Ma 4121: Introduction to Lebesgue Integration Solutions to Homework Assignment 6

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1 Ma 4: Introduction to Lebesgue Integration Solutions to Homework Assignment 6 Pro. Wickerhauser Due Thursda, April 5th, 3 Please return our solutions to the instructor b the end o class on the due date. You ma collaborate on these problems but ou must write up our own solutions. Late homework will not be accepted.. For ied c (,, deine : R R as ollows: (, de ( c /( c, i < < and < ;, otherwise. Prove that L(R and evaluate R. In act, is an upper unction. For ɛ >, deine ɛ (, de ( c /( c, i < < and < ɛ;, otherwise. Then ɛ and ɛ as ɛ. Also, ɛ is bounded and vanishes outside a bounded triangular region, so ɛ L(R or each ɛ >. Finall, use Fubini s theorem to calculate R ɛ ɛ ɛ +ɛ ( c ( c ɛ ( c [( c ɛ c c [ ɛ / ɛ ɛ c c [ ( c ( c ( c c ɛ c c ɛ c +ɛ ɛ ( c c ( ɛ( + ɛ ɛ c ɛ This evidentl converges to / as ɛ, showing that L(R with /. ( c c. Note that integration in irst, then in, gives a simple antiderivative. To ind the limits o integration, it is useul to draw a graph o the triangular region.. Suppose that S R is a measurable set with the propert that λ(s or almost ever R, where λ is -dimensional Lebesgue measure on R, and S de R : (, S}. Prove that the -dimensional Lebesgue measure o S is zero. Theorem 5.5 on p.4 o our tet. (Note: This is a partial converse to

2 Put (, de S (,. Then M(R. It suices to show that R. Suppose irst that S is bounded. Then L(R and we ma use Fubini s theorem to evaluate G(, since G( (, λ(s, a.e. R. R R Now write S n S n, where S n S [ n, n. Then or each n, S n is bounded and satisies the same hpotheses as S, so it has -dimensional Lebesgue measure. Thus the -dimensional Lebesgue measure o S is at most the sum o countabl man s, hence is zero. 3. Suppose that i : R R is deined and bounded on the compact interval [a i, b i R. I i L([a i, b i or i,..., n, prove that ( ( b bn ( n ( n d(,..., n ( d n ( n d n, Q a where Q [a, b [a n, b n R n. the same. We prove this b induction on n. It is evidentl true when n, or then both sides are Suppose the result holds or n ; let g : R n R be deined b g(,..., n ( ( n. de This g is deined and bounded on Q n [a, b [a n, b n. For an n L([a n, b n, the unction g n g(,..., n n ( n belongs to L(Q b Lebesgue s dominated convergence theorem and we have ( [ bn g(,..., n n ( n d(,..., n g(,..., n d(,..., n n ( n d n, Q Q n b Fubini s theorem. Here g(,..., n ma be removed rom the integral in n since it has no n -dependence, and the remaining unction n is Lebesgue integrable b hpothesis. The result or n now ollows rom the inductive hpothesis. 4. (a Prove that R e π b transorming the integral to polar coordinates (b Use part(a to prove that R e π. (c Use part (b to prove that R n e π n/. (d Evaluate R e t or t >, and ind t or which the value is. (a R (, (r cos θ, r sin θ : < r <, θ < π} (, }, and this mapping has Jacobian ( cos θ sin θ J r sin θ r cos θ with determinant J r r. Hence, b eq.3 on p.49 o our tet, J, R e R e r which we ma evaluate b Fubini s theorem as π [ e d(, e r r dr R r θ a n a n [ dθ π e u du π, u

3 ater substituting r u and r dr du. (b Since e e e, b using the results o eercise 3 above we get e R ( ( e e R ( e, since the two integrals are the same ecept or the name o the variable. Combine with part (a and the observation that the integrals must be positive to conclude that R e π. (c Another application o eercise 3 above gives ( n e e π n/, n where the last equalit ollows rom part (b. (d Write ( e t or t >, and put g( de (/ t e. Thus, using part (b above and the equation on p.47 o our tet gives π e g( ( t ( t e t. Thus e t π/ t, and choosing t π gives e π. 5. Let V n (a denote the volume o the ball o radius a in R n, that is, the n-dimensional Lebesgue measure o the open set R n : < a}. (a Prove that V n (a a n V n (. (b Prove that, or n 3, we have the ormula V n ( V n ( π [ (c Use the recursion in part (b to conclude that V n ( ( r n/ r dr dθ V n ( π n. π n/ Γ( n +, where Γ is the special unction deined on p.77 o our tet. In the ollowing, write + + n or R n, and let B B n B n (; R n : < } R n denote the unit open n-ball centered at. (a Observe that V n ( B ( and V n (a B (/a, n n so V n (a a n V n ( b the relation on p.47 o our tetbook. 3

4 (b Write B n (; R n : < } (, z R n : R n, z R, + z < } (, z R n : R, < ; z R n, z < } (, z R n : B (;, z B n (; }. This suggests a method to evaluate V n ( b iterated integration: V n ( V n (, B (; z B n (; B (; so using the scaling relation in part (a, we get V n ( ( (n / V n ( V n ( ( n/. B (; B (; Evaluating the integral in polar coordinates gives π θ r( r n/ r drdθ π u using elementar methods and the substitution r u. Thus V n ( π n V n (. ( u n/ du π n, (c Prove this b induction on n. First recall that Γ( and Γ( π. Also, Γ( + Γ( or all positive real, and thus Γ(n + n! or all nonnegative integers n. (Man reerences list these special values and relations, or else the ma be derived b elementar methods. Check the case n b noting that V ( Net, check the case n b observing that π π/ π Γ( π/ Γ( + V ( π π/ Γ( π/ Γ( π/ Γ( +. Now suppose that n > and that the equation holds or all k,,..., n. Then b (b, proving the inductive step. V n ( π n V n ( (ππ(n / nγ( n + πn/ n Γ( n πn/ Γ( n +, 6. Suppose that : R R is deined b e (, sin, i is rational; e, i is irrational. Prove that L(R and compute. R 4

5 Use the Tonelli-Hobson test, Th.5.8 on p.45 o our tet. Since Q has -dimensional measure zero, the set Q R has -dimensional measure zero b eercise above, so agrees almost everwhere with the unction g(, e. Now g > on R, so g g, and the iterated integral eists: [ g(, π e π, R using the results o eercise 4 above. Conclude that g L(R, so L(R, with g g π. 7. Let (, ( /( + or, <, and put (,. Prove that both iterated integrals [ (, d d, and R eist but are not equal. Conclude that / L([, [,. [ (, d d First note that or (, (,, ( + ( + ( +. We ma thus compute the iterated integrals b antidierentiation: ( [ ( + d d + d d >, + while ( [ ( + d d + d d <, + where both integrals eist because the integrands are continuous and bounded. 8. Let (, e sin sin or and, and let (, otherwise. Prove that both iterated integrals [ [ (, d d, and (, d d R R eist and are equal, but that / L(R. Eplain wh this does not contradict the Tonelli-Hobson test (theorem 5.8, p.45. I one o the iterated integrals eists, then so will the other, and the will be equal, or the unction satisies (, (, or all (, R. In Gradshten and Rzhik, Table o Integrals, Series, and Products, 5th edition, equation 3.893( on p.5 implies e sin d, i >, + so that ( e sin sin d d sin d <, + 5

6 b Lebesgue s dominated convergence theorem and comparison with /( + L(R. Now suppose toward contradiction that L(R. Let de B de t R + : sin t } B k, where B k [(k + 4 π, (k π. Then or (, B B R, we have sin sin. We deine a unction s : R R as ollows: min (s, t : (s, t Bm B s(, n }, i (, B m B n,, otherwise. Then s M(R since it is a countable sum o step unctions, and s on R. Thus L(R would impl that s L(R. But in act k s(, e (m+ 3 4 (n+ 3 4 π, i (, B m B n, and B m B n π /4 or ever n, m,,,..., so R s π 4 >,> m n e ( 3 4 +m( 3 4 +nπ π 8 m n e mnπ Finall, we ma use the integral test to show that the double series diverges, or e π ( π e d d π d +. Hence s / L(R. Conclude that / L(R, and thus that / L(R. This is not a countereample to the Tonelli-Hobson theorem precisel because / L(R, and so neither o the iterated integrals or will be inite. 9. Let I [, [,, let (, ( /( + 3 i (, I \ (, }, and let (,. Prove that / L(I b considering the integrals [ [ (, d d, and (, d d have and From Gradshten and Rzhik, Table o Integrals, Series, and Products, 5th edition, we d ( + 3 ( + ; (.(, p.68 d ( ( ; (.4(, p.69 ( + Thus or >, we have [ ( + [ ( + 3 d ( + + ( + ( +. Similar, b echanging and noting that (, (, or all,, we get ( ( + 3 d, or >. ( + 6

7 Both o these unctions have elementar antiderivatives, with which we ma compute [ d d [ [ [ +, while d d + +. Since these are not equal, the original unction cannot belong to L(I or else there would be a contradiction with Fubini s theorem.. Let I [, [, + and let (, e e i (, I. Prove that / L(I b considering the integrals [ [ (, d d, and (, d d Function is continuous on I, hence it is measurable. It is also bounded, but I is not bounded. Were L(I, it would satis Fubini s theorem and we would have ( (. But while ( ( ( (, d d ( (, d d I e [e d e [e d, e /z z [e /z dz, as shown b elementar methods, with the last equalit ollowing rom the substitution /z. But or all z (,, the continuous and bounded integrand unctions satis z e /z [e /z > z e z [e z, so the two iterated integrals eist but cannot be equal. 7