MAT Analysis II. Notes on Measure Theory. Table of Contents. Riemann, We Have a Problem. Wm C Bauldry. Henri Léon Lebesgue ( )

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1 MAT Alysis II. Notes o Mesure Theory. Wm C Buldry BuldryWC@ppstte.edu Autum, 2006 Heri Léo Leesgue ( ) Tle of Cotets Riem, We Hve Prolem. 0. Riem, We Hve Prolem. 1. Towrd Uit of Mesure 1.1 Set Algers Sider: Borel Sets 1.2 Coutly Additive Mesures 1.3 Outer Mesure 1.4 Mesured xercises 2. Leesgue Mesure 2.1 Mesurle Sets 2.2 Mesure Zero Sider: Q Is Smll 2.3 Mesurle Fuctios 2.4 Fuctiolly Mesured xercises 3. Itegrtio 3.1 Riem Itegrl 3.2 Riem-Stieltjes Itegrl 3.3 Leesgue Itegrl 3.4 Covergece s Sider: Littlewood s Three Priciples 3.5 Itegrted xercises 4. Refereces There re prolems with Riem itegrtio. { 1 if x Q 1. Defie Dirichlet s fuctio (1829) D(x) = 0 otherwise. The D(x) dx does ot exist. [0,1] 2 2 x 0 x < Set f (x) = 1 2(1 x) 2 x < 1. The 0 otherwise [0,1] lim f (x) dx lim ter Heri Leesgue i [0,1] f (x) dx.

2 A Bd Sequece of Fuctios Towrd Uit of Mesure xmple Fid f, lim f, lim f, d lim f. Defiitio The legth of itervl i R 1 is the differece of the edpoits d is give y l([, ]) =. Gol: To hve set-fuctio m : M R tht mesures the size of set where m idelly stisfies: 1. M = P(R); id est, every set c e mesured. 2. For every itervl I, ope or closed or ot, m(i) = l(i). 3. If the sequece { } is disjoit, the m( ) = m( ). 4. m is trsltio ivrit; i.e., m( + x) = m() for every d y x R. Ufortutely, this is impossile. 1 We give up the first d llow sets ot to e i the clss of mesurle sets, M P(R). 1 ve the first 3 re impossile ssumig the cotiuum hypothesis. σ-alger of Sets Defiitio Let A e collectio of sets. The A is lger of sets or Boole lger iff if A A, the A c A, if A, B A, the A B A. De Morg s lws imply tht if A, B A, the A B A. The we lso hve A d X A. Defiitio Let A e lger of sets. The A is σ-lger of sets or Borel field iff for every coutle sequece {A i } of sets from A, we hve A i A. De Morg s lws imply tht coutle itersectios sty i A. There is smllest σ-lger cotiig y collectio of sets. Sider: Borel Sets Defiitio The Borel σ-lger o R is the smllest σ-lger cotiig G, ll of the ope sets i R, d is deoted y B(R). Propositio The Borel σ-lger B(R) is (lso) geerted y ech of: F = {ll closed sets i R} {(, ] : R} {(, ] :, R} Propositio Let S δ = { S i : S i S} d S σ = { S i : S i S}. The G G δ G δσ G δσδ F F σ F σδ F σδσ B(R) P(R)

3 Coutly Additive Mesure Defiitio A coutly dditive mesure is set fuctio m such tht m is o-egtive exteded rel-vlued fuctio o σ-lger M of susets of R; tht is, m : M [0, ]. m ( ) = m( ) for y sequece of disjoit susets. xercises Let m e coutly dditive mesure o the σ-lger M. 1. If A d B re i M with A B, the m(a) m(b). 2. If there is set A M with m(a) <, the m( ) = Show tht m is coutly sudditive or tht for y sequece of sets, m( ) m( ). (Hit: B = A [ 4. Let e the coutig mesure, the umer of elemets i set. Show tht stisfies Gols 1, 3, d 4. i< A i.) Outer Mesure Defiitio The outer mesure of A is m (A) = if A S l(i ) I where I is ope d I covers A with coutle uio. Propositio The outer mesure of itervl is its legth or m (I) = l(i). I. I = [, ]. () Sice [, ] ( ɛ, + ɛ), the m (I). () Heie-Borel thm: we eed oly cosider fiite covers. Work with the fiite cover to show l(i ). II. Ay fiite itervl I. There is closed itervl J I such tht l(i) ɛ l(j) = m (J) m (I) m (I) = l(i). III. Ay ifiite itervl. Outer Mesure is Coutly Sudditive Mesured xercises Let {A } e coutle collectio of susets of R. The ( ) m A m (A ) Wolog ll A s hve fiite outer mesure. For ech A there is coutle collectio of ope itervls {I,i } coverig A such tht l(i,i ) < m (A ) + ɛ 2 i The set {I,i :, i N} covers A. Thece ( ) m A l(i,i ) = l(i,i ) <,i i ( m (A ) + ɛ 2 ) xercises 1. If A is coutle set, the m (A) = The closed itervl [0, 1] is ot coutle. 3. Show tht m (Q [0, 1]) = 0 d m (Q) = Let A = Q [0, 1] d let {I : = 1..N} e fiite collectio of ope itervls coverig A. The l(i ) Recocile 1. through Give y set A d y ɛ > 0, there is ope set G such tht A G d m (G) m (A) + ɛ. (Cofer Littlewood s Three Priciples. ) 7. Why is m trsltio ivrit?

4 Leesgue Mesure Mesurle Sets Leesgue outer mesure m stisfies gols 1, 2, d 4, ut ot gol 3, coutle dditivity; m is oly coutly sudditive. We c gi coutle dditivity y givig up gol 1 d reducig the collectio M of sets; there will e sets tht c t e mesured. This pproch is ot without difficulties, though. The existece of omesurle sets 2 leds to prolems such s Vitli s theorem which yields method of decomposig the itervl [0, 1] ito set of mesure 2. (Also see the Husdorff prdox.) We will use the defiitio of set eig mesurle tht ws give y Crthéodory. 2 See No-mesurle set for ituitive expltio. Defiitio The set is mesurle iff for ech set A we hve m (A) = m (A ) + m (A c ). Propositio If is mesurle, the the c is mesurle. Propositio If m () = 0, the is mesurle. Let A e y set. The A implies m (A ) m (). Hece m (A ) = 0. Now A c A, so m (A) m (A c ) = m (A ) + m (A c ). Properties of Mesurle A Bigger Mesurle Cup Propositio If 1 d 2 re mesurle, the so is 1 2. Let A e y set. Sice 2 M, the m (A 1 c) = m ((A 1 c) 2) + m ((A 1 c) c 2 ). From A ( 1 2 ) = (A 1 ) (A 2 1 c ), we see tht m (A ( 1 2 )) m (A 1 ) + m (A 2 1 c) So m (A ( 1 2 )) + m (A ( 1 2 ) c ) m (A 1 ) + m (A 2 c 1) + m (A ( 1 2 ) c ) = m (A 1 ) + m (A c 1) = m (A) Propositio M is lger of sets. Propositio Let A e y set d 1, 2,..., N e fiite sequece of disjoit mesurle sets. The [ N ]) N m (A i = m (A i ) ( ) Iductio o with A i = A d ( ) ( 1 ) A i c = A i.

5 A Coutle Mesurle Cup The Leesgue Mesure m. Propositio Let 1, 2,... e coutle sequece of mesurle sets. The = i is mesurle. Wolog the i re pirwise disjoit. (Otherwise defie B i = i i 1 j=1 j.) Let A e y set d set F = i. The F M d F c c. The m (A F ) = m (A i ). Hece, sice is ritrry d m (A) is idepedet of, m (A) m (A i )+m (A c ) m (A )+m (A c ) Propositio M is σ-lger of sets. Defiitio Defie Leesgue mesure to e the restrictio m = m M. The Borel sets re Leesgue mesurle. Let e set d let ɛ > 0. TFA: 1. is Leesgue mesurle 2. there is ope set G such tht m (G ) < ɛ 3. there is closed set F such tht m ( F ) < ɛ 4. there is G G δ such tht G d m (G ) = 0 5. there is F F σ such tht F d m ( F ) = 0 Mesure Zero Sider: Q Is Smll Defiitio A set S R hs mesure zero if d oly if m(s) = 0; i.e., for y ɛ > 0 there is ope cover C = {G k k N} of S such tht k N m(g k ) < ɛ. xmple 1. Ay fiite set (coutle set) hs mesure zero. 2. very itervl [, ] is ot mesure zero (whe < ). The legth of [0, 1] is 1. The rtiols cotied i [0, 1] hve mesure zero. Wht is the mesure of the irrtiols i [0, 1]? Defiitio (A..) A property tht holds for ll x except o set of mesure zero is sid to hold lmost everywhere. The rtiols re coutle. Let Q e the set of rtiol umers. The rry elow shows method of eumertig ll elemets of Q. 1/1 (1) 2/1 (2) 3/1 (4) 4/1 (7)... 1/2 (3) 2/2 (5) 3/2 (8) 4/2 (12)... 1/3 (6) 2/3 (9) 3/3 (13) 4/3 (18)... 1/4 (10) 2/4 (14) 3/4 (19) 4/4 (25) Sice ech rtiol is couted, we hve Q N where we use to idicte crdility (or size). But we kow tht N Q, so tht N Q. Hece Q = N.

6 Coverig Q The set of rtiols hs mesure zero. Let ɛ > 0. List the rtiols i order Q = {r 1, r 2, r 3,... } s give y the coutility mtrix defied erlier. For ech rtiol r k, defie the ope itervl I k = (r k ɛ/2 k+1, r k + ɛ/2 k+1 ). The the collectio C = {I k k N} forms ope cover of Q, the legth of ech I k is m(i k ) = ɛ/2 k. The m(q) m(c) which is m(q) m(c) = m(i k ) = ɛ/2 k 1 = ɛ 2 k = ɛ Mesurle Fuctios Propositio (mesurility coditio) Let f e exteded rel-vlued fuctio o mesurle domi D. The TFA: 1. For ech α R, the set {x : f(x) > α} is mesurle. 2. For ech α R, the set {x : f(x) α} is mesurle. 3. For ech α R, the set {x : f(x) < α} is mesurle. 4. For ech α R, the set {x : f(x) α} is mesurle. These imply 5. For ech β R, the set {x : f(x) = β} is mesurle. (1.) = (2.) {x : f(x) α} = {x : f(x) > α 1/} (2.) = (3.) {x : f(x) < α} = D {x : f(x) α} (3.) = (4.) {x : f(x) α} = {x : f(x) < α + 1/} (4.) = (1.) {x : f(x) > α} = D {x : f(x) α} (.) = (5.) xercise. (2 cses: β < d β = ±.) Defiitio of Mesurle Fuctio Defiitio Let D e mesurle. The f : D R is mesurle iff f stisfies the mesurility coditio. Propositio Let f d g e mesurle (rel-vlued) fuctios defied o D d c R. The f + c, cf, f ± g, f 2, d fg re mesurle. Proof (sketch). (f + c, cf): Use {x : f(x) + c < α} = {x : f(x) < α c}, etc. (f + g): If f(x) + g(x) < α, there is r Q (r = r(α) r(x)) such tht f(x) < r < α g(x). Thus {x : f(x) + g(x) < α} = ({x : f(x) < r} {x : g(x) < α r}) r is coutle uio of mesurle sets, hece is mesurle. (f 2 ): Use {x : f 2 (x)>α}={x : f(x)> α} {x : f(x)< α}. (fg): Use fg = 1 4 (f + g)2 1 4 (f g)2. Sequeces of Mesurle Fuctios Let {f } e sequece of mesurle fuctios o commo domi D. The the fuctios sup{f 1,..., f }, sup f, lim sup f re mesurle. Alogous sttemets hold for if d lim if. Set h = sup{f 1,..., f }, the {x : h(x) > α} = {x : f i (x) > α}. Hece h is mesurle. Now set g = sup f, the {x : g(x) > α} = {x : f i (x) > α}. Hece g is mesurle. Comie the ove with the defiitio lim sup f = if sup f k to fiish. k

7 Simple Fuctios re Mesurle Propositio If f is mesurle d g = f.e., the g is mesurle. Set = {x : f(x) g(x)}. The m() = 0. So {x : g(x) > α} = {x : f(x) > α} {x : g(x) > α} {x : g(x) α}. Defiitio A mesurle rel-vlued fuctio φ is simple if it ssumes oly fiitely my vlues. The φ(x) = α k χ Ak (x) where A k = {x : φ(x) = α k } If ech A k is itervl, the φ is clled step fuctio. xmple N s(x) = k 2 N 2 χ [ k 1 N, k N ](x) is step fuctio; χ Q is simple. Mesurle Fuctios re Simple Propositio Let f :[, ] R e mesurle such tht m({f(x)= ± }) is zero. Give ɛ > 0, there is step fuctio s d cotiuous fuctio h so tht f s < ɛ d f h < ɛ.e. Proof (xercise). 1. There is M such tht f M except o set of mesure < ɛ/3. 2. There is simple fuctio φ such tht f φ < ɛ except whe f > M. (Hit: (M M) ɛ.) 3. There is step fuctio g such tht g = φ except o set of mesure < ɛ/3. (Hit: look here.) 4. There is cotiuous fuctio h such tht h = g except o set of mesure < ɛ/3. (Hit: thik like splie.) Fuctiolly Mesured xercises xercises 1. Let φ 1 d φ 2 e simple fuctios d c R. Show tht. cφ is simple fuctio,. φ 1 + φ 2 is simple fuctio, c. φ 1 φ 2 is simple fuctio. 2. For set S defie { the chrcteristic or idictor fuctio to 1 x S e χ S (x) =. Show tht 0 x / S. χ A B = χ A χ B,. χ A B = χ A + χ B χ A χ B. c. χ A c = 1 χ A. 3. Let D e dese set of rel umers; i.e., every itervl cotis elemet of D. Let f e exteded relvlued fuctio o R such tht for y d D, the set {x : f(x) > d} is mesurle. The f is mesurle. Itegrtio We eg y lookig t two exmples of itegrtio prolems. The Riem itegrl over [0, 1] of fuctio with ifiitely my discotiuities did t exist eve though the poits of discotiuity formed set of mesure zero. (The poits of discotiuity formed dese set i [0, 1].) The limit of sequece of Riem itegrle fuctios did ot equl the itegrl of the limit fuctio of the sequece. (ch fuctio hd re 1 / 2, ut the limit of the sequece ws the zero fuctio.) We will look t Riem itegrtio, the Riem-Stieltjes itegrtio, d lst, develop the Leesgue itegrl. There re my other types of itegrls: Droux, Dejoy, Guge, Perro, etc. See the list give i the See lso sectio of Itegrls o Mthworld.

8 Riem Itegrl Defiitely Riem Itegrl Defiitio A prtitio P of [, ] is fiite set of poits such tht P = { = x 0 < x 1 < < x 1 < x = }. Set M i = sup f(x) o [x i 1, x i ]. The upper sum of f o [, ] w.r.t. P is U(P, f) = M i x i The upper Riem itegrl of f over [, ] is f(x) dx = if U(P, f) P xercise 1. Defie the lower sum L(P, f) d the lower itegrl f. Defiitio If f(x) dx = f(x) dx, the f is Riem itegrle d is writte s f(x) dx d f R o [, ]. Propositio A fuctio f is Riem itegrle o [, ] if d oly if for every ɛ > 0 there is prtitio P of [, ] such tht U(P, f) L(P, f) < ɛ. If f is cotiuous o [, ], the f R o [, ]. If f is ouded o [, ]with oly fiitely my poits of discotiuity, the f R o [, ]. Properties of Riem Itegrls Propositio Let f d g R o [, ] d c R. The cf dx = c f dx (f + g) dx = f dx + g dx f g R if f g, the f dx g dx f dx f dx Defie F (x) = x f(t) dt. The F is cotiuous d, if f is cotiuous t x 0, the F (x 0 ) = f(x 0 ) If F = f o [, ], the f(x) dx = F () F () Riem Itegrted xercises xercises 1. If f(x) dx = 0, the f = Show why 1 0 χ Q(x) dx does ot exist. 3. Defie +1 ( k 1 S (x) = k ) χ [ k 1 k, k k+1) (x) χ [ +1,1](x) How my discotiuities does S hve? 3.2 Prove tht S (x) = 0.e. 3.3 Clculte 1 0 S (x) dx. 3.4 Wht is S? 3.5 Does 1 0 S (x) dx exist? (See imted grph of S N.)

9 Riem-Stieltjes Itegrl Defiitio Let α(x) e mootoiclly icresig fuctio o [, ]. Set α i = α(x i ) α(x i 1 ). Set M i = sup f(x) o [x i 1, x i ]. The upper sum of f o [, ] w.r.t. α d P is U(P, f, α) = M i α i The upper Riem-Stieltjes itegrl of f over [, ] w.r.t. α is f(x) dα(x) = if U(P, f, α) P xercise 1. Defie the lower sum L(P, f, α) d lower itegrl fdα. Defiitely Riem-Stieltjes Itegrl Defiitio If f dα = f dα, the f is Riem-Stieltjes itegrle d is writte s f(x) dα(x) d f R(α) o [, ]. Propositio A fuctio f is Riem-Stieltjes itegrle w.r.t. α o [, ] iff for every ɛ > 0 there is prtitio P of [, ] such tht U(P, f, α) L(P, f, α) < ɛ. If f is cotiuous o [, ], the f R(α) o [, ]. If f is ouded o [, ]with oly fiitely my poits of discotiuity d α is cotiuous t ech of f s discotiuities, the f R(α) o [, ]. Properties of Riem-Stieltjes Itegrls Propositio Let f d g R(α) d i β o [, ] d c R. The cf dα = c f dα d f d(cα) = c f dα (f + g) dα = f dα + g dα d f d(α + β) = f dα + f dβ f g R(α) if f g, the f dα g dα f dα f dα Suppose tht α R d f is ouded. The f R(α) iff fα R d f dα = f α dx Riem-Stieltjes Itegrls d Series Propositio If f is cotiuous t c (, ) d α(x) = r for x < c d α(x) = s for c < x, the f dα = f(c) (α(c+) α(c )) = f(c) (s r) Propositio Let α = x, the gretest iteger fuctio. If f is cotiuous o [0, ], the f(x) d x = f(k) 0

10 Riem-Stieltjes Itegrted xercises xercises x dx2 2. π/2 0 cos(x) d si(x) 3. 5/2 0 x d(x x ) ex d x 5. 3/2 3/2 ex d x ex d x 7. Set H to e the Heviside fuctio; i.e., { 0 x 0 H(x) = 1 otherwise. Show tht, if f is cotiuous t 0, the + f(x) dh(x) = f(0). Leesgue Itegrl We strt with simple fuctios. Defiitio A fuctio hs fiite support if it vishes outside fiite itervl. Defiitio Let φ e mesurle simple fuctio with fiite support. If φ(x) = i χ Ai (x) is represettio of φ, the φ(x) dx = i m(a i ) Defiitio If is mesurle set, the φ = φ χ. Itegrl Lierity Propositio If φ d ψ re mesurle simple fuctios with fiite support d, R, the (φ + ψ) = φ + ψ. Further, if φ ψ.e., the φ ψ. Proof (sketch). N I. Let φ = αi χ d ψ = M Ai βi χ Bi. The show φ + ψ c K e writte s φ + ψ = (αki + β )χ kj k for the properly chose k. Set A 0 d B 0 to e zero sets of φ d ψ. (Tke { k : k = 0..K} = {A j B k : j = 0..N, k = 0..M}.) II. Use the defiitio to show ψ φ = (ψ φ) 0 = 0. Steps to the Leesgue Itegrl Propositio Let f e ouded o M with m() <. The f is mesurle iff if ψ = sup φ f ψ f φ for ll simple fuctios φ d ψ. I. Suppose f is ouded y m. Defie { k = x : k 1 M < f(x) k } M, k The k re mesurle, disjoit, d hve uio. Set ψ (x) = M k χ k (x), φ (x) = M (k 1) χ k (x)

11 SLI (cot) xmple Steps (proof cot). The φ (x) f(x) ψ(x), d so if ψ ψ = M k m( k ) k= sup φ φ = M (k 1) m( k ) k= Thus 0 if ψ sup φ M m(). Sice is ritrry, equlity holds. II. Suppose tht if ψ = sup φ. Choose φ d ψ so tht φ f ψ d (ψ φ ) < 1. The fuctios ψ = if ψ d φ = sup φ re mesurle d φ f ψ. The set = {x : φ (x) < ψ (x)} hs mesure 0. Thus φ = ψ lmost everywhere, so φ = f.e. Hece f is mesurle. xmple Defiig the Leesgue Itegrl Defiitio If f is ouded mesurle fuctio o mesurle set with m() <, the f = if ψ ψ f for ll simple fuctios ψ f. Propositio Let f e ouded fuctio defied o = [, ]. If f is Riem itegrle o [, ], the f is mesurle o [, ] d f = f(x) dx; the Riem itegrl of f equls the Leesgue itegrl of f. Properties of the Leesgue Itegrl Propositio If f d g re mesurle o, set of fiite mesure, the (αf + βg) = α f + β g if f = g.e., the f = g if f g.e., the f g f f if f, the m() f m() if A B =, the f = f + f xercise. A B A B

12 Leesgue Itegrl xmples xmples { 1 q x = p q 1. Let D(x) = Q }. The 0 otherwise { } 1 x Q 2. Let χ Q (x) =. The 0 otherwise 3. Defie +1 ( k 1 f (x) = k [0,1] [0,1] D = χ Q D(x) dx. χ Q (x)dx. ) χ [ k 1 k, k k+1) (x) χ [ +1,1](x). +2 The 3.1 f is step fuctio, hece itegrle 3.2 f (x) = 0.e f = [0,1] 1 0 f (x) dx < 3 8 xtedig the Itegrl Defiitio Defiitio Let f e oegtive mesurle fuctio defied o mesurle set. Defie f = sup h h f where h is ouded mesurle fuctio with fiite support. Propositio If f d g re oegtive mesurle fuctios, the c f = c f for c > 0 f + g = f + g If f g.e., the f g xercise. Geerl Leesgue s Itegrl Defiitio Set f + (x) = mx{f(x), 0} d f (x) = mx{ f(x), 0}. The f = f + f d f = f + + f. A mesurle fuctio f is itegrle over iff oth f + d f re itegrle over, d the f = f + f. Propositio Let f d g e itegrle over d let c R. The 1. cf = c f 2. f + g = f + g 3. if f g.e., the f g 4. if A, B re disjoit m le susets of, f = A B A f + B f Covergece s (Bouded Covergece ) Let {f : R} e sequece of mesurle fuctios covergig to f with m() <. If there is uiform oud M for ll f, the lim f = lim Proof (sketch). Let ɛ > 0. A A f 1. f coverges lmost uiformly; i.e., A, N s.t. m(a) < ɛ 4M ɛ d, for > N, x A = f (x) f(x) 2 m(). ( ) 2. f f = f f f f = + f f A A ɛ ɛ 3. f f + f + f m() + 2M 2m() 4M = ɛ

13 Leesgue s Domited Covergece (Domited Covergece ) Let {f : R} e sequece of mesurle fuctios covergig.e. o with m() <. If there is itegrle fuctio g o such tht f g the lim f = lim f Lemm Uder the coditios of the DCT, set g = sup {f, f +1,... } k d h = if {f, f +1,... }. The g d h re itegrle d k lim g = f = lim h.e. Proof of DCT (sketch). Both g d h re mootoe d covergig. Apply MCT. h f g = h f g. Icresig the Covergece (Ftou s Lemm) If {f } is sequece of mesurle fuctios covergig to f.e. o, the lim f lim if f (Mootoe Covergece ) If {f } is icresig sequece of oegtive mesurle fuctios covergig to f, the lim f = lim Corollry (Beppo Levi (cf.)) If {f } is sequece of oegtive mesurle fuctios, the f = =1 =1 f f Sider: Littlewood s Three Priciples xtesios of Covergece Joh desor Littlewood sid, The extet of kowledge required is othig so gret s sometimes supposed. There re three priciples, roughly expressile i the followig terms: every mesurle set is erly fiite uio of itervls; every mesurle fuctio is erly cotiuous; every coverget sequece of mesurle fuctios is erly uiformly coverget. Most of the results of lysis re firly ituitive pplictios of these ides. From Lectures o the Theory of Fuctios, Oxford, 1944, p. 26. The sequece f coverges to f... Defiitio (Covergece Almost verywhere) lmost everywhere if m({x : f (x) f(x)}) = 0. Defiitio (Covergece Almost Uiformly) lmost uiformly o if, for y ɛ > 0, there is set A with m(a) < ɛ so tht f coverges uiformly o A. Defiitio (Covergece i Mesure) i mesure if, for y ɛ > 0, lim m({x : f (x) f(x) ɛ})=0. Defiitio (Covergece i Me (of order p > 1)) [ ] 1/p i me if lim f f p = lim f f p = 0

14 Itegrted xercises xercises 1. Prove: If f is itegrle o, the f is itegrle o. 2. Prove: If f is itegrle over, the f f. 3. True or Flse: If f is itegrle over, the f is itegrle over. 4. Let f e itegrle over. For y ɛ > 0, there is simple (resp. step) fuctio φ (resp. ψ) such tht f φ < ɛ. 5. For = k + 2 ν, 0 k < 2 ν, defie f = χ [k2 ν,(k+1)2 ν ]. 5.1 Show tht f does ot coverge for y x [0, 1]. 5.2 Show tht f does ot coverge.e. o [0, 1]. 5.3 Show tht f does ot coverge lmost uiformly o [0, 1]. 5.4 Show tht f 0 i mesure. 5.5 Show tht f 0 i me (of order 2). Refereces Texts o lysis, itegrtio, d mesure: Mthemticl Alysis, T. Apostle Priciples of Mthemticl Alysis, W. Rudi Rel Alysis, H. Royde Leesgue Itegrtio, S. Che Geometric Mesure Theory, F. Morg Compriso of differet types of itegrls: Itegrl, Mesure, d Derivtive: A Uified Approch, G. Shilov d B. Gurevich