MATH 532 Axioms of Measure Theory Dr. Neal, WKU I. The Space Throughout the course, we shall let X deote a geeric o-empty set. I geeral, we shall ot assume that ay algebraic structure exists o X so that X is merely a collectio of objects uder cosideratio. However at times we may let X be a metric space with distace fuctio d, or let X be a ormed, liear space with orm ad distace fuctio d (x, y ) = x y. A special case of both is whe X is the real lie with the absolute value beig the orm. Whe we cosider probability applicatios, the set of all possible objects uder cosideratio is called the sample space ad is deoted by Ω. I this case, Ω also could be the set of all possible outcomes i a radom procedure or sequece of radom procedures. A sigle elemet from Ω is called a outcome ad is deoted by. After determiig Ω, we should determie its cardiality, deoted by Ω. The cardiality may be fiite, deumerable (coutably ifiite), or ucoutable. Example 1.1. Describe the sample space Ω i the followig scearios. The give Ω. (a) Pick oe studet at radom from a class of 40. (b) Roll two dice. (c) Roll a sigle six-sided die 10 times i a row. (d) Play the lottery util you wi. (e) Pick a real umber at radom from the iterval [ 2, 2]. (f) Pick a ratioal umber at radom from the iterval [0, 1]. Solutio. (a) Ω is the set of all studets i the class; Ω = 40. (b) Ω = {{1, 1}, {1, 2},..., {6, 6}} = {(x, y) : x, y {1,..., 6}}; Ω = 36. (c) Ω = { { 1, 2,..., 10 }: i {1,..., 6} }; Ω = 610 = 60,466,176. (d) Ω = {W, LW, LLW, LLLW, LLLLW, LLLLLW,... }; Ω = ℵ 0 (deumerable). (e) Ω = [ 2, 2]; Ω = c (the ucoutable cardiality of R). (f) Ω = Q [0, 1]; Ω = ℵ 0 because the ratioals are deumerable. II. The -algebra Give a set X, we shall assume the existece of a certai collectio of subsets of X, deoted by F. I order to work with these subsets, we must assume that we ca perform various set operatios, such as uios ad complemets. The followig defiitio lays out the miimum requiremets ecessary from which other operatios will follow as cosequeces. Defiitio 1.1. Let F be a collectio of subsets of a set X. The F is called a ad its elemets are called measurable sets, or evets, provided: -algebra (i) F cotais at least oe set; (ii) F is closed uder complemets; (iii) F is closed uder deumerable uios.
Note: If F is merely closed uder fiite uios, the F is called a algebra. We shall show below that every -algebra is a algebra. Propositio 1.1. Let F be a -algebra of a set X. (a) X F ad F. (b) F is closed uder deumerable itersectios. (c) F is closed uder fiite uios. (d) F is closed uder fiite itersectios. c Proof. (a) By Axiom (i), F cotais at least oe set A 1. The A 2 = A 1 F because F is closed uder complemets. Now let A i = A 1 for all i 3. Because F is closed uder c deumerable uios, we have X = A 1 A 1 = U Ai F. The = X c F as well. (b) Let A i F for i 1. Because F is closed uder complemets ad deumerable c uios, A i F for all i, ad thus c Ai F. Thus, by DeMorga s Law ad closure uder complemets, we have Ai = ( Ai c ) c F. (c, d) Let A i F for 1 i. The let A i = for i +1. Because F is closed uder deumerable uios, we have U A i = U A i F. Thus, F is also closed uder fiite uios. Now because F is closed uder complemets ad fiite uios, we have Ai = ( Ai c ) c F ; thus, F is also closed uder fiite itersectios. QED Note: (a) F is also closed uder set differece which is defied as A B = A B c. (b) Give ay set X, the power set of X, which is the collectio of all subsets of X deoted by P( X), is clearly a -algebra. The Geerated -algebra Let X be a set ad let A be a o-empty collectio of subsets of X. The we ca geerate a -algebra from A, deoted by (A), as follows: (i) P( X) is a -algebra that cotais the smaller collectio of subsets A; (ii) Let {F } be the collectio of all -algebras that cotai A; (iii) the -algebra geerated by A is give by (A) = I F.
So if F is ay -algebra that cotais A, the (A) F. That is, (A) is the smallest possible -algebra that cotais A. Of course it remais to be show that (A) actually is a -algebra; but we leave this as a exercise. Exercise 1.1. Let A be a o-empty collectio of subsets of a set X. (i) Prove that (A) cotais A. (ii) Prove that (A) is a -algebra. The Borel -algebra Let X = R (the real lie), ad let A be the collectio of all itervals of the form [a, b) where a, b R with a b. The (A) is the Borel -algebra which we shall deote as B. The B cotais R, B is closed uder complemets, ad B is closed uder both fiite ad deumerable uios ad itersectios. The followig sequece of exercises shows that the Borel -algebra also cotais various other types of itervals. Exercise 1.2. Let B be the Borel -algebra of the real lie. Let a, b, c R. Use the properties of a -algebra to prove the followig results: (a) B cotais all itervals of the form [a, ). (b) B cotais all itervals of the form (, a). (c) B cotais all itervals of the form (, b]. (d) B cotais all itervals of the form (b, ). (e) B cotais all sigleto sets {c }. (f) B cotais all itervals of the form (a, b), [a, b], ad (a, b], for a < b. The Borel -algebra ca be geerated by other types of itervals other tha [a, b). For example, we could let A 1 be the collectio of all closed itervals of the form [a, b] where a, b R with a < b, ad let B 1 = (A 1 ). The the Borel -algebra B cotais A 1 by Exercise 1.2 (f). But because B 1 is the smallest -algebra that cotais A 1, we must have B 1 B. However, as a exercise, we ca show that B 1 cotais all itervals of the form [a, b) where a, b R with a b. But because B is the smallest -algebra that cotais these half-ope itervals, we must have B B 1. Hece, B = B 1. Exercise 1.3. Prove that B 1 cotais all itervals [a, b) where a, b R with a b.
We also ca weake the coditio of the geeratig class of sets ad still geerate B. Specifically, let A q be the collectio of all itervals of the form [a, b) with a, b Q ad a < b, ad let B q = (A q ). The the Borel -algebra B cotais A q. But because B q is the smallest -algebra that cotais A q, we must have B q B. As aother exercise, we ca show that B q cotais all itervals of the form [a, b) where a, b R with a b. But because B is the smallest ope itervals, we must have B B q. Hece, B = B q. -algebra that cotais these Exercise 1.4. Prove that B q cotais all itervals [a, b) where a, b R with a b. III. The measure Give a set X with a -algebra F, a measure is a exteded real valued fuctio defied o F that satisfies the followig properties: (i) 0 ( A) for all evets A F with ( ) = 0. (ii) If { A i } is a deumerable sequece of disjoit evets, the ( i.e., is coutably additive. Ai ) = ( A i ); Collectively, the triple ( X, F, ) is called a measure space. If (A) = 0, the A is said to have measure 0 ad A is said to be a egligible or ull set. If (A) < for all evets A, the is called a fiite measure. A special case of a fiite measure is a probability measure P defied o a sample space Ω. But i this case we eed the additioal coditio that P(Ω) = 1. Some basic properties of a measure ad of a probability measure P are give ext. Ay property of also holds for P, but there are some additioal properties that hold oly for a probability measure P. Propositio 1.2. Let ( X, F, ) be a measure space ad let (Ω, F, P ) be a probability space. (a) Let { A i } be a fiite collectio of disjoit evets. The ( Ai ) = ( A i ); i.e., is fiitely additive. I particular, (A B) = (A) + ( B) for disjoit evets A ad B. (b) For all evets A F, (A) + ( A c ) = ( X ) ad P(A c ) = 1 P(A). (c) If A, B are evets with B A, the (B) ( A). If i additio (B) <, the (A B) = ( A) (B) whe B A. (d) For all evets A F, 0 ( A) ( X) ad 0 P( A) 1.
(e) Let be a fiite measure. For all A, B F, (A B) = ( A) + (B) ( A B) ad (A B) = ( A) + (B) 2 ( A B). Ai ) (A i ) for 1. be a collectio of evets that each have measure 0. The ( Ai ) = 0. be a fiite measure ad let B i be a collectio of evets such that (f) Let { A i } be a collectio of evets. The ( (g) Let { A i } (h) Let { } (B i ) = ( X) for all i. The ( Bi ) = ( X). Proof. (a) Let A i = for i >. The ifiite sequece of sets A i because ( ) = 0, we have { } is still disjoit, ad ( Ai ) = ( Ai ) = ( A i ) = ( A i ) + ( ) = (A i ). i= +1 (b) Because evets A ad A c are disjoit ad uio to X, we have (X ) = ( A A c ) = (A) + ( A c ). For a probability space (Ω, F, P ), we have i particular that 1 = P(Ω) = P(A A c ) = P(A) + P(A c ). (c) If B A, the A = B ( A B) is a disjoit uio. Thus, (A) = (B ( A B)) = ( B) + ( A B) (B), because (A B) 0. From (A) = (B) + ( A B), we obtai (A) (B) = ( A B) provided (B) <. (d) By axiom ad Part (c), we have 0 ( A) ( X) because ay evet A is cotaied i X. For a probability measure we have 0 P( A) P(Ω) = 1. (e) The evet A B ca be writte as the disjoit uio A (B A), ad the evet B ca be writte as the disjoit uio ( A B) (B A). Thus, (B) = (A B) + (B A) so that (B A) = ( B) ( A B). We the have (A B) = ( A) + (B A) = (A) + (B) ( A B).
The evet A B is the disjoit uio of A B ad B A ; hece, (A B) = ( A B) + (B A) = ( A) (A B) + ( B) (A B) = ( A) + (B) 2 ( A B). (f) We defie a collectio of disjoit evets { B i } as follows: Let B1 = A 1, B 2 = A 2 A 1, ad i geeral let B i = A i A j for i 2. The B i A i for all i ; thus, (B i ) (A i ) j < i for all i by Part (c). Moreover, Bi Ai for 1. But if Ai, the we ca choose the smallest idex i such that A i. The A i j < i A j = B i Bi. Hece, Ai Bi. So Ai = Bi for 1. Therefore, by coutable/fiite additivity, we have for 1 that ( Ai ) = ( Bi ) = (B i ) (A i ). (g) Because (A i ) = 0 for all i, we have by (f) that 0 ( Ai ) (A i ) = 0. (h) For each i, c (B i )= (X ) (Bi ) = 0. Thus by DeMorga s Law, Part (c), ad Part (g), we have ( Bi ) = (X ) ( Bi ) c = ( X ) ( c Bi ) = (X ) 0 = (X ). Note: Wheever we are subtractig a measure value, the we should assume that it is fiite. For example, let A be the Irratioals o the real lie, ad let B be the Positive Irratioals. The B A. We shall see that both of these evets have ifiite measure: (A) = (B) =. But A B is the set of egative irratioals, which also has ifiite measure. So although we ca write (B) + (A B) = (A), it does ot follow that (A B) = ( A) (B) because all three values are ifiite. Nested Evets Defiitio 1.2. (a) A sequece of evets A i { } is called ested icreasig if Aj A j+1 for all j 1: A 1 A 2 A 3... (b) The sequece is called ested decreasig if A j A j+1 for all j 1: A 1 A 2 A 3...
Propositio 1.3. (a) Let A i ( Ai ) = lim (b) Let { A i } ( Ai ) = lim ( A ). { } Dr. Neal, WKU be a ested icreasig sequece of evets. The be a ested decreasig sequece of evets with (A1 ) <. The ( A ). Proof. (a) We agai create a disjoit sequece of evets { B i } such that Bi = Ai for 1. To do so, let B 1 = A 1 ad let B i = A i A j = A i A i 1 for i 2. Because j < i { A i } is ested icreasig, the A = Ai = Bi for all <. Thus we have lim ( A ) = lim ( Bi ) = lim ( B i ) = ( B i ) = ( Bi ) = ( Ai ). (b) Because (A 1 ) <, the all measurable subsets of A 1 have fiite measure by Propositio 1.2 (c). For all 1, we have A A 1 ; thus, (A 1 A ) = (A 1 ) ( A ). But { A 1 A i } is a ested icreasig sequece with U ( A1 A i ) = A 1 I A i. Thus, by Part (a) ad Propositio 1.2 (c), we have lim ( A ) = ( A 1 ) lim ( ) = (A 1 ) U (A 1 A i ) A 1 A = ( A 1 ) A 1 I A = ( A 1 ) ( A 1 ) ( I A i ) = ( I A i ). Some Examples of Probability Measures 1. Equi-Probable Outcomes Let Ω be a fiite set with Ω = ad let F be the power set of Ω, which cosists of all 2 possible subsets of Ω. For every Ω, we ca let P({ }) = 1 / to deote that each idividual outcome is equally likely. The for ay evet A F, we have P(A) = A /. I particular, P(Ω) = Ω / = 1. Because for a fiite collectio of disjoit fiite sets we have U A i = A i, it follows that P is fiitely additive. (I this case, we oly eed P to be fiitely additive sice there is o eed to have ifiite uios of sets.)
2. Biomial Probability Let Ω = {( 1,..., ): i {W, L}}, which represets the set of outcomes i a sequece of idepedet attempts. O the i th attempt, i is either W for a wi or L for a loss. We assume that each wi occurs with probability p ad each loss occurs with probability q = 1 p. The Ω = 2. We agai let F be the power set of Ω, which cosists of the 2 2 subsets of Ω. With = 3, the Ω has 2 3 = 8 elemets, ad F has 2 8 = 256 elemets. I this case, Ω = {(W,W,W ), (W,W, L), (W, L,W ), (L,W, W), (W, L, L), (L, W, L), (L, L, W), ( L, L, L)}. Various evets i F ca be described i words. For example, A 1 = a wi o the first try; A 2 = at least oe wi; A 3 = ever the same outcome back to back. The, A 1 = {(W,W, W), (W, W, L), (W, L, W ), (W, L, L)} A 2 = {(W, W,W ), (W, W, L), (W, L, W), ( L,W,W ), (W, L, L), ( L,W, L), ( L, L, W )} A 3 = {(W, L,W ), ( L,W, L)}. I all, there are 256 possible evets. We defie a probability measure P o the set of evets F as follows: For a sigle outcome = ( 1,..., ), we let P({( 1,..., )}) = pk q k, where k is the umber of wis i the sequece ( 1,..., ). The for A F, we let P(A) = P({ }). The A P is fiitely additive as with equi-probable outcomes. Moreover, the idividual outcomes are equally likely if ad oly if p = q = 1/ 2. Now suppose we let W k be the evet that there are exactly k wis i a sequece for k = 0, 1,...,. The the evets W 0, W 1,..., W are disjoit ad uio to Ω. Because there k distict ways to have exactly k wis i a sequece, the probability of W k is k p k q k. The by the biomial expasio theorem, we have P(Ω) = P(W k ) = k = 0 k p k q k = ( p + q) = 1 = 1. k = 0 3. Geometric Probability Agai let p be the probability of a wi o ay idepedet attempt, ad let q = 1 p be the probability of a loss. But ow assume that 0 < p < 1. Let Ω = {W, LW, LLW, LLLW,...}, which represets the possible outcomes i a sequece of attempts where we stop play upo the first wi. The Ω =ℵ 0. Let F be the power set of Ω which is ucoutable with cardiality 2 ℵ0 = c. For each Ω, defie P({ }) = q 1 p, where is the total umber of attempts made i sequece. The for A F, let P(A) = P({ }). A
I this sceario, represets the sequece of outcomes up to the first wi. So if the first wi is o the th attempt, the there must have bee 1 losses followed by a sigle wi; hece, P({ }) = q 1 p. The the probability of the etire space is P(Ω) = P({ }) = q 1 p = p q = p 1 Ω =1 = 0 1 q = 1. Because P({ }) coverges absolutely to 1, the 0 P({ }) 1 for ay Ω A evet A Ω. Moreover, o rearragemet of the terms i A will affect the sum of the series. So if {A i } is a collectio of disjoit evets with A = U Ai, the P( Ai ) = P({ }) = P({ }) = P( A i ). A Hece, P is coutably additive. Although there are a ucoutable umber of evets i F, the evets of most iterest usually ca be described easily i words. For example, compute the probability of the followig evets: A i A = the first wi is withi k attempts, for k 1 B = it takes at least k attempts for the first wi, for k 1 C = it takes a odd umber of attempts for the first wi Here A is the complemet of the evet of there beig k iitial losses i a row; hece, P(A) = 1 q k. Evet B is equivalet to the evet of there beig k 1 iitial losses i a row; hece, P(B) = q k 1. Fially, P(C) = q 2 p = p (q 2 ) p = = 0 = 0 1 q 2 = 1 1 + q. 4. Uiform Probability Let Ω = [a, b], where a,b R with a < b. Let F = {[a, b] B : B B}, where B be the Borel -algebra of the real lie. For a sub-iterval [c, d ) of Ω, let P([c, d )) = d c. I this case, P measures the probability of havig a b a poit lie from c to d whe pickig a poit at radom from [a, b]. The probability measure P is iitially oly defied o sub-itervals [c, d ) [a, b]. 1 If C = U [c i, d i ) is a uio of disjoit sub-itervals, the we let P(C) = (d b a i c i ). Later we shall show that we ca use this basic defiig of P to exted P to be defied for all evets i F.
Exercise 1.5. Let { A i } be a sequece of evets i a -algebra F. Defie the limif ad limsup of this sequece by lim if {A i } = Ai limsup{a =1 i= i } = Ai. =1 i= (a) Explai why limsup{a i } ad lim if {A i } are i F. (b) Show that (limif {A i }) c c = limsup{a i } ad (limsup{ai }) c c = lim if {A i }. (c) Prove that a elemet x is i limsup{a i } if ad oly if x belogs to a ifiite umber of the sets A i. (d) Prove that a elemet x is i lim if {A i } if ad oly if x belogs to all but a fiite umber of the sets A i. (e) Show that lim if {A i } limsup{a i }. (f) If { A i } is ested icreasig, prove that lim if {Ai } = limsup{a i } = =1 A. (g) If { A i } is ested decreasig, prove that lim if {Ai } = limsup{a i } = A. =1 Exercise 1.6. Let A, B be evets such that (A B) = 0. Prove that (A) = (B). Exercise 1.7. Let A, B be evets with B A such that (A) < (B) + for all > 0. Prove that (A) = (B). Exercise 1.8. Let Ai = X. Prove that be a fiite measure ad let A i ( c Ai ) = 0 = lim ( A c ). { } be ested icreasig with Exercise 1.9. Give a couterexample usig the legth measure o the real lie to show that Propositio 1.2 (h) does ot hold for ifiite measures.