The Borel-Cantelli Lemma and its Applications

Transcription

1 The Borel-Catelli Lemma ad its Applicatios Ala M. Falleur Departmet of Mathematics ad Statistics The Uiversity of New Mexico Albuquerque, New Mexico, USA Dig Li Departmet of Electrical ad Computer Egieerig The Uiversity of New Mexico Albuquerque, New Mexico, USA Yua Ya Departmet of Electrical ad Computer Egieerig The Uiversity of New Mexico Albuquerque, New Mexico, USA October 13, 2010 Abstract We state ad prove the Borel-Catelli lemma ad use the result to prove aother propositio. 1 Defiitios ad Idetities Defiitio 1 Let { } k=1 be a coutable family of measurable subsets. The limit supremum of { } is the set lim sup ) := {x R d : x for ifiitely may k} 1

2 Propositio 1 The followig idetity holds: lim sup ) = Proof. Assume that x. So, ) ) ) x... k=1 k=2 Suppose that x / lim sup ). By defiitio, this meas that there is a positive iteger k 0 such that for all k k 0, x /. Hece, x / k=k 0. Therefore, x lim sup ). This meas that lim sup ) Coversely, assume that x lim sup ). This meas that x belogs to for ifiitely may k. That is to say, x cotiuously reappears as a elemet i a set of the sequece. The it is evidet that x k=1. It is equally evidet that x k=2, x k=3, ad so o. Thus, ) ) ) x... k=1 k=2 k=3 k=3 Therefore, Thus, lim sup ) lim sup ) = 2

3 2 The Borel-Catelli lemma ad applicatios be a coutable family of measur- Lemma 1 Borel-Catelli) Let { } k=1 able subsets of R d such that m ) < k=1 The lim sup ) is measurable ad has measure zero. Proof. Give the idetity, E = lim sup ) = Sice each is a measurable subset of R d, k= is measurable for each N, ad so is measurable as well, Stei [1]. Therefore, E is measurable. Suppose that me) = ɛ > 0. The 0 < ɛ < me) = m ) Sice for all N, E k=, by the mootoicity property, Stei [1], ) me) m By the coutable sub-additivity property, Stei [1], for all N, me) = m k= k= ) m ) By assumptio, k=1 <. It follows that the tail of the series ca be made arbitrarily small. I other words, for ay δ > 0, there is a N N such that k= 3

4 m ) < δ k=n However, if we choose δ = ɛ/2, we have 0 < ɛ < me) m ) ɛ 2 k=n Therefore, me) = 0. Propositio 2 Let {f x)} be a sequece of measurable fuctios o [0,1] with f x) < for a.e. x [0, 1]. The there exists a sequece {c } of positive real umbers such that f x) c 0 a.e. x [0, 1] Proof. Give a sequece of positive umbers {c }, cosider the set E = { f x) x [0, 1] : > 1 } c Suppose that there is o sequece of positive umbers {c } such that me ) 2. Without loss of geerally, we ca assume that {c } is a sequece of positive umbers. Fix N. The it follows that for ay N N, ma N ) = m { x [0, 1] : f x) N > 1 }) > 2 ma N ) = m { x [0, 1] : f x) > N }) > 2 4

5 So, A 1 = { x [0, 1] : f x) > 1 } A 2 = { x [0, 1] : f x) > 2 } A 3 = { x [0, 1] : f x) > 3 }. A = { x [0, 1] : f x) = } It is easy to see that this is a decreasig sequece of sets: A 1 A 2 A Sice A is a subset of each A N, A = N=1 A N. Hece, m N=1 A N) = ma ), ad so ) 2 < m A N = ma ) N=1 However, by assumptio ma ) = 0. Therefore, there is a sequece of positive umbers {c } such that me ) = m { x [0, 1] : f x) c > 1 } ) 2 Thus the series coverges by compariso to a geometric series: m ) =1 m ) 1 =1 =1 < 2 Accordig to the Borel-Catelli lemma the, lim sup E has measure zero. By defiitio, lim supe ) = {x : x E for ifiitely may } 5

6 So if x lim sup E ), the x is a umber i [0,1] such that for ifiitely may, f x) > 1 c Negatig this statemet, if x / lim sup E ), the there is a k 0 N such that f x) /c 1/ for all k 0. By compariso the, { f x) /c } would coverge to 0 sice {1/} coverges to 0. Therefore, sice mlim sup E )) = 0, the coclusio is Refereces f x) c 0 a.e. x [0, 1] [1] E. M. Stei ad R. Shakarchi. Real Aalysis. Priceto Uiversity Press, Priceto ad Oxford,