The Borel-Cantelli Group

Transcription

1 The Borel-Cantelli Group Dorothy Baumer Rong Li Glenn Stark November 14, Borel-Cantelli Lemma Exercise 16 is the introduction of the Borel-Cantelli Lemma using Lebesue measure. An approach using probability measure will be introduced later in the course. Definition 1.1. Let {E 1, E, E 3,...} be a seuence of sets. From this, we can define a decreasing seuence of sets B 1 B B 3... as follows: Let E = {x: x E k for infinitely many k} be defined to mean E = E k E k := lim sup(e n ) n Lemma 1.1. Suppose {E k } k=1 is a countable family of measurable subsets of R d that m(e k ) < k=1 Then E is measurable m(e) = 0. E = lim sup(e k ). k Proof. To show that E is measurable, note that E k. 1

2 Each set B i is the countable union of measurable sets is therefore measurable. Furthermore, E = E k = B n. The set E is the countable intersection of measurable sets is therefore measureable. It remains to be shown that m(e) = 0. Note that Define the two seuences m(e k ) = M <. k=1 n 1 s n = m(e k ) t n = k=1 m(e k ). Note that m(e k ) 0 for all k. Therefore {s n } is a monotone increasing seuence {t n } is a monotone decreasing seuence with the following properties: lim n s n = M the tail lim n t n = 0. In other words, ɛ > 0, N such that n > N t n < ɛ By the properties of measurability, E = ( m(e) = m B n B n ) m(b n ) n N E k

3 Exercise 17 ( m(b n ) = m E k ) m(e k ) = t n. 0 m(e) m(b n ) t n ɛ m(e) = 0 From Stein Sakarchi we have: Let {f n } be a seuence of measurable functions on [0, 1] with < a.e. x. Then a seuence of positive real numbers such that: 0 a.e. x Proof. First we show that k > 0 where for the set A n k = {x: > k} then m (A n k ) < ɛ Assume otherwise, then for some fixed n, ɛ > 0 such that k > 0, then when we let k, we have m {x: f (x) > k} ɛ lim k m {x: > k} ɛ, which forms a decreasing seuence. We define the seuence as follows A = {x: = } A k+1 = {x: > k + 1} A k = {x: > k} The seuence is then decreasing so that A... A k+1 A k.... Then m (A 1 ) m (A ) m (A 3 )... further define lim m (A n) = M > ɛ > 0. n 3

4 Then by Corollary 3.3 from our text which states: for a decreasing seuence where E k decreases to E the m (E k ) < for some k, then m (E) = lim N m (E N) We know this applies to our seuence because we are given that is a countable family of measurable functions x [0, 1]. Furthermore, since A k A by the measurability of each A k by the boundednes of the measure of the seuence, that is, m(a k ) <, we can apply Corollary 3.3 which gives us: lim k m (A k) = m (A k ) When we combine these results with what we were given, by the definition of almost everywhere in our text; < a.e. x, implies m (A ) = m {x: > } = 0 this gives us a contradiction. Our premise was that the measure of the set had to be ɛ for ɛ > 0, however, we have shown that: m {x: > k} < ɛ Note: m (E n ) ɛ m ( n E n) ɛ We can now define ɛ = 1 n let k = cn n. this gives us a seuence of sets, {E n} which we have shown to be a countable family of sets where: E n = { x: > 1 } n In addition, we know the m ( E n ) 1 as constructed for each of the seuences, which implies n (E n) 1. We can then define the set of E as follows { } E = lim sup E n. E = { x: n > 1 } n for i.o. n It is left to show that x appears in infinitely many E n, then by the Borel-Cantelli Lemma, m (E) = 0. If this is true, then fn(x) 0 a.e. x 4

5 Proof. Given some x, we define convergence of the seuence to be lim n 0 ɛ > 0 N > 0 s.t. n > N, < ɛ, If we assume this is not true, this seuence does not converge to zero then we need to show that ɛ > 0, s.t. N > 0 n N > N s.t. f nn (x) N > ɛ 1 If this is true then we know there is an N 1 s.t. N 1 < ɛ n N1 > N where f nn1 (x) > 1 x E nn1 n N1 We can continue in this way show that N s.t. N > n N1 further n N > N > N 1. Now if we let n N1 = N we can continue to find n Nk = N k+1 for infinitely many k. Then by the Borel Cantelli Lemma since this occurs infinitly often since from part one m (E n ) < then we can say: 0 a.e. x 3 Problem 1 Prove that the set of those x R such that there exist infinitely many fractions p/ with relatively prime integers p such that x p 1 +ɛ is a set of measure zero. Proof. For any integers p, define the interval E p It follows that m ( ) E p = +ɛ x p 1 +ɛ x E p. For every integer n N, define A p form p where p [n, n + 1]. Each A p = [ p + 1 +ɛ, p 1 +ɛ ]. to be the set of rational numbers of the 5 is a subset of Q is therefore

6 countable. For each integer, there are exactly + 1 elements of A n of the form p. For example, the set A 0 is the set A 0 = { 0 1, 1 1, 0, 1,, 0 3, 1 3, 3, 3 } 3,.... Conseuently, {E kn } kn An is a countable family of measurable subsets in R m(e kn ) < ( + 1). +ɛ k n A n =1 for infinitely many p A } n Define the sets E (n) = { x [n, n + 1] : x E p E = { x R : x E p for infinitely many p Q}. By the Borel-Cantelli Lemma, for any integer n, the set E (n) is measurable m(e (n) ) = 0. In other words, the set of those x [n, n + 1] such that there exist infinitely many fractions p/ [n, n + 1] with relatively prime integers p such that x p 1 +ɛ is a set of measure zero. Clearly, n=0 E (n) E. To prove E = n=0 E (n), it suffices to show that E n=0 E (n). Assume x R such that x E but x / n=0 E (n). There must exist some integer n such that x [n, n + 1]. In addition, x must also lie in infinitely many intervals E p, each of which must satisfy the reuirement that n 1 < p < n +. It follows that x ( ) E (n 1) E (n) E (n+1) n=0 E (n), which is a contradiction. This proves that which is the desired result. E = m(e ) = n=0 E (n) m(e (n) ) = 0 6