MATH 532 Measurable Fuctios Dr. Neal, WKU Throughout, let ( X, F, µ) be a measure space ad let (!, F, P ) deote the special case of a probability space. We shall ow begi to study real-valued fuctios defied o X. I order to do so, we eed to assume that our fuctios have the special property of beig measurable. The formal defiitio of this property depeds o the cocept of the iverse image of a fuctio which we shall discuss first. Defiitio 4.1. Let f : X! Y be a fuctio betwee spaces X ad Y. Let V! Y. The iverse image of V uder f is the subset of X give by The x! f 1 (V ) if ad oly if f ( x)!v. f!1 (V ) = {x X : f (x ) V}. It is always the case that uios, itersectios, complemets, ad set differeces are preserved by the iverse image of a fuctio. That is, (a) f!1 # & %! V ( ' =! f!1 ( V ) (b) f!1 # & %! V ( ' =! f!1 ( V ) (c) f!1 (V c ) = ( f!1 (V )) c (d) f!1 (V! U) = f!1 (V )! f!1 (U) We prove properties (a) ad (c) below, but leave the proofs of (b) ad (d) as exercises. Proof. (a) We have x! f!1 # & %! V ( ' iff f ( x)!! V iff f ( x)! V for some! iff x! f 1 ( V # ) for some! iff x!! f 1 ( V # ). # (c) We have x! f!1 (V c ) iff f ( x)! V c iff f ( x)! V iff x! f!1 (V ) iff x! ( f!1 (V )) c. Defiitio 4.2. Let f : X! be a real-valued fuctio defied o X. The f is called measurable provided f!1 ((!, b] ) #F for all itervals of the form (!, b] #. That is, {x! X : f ( x) b} must always be a measurable set by beig i the! - algebra F. We the ca measure this evet by evaluatig µ ({x! X : f ( x) b} ). For coveiece, we ofte shall abbreviate the otatio for this iverse image by writig the set as { f! b} ad by writig its measure as µ ( f! b).
Example 4.1. Let! = {1, 2,..., 20}, with F beig the power set of!, ad P({}} = 1 / 20 for all!. Let f :! # be defied by!# f () = # 2 if is prime 2 otherwise. Because F is the power set of!, it cotais all subsets of! icludig all iverse images of the form { f! b}. Thus, f is measurable. What is P( f! 10)? We simply eed to determie which values i the domai have fuctio values that are less tha or equal to 10. We see that { f! 10} = {1, 2, 3, 5}; thus, P( f! 10) = 4/20. Note: Whe (!, F, P ) is a probability space, the measurable fuctios are called radom variables ad are geerally deoted by the symbol X. I this cotext, the symbol X represets a fuctio defied o! ad does ot represet a geeric set or space. Example 4.2. Let p be the probability of a wi o ay idepedet attempt ad let q = 1! p be the probability of a loss. Let! = {(! 1,...,! ) :! i {W, L} for 1 # i # }, let F be the power set of!, ad let P({(! 1,...,! )}) = p k q k, where k is the umber of wis i the sequece (! 1,...,! ). Defie X :! # to be the umber of wis i a sequece. As i (a), X is measurable because F is the power set of!. I this sceario, X is called a biomial radom variable ad is deoted by X ~ b(, p). What is P(X = k )? The evet {X = k} is the set of outcomes i! that have exactly k wis i attempts. Moreover, because F is closed uder set differece, we have {X = k} = {X! k} {X! k 1} #F. Thus, we ca compute P(X = k ). By the defiitio of P, each idividual outcome with exactly k wis has probability p k q!k!. Because there # & distict ways to have exactly k%! k wis i a sequece, we the have P(X = k ) = # & k % p k q 'k. Whe X is a radom variable o a probability space (!, F, P ), the {X! t} is a measurable set for all t!. Thus, we always ca compute P(X! t ). This probability defies a fuctio F :!!, called the cumulative distributio fuctio or cdf, which is give by F(t) = P( X! t).
Example 4.3. (a) (Uiform Probability) Let! = [a, b] ad let X be a umber chose at radom from!. The we always have a! X! b. The cdf of X is give by # 0 if t < a % t a F(t) = P( X! t) = if a! t! b % b a & % 1 if t > b. (b) (Geometric Probability) Let p be the probability of a wi o ay idepedet attempt, ad let q = 1! p be the probability of a loss. Let X cout the umber of attempts eeded for the first wi. The for itegers! 1, the cdf gives the probability of wiig withi tries ad is give by F() = P( X! ) = 1 q (which comes from the complemet losig times i a row). We ca exted this fuctio to be defied for all t! by &( 0 if t <1 F(t) = P( X! t) = ' )( 1 q # t if t % 1. Properties of the CDF Let F(t) = P( X! t) be the cdf of a radom variable X defied o a probability space (!, F, P ). The (i) For all t!, 0! F(t )! 1. (ii) F(t) is a icreasig fuctio of t. (iii) lim F(t) = 0 ad lim F (t ) = 1. t!# t! (iv) For a < b, P(a < X! b) = F(b) F(a). (v) F(t) is right-cotiuous. That is, lim F(t ) = F(a) for all a!. t!a + Proof. (i) Because {X! t} #, we have by the axioms ad properties of probability measure that 0! P( X! t)! P() = 1; thus, 0! F(t )! 1 for all t!. (ii) If t 1 < t 2, the {X! t 1 } {X! t 2 }; thus, F(t 1 ) = P( X! t 1 )! P(X! t 2 ) = F(t 2 ).! (iii) Let {t } =1 icrease to. The the evets {X! t } =1 are ested icreasig with! {X! t } = # (assumig X is always real-valued ad ever assigs a value of + ). =1 Thus, lim F (t ) = lim ) = lim ) = P() = 1. (We leave lim F(t) = 0 as a t!!! t!# exercise.) (iv) For a < b, {X! a} {X! b}; thus, P(a < X! b) = P({X! b} {X! a}) = P( X! b) P(X! a) = F (b) F(a).
(v) Let a! ad let {a } be a sequece of real umbers that decrease to a. It suffices to show that F(a )! F(a). So let! > 0 be give. Because {a } is a decreasig sequece, the evets {X! a } form ested decreasig sequece of sets. The because P is a fiite measure, we have lim P( X # a ' ) = P! {X # a }! & ). % =1 ( Now because a! a for all, we have {X! a}! {X! a }. O the other had, if a < X(! ), the for large eough we have a < a < X(! ).! ) c which makes {X! a} {X! a } That is, we have {X! a} c ( {X! a }! ad therefore we have {X! a} =! {X! a }. Thus, # & F(a) = P(X! a) = P! {X! a } % ( = 1 ' = lim P(X! a ) = lim F(a ). ) ) Example 4.4. Let f :!! be give by f ( x) = x 3, where F is the Borel! -algebra o!. We claim that f is measurable. For all b!, we have {x : f (x )! b} = {x : x 3! b} = {x : x! b 1/3 } = (#, b 1/3 ] F, because the Borel! -algebra cotais all itervals of the form (!, c]. We ow shall prove some properties of measurable fuctios, icludig that various algebraic combiatios of measurable fuctios are still measurable. Theorem 4.1. Let f be a measurable fuctio defied o a measure space ( X, F, µ). (a) For all b!, (i) f!1 ((!, b) ) # F (ii) f!1 ((b, )) #F (iii) f!1 ([b,)) # F (iv) f!1 ({b}) F (b) For all a, b! with a < b, (i) f!1 ((a, b] ) F (ii) f!1 ((a, b) ) F (iii) f!1 ([a,b]) F (iv) f!1 ([a,b)) F Proof. (a) Because f is measurable, we kow f!1 ((!, c] ) # F for all c!. (i) We ote that (!, b) =! (!, b! 1 / ]. The because F is closed uder coutable uios, =1 we have f!1 ((!, b)) = f!1 # &! (!, b! 1 / ] % ( =1 ' =! f!1 ((!, b!1 / ] ) )F. =1
(ii, iii, iv) Because F is closed uder complemets ad itersectios, we have f!1 ((b, )) = f!1 (!, b] c f!1 ([b,)) = f!1 (!, b) c ( ) = ( f!1 ((!, b] )) c # F, ( ) = ( f!1 ((!, b) )) c #F, ad f!1 ({b}) = f!1 ((!, b] # [b,)) = f!1 ((!, b] ) # f!1 ([b,)) F. (b) Usig the results i (a), the fact that f!1 (V! U) = f!1 (V )! f!1 (U), ad the fact that F is closed uder set differece, we have (i) f!1 ((a, b] ) = f!1 ((!, b]! (!, a] ) = f!1 ((!, b] )! f!1 ((!, a] ) # F (ii) f!1 ((a, b) ) = f!1 ((!, b)! (!, a] ) = f!1 ((!, b) )! f!1 ((!, a] ) #F (iii) f!1 ([a,b]) = f!1 ((!, b]! (!, a) ) = f!1 ((!, b] )! f!1 ((!, a) ) #F (iv) f!1 ([a,b)) = f!1 ((!, b)! (!, a) ) = f!1 ((!, b) )! f!1 ((!, a) ) # F Note: Ay of the above coditios from (a) or (b) (except (a, iv)) could have bee used to defie the cocept of measurability. The followig exercise demostrates this fact. Exercise 4.1. Let f be a fuctio defied o a measure space ( X, F, µ). (a) Suppose f!1 ((!, b) ) # F for all b!. Prove that f is measurable. (b) Suppose f!1 ((a, b] ) F for all a, b! with a < b. Prove that f is measurable. Theorem 4.2. ( X, F, µ). Let f ad g be measurable fuctios defied o a measure space (a) Costat fuctios are measurable. (b) For every costat c!!, the fuctio c f is measurable. (c) For all itegers! 1, the fuctio f is measurable. (d) The fuctio f is measurable. (e) The sum f + g ad the differece f! g are measurable. (f) The product f g is measurable. (g) The fuctios max{f, g} ad mi{f, g} are measurable.
Proof. (a) Suppose h(x )! c. Let b!. If b < c, the h!1 ((!, b] )= # F. If b! c, the h!1 ((!, b] )= X # F. Thus, h is measurable. (b) If c = 0, the c f! 0 is measurable by (a). Otherwise, let b!. If c > 0, the (c f )!1 ((!, b] ) = {x #X : c f ( x) b} = {x #X : f ( x) b / c} = f!1 ((!, b / c] ) #F (c) Let b!. If is odd, the ( f )!1 ((!, b] )= {x # X : f (x ) b} = {x # X : f ( x) b 1/ } ( ) #F = f!1 (!, b 1/ ] If c < 0, the (c f )!1 ((!, b] ) = {x #X : c f ( x) b} = {x #X : f ( x) % b / c} = f!1 ([b / c, )) #F If is eve ad b! 0, the ( f )!1 ((!, b] )= {x # X : 0 f ( x) b} = {x # X :! b 1/ f (x ) b 1/ } ( ) #F = f!1 [!b 1/, b 1/ ] If is eve ad b < 0, the f! 0 > b; thus, ( f )!1 ((!, b] )= # F. (d) Let b!. If b < 0, the {x : f (x )! b} = #F. For b! 0, we have { x : f (x )! b} = {x: b! f (x )! b} = f 1 ([b, b] ) #F. (e) We shall use the result from Exercise 4.1(a) above. But we shall first show that { x : f ( x) + g(x ) < b} =! ({x : g(x ) < r}! {x : f (x ) < b r} ). r#q If g(x ) < r ad f ( x) < b! r for some r! Q, the f ( x) + g( x) < b! r + r = b. Thus, { x : f ( x) + g(x ) < b}!! ({x : g(x ) < r} {x : f ( x) < b # r} ). rq O the other had, suppose f ( x) + g( x) < b. The there exists a! > 0 such that f ( x) + g( x) < b! < b. But there also exists a ratioal umber r such that g(x ) < r < g( x) +!. But the f ( x) < b! g(x )! < b! r. Hece, { x : f ( x) + g(x ) < b}!! ({x: g( x) < r}! {x : f ( x) < b r} ). r #Q So for every b!, {x : g(x ) < r}! {x : f (x ) < b r}! F because f ad g are both measurable. Thus, the deumerable uio of these evets over all ratioal umbers is also i F. Hece, { x : f ( x) + g(x ) < b}! F for all b!, which makes f + g a measurable fuctio. From this result ad (b), it the follows that f! g = f + (!1)g is measurable.
(f) f g = 1 ( 4 ( f + g)2! ( f! g) 2 ) is measurable by Parts (b), (c), ad (e). (g) The result follows from (b), (d), (e), ad the followig idetities: max{f, g} = 1 2 ( f + g + f! g ) ad mi{f, g} = 1 2 ( f + g! f! g ). Exercise 4.2. Let f be a measurable fuctio defied o a measure space ( X, F, µ). Prove: (a) If m is a odd iteger, the f 1/m is measurable. (b) If m is a eve iteger ad f! 0, the f 1/m is measurable. (c) Let, m be positive itegers with gcd(, m) = 1. Explai the coditios that make f /m measurable. Defiitio 4.3. Let f be a measurable fuctio defied o a measure space ( X, F, µ). The kerel of f is defied by ker f = {x : f ( x) = 0}. We ote that ker f! F by Theorem 4.1 (a) (iv). If we exclude ker f from the domai, the we ca evaluate the reciprocal fuctio 1 / f. Theorem 4.3. Let f be a measurable fuctio defied o a measure space ( X, F, µ). Let 1 f : X! ker( f ) # be give by 1 f ( x) = 1 f (x ). The 1 is measurable. f Proof. Let b!. If b > 0, the 1 / f ( x)! b iff f ( x) < 0 or ( f ( x) > 0 ad 1 / b! f (x )). Thus, {1 / f! b} = { f < 0} ({1 / b! f } # {f > 0} )!F. If b = 0, the 1 / f ( x)! b iff f ( x) < 0 ; thus, {1 / f! b} = { f < 0}!F. If b < 0, the {1 / f! b} = ({1 / b! f } {f < 0} )!F. Hece, 1 / f is measurable. QED Exercise 4.3. Let f ad g be measurable fuctios defied o a measure space ( X, F, µ). Prove that f : X! ker(g) # is also measurable. g Defiitio 4.4. Give a measurable fuctio f, the positive part of f is defied by f + ( x) = # f ( x) if f (x )! 0 0 if f ( x) < 0 = max{f, 0}. The egative part of f is defied by f! #! f ( x) if f ( x) < 0 ( x) = =! mi{f, 0}. % 0 if f ( x) 0
Both f + ad f! are o-egative fuctios. It is easy to verify that f = f +! f! ad f = f + + f!. Because f + ad f! ca be expressed i terms of the max ad mi of the measurable fuctios f ad 0, the ext result follows from Theorem 4.2(g): Theorem 4.4. Let f be a measurable fuctio defied o a measure space ( X, F, µ). The f + ad f! are measurable. The Idicator Fuctio Let ( X, F, µ) be a measure space. For every A!F, we defie the idicator fuctio 1 A : X! by # 1 if x!a 1 A (x ) = % 0 if x A. Idicator fuctios ca be used to write fuctios that are piecewise defied over evets that form a partitio of the space. Example 4.5. Let X be a set of college studets with F cosistig of all subsets of X. Let A be the subset of all udergraduate males, let B be the subset of all udergraduate females, ad let C be the subset of all other studets. The A, B, C are mutually disjoit, ad X = A! B! C. Defie f : X! by f ( x) = # % 5 if x! A 10 if x!b 15 if x!c The f ca be writte i oe lie as f = 51 A +101 B + 151 C. We ote that the variable x is ofte suppressed whe usig the idicator otatio. Techically, we could write f ( x) = 51 A (x ) + 101 B ( x) + 151 C ( x). Because ay specific elemet x must be i oe ad oly oe of the subsets, oly oe idicator fuctio ca be 1 at a time while the others are 0.
Theorem 4.5. Let ( X, F, µ) be a measure space. For every A!F, the idicator fuctio 1 A is measurable. Proof. Let b!. Suppose first that b! 1. Because 0! 1 A ( x)! 1! b for all x, the 1 A!1 ((!, b]) = X! F. If 0! b < 1, the 1A!1 ((!, b]) = A c! F. If b < 0, the 1 A!1 ((!, b]) =!! F. QED Theorem 4.6. Let ( X, F, µ) be a measure space ad let A, B!F. The (a) 1 A!B = 1 A 1 B (b) 1 A!B = 1 A + 1 B 1 A 1 B. Proof. (a) For x!x, the value 1 A!B (x ) is either 1 or 0, ad the product 1 A (x ) 1 B (x ) ca oly be 1 or 0. But 1 A!B (x ) = 1 if ad oly if x!a B if ad oly if x!a ad x!b if ad oly if 1 A (x ) = 1 ad 1 B (x ) = 1 if ad oly if 1 A (x ) 1 B (x ) = 1. (b) Let x!x. We cosider four cases. (i) If x!( A B) c, the x!a c ad x!b c ; thus, 1 A!B (x ) = 0 = 1 A (x ) + 1 B ( x) 1 A (x )1 B (x ). (ii) If x!a B, the 1 A!B (x ), 1 A (x ), 1 B (x ), ad 1 A (x )1 B (x ) are all equal to 1 ad the result follows. (iii) If x!a B c, the 1 A!B (x ) ad 1 A (x ) are equal to 1 while 1 B (x ) is 0, ad the result follows. (iv) The case of x!a c B is similar to (iii). QED Example 4.6. Let f be a measurable fuctio ad let A = {f! 0}. The f + = f 1 A ad f! =! f 1 A c = f 1 A c. Defiitio 4.5. Let { A i } i=1 be a fiite collectio of disjoit evets i a measure space ( X, F, µ). A fuctio f : X! is called a simple fuctio if f is of the form f =! a i 1 Ai = a 1 1 A1 +...+ a 1 A, i =1 where a 1,..., a are real umbers. Notes: (i) Wheever coveiet, we ca assume that a 1,..., a are distict real umbers. For if a i = a j = a, the because A i ad A j are assumed to be disjoit, we the ( ) = a1 Ai! A j. That is, we ca replace the sum of the two have a i 1 Ai + a j 1 Aj = a 1 Ai + 1 Aj terms a i 1 Ai + a j 1 Aj with the sigle term a1 Ai!A j, ad A i! A j is still i F ad is disjoit from the other evets A k.
(ii) By re-labelig the evets, we ca always assume that a 1 < a 2 <... < a. From Theorems 4.5 ad 4.2, we obtai: Theorem 4.7. Simple fuctios are measurable. Exteded Real-Valued Fuctios Suppose f : X! [#, #] is a fuctio that is allowed to take ifiite values. We ca still say that f is measurable provided { f! b}!f for all b!. If f is measurable,! the {x : f (x ) <!} =! {f }!F ad {x : f (x ) =!} = {x : f (x ) <!} c F. =1 More Exercises Exercise 4.4. (a) Let f :!! be give by f ( x) = x 2. Use the formal defiitio of measurability to prove that f is measurable with respect to the Borel! -algebra o!. (b) Prove that the followig real-valued fuctios are Borel measurable: (i) g(x ) = ta x (ii) h(x ) =! x Exercise 4.5. Let X be a set of college studets with F cosistig of all subsets of X. Let A be the subset of all udergraduate males, let B be the subset of all udergraduate females, ad let C be the subset of all other studets. Assume P({!}) = 1 / X for all! X, ad that P(A) = 0.50, P(B) = 0.30, P(C) = 0.20. Defie f : X! by f ( x) = # % 5 if x! A 10 if x!b 15 if x!c Fid the cdf fuctio F(t) = P( f! t ) as a piecewise defied fuctio of all t!. Exercise 4.6. Prove that lim F(t) = 0 for ay cdf fuctio F. t!# Exercise 4.7. Let f :!! be a cotiuous fuctio. Prove that f is measurable with respect to the Borel! -algebra o!. Exercise 4.8. Let f : X! [#, #] be a exteded real-valued fuctio defied o a measure space ( X, F, µ). Prove that {x : f (x ) >!}!F ad {x :! < f ( x) < }!F.