ACM 105: Applied Real and Functional Analysis. Solutions to Homework # 1.

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1 ACM 15: Applied Real ad Fuctioal Aalysis. Solutios to Homewor # 1. Ady Greeberg, Alexei Noviov Problem 1. Beroulli sequeces. a. I = {x l : x = a}. But by costructio of Beroulli sequeces, a ca oly be ±1. Moreover, if for I, a = 1, the Ĩ, the complemet of I, has the same structure with a = 1. Therefore the properties of I are iherited by Ĩ. Note that if we have a sequece x I, the lim x also lies i I, so I is closed. But by the argumet above, Ĩ is also closed. However the complemet of a closed set is ope, so I is also ope. It thus follows that I is both ope ad closed. b. We use the fact that for each, the umber of elemetary evets is fiite. Therefore the power set, i.e., the set of all possible evets depedig o the outcome of elemetary evets up till, is i itself fiite ad will be our σ-algebra F. Sice all sets are icluded, closeess uder all ids of uios ad itersectios is immediate. Note that usually (i.e., whe it is ot fiite) this set is too large to cosider as the σ-algebra of measurable evets. Now cosider F = F i. Sice X 1, X F such that X 1, X F (by defiitio of the uio of sets) ad each F is a σ-algebra, F has to be closed uder fiite uios ad itersectios, i.e., F is a algebra. However the set = I i = (1, 1,..., 1,...) / F i=1 sice does ot belog to ay F for ay fiite. Therefore, F is ot a σ-algebra. c. Usig Carathéodory s extesio theorem we ca defie the miimal σ-algebra that cotais uios of elemetary evets ad a measure o this σ-algebra. It will cotai more elemets tha F, sice, as we have just established, F is ot closed uder ifiite itersectios, while a σ-algebra always is. Problem. Probability of log leads. The arc-sie law. a. Cosider the set G α = {x : g (x) > α}. Obviously, G α is a uio of a umber of elemetary evets (amely, of those whose total umber of votes is greater tha α, coutig a vote for Al as +1 ad a vote for Bill as -1). Sice each elemetary evet is measurable by costructio ad there is at most coutably ifiite of them, G α has to be a measurable set by defiitio of the σ-algebra (of measurable sets). b. Obviously, f ( ) 1 = A i () where A i () = Al is wiig i exactly i out of ( tossigs. ach of the A i is a uio of )(at most, coutable) evets of the form {x : f (x) > } {x : f (x) = } {x : f 1 (x) > }, each of which is measurable, because so is f. Therefore, as a uio of a coutable umber of measurable evets, f(/) is measurable. c, e. See attached plot. We used M = N = 1 iteratios, sice for a larger umber of rus ad simulatios, the covergece is so good that you caot tell the two curves apart. The solid curve is the simulatio results; the dashed curve is the actual arc-sie. i= 1

2 Figure 1: Log leads: simulatios vs. limit distributios d. Here are the approximate umerical values: ( ( ) ) P f <. 11.3%; ( ( ) ) P.49 < f < %; ( ( ) ) P f >. 88.7%; {( ( ) ) ( ( ) )} P f >.976 f < %; {( ( ) ) P f <.54 (wi) ( ( ) ) } f >.476 (loss) 3.5% It turs out that it is much more liely for the wier to wi overwhelmigly tha by a slim margi. f, g. See plot o ext page. Problem 3. Measurable Sets ad Ope/Closed Sets. We first set off to prove that measurability implies existece of the specified ope ad close sets (i.e., ecessity). The coverse statemet (sufficiecy) follows after that. Necessity. Recall (Royde, p. 53) the defiitios of two classes of sets, G δ, which icludes coutable uios of ope sets, ad F σ, which icludes coutable itersectios of closed sets. By Carathéordory s theorem, both of these types of sets are measurable. We first prove that there exist sets F F σ ad

3 Figure : Number of returs: simulatios vs. limit distributios G G δ, such that F G ad µ(g\f) =. Recall the pricipal defiitios. Outer measure of a set is µ () = if µ(p ) P where P are disjoit ope itervals with the atural measure assiged to them. (This implies that P = P G δ, is measurable ad µ(p ) = µ(p ), so that µ () = if µ(p ).) Furthermore, a set is measurable if for ay set A µ (A) = µ (A ) + µ (A Ẽ) (1) where Ẽ is the complemet of. If we deote X Ẽ (i.e., the whole set ), the by substitutig X ito (1) for A we get a simple implicatio, amely, if is measurable, the µ () + µ (Ẽ) = µ (X) = µ(x) () Now proceed to the proof of the itermediate claim. Sice µ () is the ifium over all possible ope coverigs of, there exists a coverig O = P, O G δ, such that µ(o ) µ () + 1 Applyig the same argumet to Ẽ we fid a coverig Ẽ O G δ such that If we deote F = implies µ(o ) µ (Ẽ) + 1 O, the F F σ, is measurable ad F O. Moreover, the last iequality µ(x) µ(f ) µ (Ẽ) (3)

4 Sice is measurable, () holds, so we have µ(x) µ(f ) µ(x) µ () + 1 or µ(f ) µ () 1 Subtractig (4) from (3) ad usig the iclusio F O we obtai (4) µ(o ) µ(f ) = µ(o \F ) 1 Sice the right-had side teds to as, we have foud the desired F = lim F F σ ad G = lim O G δ. It is ow easy to fid the desired ope set O ad closed set F, followig from the measurability of G ad F. Sice G G δ, G = P, where P are ope sets. Without loss of geerality, assume that P are disjoit. The, sice G is measurable, µ(g) = µ(p ) <, so there exists a positive iteger N such that We deote µ(p ) < ε >N O = N =1 Note that as a fiite uio of ope sets, it is itself a ope set ad O with µ(o) µ() < ε/. Taig complemets or repeatig the argumet with F we fid the set F = N =1 where Q are closed, which is a closed set ad F with µ() µ(f ) < ε/. Together, the two iequalities give µ(o) µ(f ) = µ(o\f ) < ε as required. Sufficiecy. Let there exist a ope set O ad a closed set F such that F O ad µ(o\f ) < ε. Let us prove that is measurable, i.e., that (1) holds for ay set A. Sice both ope ad closed sets are measurable, for ay set A we have P Q µ (A) = µ (O A) + µ (Õ A) µ (A) = µ (F A) + µ ( F A) (5) Addig these two up produces, after some rearragemet, µ (A) = µ (O A) + µ ( F [ ] A) µ (A) µ (Õ A) µ (F A) The terms outside the square bracets ca be bouded from below usig the iclusios F O ad Õ Ẽ F ad mootoicity of the outer measure: ] µ (A) µ ( A) + µ (Ẽ [µ A) (A) µ (Õ A) µ (F A) (6) We ow perform some slic maipulatio of the terms iside the square bracets. Add ad subtract µ (O A): µ (A) µ (Õ A) µ (F A) = µ (A) µ (Õ A) µ (O A) }{{} +µ (O A) µ (F A) 4

5 The first three terms combie to because of the first of the equalities (5). Note also that sice O F, it follows that O = F (O\F ), so that the last three terms reduce to µ (O A) µ (F A) = µ (( F (O\F ) ) A ) µ (F A) µ (F A) + µ ((O\F ) A) µ (F A) = µ ((O\F ) A) µ (O\F ) < ε by subadditivity ad mootoicity of outer measure. Thus we bouded the term i the square bracets i (6) by ε, so it follows that µ (A) µ ( A) + µ (Ẽ A) ε Sice ε is arbitrary, this implies µ (A) µ ( A) + µ (Ẽ A) for ay A. Subadditivity implies that always µ (A) µ ( A) + µ (Ẽ A). The two iequalities ca hold simultaeously oly if i.e., (1) holds, so is measurable. µ (A) = µ ( A) + µ (Ẽ A) Note. The quatity µ(x) µ (Ẽ) is called the ier measure of set. A more geeral theorem (cf. Royde, Propositio 1.4) says that () is ot oly a ecessary, but also a sufficiet coditio of measurability: is measurable (i.e., (1) holds) if ad oly if µ () = µ (). It is ow easy to establish the results µ() = if µ(o) = sup µ(f ) O F Recall that a = if{x A} if (a) for ay x A, x a; ad (b) for ay ε > x A such that x < a + ε. Similarly, b = sup{x B} if (a) for ay x B, x b; ad (b) for ay ε > x A such that x > b ε. Sice we oly cosider F O, mootoicity of the measure implies µ(f ) µ() µ(o) for all such sets F ad O. Moreover, the property that has just bee established esures existece of such ope set O ad closed set F that µ(f ) < µ() + ε ad µ(o) > µ() ε Therefore both criteria for the sup (if) are satisfied. Problem 4. gorov s Theorem. Theorem (D.F.gorov). If {f } is a sequece of measurable fuctios that coverge to a real-valued fuctio f almost everywhere o a measurable set with fiite measure µ() <, the for ay ε > there is a subset A with µ(a) < ε, such that f coverges uiformly o \A. Remar. Rather tha usig the hit ad citig Propositio 3.4, we choose to prove the gorov s Theorem from the groud up. I fact, you will see that the proof follows the same lie of thiig as i Royde s text. Proof. Sice f f a.e., f is measurable o. Defie the collectio of sets m = { x : f i (x) f(x) < 1 } m For a fixed value of m, m i> is simply the set of all poits x such that f i (x) f(x) < 1 m holds for all i >. Obviously, m 1 m... m..., ad we ca defie m = =1 m. By the cotiuity property of measure (which ca be show to follow from coutable additivity), lim µ(m ) = µ( m ), ad therefore give ay m > ad ε >, there exists (m) such that µ( m \ m ) < ε (m) m 5

6 Now let ε = m=1 m (m). We claim that the desired set A is just \ ε. To see this we eed to prove that f f uiformly o ε ad that µ(\ ε ) < ε. Sice x ε we have f i (x) f(x) < 1 m for ay m 1 ad all i > (m), it follows that sup f i (x) f(x), ad thus uiform covergece o ε is established. To estimate the measure of the set \ ε, we first ote that µ(\ m ) = for every m. Ideed, if x \ m, the for ay i, o matter how large, the iequality f i (x ) f(x ) 1 m holds, so that f caot coverge to f o \ m. But sice f f a.e., the measure of this set has to be. Thus µ(\ m ) = (m) µ(m \ m ) < ε (m) m Therefore ( ) ( ) µ(\ ε ) = µ \ m (m) = µ (\ m ) (m) µ(\ m ) < ε (m) m = ε m=1 so that µ(\ ε ) < ε, ad we are doe. m=1 m=1 m=1 Problem 5. Luzi s Theorem. Theorem (N.N.Luzi). Let f(x) be a measurable real-valued fuctio o [a, b] ad let ε > be arbitrary. The there is a cotiuous fuctio g(x) such that the measure of the set {x : f(x) g(x)} is less tha ε. Proof. We first outlie the followig steps i the proof. 1. Based o Propositio 3. from Royde s boo (proved below), fid a sequece of cotiuous fuctios g such that g f < ε = ε/(3 ) o a set [a, b]\ with µ( ) < ε.. Coclude that there exists a sequece of cotiuous fuctios g f o [a, b]\ with µ() < ε/3. 3. Fid a subset A = [a, b]\ with µ( ) < ε/3 such that g f uiformly o A. 4. Fid a closed set F A with µ(a\f ) < ε/3 ad coclude that we have built a sequece of cotiuous fuctios g which coverges uiformly to f o a closed set F [a, b] with µ([a, b]\f ) < ε. Sice uiform covergece preserves cotiuity, lim g = g is cotiuous o F ad g = f o F. 5. xted g cotiuously to the whole iterval [a, b]. Thus we have a fuctio g which is cotiuous o [a, b] ad µ({x : g(x) f(x)}) = µ([a, b]\f ) < ε, Q..D. We first prove the followig claim, which is part of Propositio 3.. Claim. Give a measurable fuctio f o the iterval [a, b], for ay δ >, there exists a set A with µ([a, b]\a) < δ ad a cotiuous fuctio h such that f h < δ o A. Proof. Defie f (x) = m m (m + 1) if f(x) < with m ad > itegers. The the fuctios f are simple ad f f uiformly, sice f (x) f(x) = = 1/. Therefore for ay δ > there exists a idex N such that f (x) f(x) < δ for all N. Thus we have a simple fuctio (ϕ = f N, say), which is at most δ-close to our measurable fuctio f. Moreover, sice ϕ is simple, there exist a collectio of measurable sets i such that ϕ c = cost o. By the argumets used i solvig Problem 3 of this homewor, we ca approximate each with a fiite uio of disjoit ope itervals I = K (a, b ), so that µ(i \ ) < δ/ +1. By defiig g(x) = c =1 6

7 o I, we arrive at a step fuctio g (i.e., a fuctio which assumes costat values o itervals), such that g(x) = ϕ(x) except o the set = ( \I ) with µ() < µ( \I ) = δ +1 = δ The desired cotiuous fuctio h is simply the liear iterpolat of g: if, e.g., g = c 1 o the iterval (a 1, a ) ad g = c o the iterval (a, a 3 ), the we defie c 1, o ( ) a 1, a δ + h = +1 c c 1 δ x + c 1 +c, o ( ) a δ, a + + δ + c, o ( ) a + δ, a + 3 (these are othig else tha formulas for liear iterpolatio). The h is differet from g oly o the uio of the small itervals aroud each edpoit. The measure of this set is less tha δ/+1 = δ/. Thus for ay give δ we have the followig: a simple fuctio ϕ such that f ϕ < δ a step fuctio g such that g = ϕ except o a set of measure < δ/; this implies that g f < δ except o a set of measure < δ/ a cotiuous fuctio h such that h = g except o a set of measure < δ/, implyig that h f < δ except o a set of measure < δ/ + δ/ = δ The Claim is thus established. It is ow easy to fill i the details of the proof of Luzi s theorem. We shall stic with umbers i the outlie above. 1. For each simply tae δ = ε = ε/(3 ) for a give ε ad use the Claim above to come up with the correspodig g.. Follows immediately, sice g f as. 3. xistece of such a set A is guarateed by gorov s theorem. 4. xistece of such a closed set F A is proved i Problem 3 of this homewor. 5. xtesio is possible due to the result i Problem.4 i Royde s text. We are doe. Problem 6. Cotiuity of the Lebesgue Itegral. We eed to prove that for ay ε > there exists δ > such that y F (x) F (y) = f(x) dx < ε if oly x y < δ. This fact is clear if f(x) is bouded. Ideed, i this case, y y F (x) F (y) = f(x) dx f(x) dx sup f(x) x y < ε x x x 7

8 if we tae δ < ε/ sup f(x). I the geeral case, cosider the followig sets: A = {x : f(x) < + 1} N B N = = A C N = R B N Thus costructed, A is a disjoit partitio of the real lie; therefore, by the properties of the Lebesgue itegral, f(x) dx = f(x) dx Thus there exists N such that R =N +1 Now tae < δ < (N. The +1) y F (x) F (y) = f(x) dx ad we are doe. ε x y x A f(x) dx = f(x) dx = = A C N f(x) dx < ε [x,y] B N f(x) dx + (N + 1) x y + [x,y] C N C N f(x) dx < f(x) dx ε + ε = ε Problem 7. Covergece Theorems for Covergece i Measure. Theorem (bouded covergece ). Suppose the sequece of fuctios f measurable o the set with fiite measure is bouded above, i.e., f M. The if f f i measure, the f f. Proof. Covergece i measure meas that for ay ε > there exists a δ > such that the measure of the set S = {x : f (x) f(x) > δ} is less tha ε. (Note that the sets S may be very differet for each fuctio f, depedig o the structure of the sequece.) Therefore f f = (f f) f f = f f + f f S f f <=δ By choosig a sufficietly small δ, the measure of the set {x : f (x) f(x) > δ } ca be made less tha ε/(4m). Sice f < M ad f f i measure, f is also bouded by M, so that f f M, ad thus the first term i the expressio above is less tha ε/. As for the secod term, it is clearly bouded above by δµ(). Choosig δ = mi {δ, ε/(µ())}, obtai f f ε + ε = ε Sice ε is arbitrary, covergece of the itegrals is thus proved. Havig established the bouded covergece theorem for covergece i measure, we ca ow proceed exactly as i the covergece a.e. case, to prove Fatou s lemma ad the mootoe ad Lebesgue covergece theorems (cf. Royde s text). 8

9 Lemma (Fatou ). If f is a sequece of oegative measurable fuctios ad f f i measure o set, the f lim Proof. Defie a measurable fuctio h supported o a set with fiite measure, with { f, f M h =, f > M for some M >. Obviously, h M ad h f o. Now defie the sequece h (x) = = mi{h(x), f (x)}. Clearly, h h i measure, so by the bouded covergece theorem h = h = lim h lim But h f, so h f, ad taig the supremum over h we get f lim as required. Theorem (mootoe covergece ). Let f be a icreasig sequece of mesurable fuctios such that f f i measure. The f = lim f f f f Proof. By Fatou s Lemma, f lim f But sice for each we have f f, it follows that f f, which implies lim f f For both iequalities to hold, we must have lim f = lim f = lim f = f as required. Theorem (Lebesgue s covergece theorem). Let f be a sequece of fuctios measurable o the set with fiite measure ad suppose there exists a measurable fuctio g(x) such that f g. The if f f i measure, the f f. Proof. Obiviously, f g, ad therefore f is also measurable. Cosider the fuctio g f. It is oegative ad coverges to g f i measure; therefore, by Fatou s Lemma, (g f) lim (g f ) g lim which implies f lim f f (7) Now tae the fuctio g + f, which is also oegative ad coverges i measure to g + f. Applyig Fatou s Lemma agai, we obtai (g + f) g + lim 9 f

10 implyig f lim f (8) But both (7) ad (8) ca be true oly if f = lim f which proves the theorem. Problem 8. Gree s Fuctio for Poisso s quatio o a Uit Dis. a. Sice the problem is set i a ball, spherical coordiates are appropriate. Recall that if x 1 = r cos θ cos φ, x = r cos θ si φ, x 3 = r si θ ad y 1 = ρ cos θ cos φ, y = ρ cos θ si φ, y 3 = ρ si θ, the ( ) x y = r + ρ rρ cos θ cos θ + si θ si θ cos(φ φ ) ad the Gree s fuctio is 1/(4π x y ). We mae use of spherical symmetry to claim that the value of the itegral at ay poit x lyig o a sphere aroud the origi is the same. Therefore it suffices to evaluate the itegral at oe coveiet value of θ, say, θ =. I this case x y taes a particularly simple form, x y = r + ρ rρ cos θ ad we have u(x) = 1 4π D dy x y = 1 4π π π 1 ρ si θ dρ dθ dφ r + ρ rρ cos θ There is o depedece o φ, so the outer itegral reduces to simple multiplicatio by π. Itegratio dθ (e.g., i Maple) yields u(r) = 1 1 ρ ( ) ρ + r (ρ r) sg(ρ r) dρ r To itegrate the latter expressio, we split the iterval up i two parts: [, 1] = [, r] + [r, 1] ad use liearity of the itegral to get = 1 r u(r) = 1 1 ρ r dρ 1 1 which is the required expressio, sice r = x. b. Sice r ρ ( ) ρ + r (ρ r) sg(ρ r) dρ = r ρ dρ = 1 u(x) = 1 4π D ) ( r = r 6 1 f(y) x y dy ad the domai of itegratio (i.e., the uit dis D) is fixed, the oly x depedece comes from the deomiator of the itegrad. For ay h >, h is a simple liear operator, so it commutes with itegratio. Thus we have h u(x) = h 1 4π D f(y) x y dy = 1 4π D 1 f(y) h x y dy = D f(y) h G(x y) dy 1

11 c. For ay fixed y, x G(x y) = x G(x). We use the fact that G(x) oly depeds o x, so we use spherical coordiates agai: G(x) = 1 ( ρ ) ρ G ρ ρ + 1 ρ si θ G 1 ( φφ + si θgθ ρ ) θ = 1 ) (ρ 1 si θ ρ 4πρ = ρ This is true for ay fixed y; for y = x, this does ot hold. d. I b. above, put h = 1/. The we have a sequece of oegative measurable fuctios g (x) = 1/ G(x y). Sice lim g = g = a.e. (as proved i c.), it follows that g =. But we ow that lim g = 1. Therefore we see that Fatou s lemma still holds. However we ca easily establish that g is either bouded or mootoe, so oe of the covergece theorems implyig equality of the itegrals ca be used. Refereces [1] Royde H.L., Real Aalysis, 3rd ed., Pretice-Hall, [] Kolmogorov A.N., Fomi S.V., Itroductory Real Aalysis, Dover,