)dx. a relatively easy one. It’s enough to observe that [a. the result follows.