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# Real Variables 2 MATH 651

Rae Kutch
WVU
GPA 3.8

Staff

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COURSE
PROF.
Staff
TYPE
Class Notes
PAGES
11
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 11 page Class Notes was uploaded by Rae Kutch on Saturday September 12, 2015. The Class Notes belongs to MATH 651 at West Virginia University taught by Staff in Fall. Since its upload, it has received 21 views. For similar materials see /class/202660/math-651-west-virginia-university in Mathematics (M) at West Virginia University.

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Date Created: 09/12/15
Real Analysis 2 Math 651 Spring 2005 April 26 2005 1 Real Analysis 2 Math 651 Spring 2005 Krzysztof Chris Ciesielski 11205 sec 31 and my article How good is the Lebesgue measure Math Intelligencer 112 1989 5458 N 123 Measure on R is a function m from a family M of subsets of R into 07 oo Desired properties for a measure m U En Zn MEL iii m is nitely additive lf EF 6 M7 then mE U E iv m is translation invariant If E E M and z E R then x E E M and mz E iv m is isometrically invariant If E E M and and 239 is an isometry of R then 6 M and Facts discussed 0 There is no measure satisfying i ii iii iv To be shown later 0 Lebesgue measure we will construct will satisfy i ii iii iv We will construct it only for n 17 but it works also for any 0 There are many measures satisfying i iii iv Eg m E 07 m E 007 m being counting measure 0 There are measures satisfying i ii iii iv Eg m E 07 m E 007 m being counting measure Real Analysis 2 Math 651 Spring 2005 April 26 2005 2 0 There are measures satisfying i ii iii iv for 71 S 27 but there are no such measures for 71 2 3 by Banach Tarski Paradox o The existence of measures satisfying i ii iii cannot be easily decided with the framework of usual axioms of set theory 11305 sec 32 Homework EX 1 Prove that for any measure 771 M a 07 oo satisfying i ii iii iv we have also ii every interval J belongs to M and 771J is equal to its length J Hint Prove the statement showing the following steps H Every interval J belongs to M D 0 for every x E R 3 77102 2 for every 71 012 4 771ab Zab for every ab 6 R a lt b 5 771J J for every interval J We proved de ned outer measure 771 and proved Proposition 17 page 56 that 771J J for every interval J 11905 sec 32 and 33 Proposition 2 page 57 Outer measure 771 is subattitive for every sets An C R we have 771 An anAn Corollary 3 page 58 If A C R is countable7 then 771A 0 Corollary 4 page 58 The interval 01 is not countable Real Analysis 2 Math 651 Spring 2005 April 26 2005 3 Proved Proposition 5 Asked to try Exercise 8 for the next day7 A set E C R is measurable E E M provided for every A C R 771A 771A E 771A E or7 equivalently7 771A 2 771A E 771A A Lebesgue measure 771 is equal to 771 restricted to M We need to show that M is a U algebra containing all intervals This will be done in the following steps 1 If E 6 M7 then E 6 M7 as the de nition is symmetric in E and E 2 MR 6 M 3 Lemma 6 If 771E 07 then E E M 12005 sec 33 4 Lemma 7 If E17 E2 6 M7 then E1 U E2 6 M 5 Corollary 8 M is an algebra on R 6 Lemma 9 771 is nitely additive Moreover7 if E17 En E M are pair wise disjoint and A C R7 then 771 A UL1 21 771 A 7 Theorem 10 M is a U algebra on R containing all sets of outer mea sure 0 8 Lemma 11 aoo E M 9 Theorem 12 M contains all Borel sets 10 Exercise 9 M is translation invariant Proposition 13 771 is countably additive Homework Exercises 7 X5 8 page 58 and exercise 97 page 64 Real Analysis 2 Math 651 Spring 2005 April 26 2005 4 Solution to Exercise 1 Prove that for any measure m M a 0 oo satis fying i ii iii iv we have also quotevery interval J belongs to M and mJ is equal to its length 6U PROOF Step 1 Every interval J belongs t0 M By iv for every x E R we have xz 1 z 01 6 M and so by the property iii the sets L 00 UO n z n 1 and foo a Uf h 7 n z 7 n 1 belong to M Thus by i R foo oo foo 0 U 0 00 E M and so for every x E R the intervals 700 R zoo and Loo R foox are in M We proved that every unbounded interval is in M Since every bounded interval is an intersection of two of such intervals the claim of Step 1 follows Step 2 0 for every x E R Let a 6 000 By iv for every x E R we have mx 1 Since Uf12 C 01 by ii and rnonotonicity of m we get 21 0 21 m2 mU12 S 77100710 1 Thus a 0 and the claim of Step 2 follows Step 3 m02 2 for every n 012 For n 0 this is true since by iii m01 m01 0 m01 m01 1 So assume that for some n we have m0 2 2 We will prove this for n 1 Indeed by iv additivity and the inductive assumption we have 2 m 02 lt 1gt m 0 2 lt 1gt mlt27ltn1gt 02 lt 1gt M We mlt2lt 1gt72rgtgt m f 721M U 2 172 m 02 2 0 i 0 Thus m 02 1 Tm and by induction we conclude that the claim of Step 3 holds true Step 4 m0 z for every x 6 01 Let 1122 be a binary represen tation of a that is in 6 01 for every n and z 2f1ln2 Put 0 0 and for every k 12 put xk U1ln2 and k xk1k Notice that the intervals k are pairwise disjoint and that 0 a U2 1k Note also that may mk1 QM211 m0lk2k lkQk where for lk 1 this follows from Step 3 and for lk 0 it is obvious since then 1k 0 Thus by in we have mam m 01m 221mm 2 2W z concluding the argument for Step 4 Real Analysis 2 Math 651 Spring 2005 April 26 2005 5 Step 5 mab b 7 a Zab for every ab E R a lt b First note that m0 z for every x gt 0 If x lt 1 this follows from Step 4 So assume that z 2 1 and let n 6 123 be maximal such that n S x Then z 7 n 6 01 and 0a 71z U Uz lk 7 1k and all the intervals in this representations and disjoint So by Step 4 m0x m pm u Um 7110 71k H i F 3 M 7 mltn0z7nZmltk71071 k1 z7nn1 The general case follows as mab ma0 b7a m0 b7a b7a Step 6 mJ J for every interval J Let a lt b be real numbers Then mltlta7bgtgt 7 0 mltlta7bgtgt 7 we 7 mltlta7bgtgt 7 mltla7b 7 b 7 a and mab mab mab b 7 1 Since for every bounded interval J there are a lt b such that ab Q J Q ab for such J we have b 7 a mab S mJ S mab b 7 a and so mJ b 7 a J Thus mJ J holds for every bounded interval J To prove that it also holds for any unbounded interval notice that for every 1 E R we have mltltaoogtgt 2 mU1la 7w n 71 7 21m a n 071 7 oo and mltlt7ooagtgt 2 mU1la7n7a7n1 7 21mlta7n 071 7 oo Since any unbounded interval J contains an interval of the form aoo or 7ooa we conclude that for such J we have mJ 2 00 Therefore we have mJ oo J Exercise 8 page 58 Prove that ifmA 0 then mA U B mB PROOF Too complicated Clearly mB S mA U B Fix an 8 gt 0 Then there exist the families 12 and 2n1 of open intervals such that A C U 2 and B C U 2ndL with the property Zn 1 S mA 8 8 and Zn 12n1 S mB 8 Then A U B C U In and quotNA U B S Zn 5In Zn 1 Zn512n13 8 7703 8 So mB S mA U B S mB proving the result Rea Analysis 2 Math 651 Spring 2005 April 26 2005 6 12605 sec 33 Proposition A Let in be art arbitrary measure satisfying arid iii Let E1 Q E2 Q be measurable Theri m En lirniH00 PROOF Let E0 Q and Fk Ek Ek1 for every k 123 Then in En m F 221 in lirnH00 21 m Therefore7 m En lirnH00 21 m lirnH00 m U221 F lirnH00 m nishing the proof I Proposition B version of Proposition 14 page 62 Let in be art arbi trary measure satisfying arid iii Let F1 2 F2 2 be measurable IfmF1 lt 00 theri m F lirniH00 PROOF Note that if A Q B are measurable and mB lt 00 then WE A 7713 WA since mB mA U B A mA mB A Let En F1 Then En7s satisfy Proposition A Thus WWW Woe lirn lirn mF1 F lirn 7 mF1 7 lirn So7 m F lirniH00 l EX 11 If E17 E2 6 M then mE1 U E2 mE1 E2 mE1 Homework Exercise 117 page 64 Proposition 157 with the proof iigtivgti Real Analysis 2 Math 651 Spring 2005 April 26 2005 7 12705 sec 33 Proof of Proposition 15 iilt iii7 ivlt v7 and i gtvi Proof of igtvi Fix 8 gt 0 Let Lg n E N be a family of open intervals such that E C U In and Zn In lt mE 82 Let V U In Then mV m1 3 Zn mIn Zn In lt 82 As mE lt 007 we conclude that mV E mV 7 lt 82 By Proposition A we have lirnNH00 m U2 In m U21 In Thus7 there is an N for which mV 7m US In lt 82 Let U US In Then U is a nite union of open intervals We will show that it satis es vi For this rst note that MV lt 00 implies mV U mV 7 mU lt 82 Thus7 since V contains U and E7 mme mUEmEU mVEmVU e2e2 a Proof of vigti ln fact7 it is enough to show that vi irnplies ii So7 x an 8 gt 0 By vi there exists an open set U which is a nite union of intervals such that mUAE lt 84 Then7 mE U S mUAE lt 847 since E U C UAE Thus7 there exists an open set W D E U such that mW lt mE U 84 lt 64 84 82 Let V U U W Then V is open and contains E Moreover7 mVE mUE UWE mUE mWE mUAE mW 8482 lt 8 ll So7 V satis es ii I Solved Ex 14a Show that Cantor ternary set has rneasure zero Homework Exercise 14b7 page 64 Real Analysis 2 Math 651 Spring 2005 April 26 2005 8 2205 sec 34 and 35 Construction of a non measurable Vitali set as in How good is the Lebesgue measure page 55 see httpJacobimathwvuedu kciesOtherElectronicReprints12pdf For x E R let Em y E Ryiz E Q xQ Notice that E1 is an equivalence class of x with respect to the equivalence relation on R de ned as z N y if and only if y 7 x E Q Thus 8 Ema E R is a family of non empty pairwise disjoint sets and so is 80 Em 01z E R So by the Axiom of Choice there exists a set V such that V E contains precisely one element for every E E 80 Step 1 Exercises 15 page 66 If E C V is measurable then 0 Indeed as ZqEQ O1mq E m quQ pJ nL E S m02 2 and mq E mE we must have 0 as set Q 01 is in nite Step 2 V cannot be measurable By way of contradiction assume V E M Then by Step 1 mV 0 and we would have mRmltUqVgt ZmqV0 qu qEQ a contradiction Solve Exercise 16 page 66 by noting that if mA gt 0 then one of the sets A q V q E Q must be non measurable Homework bonus Exercise 17 page 66 Proposition 18 page 66 Note that is equivalent to f 1U E M for every open set U C 700 00 Let E E M and f E a 70000 We say that f is measurable if one of the above conditions hold Solve Exercise 19 page 70 Real Analysis 2 Math 651 Spring 2005 April 26 2005 9 2305 sec 35 Exercise 24 page 71 If fE a 70000 is measurable then f 1B E M for every Borel set E C 700 00 Let f S C 70000 f 1B E Then f contains all open sets and is a U algebra So it contains all Borel sets Proposition 19 page 67 Theorem 20 page 68 Discuss Exercises 25 and 28 page 71 De ne almost everywhere abbreviated as ae Proposition 21 page 69 Proposition 22 page 69 Homework project Exercise 23 page 71 prove Prop 22 page 69 De ne simple function Homework Exercise 20 page 70 Real Analysis 2 Math 651 Spring 2005 April 26 2005 10 2905 sec 36 and 42 Three Littlewood s Principles Every measurable E C R of nite measure is nearly a nite union of intervals Proposition 15Vi For every 8 gt 0 there is an U C a7b which is a nite union of open intervals such that mUAE lt 8 Also7 other parts of Proposition 15 Eg7 For every 8 gt 0 there is a closed FCEwithmEFlt6 Every measurable function f a7b gt 00700 is nearly continuous Proposition 22 part For every measurable function f a7b a R and 8 gt 0 there is a continuous function h a7b a R such that max 6 am W 7 WM 2 8 lt e Lusin7s Theorem Exercise 317 page 74 For every measurable function f a7b a R and 8 gt 0 there is a continuous function h a7b a R such that max 6 am M he lt a Every convergent sequence fn a7b gt 00700 of mea surable functions is nearly uniformly convergent Egoroff7s Theorem Exercise 307 page 73 Let E be measurable of nite measure and assume that the sequence fn E a R of measurable functions converges to f E a R ae on E Then for every 77 gt 0 there is an A C E with mA lt 77 such that fn converges uniformly to f on E A

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