The Boolean Ring of Intervals

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1 MATH 532 Lebesgue Measure Dr. Neal, WKU We ow shall apply the results obtaied about outer measure to the legth measure o the real lie. Throughout, our space X will be the set of real umbers R. Whe ecessary, we shall assume that R is a metric space with the distace betwee poits give by d (x, y ) = x y. The Boolea Rig of Itervals We ow let R(I) be the collectio of all fiite uios of disjoit itervals of the form [a, b) o the real lie where a b. To show that R (I) is a rig, we must argue that R (I) is closed uder fiite uios ad set differeces. We shall first use iductio to show that ay fiite uio of itervals ca be re-writte as a uio of at most disjoit itervals. First, cosider [a 1, b 1 ) [a 2, b 2 ), where a 1 a 2. If b 1 a 2, the [a 1, b 1 ) [a 2, b 2 ) is already a disjoit uio, ad if b 1 = a 2 the it actually ca be re-writte as [a 1, b 2 ). If a 1 a 2 < b 1, the [a 1, b 1 ) [a 2, b 2 ) is the sigle iterval [a 1, max{b 1, b 2 }). So ay two itervals are either already disjoit or ca be re-writte as a sigle iterval. Now assume that for 2 k, ay set of k itervals ca be re-writte as a uio of at most k disjoit itervals, ad cosider a uio of + 1 itervals. By the iductio hypothesis, we ca assume that the first itervals reduce to at most disjoit itervals. If the ( + 1)st iterval is already disjoit from all the others, the we have at most + 1 disjoit itervals. But if the ( + 1)st iterval itersects say the j th iterval, the those two ca be re-writte as oe iterval as discussed i the achor step above with just two itervals. So ow we are reduced to at most itervals, which by the iductio hypothesis, ca be re-writte as a uio of at most disjoit itervals. m Thus, if A = U [a i, b i ) R (I) ad B = U [c j,d j ) R (I), the A B is still a fiite j =1 uio of disjoit itervals of the form [a, b); thus, A B R (I). Note that = [a, a) R (I). Next cosider C = [a 1, b 1 ) [a 2, b 2 ), where a 1 a 2. If b 1 a 2, the C = R (I). If a 1 a 2 < b 1, the C = [a 2, mi{b 1, b 2 }) R (I). That is, C = [a 1, b 1 ) [a 2, b 2 ) always reduces to a sigle iterval of the form [a, b). So if m m A = U [a i, b i ) ad B = U [c j,d j ) are i R (I), the A B = U U [a i, b i ) [c j, d j ) is j =1 j =1 still i R (I) because it is a fiite uio of itervals of the form [a, b), which the reduces to a fiite uio of disjoit itervals of the form [a, b). So R (I) is closed uder fiite itersectios. Now cosider D = [a 1, b 1 ) [a 2, b 2 ). If a 2 a 1, the D = [b 2, b 1 ) which is empty if b 1 b 2. If a 1 < a 2, the D = [a 1, a 2 ) if b 1 b 2 ad D = [a 1, a 2 ) [b 2, b 1 ) if b 2 < b 1. I m ay case, we have D R (I). Now if A = U [a i, b i ) ad B = U [c j,d j ) are i R (I), j =1 m the we have A B = U I ([a i, b i ) [c j, d j )) R (I). Thus, R (I) is a Boolea rig. j =1

2 The Legth Measure Agai let R (I) be the collectio of all fiite uios of disjoit itervals of the form [a, b). Let : R (I) [0, ) be defied as follows: If [a i, b i ) is a collectio of { } disjoit itervals with A = U A i, the let ( A) = (b i a i ). I particular, let i= 1 ([a, b) ) = b a, which gives the legth of the iterval. Clearly, 0 ( A) < for all A R(I). We assert that is a measure o R (I). Let A, B R (I) with A B =. The A ad B are each fiite uios of disjoit itervals. Thus we ca write A = U [a i, b i ) where a 1 < b 1 <... < a < b, ad +m B = U [a i, b i ) where a +1 < b +1 <... < a +m < b +m. Because A B =, there is o i = +1 +m overlap amog ay of the itervals. Therefore, A B = U [a i,b i ) is still i the form of a disjoit uio of itervals; thus, +m + m ( A B) = (b i a i ) = (b i a i ) + (b i a i ) = (A) + ( B). i= +1 That is, is fiitely additive. Also ( ) = ([a,a)) = 0. But i order for to be a measure, it must be coutably additive. We develop this result ext. Lemma 3.1. Let B, C R (I) with B C. The (B) (C). Proof. We write C = B (C B) as a disjoit uio i R (I). The by the fiite additivity of, we have (C) = (B) + (C B) (B) because (C B) 0. Lemma 3.2. Let { A i } be a sequece of disjoit sets from R (I) such that A = U Ai is also i R (I). The (A i ) ( A). Proof. By the fiite additivity of ad Lemma 3.1, we have for all 1 that (A i ) = U A i ( A ). We obtai the result by takig the limit as.

3 Lemma 3.3. (a) Let [c, d ) be partitioed by c = e 1 < f 1 = e 2 < f 2 =... = e < f = d so that [c, d ) = U [e i, f i ) is a disjoit uio with strictly icreasig edpoits. The i= 1 ( f i e i ) = d c. (b) Let [c, d ) be partitioed by c = e 1 < f 1 = e 2 < f 2 =... = e < f =...< d so that [c, d ) = U [e i, f i ). The d = lim f ad ( f i e i ) = d c. i= 1 Proof. Part (a) follows from the fiite additivity of or by a telescopig sum usig the assumptio that f i = e i+1 for 1 i 1, e 1 = c, ad f = d. For Part (b), the sequece { f } is strictly icreasig ad bouded above by d ; thus lim f exists ad c lim f = sup{f } d. But if sup { f } < d, the there exists a poit x [c, d ) such that sup{ f } < x < d. But this x cotradicts the assumptio that [c, d ) = U [e i, f i ). Hece, lim f = d. The by telescopig series, we have i= 1 ( f i e i ) = lim ( f i e i ) = lim (f e 1 ) = d c. QED Lemma 3.4. Let { A i } be a sequece of disjoit sets from R (I) such that A = U Ai is also i R (I). The, ( A) ( A i ). i Proof. For each i, let A i = U [a ij, b ij ) be a fiite uio of disjoit itervals with strictly j =1 i icreasig edpoits. The A = U U [a ij, b ij ) is a deumerable uio of disjoit j =1 itervals. But because A is i R (I), A also ca be writte as a fiite uio of disjoit N itervals A = U [c k, d k ), also with strictly icreasig edpoits: c 1 < d 1 <... < c N < d N. k = 1

4 i Dr. Neal, WKU N Next we ca write A as A = U U U ([a ij, b ij ) [c k, d k )). Because we still have a j =1 deumerable uio of disjoit itervals [a ij, b ij ) [c k, d k ), we ca rewrite this uio further as A = U [e i, f i ) where ow all e i ad all f i are bouded withi [c 1, d N ), all itervals are disjoit, ad each iterval is completely cotaied withi [c k, d k ) for some k. For 1 k N, we ow partitio each [c k, d k ). Because c k A, let e k1 be the first edpoit such that c k [e k1, f k1 ). The we must have e k1 = c k ad [e k1, f k1 ) [c k, d k ). If f k1 = d k, the we stop. Otherwise, because f k1 A, we let e k 2 be the first edpoit such that f k1 [e k2, f k 2 ). Because all [e i, f i ) are disjoit, we must have f k1 = e k 2 ad still [e k 2, f k2 ) [c k, d k ). We cotiue as ecessary. We thereby obtai either a fiite or ifiite sequece c k = e k1 < f k1 = e k 2 < f k2 =... = e k < f k =... d k such that [c k, d k ) = U [e kl, f kl ). l By costructio, each [e i, f i ) is also cotaied i some particular iterval [a ij, b ij ). Thus, by Lemma 3.3, we have N N N ( A) = ([c k, d k )) = (d k c k ) = ( f kl e kl ) l i (b ij a ij ) = (b ij a ij ) = ( A i ). i, j j=1 QED Combiig Lemmas 3.2 ad 3.4, we obtai: Theorem 3.1. The legth measure is coutably additive o R (I). The Hereditary -algebra The hereditary -rig of R(I) is the collectio of all subsets of the real lie that have a coutable coverig of sets from R (I): C(R(I)) = A (, ) : A U A i, for some, where 1, ad all A i R (I). I particular, ay bouded subset of the real lie is cotaied i a sigle iterval of the form [, ) ; thus, C(R(I)) cotais all bouded sets. Of course, C(R(I)) the cotais every subset of a bouded set as well.

5 We have see that the hereditary -rig actually is a -rig. However, the etire real lie R is covered the sequece of sets {[, ) } = 1. Thus, R C(R (I)), which makes C(R(I)) a -algebra. The ay subset of R is covered by the same coverig as R; hece, C(R(I)) is actually the power set of R. The Borel -algebra ad the Lebesgue -algebra Previously, we defied the Borel -algebra B to be the smallest -algebra cotaiig all itervals of the form [a, b), where a b. But (R(I)), the smallest -algebra cotaiig R (I), also cotais all itervals of the form [a, b); so B (R(I)). However, B also cotais R (I), so (R(I)) B. I other words, (R(I)) is the Borel -algebra. I this cotext, the measurable sets M (C(R (I))) are called the Lebesgue measurable sets ad are deoted by L. So the rig of itervals is cotaied i the Borel -algebra, which is cotaied i the Lebesgue -algebra of measurable sets, which is cotaied i the hereditary - algebra which equals the power set of the real lie: R (I) (R(I)) M (C(R(I))) C(R(I)) = P(R) or R (I) B L P(R) The outer measure is defied for every set i C(R(I)) = P(R), which meas that it is defied for every subset of the real lie. Also, ( A) = ( A) for all A R(I). I particular, ([a, b) ) = ([a, b) ) = b a for all itervals [a, b). Moreover, is a coutably additive measure o the set of Lebesgue measurable sets L. Whe we are specifically evaluatig the outer measure of a arbitrary subset, the we will use. However, at this poit, if we are specifically evaluatig the measure of a Lebesgue measurable set, the we will simply use. Recall that E L if ad oly if ( A) = (A E) + ( A E c ) for all A P(R), which is satisfied by every set E that has outer measure 0 (Theorem 2.1). Ad every subset of a set of measure 0 is Lebesgue measurable ad also has measure 0. The sets of measure 0 are very importat; i fact, we will show that every Lebesgue measurable set differs from a Borel measurable set by a set of measure 0.

6 The Borel measurable sets are essetially the subsets of the real lie that ca be obtaied through a coutable umber of operatios o itervals of the form [a, b) that ivolve uios, complemets, or itersectios. For istace, (, ) = U [, ) Primes = U I [ p, p +1 / ) =1 p =1 Cator Set: C = I C, where = 0 C 0 = [0, 1], C 1 = C 0 (1 / 3, 2 / 3) = [0, 1/ 3] [2 / 3, 1], C 2 = C 1 (1 / 9, 2 / 9) (7 / 9, 8 / 9) = [0, 1/ 9] [2 / 9, 3/ 9] [6 / 9, 7 / 9] [8 / 9, 1], etc. Theorem 3.2. Every ope subset of the real lie is Borel measurable. Proof. Let E be ope. If E =, the E B. So assume that E. The there is some x E. Because E is ope, there is a eighborhood (x, x + ) E. Now there must be a ratioal umber withi (x, x + ). So E must cotai a ratioal umber. For every ratioal umber q E, there is ope iterval (q r q, q + r q ) E. By takig a smaller radius if ecessary, we ca assume that r q is ratioal. Now for each q E, let R q be the set of ratioal radii r such that (q r, q + r) E. Because the ratioals are deumerable, S q = U (q r, q + r) is a deumerable uio ad therefore r R q is Borel measurable because each iterval (q r, q + r) is i B. Moreover, S q E for all q E. The S = U S q = U U (q r, q + r) is also a Borel measurable set cotaied q E q E r R q i E, ad S clearly cotais all ratioal umbers i E. But if x is a irratioal umber i E, the there is a ope iterval (x, x + ) E for which is ratioal. The there is a ratioal umber q such that x < q < x + / 2. But the x (q / 2, q + / 2) (x, x ) E. So x S q S. Thus, E = S ad E is Borel measurable. QED Corollary 3.1. Let (O) be the -algebra geerated by the ope sets of the real lie. The (O) is the Borel -algebra. Proof. Because B is a -algebra that cotais all ope sets, B cotais the -algebra geerated by the ope sets (O). O the other had, (O) cotais all itervals of the form [a, b) = I (a 1/, b). Thus, (O) cotais the -algebra geerated by =1 itervals of the form [a, b) which is B. QED We ow show that a Lebesgue measurable set is always withi a set of measure 0 of two Borel measurable sets:

7 Theorem 3.3. Let E L. There exist Borel measurable sets B, A such that B E A ad ( A E) = 0 = (E B). Proof. First, assume that (E ) <. By the defiitio of outer measure, for every iteger 1, there exists a coutable collectio of sets { A i } from the rig of itervals R (I) such that E U A i ad ( A i ) < ( E) +1 /. Now let A = I U A i. The i i = 1 i A B ad E A U A i for all 1. Also, (E ) = (E ) by covetio. Hece, for i all, (E ) ( A) U A i i i (A i ) < ( E) +1 /. By lettig, we have (E ) ( A) (E) ; or (E ) = (A). Now we have the disjoit uio E (A E) = A, which gives (E ) + (A E) = ( A). Because (E ) = (A) <, we have ( A E) = 0. Now for E L, let E k = E [ k, k) for all itegers k 1. The E k L for all k ad (E k ) ([ k, k )) <. So for each k, there exists a Borel measurable set A k ad a set of measure 0, give by N k = A k E k, such that E k A k ad E k N k = A k. Now let A = U A k B. The E = U E k A so that E (A E) = A. Also, k = 1 A E U ( A k E k ) = U N k. By Propositio 1.2 (g), U N k still has measure 0, so ( A E) = 0. Because E c L, there exists C B, with E c C ad 0 = (C E c ) = (C E ). But the C c B, C c E, ad (E C c ) = ( E C) = 0. Lettig B = C c, we have B E, ad B ( E B) = E with (E B) = 0. QED The precedig theorem gives two way to characterize the Lebesgue measurable sets: L = {B N : B B ad ( N) = 0} ad L = {A N : A B ad (N ) = 0}. Lebesgue Measure The Lebesgue measure : L [0, ] is a extesio of the legth measure for which ([a, b) ) = b a. So gives a way of measurig the legth of other subsets of the real lie i a coutably additive way. Next are some immediate legth properties:

8 Theorem 3.4. For all a, b R with a < b: (a) ({a}) = 0 (b) ((a, b) ) = ((a, b] ) = ([a, b] ) = b a (c) ([a, ) ) = ((a, ) ) = ((, b) ) = ((, b] ) = + (d) ((, ) ) = + (e) If A is a coutable subset of R, the ( A) = 0. Proof. We simply use the properties of a measure o a -algebra. (a) Because {[a, a + 1/ i) } is a ested, decreasig sequece of measurable sets with ([a, a +1) ) <, we have by Propositio 1.3 (b), ({a}) = I [a, a + 1/ i) = lim ([a, a +1/ ) ) = lim 1 / = 0. (b) ((a, b) ) = ([a, b) {a}) = ([a, b) ) ({a}) = b a ((a, b] ) = ((a, b) {b}) = ((a, b) ) + ({b} ) = b a ([a, b] ) = ([a, b) {b}) = ([a, b) ) + ({b}) = b a (c) Because { [a, a + i) } is a ested, icreasig sequece of sets, we have ([a, ) ) = U [a, a + i) i= 1 = lim ([a, a + ) ) = lim = +. The other results of Part (c) are simple exercises. (d) Because [0, ) (, ), we have ((, ) ) ([0, ) ) = +. (e) Because is A coutable, we have A = {a i }, where 0, ad all a i R are distict. Thus, ( A) = U{a i } i= 1 = ({a i }) = 0. i= 1 Applyig (d) ad (e), we have: Corollary 3.2. O the real lie, the ratioals Q have Lebesgue measure 0 ad the irratioals I have ifiite Lebesgue measure.

9 Uiform Probability Let a < b, let Ω = [a, b], ad let F be the -algebra of all Lebesgue measurable subsets of [a, b]. ( F = {E [a, b]: E L}.) Defie P : F [0, 1] by P(A) = ( A) b a. I particular, for [c, d ] [a, b], we have P([c, d ]) = d c ad also P(Ω) = 1. Because b a is coutably additive, it follows that P is a probability measure. This measure is used to compute probabilities whe choosig values at radom from the iterval [a, b]. Example 3.1. Pick a value x at radom from [0, 20]. What is the probability of it beig (a) at least 5? (b) at most 8? (c) at least 3 away from 10? (d) ratioal? Solutio. (a) P([5, 20]) = 15 / 20 (b) P([0, 8]) = 8 / 20 (c) P([0, 7] [13, 20]) = 14 / 20 (d) P(Q [0, 20]) (Q) = 0. The Cator Set We previously described the Cator Set as C = I C, where C 0 = [0, 1], = 0 C 1 = C 0 (1 / 3, 2 / 3) = [0, 1/ 3] [2 / 3, 1], C 2 = C 1 (1 / 9, 2 / 9) (7 / 9, 8 / 9) = [0, 1/ 9] [2 / 9, 3/ 9] [6 / 9, 7 / 9] [8 / 9, 1], etc. The C is Borel measurable beig the deumerable itersectio of Borel sets. Each set C is the uio of 2 disjoit closed itervals each of which has legth 1 / 3. Give C, set C +1 is created by removig the middle ope third of each closed iterval i C. The Cator set ca be uderstood easily by cosiderig the tertiary (base 3) expasio of the umbers x i [0, 1]. Each x [0, 1] ca be writte as x = 0. a 1 a 2 a 3... = a 1 / 3 + a 2 / 9+ a 3 / where a i {0, 1, 2}. The expasio is ot uique; for istace, 1/3 = = The Cator Set cosists of those x [0, 1] that ca be writte with a i {0, 2} for all i. So if a 1 = 0, the the largest x ca be is = 1/3. If a 1 = 2, the the smallest x ca be is = 2/3. So havig a 1 {0, 2} meas that x C 1 = [0, 1 / 3] [2 / 3, 1]. Next we have x = , or x = , or x = , or x = , which meas that x [0, 1/ 9] [2 / 9, 3 / 9] [6 / 9, 7 / 9] [8 / 9, 1].

10 So C = {0.a 1 a 2 a 3... : a i {0, 2} for all i}. Thus C is i oe-to-oe correspodece with the set B = {0. b 1 b 2 b 3... : b i {0,1} for all i}, which gives the biary expasios of all x [0, 1]. So C ~ B ~[0, 1] ~ R, which meas that the Cator set is ucoutable. But C C for all ; thus, (C) (C ) = 2 ( 1/ 3) = (2 / 3) for all 1. By lettig, we see that (C) = 0. Thus, the Cator Set gives a example of a ucoutable set that has Lebesgue measure 0. Moreover, ay subset of the Cator Set is also Lebesgue measurable ad also has measure 0. Note: With some more advaced set theory, it ca be show that the collectio of Borel sets is also i oe-to-oe correspodece with the real lie. Thus, B = C = R = c. But every subset of C is Lebesgue measurable, ad the cardiality of the umber of subsets of the Cator Set is 2 c > c = B. So there must be more Lebesgue measurable sets tha Borel measurable sets. I other words, there exist Lebesgue measurable sets that are ot Borel measurable. Exercises Exercise 3.1. Let (Cl) be the -algebra geerated by the closed sets of the real lie. Prove that (Cl) is the Borel -algebra. Exercise 3.2. Let N = { A L : (A) = 0} be the collectio of all sets of Lebesgue measure 0. Prove that N is a -rig. Exercise 3.3. Cosider the followig sets o the real lie: (i) I (10 1/ i,15 + 1/ i) (ii) U [20 +1 / i, 30 1/ i] For each: (a) Derive its Lebesgue measure by usig a limit theorem for ested sets. (b) Evaluate the set as oe iterval ad give its Lebesgue measure directly. Exercise 3.4. Cosider the base 10 decimal expasio of values i [0, 1]. (The expasio is ot uique: 1/2 = = ) Fid the Lebesgue measure of the followig sets: A 1 = {x : x = 0.4 a 2 a 3...} A 2 = {x : x = 0.53 a 3 a 4...} A 3 = {x : x = a 4 a 5...} Exercise 3.5. Cosider the base 3 expasio of values i [0, 1]. Fid the Lebesgue measure of the followig sets: B 1 = {x : x = 0.1 a 2 a 3...} B 2 = {x : x = 0.21a 3 a 4...} B 3 = {x : x = a 4 a 5...}