REAL ANALYSIS I HOMEWORK 3 CİHAN BAHRAN The quesions re from Sein nd Shkrchi s e. Chper 1 18. Prove he following sserion: Every mesurble funcion is he limi.e. of sequence of coninuous funcions. We firs suppose h f : E R is mesurble funcion (finie vlued) wih m(e) <. Then for every n N, by Lusin s heorem here eiss closed se F n E such h m(e F n ) 1/n nd f Fn is coninuous. For ech n N, wrie C n = n k=1 F n nd define g n = f Cn. By he psing lemm every g n is coninuous (he coninuous f Fk s re psed on finiely mny closed ses). Now since ech C n is closed, by he Tieze eension heorem here eiss coninuous funcion h n : E R such h h n Cn = g n. We clim h h n f.e. For, leing B = C n, since C n s re incresing, every B evenully flls in he C n s (h is, here eiss N N such h C n whenever n N); herefore lim n h n () = f(). So h n f on B nd m(e B) = m( (E C n )) m( (E F n )) =.. Show h here eiss closed ses A nd B wih m(a) = m(b) =, bu m(a + B) > : () In R, le A = C (he Cnor se), B = C/. Noe h A + B [, 1]. (b) In R, observe h if A = I {} nd B = {} I (where I = [, 1]), hen A + B = I I. () Le [, 1]. We know h hs ernry epnsion = n 3 n. Then if we define nd n if n 1 b n = oherwise if n = 1 c n = oherwise 1
REAL ANALYSIS I HOMEWORK 3 hen we hve = b n 3 n + c n 3 n = b n 3 n + 1 c n 3 n C + C/ since (b n ) nd (c n ) re sequences of s nd s. As ws rbirry bove, we obin [, 1] A + B. Hence m(a + B) 1, bu A nd B re closed ses of mesure zero. (b) Given (, y) I I, (, y) = (, ) + (, y) A + B. So I I A + B nd he reverse coninmen is similr so we hve A + B = I I. Therefore m(a + B) = 1, however boh A nd B cn be covered by recngles of re ε for ny given ε >, herefore m(a) = m(b) =. 1. Prove h here is coninuous funcion h mps Lebesgue mesurble se o non-mesurble se. [Hin: Consider non-mesurble subse of [, 1], nd is inverse imge in C by he funcion F in Eercise.] We know h [, 1] conins non-mesurble se N. So if we le F : C [, 1] o be he funcion in Eercise, since A := F 1 (N ) C, we hve m (A) = nd hence A is mesurble. Bu s F is surjecive, F (A) = N, ye N is no mesurble.. Le χ [,1] be he chrcerisic funcion of [, 1]. Show h here is no everywhere coninuous funcion f on R such h lmos everywhere. f() = χ [,1] () Wrie g = χ [,1] nd suppose such n f eiss. Since g = on se of posiive mesure (nmely R [, 1]), here eiss R such h f() =. Similrly, here eiss b R such h f(b) = 1. WLOG we my ssume < b. Consider he se S = { [, b] : f() = 1}. Firs, S Ø since b S. And S is bounded below from, hence hs n infimum, sy d. So here eiss sequence ( n ) in S such h f( n ) = 1 nd lim n n = d. Since f is coninuous, we ge f(d) = 1. So < d b nd by he definiion of c, for every [, d) we hve f() 1. In similr fshion, if we define c o be he supremum of he se { [, d] : f() = } hen f(c) =, c < d nd for every (c, d] we hve f(). Therefore, 1 / f((c, d)). Bu g is lwys eiher or 1, so is conrdicion. m ({ R : f() g()}) m((c, d)) = d c > 6. Suppose A E B, where A nd B re mesurble ses of finie mesure. Prove h if m(a) = m(b), hen E is mesurble. Noe h m(b) = m(a)+m(b A) nd since m(a) = m(b) < we hve m(b A) =. Now E A B A, so E A is null se nd in priculr mesurble. Thus
E = A (E A) is mesurble. REAL ANALYSIS I HOMEWORK 3 3 8. Le E be subse of R wih m (E) >. Prove h for ech < α < 1, here eiss n open inervl I so h m (E I) αm (I). Loosely speking, he esime shows h E conins lmos whole inervl. [Hin: Choose n open se O h conins E, nd such h m (E) αm (O). Wrie O s he counble union of disjoin open inervls, nd show h one of hese inervls mus sisfy he desired propery.] Since m (E) > nd 1 α > 1, we hve m (E) < 1 α m (E). So m (E) = inf{m(o) : E O-open} < 1 α m (E) herefore here eiss n open se O R conining E such h m(o) < 1 α m (E). Wrie O = I n where I n s re disjoin (nonempy) open inervls. Then E = E O = (E I n ) so we ge αm(i n ) = αm(o) < m (E) m (E I n ). Hence he semen for every n N, αm(i n ) m (E I n ) would yield conrdicion. Thus here eiss n N such h αm(i n ) < m (E I n ). 37. Suppose Γ is curve y = f() in R, where f is coninuous. Show h m(γ) =. Noe h since he mp preserves res of recngles, Γ hs he sme mesure wih he curve given by y = f(). Therefore we my ssume f is nonnegive. Also since Γ = {(, f()) : [n, n + 1]} nd mesure is counbly subddiive, i suffices o show h ech erm in he bove union hs mesure zero. Thus we my ssume h f : [, b] R where [, b] R is finie inervl. Moreover by replcing f wih f + 1, we my ssume h f() 1 for every [, b]. Then given < ε < 1, he se E ε = {(, y) : b, f() ε y f() + ε} conins Γ. Bu since f 1 > ε, boh f + ε nd f ε re nonnegive nd coninuous, herefore he mesure of E ε cn be clculed by definie Riemnn inegrl s m(e ε ) = (f() + ε)d (f() ε)d = ε d = ε(b ),
REAL ANALYSIS I HOMEWORK 3 4 So m(γ) ε(b ) for rbirrily smll ε. As, b is independen from ε, his shows h m(γ) =. 4. Suppose f is inegrble on [, b], nd Chper f() g() = d for b. Prove h g is inegrble on [, b] nd g()d = f()d. Firs, we show he clim ssuming f is non-negive. Le E = {(, ) R : < b }. Noe h E is mesurble nd conined in [, b] (, b]. Now define h : [, b] (, b] R (, ) χ E(, )f() which is non-negive nd mesurble (i is obined by lgebric operions on mesurble funcions). Using Tonelli s heorem we ge he following: firs, for lmos every [, b], he funcion is mesurble. Second, he funcion h : (, b] R [, b] R h(, ) h (,b] is mesurble. Noe h his funcion is nohing bu g becuse f() h = h(, )d = d. (,b] Third, we hve [,b] (,b] h = [,b] g = g()d. Also using Tonelli for he oher vrible, we ge he following: Firs, for lmos every (, b), he funcion is mesurble. Second, he funcion h : [, b] R (, b] R h(, ) h [,b]
REAL ANALYSIS I HOMEWORK 3 5 is mesurble. Noe h his funcion is nohing bu f, since Third, we hve Thus we ge [,b] h = [,b] (,b) h(, )d = h = (,b] g()d = f = f() d = f(). f()d. f()d. As f is inegrble on [, b], he inegrl bove is finie; hence g is lso inegrble on [, b]. For he generl cse, le P = { (, b] : f() } nd N = (, b] P nd define f + : (, b] R f() if P if N, f : (, b] R if P f() if N. Observe h f + nd f re inegrble non-negive funcions such h f = f + f. Now define g + : [, b] R g : [, b] R By he firs pr g +, g re inegrble nd g + ()d = g ()d = f + () d, f () d. f + ()d, f ()d. Observe h g + () = g () = [,b] P [,b] N f() d, f() d.
REAL ANALYSIS I HOMEWORK 3 6 Thus f() f() g + () g () = d + d [,b] P [,b] N f() = d = g() [,b] nd hence since everyhing is inegrble we cn do subrcion o ge g()d = = = g + ()d f + ()d f()d <. g ()d f ()d 6. Inegrbiliy of f on R does no necessrily imply he convergence of f() o s. () There eiss posiive coninuous funcion f on R so h f is inegrble on R, bu ye lim sup f() =. (b) However, if we ssume h f is uniformly coninuous on R nd inegrble, hen lim f() =. [Hin: For (), consruc coninuous version of he funcion equl o n on he segmen [n, n + 1/n 3 ), n 1.] () To pch he holes of he hined funcion o eend i o coninuous funcion in nice wy, we use lemm. Lemm 1. Le, b, c, d R such h < b nd c, d >. Then inf f()d f : [, b] R+ is coninuous, f() = c, f(b) = d =. Proof. Le ε > such h ε < min{(b )/, c, d}. Then he funcion f : [, b] R ε c ( ) + c if + ε ε ε if + ε b ε d ε( b) + d if b ε b ε is well-defined, posiive nd coninuous. Is grph looks like he following: So we see h f()d = 1 ε(c ε) + ε(b ) + 1 ε(d ε) Ç c ε = ε + b + d ε å Ç å c + d = ε + b ε ε(c + d + b ).
REAL ANALYSIS I HOMEWORK 3 7 y d y = f() c ε + ε b ε b Hence indeed, he inegrl of posiive coninuous funcion f wih f() = c nd f(b) = d cn be mde rbirrily smll. Corollry. Le C be closed subse of R nd f : C R be posiive coninuous funcion. Then given ε >, f cn be eended o posiive coninuous funcion g : R R such h g C f + ε. Proof. Since U := R C is n open subse of R, i is (counble) union of disjoin open inervls, h is, here eiss sequences ( n ), (b n ) such h nd 1 < b 1 < < b < 3 < b 3 < U = ( n, b n ). Therefore ech n nd b n belongs o C. So by he bove proposiion, for every n N here eiss posiive coninuous funcion g n : [ n, b n ] R wih g n ( n ) = f( n ), g n (b n ) = f(b n ) such h g n ε [ n,b n]. n So he funcion is well-defined nd coninuous. And g = = g : R R f() if C g n () if [ n, b n ] C C C g + U f + f + ε. g g n [ n,b n]
Now le REAL ANALYSIS I HOMEWORK 3 8 C = ñn, n + 1n ô 3 n= which, being union of uniformly disn closed ses, is closed. And he funcion f : C R n is well-defined nd coninuous, wih f = C n= if î n, n + 1 n 3 ó 1 n n = 1 3 n <. So by he corollry, f cn be eended o coninuous funcion g : R R such h g <, h is, g is inegrble. However, since lim sup f() = we hve lim sup g() =. (b) We prove he conrposiive, h is, ssuming h f is uniformly coninuous nd lim f() we show h f is no inegrble (i follows h f is no inegrble). By uniform coninuiy, here eiss < δ < 1 such h f() f(y) ε/ whenever y < δ. Also since lim f() here eiss 1 such h f( 1 ) ε. Agin since lim f(), we my choose > 1 + 1 such h f( ) ε. Then we choose 3 > + 1 such h f( 3 ) ε nd proceed in his wy. Since δ < 1, by our consrucion he collecion of inervls I n := ( n δ, n + δ) -s n vries- is disjoin. Noe h for fied n, if y I n hen f( n ) f(y) ε/. Bu f( n ) ε, so f(y) ε/ by ringle inequliy. Therefore f f = f ε/ =, In I n s desired. (b) For every k N wrie E k = { R : f() > 1/k}. I suffices o show h ech E k is bounded. Suppose no, h is, for some k he se E k is unbounded. Then we cn find sequence { n } in E k such h n+1 > n + 1 for every n. Now by uniform coninuiy here eiss δ (, 1) such h y < δ implies f() f(y) < 1/k. Noe h he inervls I n := ( n δ, n + δ) re disjoin. And given y I n since f( n ) > 1/k we hve f(y) > 1/k. Thus f conrdicing h f L 1 (R). In f = n= f 1/k =, I n 8. If f is inegrble on R, show h F () = f()d is uniformly coninuous. Given, y R wih y, by ddiiviy of he Lebesgue inegrl we hve f + f = f (,] f()d + y [,y] f()d = (,y] y f()d.
REAL ANALYSIS I HOMEWORK 3 9 Since f is inegrble on R, he bove inegrls re ll finie. Therefore we cn perform usul lgebr o ge y f()d = y f()d = F (y) F (). f()d Given ε >, by Proposiion 1.1 pr (ii) in Sein & Shkrchi s e, here eiss δ > such h y y F (y) F () = f()d f() d < ε whenever y < δ (king E = [, y] in he semen of he proposiion). This is precisely uniform coninuiy for F. 11. Prove h if f is inegrble on R d, rel-vlued, nd E f()d for every mesurble E, hen f().e.. As resul, if E f()d = for every mesurble E, hen f() =.e. Wrie E n = { R d : f() < 1/n}. E n s re mesurble. Noe h { R d : f() < } = E n, So i is enough o show h every E n hs mesure zero. Suppose no, so m(e n ) > for some n. So using he ssumpion on E n, we hve 1 f E n E n n = 1 n m(e n) <, conrdicion. Now le s do he second pr. By he firs pr, we hve f.e. Wriing g = f, since E g = E ( f) = for every mesurble E, gin by he firs pr we deduce h f = g.e. Thus f.e. nd hence f =.e. 1. Show h here re f L 1 (R d ) nd sequence {f n } wih f n L 1 (R d ) such h bu f n () f() for no. f f n L 1. [Hin: In R, le f n = χ In, where I n is n ppropriely chosen sequence of inervls wih m(i n ).] For every n N here eiss unique k, j N such h n + k + j. Noe h k s n. So defining I n = î j, j+1 ó k, by binry epnsions every [, 1] belong k in infiniely mny I n s nd lso every [, 1] belong in he complemen of infiniely mny I n s. So given [, 1], if we define f n = χ In, he limi lim f n () does no eis. However f n = f n = m(i n ) = k s n. So king f = gives couneremple.