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by: Golden Bernhard


Marketplace > University of Florida > Statistics > STA 6466 > PROBABILITY THEORY 1
Golden Bernhard
GPA 3.83


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This 3 page Class Notes was uploaded by Golden Bernhard on Friday September 18, 2015. The Class Notes belongs to STA 6466 at University of Florida taught by Staff in Fall. Since its upload, it has received 19 views. For similar materials see /class/206580/sta-6466-university-of-florida in Statistics at University of Florida.




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Date Created: 09/18/15
STA 6466 Probability Theory Details and Comments on Vitali s Construction of a Nonmeasurable Set Following the notation of Billingsley we let 3 represent the class of subintervals a b of the unit interval 07 l and we let 3 represent the Borel a eld on 07 1 For x y C 07 1 let x 63 y denote addition modulo 1 in 07 l ie m if 0 lt m g l as e y y y 1 myillf1ltmy 2 Note that this can be thought of as addition of angles if we identify z with the angle 27m and agree to call the zero angle 27139 For any A C 07 land z E 07 1 let Amaxa A and let L A C B A6995 CB andAGBx AVx C 07 Then L is a system containing 3 see A2 and thus 3 C L by the 7r Theorem In other words A for any A C B and z E 07 l we have A 63 z E B and A 69 A Now for z y C 07 1 de ne z to be equivalent to y and write x N y if there eXists a rational number r C 07 1 such that z 63 r y This de nes an equivalence relation see A3 and thus 07 1 can be partitioned into the corresponding equivalence classes Let H C 07 1 be contain exactly one point from each equivalence class this is possible by the Axiom of Choice Then we claim that H M To show this we proceed as follows Let r1 r2 be an enumeration of the rationals in 07 l and consider the sets H 63 n 2 12 These sets form a partition of 07 l ie B the sets H 63 n are disjoint and their union is 07 1 see A4 Now if H C B then by A above H 63 r C B and H 63 n H for all i l7 2 By B and countable additivity of we then have 1 01 E H em ZMH i1 i1 But if H 0 then the right hand side of this equation is 0 while if H gt 0 then the right hand side is in nity In either case we have a contradiction and thus H 3 Remark 1 It is possible to prove the stronger and more interesting result that H M ie H is not a Lebesgue measurable set The only signi cant additional dif culty is in proving that A C M implies A 63 z E M but this becomes simpler once we have eXtended Lebesgue measure to 7007 00 Note that the argument used above invoking the 7r Theorem suf ces to show that A C 3 implies A 63 z E B but does not work for the larger class M of Lebesgue measurable sets A Further Details A1 Basics Facts We begin by proving a few basic facts concerning the binary operation 69 F11fmy E 071andzmythenzz1iy To see this note that ifO lt z y g 1 then zzyxy gt 217ym1gt z17yz whileif1ltmy 2then zmymy71gt 217ym gt 26917ym Thus in either caseze liy m F2 For 73472 6 071 yz zyz if0ltyz 1 m y271 if1ltyz 2 To see this note that To lt1 569wa m ByHA ltx yzi m69y2717 1f1ltmyz 2 myz if0ltxy 1and0ltzyz 1 7 my71z if1ltxy 2and0ltzy712 1 7 my271 if0ltxy 1and1ltzyz 2 my71271 if1ltxy 2and1ltzy712 2 myz if0ltmyz 1 my2717 if1ltmyz 2 my2727 if2ltmyz 3 the last equality results from combining the middle two cases in the previous step Nowif0lty2 1theneither0ltzyz lor1ltmyz 2andwe obtain y27 zyz if0ltmyz 1 zyz71 if1ltmyz 2 m yz Ifonthe otherhand l lt yz 2theneitherl lt myz 2or2 lt myz 3 andwe obtain 69 27 zyzil if0ltmyzil l y zyzil7l if1ltzy271 2 xEByzil This proves the result F3 Ifmy17 yg E 07 l and y1 63 z yg 63 m then yl yg This follows from F1 and F2 since yl 63 z yg 63 z implies 241 242 69 90 69 1 i 90 by F1 y2691 byF2sincezlizl yz bylsincellty2l 2 F4 If A7 B C 07 l are disjoint then so are A 63 z and B 63 m This follows immediately from F3 since if yl E A and yg E B are such that yl 63 z yg 63 m then yl yg contradicting A O B 0 F5 IfA1A2 c 0 1 then UnAn ea 25 04A ea z This follows from simple set theory For example if we de ne the function y y 63 m then this says that fUAn UfAn which was proven in the rst assign ment F61fA c 0 1 then A any A0 ea 35 That A 63 z and Ac 63 z are disjoint follows from F4 and the fact that A and A0 are disjoint while from F5 we have A 63 U A0 63 A U Ac 69m 07 l 69 07 1 A2 Translation Invariance TEXT DELETED A3 The Equivalence Relation TEXT DELETED A 5 The Partition TEXT DELETED


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