s = and t = with C ij = A i B j F. (i) Note that cs = M and so ca i µ(a i ) I E (cs) = = c a i µ(a i ) = ci E (s). (ii) Note that s + t = M and so

Transcription

1 3 From the otes we see that the parts of Theorem 4. that cocer us are: Let s ad t be two simple o-egative F-measurable fuctios o X, F, µ ad E, F F. The i I E cs ci E s for all c R, ii I E s + t I E s + I E t, iiiif s t o E the I E s I E t, iv If F E the I F s I E s. Proof As i Lemma 3.7 write s a i χ Ai i N a i χ Cij ad t i j N b j χ Bj j N b j χ Cij i j with C ij A i B j F. i Note that cs M i ca iχ Ai ad so I E cs c ca i µa i i a i µa i ci E s. i ii Note that s + t M i N j a i + b j χ Cij ad so N I E s + t a i + b j µc ij E i i j N a i µc ij E + j N b j µc ij E i N a i µ C ij E + j N M b j µ C ij E i j j i sice µ is additive N a i µa i E + b j µb j E i j I E s + I E t.

2 iii Give ay pair i, j : i M, j N for which C ij E φ, we have for ay x C ij E that a i sx sice x C ij A i tx sice s t b j sice x C ij B j. So I E s N a i µc ij E i j N b j µc ij E I E t. i j iv By the mootoicity of µ we have I F s 33 a Start with a i µa i F i log + t + t a i µa i E I E s. i dy t y is a icreasig sequece of o-egative, Lebesgue i- So by Lebesgue s Mootoic Covergece Theorem we fid that lim log + t t dx t lim + x dx lim + x t. Now { } +x tegrable fuctios with limit fuctio. dx + x t dx + x. b Give m > the for x > we have m + mx > m + x ad so or Thus m log + t t m m m + x > + x. mdx t m + x > + m t m > + t. dx + x log + t,

3 Defiig g x { + x e x for x for < x we have a icreasig sequece of o-egative Lebesgue measurable fuctios. The limit fuctio is gx e x e x e x for all x havig used part a. So by Lebesgue s Mootoic Covergece Theorem we fid that lim + x e x dx e x dx. 34 a Startig with x + x + 3x x, gives log x + x log x. x Apply Corollary 4.3 with f x + x log x. These are cotiuous fuctios o [, ] ad so are Lebesgue measurable. Obviously oegative so log x dx x + Itegrate by parts twice to see that x log x dx. Hece x log x dx + 3. log x dx x b As i a use Corollary 4.3 to justify Itegratig by parts gives x p log x x dx + x p+ log xdx. π 3. [ ] x x p+ p++ log xdx p + + log x x p+ p + + dx. To deal with the first term at x we require p++ > for all, that is, p >. I which case 3

4 Hece x p+ log xdx as log as p >. [ x p++ ] p + + p + + x p log x x dx p + + p + +. p + 35 Lookig first at those x for which f takes o the values, ad we fid fx for either x Q or. x, fx for x / Q ad. x <., fx for x / Q ad. x <., etc. There might be cocer that a umber such as x. ca also be writte as for which f would give a differet value. We eed ot worry, the collectio of such poits is such a small set, i.e. oe of measure zero, that it would ot effect the value of the itegral. For our sequece of simple fuctios simply choose { fx f N x N x elsewhere. Sice fx this is a icreasig sequece of fuctios. Each f N is simple, takig oly itegral values betwee ad N. Also the fuctios are measurable, with ad, for each N, {x : fx } Q [, {x : fx } Q c [ +,, which are all measurable sets. We ca quickly evaluate the itegral of these simple fuctios, I [,] f, as 4

5 N j [ ] jµ Q, j+ j N j N j j j j+ j N j N + N j j j j j j j j N N N j. N By Lebesgue s Mootoic Covergece Theorem we get fdµ lim N N j N j N j j Usig the defiitio of φ give i the questio ad the represetatio of f as a series we see that φ E Hece φ is σ-additive. S E m m fdµ S E f m dµ E m fdµ E m φe m. m m S E f m dµ m f m dµ by Corollary 4.3 sice f m o E whe m sice f m f o E m 37 Let x R be give. The { } lim if g x lim if g rx. r 5

6 But g r x ad g r x for all r x see defiitio of g r, so if r g r x. True for all x implies lim if g ad so R lim if g dµ. Yet R g dµ for all ad so lim if R g dµ. Thus we have strict iequality. 38 a Sice we have both maxa, ad a maxa, it follows that ad so + maxa, + maxb,, a + b maxa, + maxb,, maxa + b, maxa, + maxb,. b Sice we have both mia, ad mia, a it follows that ad so mia, + mib,, mia, + mib, a + b, c mia, + mib, mia + b,. f + g + x maxf + gx, maxfx + gx, maxfx, + maxgx,, by part a, f + x + g + x. d f + g x mif + gx, mifx, migx,, by part b, f x + g x. 39 If we have a b a + b where a, b we ca square both sides to get a b a + b, i which case, ab ab. Thus ab ad so either 6

7 a or b. It should be oted that with either of these possibilities we have equality as required. 4 i Let f x { fx if x if x >. Recall from otes that f is Lebesgue itegrable if, ad oly if, f is Lebesgue itegral. Sice f f this first gives us that f ad thus f are Lebesgue itegrable. But further, lim f x fx with f is itegrable ad f f meas that we ca apply Lebesgue s Domiated Covergece Theorem ad deduce that fdµ lim lim lim f dµ f dµ sice f x for all x >, fdµ. sice f x fx for all x. ii We ow assume that f is Lebesgue measurable which is weaker tha Lebesgue itegrable ad o-egative. The sequece of fuctios f defied i part i form a icreasig sequece of Lebesgue measurable o-egative fuctios ad so we ca apply Lebesgue s Mootoic Covergece Theorem to deduce that fdµ lim f dµ lim f dµ lim fdµ. * 4 The fuctio ft e t t x is cotiuous over [, ad so is Lebesgue measurable over that iterval. We have to show that it is Lebesgue itegrable by showig that it s itegral over [, is fiite. We do this by applyig 4ii ad calculatig lim fdµ. I fact all we do is show that this limit is fiite by boudig the itegrad e t t x i terms of fuctios that are easier to itegrate. For example, for t [, ], we have e t t x e t x ad so [ t e t t x dµ e t x dµ e x x ] e x <. For t we ca fid κ κx such that t x κe t/ i which case e t t x κe t/ ad so 7

8 e t t x dµ κ e t/ dµ < κ. So the limit i * is of a icreasig sequece of values of itegrals bouded above, hece the limit exists. Thus e t t x is Lebesgue itegrable. Let N be a iteger > x. It suffices to fid κ such that t N κe t/. If N the κ will suffice. Assume N. Expadig, ad takig the j N term, e t/ j j! sice t. So choose κ N N!. j t > N! N t, 4 The restrictio of t i questio 33a was ot ecessary so that result holds also with t <, or equivaletly, lim log t t, i.e. lim t e t for t. Now defie f t { t t x if t < otherwise. As i questio 33b this is a icreasig sequece. First ote that log t t dx x. Assume m > the m mx < m x so m m x < x which is well defied sice x t <, ad so or m log t t > m mdx t m x > m t m > t. dx x > log t 8

9 Thus we ca apply Lebesgue s Mootoic Covergece Theorem to justify the iterchage of itegratio ad limit i Γx lim e t t x dt lim f tdt lim f tdt sice f t for all t >. f tdt By chages of variables ad itegratig by parts we see that, as log as x + j for all of j, we have Hece f tdt t t x dt x x { [ y y x x x x ] + x y y x dy... x! xx +...x +! x xx +...x +.! x Γx lim xx +...x +, y y x dy } y y x dy y x+ dy as log as x + j for all j N {}, i.e. x / N {}. 43 Startig with e x e x e x e x j e x j we fid that x a e x x a e x, for x >. This is a sum of o-egative Lebesgue measurable fuctios. So we ca use Corollary 4.3 to justify the iterchage i 9

10 x a e x dµ x a e x dµ a x a e x dµ a Γa t a e t dt by the defiitio i Questio 4. Note that the equality justified by Corollary 4.3 meas that if oe side is fiite the so are both with the same value ad if oe side i ifiite the so is the other. I our case if a > the a is coverget ad so both sides of our result are fiite. 44i Apply Lebesgue s Domiated Covergece Theorem. From Questio 33 we have that lim + x x si e x for all x >. For a domiatig fuctio, h, choose a boud o the 3 itegrad: + x x si + x + x 3 3 for all 3. Havig used the fact proved i questio 33 that { + x } is a icreasig sequece. The fuctio hx + 3 x 3 i itegrable over [, ] ad so the Domiated Covergece Theorem justifies lim + x x si dx lim dx. + x x si dx ii Start from + x + x > + + x to see that + x > ++x +x ad so + x + + x > + x + x +.

11 { Thus the sequece of fuctios +x +x } is decreasig. Though the terms are o-egative ad measurable we caot use Lebesgue s Mootoic Covergece Theorem directly sice it is cocered with icreasig sequeces. Istead we might hope to use the Domiated Covergece Theorem. But for this we eed to kow the limit, that the fuctios are itegrable ad domiated by a itegrable fuctio. The limit is easily see. If x the all terms i the sequece equal so the limit is. If x > we start with the observatio that the biomial expasio gives + x + x + x ad so + x + x + x + x + x as. Thus the limit is if x ad elsewhere, that is, a.e.µ o [,. We could choose the domiatig fuctio to be the 3 term, i.e. hx + 3x/ + x 3. It is easily show that hxdx ad so h is itegrable. But also sice the sequece of fuctios is decreasig each fuctio, at least for 3, is itegrable. Hece, usig the Domiated covergece Theorem we ca justify the iterchage i lim 45i Start from + x + x dµ lim + x + x dµ dµ. sech x e x + e x e x + e x e +x. We caot use Corollary 4.3 directly because of the alteratig sig. Istead we write the sum as m e 4m+x e 4m+3x, which is ow a sum of o-egative measurable fuctios. So by Corollary 4,3 we ow get

12 sech x dx m π 4m + m π 4m + π m e 4m+x e 4m+3x dµ +. π 4m + 3 4m + 3 Here the hit give i the questio has bee used to get the secod lie. Also the last lie is oly a coditioally coverget series coverget by the alteratig sig test ad so the order of summatio is importat ad that is give by the bracketig i the lie before. ii As so ofte see start with cos x e x + e +x cos x. We hope to use Theorem 4.9 with g N the N th -partial sum, so g N N e +x cos x N e x e x e x N+ e x e x e x + e x + e sice x ex for x >, e x. So we ca apply Lebesgue s Domiated Covergece Theorem with hx e x. Thus cos x e x + dx e +x cos xdx e +x cos xdx.

13 Settig I e x cos xdx ad itegratig by parts twice shows that I I, that is, I. Thus + 46 Start from cos x e x + dx I +. sih bx sih ax ebx e bx e ax e ax ebx e bx e ax e ax e bx e bx e +ax. For x > this is a sum over o-egative measurable fuctios ad so we ca apply corollary 4.3 to deduce sih bx sih ax dx e bx e bx e +ax dx e bx e bx e +ax dx [ e bx +ax b + a ] e bx +ax b + a b + a b + a b + a b. Note that the coditio b < a esures that oe of the deomiators of terms i the sum are zero. 3