f(1), and so, if f is continuous, f(x) = f(1)x.

Transcription

1 2.2.35: Let f be a additive fuctio. i Clearly fx = fx ad therefore f x = fx for all Z+ ad x R. Hece, for ay, Z +, f = f, ad so, if f is cotiuous, fx = fx. ii Suppose that f is bouded o soe o-epty ope set. The there is a a R, a δ > 0, ad a r > 0 such that fx r for x a δ, a + δ. Hece, fx R fa + r for x δ, δ. Give x R, either x = 0, ad therefore, sice f0 = f0 + f0, f0 = 0, or x 0 ad there exists a ratioal uber q such that δ x q < 2δ x. I the latter case, x q x δ, δ ad therefore fx = q f q qr 2 x δ. Hece f is cotiuous at 0, ad therefore, sice fy fx = fy x, it is cotiuous everywhere. Now apply i to coclude that f is liear. iii Now assue that f is B R λ R-easurable. Choose R > 0 so that λr > 0 whe = { x : fx R 2 }. The, by Lea 2.2.6, there exists a δ > 0 such that δ, δ. I particular, if x < δ, the there exist y, z such that x = y z ad therefore fx fy + fz R. Thus, by ii, f is liear. 3..2: There is othig to do whe f is easurable. Now suppose that, for every a fro a dese subset D of R, {f > a} B, ad ote that this assuptio is equivalet to sayig that f B for every C {a, ] : a D}. Sice the set of B R for which f B is a σ-algebra, all that we have to do is check that B R = σc, ad this coes dow to checkig that x, y σc for every x < y. But if < x < y < are give, the oe ca fid {a : } x, D ad {b : }, y D so that a x ad b y, which eas that x, y = a, ] b, ] B. The cases whe x = or y = ca be hadled i the sae way. If {f a} B for each a D, let b R be give, choose {a : } D b, so that a b, ad coclude that {f > b} = {f a } B. Hece, this case reduces to the precedig oe. The cases whe the iequalities are reversed follow whe oe replaces f by f. To hadle the secod part of the proble, first ote that it suffices to show that {f < g} B. Ideed, {f g} = {g < f}, {f = g} = {f g} {g f}, ad {f g} = {f = g}. But, if Q deotes the set of ratioal ubers i R, the {f < g} = a Q{f < a} {g > a} B. 3..3: To see that µ is a fiite easure o E, B, first observe that µe = f dν < ad that µ = 0. Secod, give a sequece { : } B

2 2 of utually disjoit sets, set B = = ad defie ϕ = =. The 0 ϕ f B f, ad so, by the Mootoe Covergece Theore, µb = li ϕ f dν = li = µ = µ, which copletes the proof that µ is a fiite easure o E, B. Fially, to see that µ ν, suppose B B satisfies νb = 0. The, by Markov s Iequality, νf B 0 = 0 ad so µb = B f dν = 0. Kowig this, the reaiig assertio is a applicatio of Exercise ex3..4: Suppose that f, g L µ; R. If µf > g = 0, the, for every B, g f = g f 0 µ-alost everywhere ad therefore = g dµ f dµ = g f dµ 0. Coversely, if g dµ f dµ for every B, set B = {f g + } ad coclude that 0 B g f dµ µb. Hece, µb = 0. Sice B {f > g}, it follows that µf > g = li µb = 0. Give these facts, the rest of the exercise is trivial. 3..5: First ote that ωi = if ad oly if there are exactly i {,..., } for which ωi =. Hece, β p {ω : ωi = } equals p q ties, which is the uber of η {0, } with ηi = for exactly i s. To verify the secod equality, siply use the first part to justify exp λ ωi p β p dω = e λp = e λp pe λ + q = pe λq + qe λp. =0 pe λ q Next set fλ = log pe λq + qe λp, ad apply eleetary calculus plus the fact that p + q = to see that f0 = f 0 = 0 ad f λ = pqe λq p pe λq + qe λp 2. Because, for ay a, b > 0, ab a + b 2 = 4 [ a + b 2 a b 2 ] a + b 2 4, it follows that f0 = f 0 = 0 ad f λ 4, which, by eleetary calculus, eas that fλ λ2 8 ad therefore that * exp λ ωi p β p dω e λ2 8 for all λ R.

3 3 Now apply Markov s Iequality ad * to see that, for ay λ > 0, β p ωi p R = β p exp λ ωi p e λr e λr exp λ ωi p β p dω e ad therefore, after iiizig over λ > 0, that β p ωi p R e 2R2 The sae auget, oly with λ < 0, shows that. β p ωi p R e 2R2 λ2 λr+ 8, ad so we ow kow that β p ωi p R = β p ωi p R 2e 2R2. Fially, apply this with R = 4 to coclude that β p ωi p 4 <, = ad therefore, by the Borel Catelli Lea i Exercise 2..26, that for β p - alost all ω Ω, there exists a such that ωi p for all : There is very little to do here. Ideed, sice µ p B p = β p Ω p ad β p Ω p is 0 or oe depedig o whether p p or p = p, it is obvious that µ p µ p wheever p p. I particular, sice, by , µ = λ [0,], µ 2 p λ [0,] whe p 2. Thus, if F px = µ p 0, x + ] is the distributio fuctio deteried by µ p, the, by Exercise , F p is a sigular, o-decreasig fuctio. Moreover, sice β p {ω} = 0 for all ω Ω ad Φ x cotais at ost two eleets, F p x F p x = µ p {x} = β p {Φ x} = 0, it follows that F p is cotiuous. Fially, for ay ad 0 < 2, part i of Exercise shows that µ p [2, + 2 = β p ωi = = ad so F p is strictly icreasig o [0, ]., p q > 0,

4 : Suppose f f i µ-easure. The, because µ is fiite ad therefore f f L µ; R, Lebesgue s Doiated Covergece Theore iplies that f f dµ 0. Coversely, if f f dµ 0, the, by Markov s Iequality, for ay ɛ > 0, µ f f ɛ µ f f ɛ ɛ f f dµ : i To prove the first part, let C deote the collectio of N rectagles which oe obtais by subdividig the sides of ito equal parts ad coectig all the subdivisio poits by lies. Next, for each I C, choose of poit ξ I I, ad defie f to be the fuctio o such that f x equals the average of {fξ I : I x}. The it is easy to check that, for each I C, f I is Riea itegrable o I ad R I f x dx = fξ I voli. Thus R f x dx = R f x dx = fξ I voli = f dλ R N, I C I I C ad therefore, sice f f uiforly o, it follows that R fx dx = f dλ R N. Fially, suppose f is a cotiuous eleet of L λ R N ; R. The, by the precedig ad Lebesgue s Doiated Covergece Theore, f dλ R N R fx dx = f dλ R N R N \ f dλ R N 0 as R N. R N \ ii This part is essetially the sae, oly easier : i If K is uiforly µ-absolutely cotiuous ad M sup f K f L µ;r <, choose, for a give ɛ > 0, δ > 0 so that sup f K f dµ < ɛ wheever B satisfies µ < δ, ad choose R 0, so that M R < δ. The, by Markov s iequality, sup f K µ f R δ ad so sup f K f dµ ɛ. { f R} Next, suppose that K is uiforly µ-itegrable, ad defie AR sup f K f dµ for R 0,. Clearly { f R} sup f dµ Rµ + AR f K for ay B ad R 0,. Hece, if, for give ɛ > 0, we choose R 0, so that AR < ɛ 2, the ɛ f dµ < ɛ for all f K ad B with µ < 2R ; ad so K is uiforly µ-absolutely cotiuous. I additio, whe µe <, by choosig R 0, so that AR, we see that f L µ;r RµE+ < for all f K.

5 5 ii Note that, for ay R 0,, f dµ R δ f +δ dµ R δ { f R} f R f +δ dµ. iii Suppose that f f i L µ; R. Clearly f L µ; R ad sup Z + f L µ;r <. For ɛ > 0, choose Z + so that f f L µ;r < ɛ 4 for, ad, apply Exercise 3..3 to fid a δ > 0 so that f f dµ < ɛ 4 for all B with µ < δ. Sice ax sup f dµ sup f f L µ;r + f dµ, it follows that, for this choice of δ > 0, µ < δ = sup f dµ < ɛ. Coversely, if µe <, f f i µ-easure, ad {f : } is uiforly µ-itegrable, the, by i ad Fatou s Lea, f is µ-itegrable, ad therefore, for ay ɛ > 0, there is a δ > 0 such that µ < δ = for all Z +. Hece, if = µ < δ for all, the f f L µ;r = f f dµ f dµ + f dµ < ɛ { } f f ɛ µe ad is chose so that f f dµ + f f dµ 2ɛ for. iv Tightess eables oe to reduce each of these assertios to the fiite easure situatio, i which case they have already bee established.