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# REAL ANALYSIS MATH 532

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This 55 page Class Notes was uploaded by Sadie Schroeder on Wednesday September 30, 2015. The Class Notes belongs to MATH 532 at Western Kentucky University taught by David Neal in Fall. Since its upload, it has received 38 views. For similar materials see /class/216738/math-532-western-kentucky-university in Applied Mathematics at Western Kentucky University.

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Date Created: 09/30/15

Dr Neal Spring 2009 MATH 532 Measurable Functions Throughout let X F p be a measure space and let Q F P denote the special case of a probability space We shall now begin to study realevalued functions defined on X In order to do so we need to assume that our functions have the special property of being measurable The formal definition of this property depends on the concept of the inverse image of a function which we shall discuss first De nition 41 Let f X gt Y be a function between spaces X and Y Let Vg Y The inverse image of V under f is the subset of X given by f 1V x eXfxe V Then x ef 1V if and only if fx 6 V It is always the case that unions intersections complements and set differences are preserved by the inverse image of a function That is a f ILU VuJ U lm an r1 n VuJ n rm c f 1Vc f 1VC d f 1V U f Rm f lw We prove properties a and c below but leave the proofs of b and d as exercises Proof a We have x e fTIEU Va iff fxe U Va iff fxe Va for some 1 iff X I x e f 1Va for some a iffx e U f 1Va l c We have x e f 1VCifffx e V0 ifffx e V iffx e f1V iffx 6f1Vc Definition 42 Let f X gt SR be a realevalued function defined on X Then f is called measurable provided f 1 oob e F for all intervals of the form oob g SR That is x EX fxS b must always be a measurable set by being in the of algebra F We then can measure this event by evaluating px EX f x S b For convenience we often shall abbreviate the notation for this inverse image by writing the set as fS b and by writing its measure as f S b Dr Neal Spring 2009 Example 41 Let Q 1 Z 20 with F being the power set of Q and Pn 1 20 for all n 69 Let f Q gt SR be defined by 2 n if n is prime fnn2 otherwise Because F is the power set of Q it contains all subsets of Q including all inverse images of the form fS b Thus f is measurable What is P f S 10 We simply need to determine which values in the domain have function values that are less than or equal to 10 We see that f S 10 1 2 3 5 thus Pf s 10 420 Note When Q F P is a probability space then measurable functions are called random variables and are generally denoted by the symbol X In this context the symbol X represents a function defined on Q and does not represent a generic set or space Example 42 Let p be the probability of a win on any independent attempt and let q 1 p be the probability ofa loss Let Q m1 con 11 eW L for 1 S z39S n let F be the power set of Q and let Po1 wn pk qnik where k is the number of wins in the sequence 11 con Define X 9 SR to be the number of wins in a sequence As in a X is measurable because F is the power set of Q In this scenario X is called a binomial random variable and is denoted by X bn p What is PX k The event X k is the set of outcomes in Q that have exactly k wins in n attempts Moreover because F is closed under set difference we have XkXSk XSk 1EF Thus we can compute PX k By the definition of P each individual outcome with exactly k wins has probability pk qnik Because there Z distinct ways to have exactly k wins in a sequence we then have PX k pk qnik When X is a random variable on a probability space Q FP then X St is a measurable set for all t 6 SR Thus we always can compute PX S t This probability defines a function F SR gt SR called the cumulative distribution function or cdf which is given by Ft PXS t Dr Neal Spring 2009 Example 43 a Uniform Probability Let Q a b and let X be a number chosen at random from Q Then we always have a S X S b The cdf of X is given by 0 iftlta FtPth ifaStSb 1 iftgtb b Geometric Probability Let p be the probability of a win on any independent attempt and let q 1 p be the probability of a loss Let X count the number of attempts needed for the first win Then for integers n 2 1 the cdf gives the probability of winning Within n tries and is given by Fn PX S n 1 qquot which comes from the complement losing n times in a row We can extend this function to be defined for allt 6 SR by 0 if I lt1 FtPXSt 0 1 97m iftZl Properties ofthe CDF Let F t PX S t be the cdf of a random variable X defined on a probability space Q FP Then i For allt GER 0 S Ft S 1 ii Ft is an increasing function of t iii lim Ft 0 and lim Ft 1 iv For a lt b Palt XS 1 Fb Fa 9700 00 v Ft is rightecontinuous That is 1im Ft Fa for all a GER Igta Proof 1 Because X St g Q we have by the axioms and properties of probability measure that 0 SPXS t S PQ 1 thus 0 SFt S 1 for all t GER 3911 If t1 lt Q then X sq g XS t2 thus F01 PXS t1 3 PX 3 t2 F02 iii Let In ll increase to 00 Then the events X S In ll are nested increasing with U X S tn Q assuming X is always realevalued and never assigns a value of 00 nl Thus lim Ft lim Fan lim PXS tn PQ 1 We leave lim Ft 0 as an 1400 new new raise exercise iv For a lt b X S agX S b thus PaltXSbPXS b XSaPXS b PXSaFb Fa Dr Neal Spring 2009 V Let a GER and let an be a sequence of real numbers that decrease to a It suffices to show that Fan gtFa So let 1 gt0 be given Because an is a decreasing sequence the events X S an form nested decreasing sequence of sets Then because 00 P is a finite measure we have lim PXS an P H XS 613 n 00 nl Now because a S an for all n we have X S a g H X S an On the other hand if n altXu then for n large enough we have altanltXn That is we have X S ac g X S anc which makes X S a 2 H X S an and therefore we have n n X s a XS an Thus n FaPXS a P XS cw lim PX San lim Fan nl new ngtoo Example 44 Let f SR gt SR be given by fx x3 where F is the Borel o algebra on SR We claim that f is measurable For all b 6 SR we have xfxSb x x3 Sbx be13 oo b13 eF because the Borel o falgebra contains all intervals of the form 00 c We now shall prove some properties of measurable functions including that various algebraic combinations of measurable functions are still measurable Theorem 41 Let f be a measurable function defined on a measure space X F p a For all b GER i f 1 oobeI39 n f 1boo eF in f 1booeI39 iv f 1beF b Forall a b GER witha lt b 1 f71abeF 3911 f71abeF iii f 1aber 1V f71abeF Proof a Because f is measurable we know f71 ooc e F for all c GER i We note that oob U 00 b 1 n Then because F is closed under countable unions n l we have f1 oobf1G oob 1nJ jf 1 oob 1neF 711 nl Dr Neal Spring 2009 ii iii iv Because F is closed under complements and intersections we have f 1boo f 1 oo bicf 1 oobic e F f 1boof 1 oobcf 1 oobc eF and f 1bf 1 oobi ob 0 f 1 oo we f 1booe F b Using the results in a the fact that f71V U f71 V f71U and the fact that F is closed under set difference we have i f 1abif 1 oobi 0001 f 1 oobi f 1 ooai e F ii f 1abf 1 oob ooai f 1 oob f 1 ooai e F iii f lqabif 1 oo b ooa f 1 oobi f 1 ooa e F iv f 1abf 1 oob ooaf 1 oob f 1 ooa e F Note Any of the above conditions from a or b except a iv could have been used to define the concept of measurability The following exercise demonstrates this fact Exercise 41 Let f be a function defined on a measure space X F u a Suppose f 1 oob e F for all b GER Prove that f is measurable b Suppose f71a b e F for all a b 6 ER with a lt b Prove that f is measurable Theorem 42 Let f and g be measurable functions defined on a measure space X F M a Constant functions are measurable b For every constant c 6 SR the function cf is measurable c For all integers n 2 1 the function f n is measurable d The function fl is measurable e The sum f g and the difference f g are measurable f The product f g is measurable g The functions max f g and min f g are measurable Dr Neal Spring 2009 Proof a Suppose hxE c Let b GER lf b lt c then h71 oob Q6 F If b 2c then h71 oo b X e F Thus h is measurable b If c 0 then cf E 0 is measurable by a Otherwise let b GER lfcgt0then lfclt0then cf 1 oob x eX cm s b cf 1 oob x eX cm s b xeXfxSbc xeXfxZbc f 1 oobc eF f 1b cooe F c Let b GER If n is odd then If n is even and b 2 0 then f 1oob x eXf1xS b f 1 oob x ex 0 s f x s b xexfxsb xeX b sfxsb flwbln E flblnbln El lfn is even and b lt 0 then fquot 2 0 gt b thus fn 1 oob Q e F d LetbeSR lfblt0thenxfx b eF Forb20wehave xfx b x b fxs b f 1 bb 6F e We shall use the result from Exercise 41 a above But we shall first show that x1fxgxlt 13 UQx1gxltV x1fxlt 13 r lfgxltr and fxltb r for some re Q then fxgxltb rr b Thus x1fxgxltb2 UQx1gxltV x1fxltbV On the other hand suppose f x g x lt b Then there exists an s gt 0 such that fxgx lt b e lt b But there also exists a rational number r such that gxlt rlt gx 5 But then fx lt b gx a lt b r Hence x fxgxlt bg U xgxltrnxfxltb r rEQ So for every b GER xgxlt rnx fx lt b r e F because f and g are both measurable Thus the denumerable union of these events over all rational numbers is also in F Hence xfxgxlt b e F for all b GER which makes fg a measurable function From this result and b it then follows that f g f 1g is measurable Dr Neal Spring 2009 f fg f g2 f g2 is measurable by Parts b c and e g The result follows from b d e and the following identities 1 1 maXfg3fgIfgl and m1nfg 5fgfgl Exercise 42 Let f be a measurable function defined on a measure space X F u 1 m is measurable b If m is an even integer Prove a If m is an odd integer then f and fZO then flm gcd n m 1 Explain the conditions that make f is measurable c Let nm be positive integers with n m measurable De nition 43 Let f be a measurable function defined on a measure space X F u The kernel of f is defined by kerf x fx 0 We note that ker f e F by Theorem 41 a iv If we exclude ker f from the domain then we can evaluate the reciprocal function 1 f Theorem 43 Let f be a measurable function defined on a measure space X F u l l l 1 Let X kerf gt SR be given by x Then is measurable f f f x f Proof Let beSR lfbgt0 then 1fxSb iff fxlt0 or fxgt0 and 1bex Thus1fsbflt0o1bsfnfgt0 eF lfb0then1fx b ifffxlt0thus1beflt0 eF lfblt0 then 1be1b fnflt0 eF Hence lfis measurable QED Exercise 43 Let f and g be measurable functions defined on a measure space X F u Prove that j X kerg gt SR is also measurable g De nition 44 Given a measurable function f the positive part of f is defined by fx iffx20 f x 0 iffxlt0 maXf 039 The negative part of f is defined by fx iffxlt0 700 0 iffx20 Z minf 039 Dr Neal Spring 2009 Both f T and f T are nonenegative functions It is easy to verify that ff f and ff f Because f T and f T can be expressed in terms of the max and min of the measurable functions f and 0 the next result follows from Theorem 42 g Theorem 44 Let f be a measurable function defined on a measure space X F 1 Then f T and f7 are measurable The Indicator Function Let X F p be a measure space For every A e F we define the indicator function 1anmw 1 1 xeA A 0 xeA lndicator functions can be used to write functions that are piecewise defined over events that form a partition of the space Example 45 Let X be a set of college students with F consisting of all subsets of X Let A be the subset of all undergraduate males let B be the subset of all undergraduate females and let C be the subset of all other students Then A B C are mutually disjoint and X A U B U C Define fX gt SR by 5 ifxeA fx 10 ifx EB 15 ifxeC Then f can be written in one line as f 51A 1013 151C We note that the variable x is often suppressed when using the indicator notation Technically we could write f x 51Ax1013x151cx Because any specific element x must be in one and only one of the subsets only one indicator function can be 1 at a time while the others are 0 Dr Neal Spring 2009 Theorem 45 Let X F p be a measure space For every A e F the indicator function 1A is measurable Proof Let b GER Suppose first that b 2 1 Because 0 S 1Ax S 1 S b for all x then 1A 1 oob X e F If 0Sblt1 then 1A 1 oob Ace F If blt0 then 1A 1 oob o e F QED Theorem 46 Let X F p be a measure space and let A B e F Then a 1AmB 1A 13 b 1AoB 1A 13 1AlB Proof a For x eX the value lA B x is either 1 or 0 and the product 1A x 13 x can only be 1 or 0 But lA Bx l ifand only if x EA nB ifand only ifx EA and x EB if and only if1Ax l and 13x l ifand only if 1Ax 13x l b Let x eX We consider four cases i If x eA UBC then x 6A0 and x 630 thus 1AuBx 0 1AxlBx 1AxlBx 391391If x EA n B then 1AuBx 1Ax 13x and 1A x13x are all equal to l and the result follows iii If x eAnBc then IAUE x and 1A x are equal to 1 while 13 x is 0 and the result follows iv The case of x eAc n B is similar to iii QED Example 46 Let f be a measurable function and let A f 2 0 Then f flA and f7f1Ac f1Ac Definition 45 Let AZH11 be a finite collection of disjoint events in a measure space X F u A function f X gt SR is called a simple function iff is of the form n f 211691141 a11A1 anlAn 1 where all an are real numbers Notes 1 Whenever convenient we can assume that al an are distinct real numbers For if 611 611 a then because AZ and A are assumed to be disjoint we then have ell141 611 1A a1Ai IAJ alAiUAj That is we can replace the sum of the two terms ell141 aJIAJ with the single term alAiUAj and AZ UAJ is still in F and is disjoint from the other events Ak Dr Neal Spring 2009 ii By reilabeling the events we can always assume that all lt a2 lt lt an From Theorems 45 and 42 we obtain Theorem 47 Simple functions are measurable Extended RealiValued Functions Suppose f X gt oo 00 is a function that is allowed to take infinite values We can still say that f is measurable provided f S b e F for all b GER If f is measurable then xfxltoo glfSn eFandxfxooxfxltooc eF More Exercises Exercise 44 Let f SR gt SR be given by f x x2 Use the formal definition of measurability to prove that f is measurable with respect to the Borel 0 algebra on SR Exercise 45 Let X be a set of college students with F consisting of all subsets of X Let A be the subset of all undergraduate males let B be the subset of all undergraduate females and let C be the subset of all other students Assume Pm 1 X for all u EX and that PA 050 PB 030 PC 020 Define f X gt SR by 5 ifxeA fx 10 ifx EB 15 ifxeC Find the cdf function Ft Pf S t as a piecewise defined function of all t 6 SR Exercise 46 Prove that lim F t 0 for any cdf function F raise Exercise 47 Let f SR gt SR be a continuous function Prove that f is measurable with respect to the Borel o algebra on SR Exercise 48 Let f X gt oo 00 be an extended realivalued function defined on a measure space X F u Prove that x fx gt oo 6 F and x 00 lt fx lt oo 6 F Dr Neal Spring 2009 MATH 532 Sequences of Functions Throughout let X T u be a measure space and let fn 5101 be a sequence of reali valued functions defined on X We now shall study some properties of this sequence of functions including various types of convergence Theorem 51 Assume fn are all measurable functions Let f X gt oo co and g X gt oo oo be defined by fx SUP fnx and gx inf fnx n n Then f and g are measurable Proof Fix x EX lf fn x01 is bounded above respectively bounded below then fx sup fn x respectively gx inf fnx exists as a finite number Otherwise n n we let fx 00 respectively gx oo Now let b GER Because sup fnx S b if and only if fnx S b for all n we have n x fx S b xfnx S b e 9quot Thus f is measurable nl Also inf fnx 2 b ifand only if fnx2 b for all n So g2 b fn 2 b e T n nl Thus glt b g 2 bc e T for all b GER which implies that g is measurable see Exercise 41 a QED Two Modes of Convergence Again let fn 3101 be a sequence of realivalued functions defined on X and let E gX For each x EX fn x01 is a realivalued sequence that may or may not converge with respect to the absolute value metric of the real line I Pointwise Convergence 7 A sequence of functions fn converges pointwise on E provided fn x01 converges in SR for every x eE Then fx lim fn x is a welli ngt00 defined function on E because limits are unique We then write fn gt f on E Then given a fixed x eE we can say For every 2 gt 0 there exists an integer Nx Z 1 that depends on the x being used such that ifn Z Nx then I fn x fx lt s II Uniform Convergence 7 A sequence of functions fn converges uniformly on E to the function f x provided For every 5 gt 0 there exists an single integer N Z 1 that works for every x eE such that if n ZN then lfnx fx I lt e for all x eE In this case we again write fx lim fn x but also write fn 3 f on E ngt00 Dr Neal Spring 2009 Theorem 52 a Uniform convergence always implies pointwise convergence b If fn are measurable and fn gt f on X then f is measurable Proof a Follows directly from the definitions of convergence b Let b GER We shall show that x fxS b is a measurable set To do so we assert that Xi003 U1 0Nxfnxltbs EEQ N which is a measurable set because it consists of denumerable intersections unions of measurable sets Fix x EX lf fx S b then for every 5 gt 0 there exists an integer N Z 1 such that fnxltfxsSb foralanN Thusxe H U H xfnxltbs 56QNl nN On the other hand suppose x 6 H U H xfnxltbe Let s39gt0 be 6Q N1 nN given Then there exists an e eQ with 0 lt e S 539 And for this s eQ there is an integer N 21 such that fn x lt b 2 for all n ZN But this means that fx lim fnxSbe S be39 for all e39gt0 Because fxS b 39 for all e39gt0 we ngt00 must have f x S b QED Example 51 Let fn 0 oo gt SR be defined by fnx10i for n 2 1 We assert nx that fn converges pointwise on 0 00 to the function f x 10 but does not converge uniformly So let x 60 00 and let 8 gt 0 be given Let N be the smallest integer such that 1xeltNsothat1Nltxe Thenfor nZNwehave l Ifnxfx10 10 nx Thus fn x gt fx pointwise for all x 60 00 But as we can see the choice of N depends on the specific x in the domain But suppose fnx fx i lt a for all x 60 00 if n 2N0 for some N0 2 1 nx 1 Then we would have 0 lt N lt x for all x 60 00 which is a contradiction because we 0 E 1 can always find a real number between 0 and N 0 6 Dr Neal Spring 2009 Using properties of realivalued sequences we obtain the following results Proposition 51 Let fn gt f on E and let gn gt g on F Then a ForallceSRcfn gtcfonE b fngn gtfgonEnF c fngn a fg on MR d fnx gnx gt fxgx for x eEnF provided gx 0 and gnx 0 for all butafinite number of n Properties a and b also hold for uniform convergence but Property c does not hold for uniform convergence without additional boundedness conditions We next prove a Cauchy criterion that is equivalent to uniform convergence Theorem 53 A sequence of functions fn converges uniformly on E if and only if for every 5 gt 0 there exists an integer N21 such that fnx fmxlt e for all x eE whenever mm 2 N Proof Suppose fn 3 f Let s gt 0 Then there exists an integer N 21 such that Ifnx fx I lt a 2 for all x eE whenever n ZN Then for mm 2 N we have for all x eE thatlfnx fmxlslfnx fxllfx fmxllt e 2 5 2 a Conversely suppose the Cauchy criterion holds Then for each x eE fn x is a Cauchy sequence and therefore has a realivalued limit f x because SR is complete We assert that fn 3 f on E Let s gt 0 Then there exists an integer N 21 such that Ifnx fmxlt 82 for all x eE whenever nm ZN Let n ZN and let x eE For this x we can choose an integer MZN such that IfMx fxlte 2 Then IfnxfxSfnxfMxfMxfxlt82828 QED It is also the case that the uniform limit of continuous functions is continuous Theorem 54 Let X dX be a metric space and let EgX Let fn E gt SR be a continuous function for all n 2 1 and suppose that fn 3 f on E Then f is continuous on E Proof Let a GE and let a gt 0 Then there exists an integer N Z 1 such that if n 2 N then fnx fxlt e 3 for all x eE Because fN is continuous at a there exists 5 gt 0 such that if x eE and dXxa lt6 then IfNx fNalt 23 So let x eE with dXx a lt 6 Then fxfaSfxfNxfNxfa SI MfNOCfNxfNafNafa lte3e3e3e Thus f is continuous at a QED Dr Neal Spring 2009 Theorem 54 gives an interesting interchange of limits result in the case when a is also a limit point of E Because f is continuous at a and fn x gt fx for all x we have fa lim fx lim lim fnx xgta xgtangtoo Then because each fn is continuous ata we have lim lim fnx lim fnafa new xgta 7100 Hence lim lim fnx lim lim fnx new xgta xgta new The next result gives conditions for when pointwise convergence implies uniform convergence Theorem 55 Let fn a b gt SR be a sequence of realevalued functions defined on a closed bounded interval ab g SR Assume a Each fn is continuous on 61b b fn gt f pointwise on 6113 and that f is also continuous on 61b c For each x ea b fnx Z fn1x for all n Then fn converges uniformly to f on ab Proof Let gn fn f for all n Because f1 Zfz Z an anH 2 we must have that fn x 2 lim fnx fx for all x ea b Thus gn Z 0 for all n But we also have new gn Z gn 1 for all n because of the decreasing assumption of fn Because f and each fn are continuous on ab then gn is also continuous on 6113 for all n and we also have that gnx fnx fx gt 0 for all x ea b Let a gt 0 be given For each n let Kn x eab gnx2 a If gn1x2 a then gn x Z s also because gn Z gnH thus KnH g Kn Now suppose Kn i Q for all n Then Kn is actually the set Kn g1e max gn a x b which is the inverse image of a closed set under a continuous function and therefore must be closed But Kn is also a subset of ab so Kn is a closed bounded set and therefore must have the form an bn So Kn is a nested decreasing set of none empty intervals and therefore Kn must contain a point x0 by the Nested Interval n Theorem However gn x0 gt 0 thus there exists an integer N Z 1 such that gn x0 lt s for all n ZN That is x0 sEKn for n ZN which contradicts that x0 6 Kn So we must n have that KM Q for some M which makes Kn Q for all n 2M Thus for all n 2M we have Ifnx fx gnx lt s for all x ea b Hence fn 3 f on ab QED Dr Neal Spring 2009 Note The functions fn x 10 1 n x are all continuous on 0 00 they satisfy an S fn for all n and they converge pointwise to the continuous function fx 10 However they do not converge uniformly as shown in Example 41 This example illustrates why we need the additional condition of working on a closed bounded interval a b De nition 51 a A function f X gt SR is bounded if there exists a positive real number M such that I fx S M for all x EX b A sequence of realivalued functions fn is uniformly bounded if there exists a positive real number M such that for all n Ifnx SM for all x EX Exercise 51 Let fn 5101 be a sequence of realivalued functions defined on a set X Assume that each f n is a bounded function Prove a If fn converges uniformly to f on X then f is bounded b If fn converges uniformly on X then fn is uniformly bounded Exercise 52 Let fn and gn be sequences of realivalued functions defined on X that converge uniformly on E g X a Prove that fn gn converges uniformly on E b Assume also that fn and gn are sequences of bounded functions Prove that fn gn converges uniformly on E Exercise 53 Construct sequences fn and gn that converge uniformly on some subset of SR but such that fn gn does not converge uniformly Exercise 54 Let X dX be a metric space and let EgX Let fn E gt SR be a continuous function for all n 2 1 and suppose that fn 3 f on E For a GE prove that In fnxnfa ngt00 for every sequence xn from E such that xn gt a in X More Modes of Convergence The definitions of pointwise and uniform convergence do not require any conditions on the space X However the next two forms of convergence do require that we are working on a measure space and that all functions are assumed to be measurable Dr Neal Spring 2009 III Convergence Almost Everywhere 7 Let X T p be a measure space and let fn be a sequence of measurable functions Then fn converges almost everywhere to the measurable function f provided fn x01 converges in SR to f x for every x e X A where A is a set of measure 0 In this case we write fn gt f ae IV Convergence in Measure 7 Let X T p be a measure space A sequence of functions fn converges in measure to the measurable function f provided for every 1 gt0wehave lim Ifn fl Za0 Inthiscasewewrite fn gt f ngt00 1L Notes a If fn gt f then fn gt f ae using the set A Q b On a probability space we use the alternate phrase converges almost surely and write fn gt f as in place offn gt f ae c If f a f ae then we fnoc 4 fx 0 d fn gt f if and only if for every 8 gt 0 and every 839 gt 0 there exists an integer N Z 1 suchthat Ifn f28lt839f01 alanN e If fn converges in measure to f in a probability space then we say instead that fn converges in probability to f and write fn gt f P Example 52 a For n 2 1 let fn 0 00 gt SR be defined by ix ifxex fnx n1 n ifxex For instance f1 x x2 provided x ex but f1x l for all x ex Likewise f20x 20x21 for x ex but f20x 20 for all x ex Then for all x ex we have n 1 Now define f 0 00 gt SR and f 0 00 gt 0 00 by lim fnx lim x x Butforall x ex lim fnx lim n oo ngtoo ngtoo 7 ngt00 new d N x if x ex fxx an fx oo ifxex Then fn gt f everywhere on 0 00 However with respect to Lebesgue measure A fn gt f ae on 0 00 That is fn gt f for all x e 0 oo x and hx 0 Dr Neal Spring 2009 b For n 2 1 let fn 0 00 gt SR be defined by 10 ifoln fquotx 0 ifOSxlt1n and let f 0 co gt SR be the constant function fx 10 For every 0 lt 1 S 10 the set on which fn and f differ by at least 8 is always x e0oolfnx fxI22 01n And with regard to Lebesgue measure lim A0 1 n lim 1 n 0 For 8 gt10 naoo ngt00 fn fZe and A 0 Thus fn gt f It is also the case that fn gtf ae 7x for lim fnx 10 for all x 60 00 new Proposition 52 Let fn gt f ae and let gn gt g ae Then a ForallceSRcfn gtcfae b fngn gtfg ae c fn gn gt fg ae d fnx gnx gt fxgx aeon the set g 0 The above proposition follows immediately from the properties of realivalued sequences with convergence occurring on the set x ifnx 4gt fxUx1gnx4gt gxc The following results show that the limiting functions with regard to convergence almost everywhere and convergence in measure are unique up to a set of measure 0 Theorem 56 Let X T p be a measure space a If fn gt f ae and fn gtg ae then fg ae b lffn gt fand fn gt gthenfg ae u u c If fn gt f ae and fg ae then fn gtg ae d lffn gt fandfg aethen fn gt g u u Proof a Let Al x fnx fx and A2 x fnx w4gtgx Because fn gt f ae and fn gtg ae we have 141 0 142 Thus A1 U142 0 also Then for all x EX AIUAz we have fnx gt fx and fnx gtgx By uniqueness of limits we have fx gx for all x eX A1 U142 Thus f g ae Dr Neal Spring 2009 b Let A x eXfx gxgt 0 We mustshow that MA 0 lffx fnxlt 2 and Ifnx gxlts2thenwe have fxgxS fxfnx fnxgx lt8 Thus xfxfnxlt82ox1fnxgxlltE2gxfxgxlt By taking complements we have xfxgx28gx1fxfnx282Ux1fnxgxZE2 Let s39 gt 0 be given Because fn gt f and fn gt g there exists N Z 1 such that M u uf fNZe2lte392 and IfN gIZe2lts392 Hence If gZ SpLf fNZs2pfN gZs2lts39 for every s39gt0 Thusuf gZs0 foreverya gt0 Butthen A Ufg28 EEQ which is a denumerable union of sets of measure 0 Thus 1 A 0 c Let A1 xfnx fx and A2 xfx gx Then A1 UA20 and for x 6 A1 UA2cwe have fnx gt fx gx Thus fn gtg ae d LBAn xfnxgx25 Then AnAn fgUAn f g50 that MAnuAnof guAnof g Sllltxifnx gx213gt fg Urig Mxilfnx fx28 fg0 S x Ifnx fx Ze gt0 as n gt 00 Thus fn gt g QED u Of the four modes of convergence that we have defined uniform convergence is the strongest If fn 3 f then fn gt f which in turn implies fn gt f ae The next result shows that fn 3 f also implies fn gt f l1 Dr Neal Spring 2009 Theorem 57 Let X T p be a measure space If fn are measurable functions that converge uniformly to f on X then fn converges in measure to f Proof Let s gt 0 be given Because fn 3 f there exists an integer N 21 such that fnx fxlte forallx EX whennZN Thus for nZN fn fZe which makes Ifn fZx0 Hence lim Ifn f220foralle gt0and fn gt f ngt00 ll QED Convergence in measure does not imply uniform convergence nor does it imply convergence ae However we can say the following Theorem 58 If fn gt f then there exists a subsequence such that fnk gt f ae 1 Proof Because fn gt f there exists an increasing sequence of integers nk such that Ifn flz 27klt27k foralln an Now let Ek xeXfnkx fx2 27kand letA UEk Suppose x6140 U E U fnk flt27k and let gt0 ilki lki ilki 139 be given 00 Because x 6146 there is some 13921 such that x 6 Ifnk flt 27k Then for ki K 2139 large enough we have 27K lt a so for k2 K we have Ifnk x fxlt 27k lt 6 Thus fnk gt f for all x 6146 Moreover we assert that uA 0 For each 139 Z 1 we have 12H 2 1 12 uA M UEkJSP UEkJS Z uEkS Zf il ki ki ki ki Because 0 S uA S1 21 for all 139 Z 1 we must have uA 0 Thus fnk gt f ae QED Finally convergence almost everywhere does not imply uniform convergence nor pointwise convergence However it will imply convergence in measure provided we have a finite measure space Dr Neal Spring 2009 Theorem 59 Egorov Let X T u be a finite measure space If fn gt f ae then fn gt f u Proof Let 8 gt0 be given We must show that lim Ifn fIZs0 Let ngt00 A x eXfnx fx Because fn gtf ae we have pA 0 and thus uAc uX ultAgt MX For all n21 fZ flt g fl flte thus the sets fZ flte in inl in form a nested increasing sequence of events as n increases Therefore 00 00 00 u U flt8 11m u f1flte n l in 7 139 n However if x 6146 then fnx gt fx thus x e U fl flts Hence nl in 0 Ac U n Ifi fllteandtherefore l in uXZn139nwu I flt5u Li fiflt5ZMAcMX Because uX ltoo we must have lim u fi flte 0 Therefore quot00 in hm ultlfn fI22gts1im uUfifZEJ0 QED new new 171 The following definition gives another version of the concept of Cauchy Definition 52 A sequence of measurable functions fn is called Cauchy in measure or fundamental in measure provided for every e gt 0 we have lim I fn fm I Z a 0 nm gtoo That is for every 8 gt 0 and every 239 gt 0 there exists an integer N 21 such that pfn fmZ elt 839 whenever mm 2 N Theorem 510 If fn gt f then fn is Cauchy in measure 1 Dr Neal Spring 2009 Proof Let s gt 0 and s39 gt 0 be given Because fn gt f there exists an integer N 21 u suchthatufn fZe2lte392 foralanN Thenforallnm ZNwehave Ifnflt82 ffmlt 2 fnfmlt8 thus bytakingcomplements Ifn fmIZegfn f282Uf fmIZeZ So for mm 2 N we have lenfmIZESMfnfZE2HIffmIZS 2lte392e392e39 Hence fn is Cauchy in measure QED Convergence in Distribution Let Q T P be a probability space and let X n 1 be a sequence of random variables defined on Q We say that Xn converges in distribution to the random variable X provided lim PXn S t PX S t for all t GER new It follows that lim Ps S Xn S t Ps S XS t for all s t GER Although this new concept can be generalized to any measure space it is used mostly in probability and statistics to approximate probabilities Example 53 a Let X be any random variable and let Xn be the truncation of X defined by X ifXSn XHZXlngn n ifXgtn for n 2 1 Then Xn converges in distribution to X To see this let I 6 SR Then let tlt n leS t then XS n so Xn X and Xn St also Moreover if Xn St then Xn lt n so we cannot have Xgt n Thus XXn and XSt also That is XSt iff Xn St for tlt n So for n gtt we have PXn S t PXS t b As a specific example of truncation let X N geo p be a geometric random variable that counts the number of attempts needed for the first win where p is the probability of a win on any independent attempt and q 1 p is the probability of a loss If you do not win within n attempts then you quit so that you make a maximum of n attempts Then Xn X 1391 gives the number of attempts needed to win if it takes no more than n tries or gives n when it takes more than n tries to win The probability distribution functions are given by k4 qkilp iflSkSn l PXkq pforkZl and PXnk qn l ifk n Dr Neal Spring 2009 If you have n 1 losses in a row with probability qnil then it automatically takes at least n tries thus Xn n The cumulative distribution functions are given by 0 HkSO 0 ika0 k PXSk and PX Sk 1 iflSkSn l l qk ikaI 1 q 1 1kan For t GER PXStPXSLtJ and PXn StPXn SLtJ So for t eiRby choosing ngtt we then have LtJSn l thus PXnSt1 qmPXSt for tzl and PXnSt0PXStfortlt1 c Let Xn and X be random variables such that 0 SXn S XnH S X for all n 2 1 and lim Xn on X 1 for all u e Q We assert that X n converges in distribution X n 00 lf Xu S t then Xn1u S t which in turn implies that Xn 1 S t Thus we have X S t g XnH S t g Xn S t for all n 2 1 which makes Xn S t a nested decreasing sequence of sets Because a probability space is a finite measure space we then have 00 thatP norms lim PXnSt n71 new But ifo 6 Xn S tthen XnoS t for all n thus Xru lim XnoS t Hence nl new XnStXSt Thus PXStP XnSt lim PXnSt for all teSR ngt00 nl hence X n converges in distribution X The Truncation of a Function into Simple Functions This next construction gives a further generalization of truncation and is crucial to the development of integration theory Initially let f X gt 0 00 be a nonnegative realivalued function We will construct a sequence of simple functions fn such that for all x EX 0 Sf1xSf2xS anxS Sfx and quotlinwfnbo fx That is the sequence fn increases pointwise to f For all n 2 1 we shall let fn x n whenever fx 2 n Next suppose 0 S fx lt n Then k S f x lt k 1 for some integer k with 0 S k lt n 1 Now we divide this interval i l 39 10 k k 1 into 10quot intervals of equal length Then we have k S f x lt k for some integer z39 with 1 Si S 10quot Dr Neal Spring 2009 To define the required simple functions we let En x fx 2 n for n 2 1 Next 39 1 welet Elan xk117Sfxltk for k0 n 1 andz391 10quot and let 4 10 mil 10quot l1 fn Z Z k n1Ekin n1En k0 11 10 For example suppose 4 S fx lt 5 Then f1 x 1 f2 x 2 f3x 3 and f4 x 4 For f5x we divide the interval 4 5 into 105 100 000 subintervals of the form 4 i 1105 4 i 105 with the endpoints being 400000 400001 400002 400009 400010 400199 499999 5 Then f x is in exactly one of these subintervals and we round f 5x down to the left endpoint 4 139 1 105 So f5x 4611612 a3 a4 a5 which is the truncation of the decimal expansion of fx to 5 decimal places Then for n 2 5 fn x 4611612 a3 a4 an which is the truncation of the decimal expansion of fx to n decimal places Then 0 S fn x S fn1x S fx for all x eX and lim fnx fx new Note a If f is measurable with respect to a measure space X T u then each set En and ElmJ are in T thus each indicator function in fn will be measurable and thus each fn will be measurable b For an arbitrary f we can apply the construction to each of f T and f T obtaining sequences of simple functions kn and gn for each respectively Then let fn 1111320 gn1flt0 Then fn are still simple functions such that for all x eX lim fnx fx Moreover when fxZ 0 then 0 anxan1x Sfx and new when fxlt0then 0 anx2fn1x Zfx c If f 2 0 and we are working on a finite measure space X T u in particular a probability space then as in Example 53 c we would have that lim fn St fSt forallteSR new The preceding construction gives us the following result Theorem 511 Let X T p be a measure space and let fX gt SR be a realivalued function defined on X There exists a sequence of simple functions fn that converges pointwise to f on X If f is measurable then the functions fn may be chosen to be measurable lffZ 0 then fn maybe chosenso that 0 Sfl S f2 S an S Sf Dr Neal Spring 2009 Additional Exercises Exercise 55 Let X T u be a finite measure space Assume fn ugt f and gn gt g a Let a b be nonizero constants Prove that afn bgn E af b g b Prove thatfn Ifl c Prove that f3 Mgt f2 d Prove that f n gn ugt f g Hint Apply the identity that expresses a product in terms of sums and squares Exercise 56 Let X T p be a measure space For measurable functions f g say f N g provided f g ae Prove that is an equivalence relation Dr Neal Spring 2009 MATH 532 Outer Measure Throughout let X be a noniempty set let R be a noniempty collection of subsets of X that is closed under finite unions and set differences and let p be a finiteivalued countably additive measure defined on R We now shall see how to measure a larger class of sets generated by R De nition 21 Let R be a noniempty collection of subsets of a noniempty set X Then R is called a Boolean ring if the following condition holds lf AB e R then A U3 6 R and B A e R That is R is closed under finite unions and set differences Lemma 21 A Boolean ring R is closed under symmetric differences and finite intersections Proof Let AB e R Then A AB A BU B A e R because R is closed under finite unions and set differences Then A n B A U B A A B e R also i A ring is not necessarily closed under complements and the set X may or may not be in R ii lee Rthen foranysetA e R we have Ac X A e R Thus ier R then R is closed under complements and R is then called an algebra iii If R is closed under denumerable unions then R is called a o ring iv If R is a a ring and X e R then R will be a o ialgebra V We have seen previously that a o algebra is a Boolean ring because a o algebra is closed under finite unions and intersections as well as complements Thus a o algebra is also closed under set differences using B A B n Ac vi Because R is noniempty there is some set A e R Thus Q A A e R vii The power set of X TX is a o ring that contains R The intersection of all 0 7 rings that contains R is a o iring that contains R and is denoted by EUR viii A ring R is not necessarily closed under denumerable unions But a measure p on R is still assumed to be countably additive That is if A1 is a denumerable sequence of disjoint events from R and SIAZ e R then 814 1uAl Because 1 z 139 Q e R 1 also will be finitely additive We now give some results about u that were previously proven for a countably additive measure on a oialgebra but the proofs of which require only the weaker assumptions about a ring R Dr Neal Spring 2009 Lemma 22 Let R be Boolean ring and let p R gt 0 00 be a measure a lfAB e R with B Q A then uB S MA n n b HA 6 R andAg UAZ where 1 SnSooW39lthallAl e Rthen pAS 2 AZ 1quot z391 Proof a Because B Q A set A can be written as the disjoint union A BU A B Thus uA uB MA B Z uB because uA B Z 0 iil b Let B1 A1 B2 A2 A1 and in general let B1 AZ for 139 Z 2 Because R is J a ring B1 6 R for all z39 The sets B1 are mutually disjoint and B Q Al for all 139 thus n n n B1 S A0 for all 139 by Part a Moreover U B U Ai so A Q U Bi il il iil Now consider the disjoint sequence of sets AnBlhil all of which are in R because R is closed under finite intersections and for which we have n n U A n B A n UBZ A e R By countable additivity we have il il MA EIMA Bi 1HBis1l 14i QED n n Part c also implies that M U AiJ S 2 A10 whenever A1 An e R But when il il 00 n 00 then U Al may not be in R so we may not be able to compute its measure So 139l 00 00 we can only say that for A e R A Q U Al with all AZ e R then pAS Z uAZ iil 139l The Hereditary airing De nition 22 Let R be a Boolean ring of a noniempty set X and let A gX A collection of sets is a covering of A if A Q U A0L The hereditary o fring of R is a n CRAgXAg UAi forsomen where lSnSoo andallAl ER i l In other words C R is the collection of subsets of X that have a countable covering of sets from R The covering AZH11 could be finite when n lt 00 or it could be denumerable when n 00 Dr Neal Spring 2009 Lemma 23 CiR is a 0 ring that contains ZR Proof For all A e R A is covered by the countable covering A thus A 6 C91 If B C 6 C91 then C B Q C so the covering of C is also a covering of C B Hence C B 6 C91 and C91 is closed under sets differences n J Now let CJO1 be a sequence of sets in CiR such that for all j C Q UAJZ with il oo 00 quotj 1 S n S 00 and all A e R Then U C Q U U AJZ which is still a countable covering jl jl il 00 of sets from R so U C 6 C91 Thus CiR is closed under denumerable unions J391 and also finite unions hence C R is a 0 fring QED Note lfA 6 C91 and B Q A then B 6 C91 also because B is covered by the same covering as A Outer Measure De nition 23 Let u R gt 0 00 be a measure defined on a Boolean ring R For any set A in CiR we define the outer measure 11 of A by n n 1 Ainf Z pAZAg UAi lSnSoo allAZ ER il il n Essentially uquot A is the greatest lower bound of the sums 2 A0 over all 139l countable coverings A1 of A by sets from R We note that for 1 S n S 00 the sum n n 2 A0 is always defined as a value in 0 00 Moreover for n lt oo 2 A0 lt oo 139l 39l n n n U Al is in R and p U Ai S 2 uAZ with equality holding if A are mutually zquot z391 il disjoint The next result shows that pquot agrees with p on R Lemma 24 Let 11 be the outer measure induced by a measure 11 on a Boolean ring R Then pquot A pA for all A e R In particular uquot Q 0 Dr Neal Spring 2009 1 Proof Let A e R Using n 1 and A1 A in the definition of pquot we have A g UAZ 39 l 1 1 thus uquot AS Z AZ MAI pA because uquot A is a lower bound of all such 139l n sums 2 A0 On the other hand for any collection AZl from R with 1 S n S 00 139l n n and Ag UAi we have AS 2 AD by Lemma 22 b Because pquot A is the i l il n greatest of the lower bounds of such sums Z pAZ we must have that pA S uquot A 139l Thus pquot A uA QED An outer measure is not necessarily countably additive nor even finitely additive But there are various properties that hold that we shall need Lemma 25 Let pquot be the outer measure induced by a measure u on a ring R a lfAB 6 C91 with A g B then uquot A S pquot B 8 b For any sequence of subsets BJO1 from CiR pquot I B S 21 pquot Bj J Proof a For A g B any covering AZEli from R of set B is also a covering of set A n thus pquot A S 2 uAZ But because 11 B is the greatest lower bound of the set of 139l n sums 2 A0 taken over all coverings from R of B we must have uquot A S Mquot B 139l b lf 2 pquot Bj oo then we are done So we may assume that Z 51 Bj lt 00 J391 F1 and therefore pquot Bj lt 00 for all j Now let a gt 0 be given For all j the value n 1 Bj s 2 cannot be an upper bound of the set of sums 2 A0 taken over all il coverings from R of Bj So for all j B has a covering A 11 of sets from R such n J that Z AJilt m BJe2 il Dr Neal Spring 2009 00 00 nj 00 Then U 31 Q U U A which is a denumerable covering of U B by sets from R jl jl il j1 Therefore 00 00 quotj 00 00 uUBJJs Z Z MAJi Z uBJe2J7 2 M02 2 J jl jl Because a gt 0 was arbitrary we must have pquot U BJJ S 2 uquot Bf QED J391 F1 Note Because pquot Q 0 by taking 31 Q for j Z 3 we also have for all Bl 32 in C91 that Mquot 31 U32 3 Mquot Bl Hquot 32 u 7 measurable Sets Although we can evaluate 11 A for every set A in CiR we want to restrict our measurements to subsets that behave in a certain way We will show that on these sets that pquot is actually countably additive De nition 24 Let uquot be the outer measure induced by a measure p on a ring R A set E 6 C91 is called pquot 7measurable provided uquot A pquot An E Mquot An EC for all A 6 C91 The set of all pquot 7 measurable sets is denoted by rlkC R Note For all sets A we have A A EUA EC Thus we already have that 1 A S 1quot An E pquot An EC for all AE in CiR Thus we need only show that uquot AZ pquot AnE uquot AnEc for all A 6 C91 in order to show that E is 7 measurable Theorem 21 rlkC R contains all sets with outer measure 0 Proof Suppose pquot E 0 Then for all A 6 C91 we have by Lemma 25 a that uquot AnE uquot AnEc s uquot E m A 7 uquot A Thus E is pquot 7 measurable by the preceding note Dr Neal Spring 2009 Theorem 22 rlkC R is a ring Proof For A 6 C91 we have pquot An p An QC pquot u A pquot A thus Q e rlkC R and rMC R is a noneempty class of sets Next let E F e rlkC R We must show that EUF and E F are in rlkC R So let A 6 C91 Because E F e rMC R we have pquot Dp DnEu DnEc and uquot Dp DnF pquot DnFc for all D 6 C91 Moreover because CiR is a of ring DnE and DnEc are in CiR So for all D 6 C91 we have i uDuDoEMDoEC ii uDnE pDnEnF DnEnFC iii in DnEC m DnEan w DnEanc By substituting ii and iii into i we have iv uD pDmEnFuDnEnFCuDnECnFuDnEcch Now let D A n E U F in iv The last term drops out due to a null intersection But E UF contains each of E nF E nFC and EC nF which leaves v m AnEUF m AnEnFpi AnEnFcp AnEan Now let D A in iv We then have pApAnErFpAmEnFc AmEanuAnEanc vi pArEnFpArEnFcp AnEanpAnEuFc u AoEuFMAoEuFC thus EUF e MC R Next let D An E Fc A nEnFcc A nEc UF in iv Now the second term drops out Moreover EC UF contains EnF EC nF and EC nFC which leaves 11AmE FcuArErFpAnEanuAmEcch Now vi becomes uAuAoEoFCuAoEFcuAoEFuAoEFC hence E F e Mam QED Dr Neal Spring 2009 Theorem 23 For 1 S n S 00 let BlELI be a sequence of disjoint sets in rlkC R with n n B U Bi Then 11 AnB Z AnBi forallA 6 C91 1 il 1 Proof For n 1 the result is obvious In line V of the previous proof we had w AnEUF m AnEnFu AnEn Fcu AnEc nF for all A 6 C91 if E F e rMC R So if E F are disjoint then E gFC and F g E0 and we obtain pf AnEUF pf AnE pf AnF Thus the result holds for n 2 n1 Assume now that the result holds for some n 2 2 and let B U B Then by the 11 result for n 2 and the induction hypothesis we have n n MAoBM An U12 uBM u An U12 uAan1 139l il n n1 Z uAnBuAan1 Z uAnB ii il Thus by induction the result holds when 1 S n lt 00 Now for B U Bi we have by the result for finite n and Lemma 25 a that i l uAoBZ uAoOBZDS uAolszD il il il By taking the limit as n gt co and applying Lemma 25 b we obtain Emma3 uAnijD mm AnaJ Emma il il il il QED Dr Neal Spring 2009 Corollary 21 rlkC R is closed under denumerable unions of disjoint sets Proof Let B 1 be a sequence of disjoint sets in rMC R with B 81B We must 1 show that B is 51 measurable For n 2 1 let Fn OIBZ which is in rlkC R because 1 rlkC R is a ring Because Fn g B we have B0 Q F5 So for all A 6 C91 we have by Lemma 25 a and Theorem 23 that uAuAnFnuAan 2uAnFnuAnBc n M An U3 uAnBc il n 2 u AoBzHM AnBc il By taking the limit as n gt co and applying Theorem 23 again we obtain in 202 2 u AnBi m AnBC uquot AoBu AnBc il It follows that B is pquot rmeasurable by the Note after Definition 24 QED Corollary 22 rlkC R is a 0 ring Proof Let Egg1 be a sequence of sets in rlkC R We must show that their union is in rlkC R But we simply disjointify the sets by letting B1 E1 B2 E2 E1 and nil in general let Bn En UEZ for n 22 Because rlkC R is a ring B1 6 rMC R il for all z39 The sets B1 are mutually disjoint and U EZ UBZ e rlkC R by il il Corollary 21 QED We conclude with the result that shows that pquot is actually a measure that extends u from being defined only on a ring R to being defined on the a ring rlkC R Dr Neal Spring 2009 Theorem 24 Let 11 be the outer measure induced by a measure 11 on a ring R Then 11 MC R gt 0 00 is a countably additive measure Proof Let Bl 1 be a sequence of disjoint sets in rMC R with B U Bl Then B is 139l in rMC R by Corollary 22 Thus by Theorem 23 m B u BoBuBo U3 Z uBnB Z uB il 139l il Thus 11 is countably additive on rlkC R which makes it a measure QED Note By Theorem 21 and Lemma 25 a the measure 11 is a complete measure on Mam That is if B e MC R with 11 B 0 and A g B then A e MC R and 11 A 0 In other words rMC R contains all subsets ofsets ofmeasure 0 Finally let 6R be the 0 iring generated by R which is the intersection of all 0 7 rings that contain R We assert that 66R is contained in rMC R Theorem 25 rMC R contains EUR Proof We first show that rMC R contains ZR So let E e R Let A 6 C91 and let a gt 0 be given By the definition of C R and outer measure there exists a sequence of sets A from R such that A g UAZ and Z 11Ai lt 11 A a Then because 139 139 A nE g UAZ n E and A nEc g UAZ n EC are coverings from R we have 139 139 AoEHMAoEcSZMAioEZMAioEc ZP Ai EH 1414139 ED ZuAiltuA8 Because a gt 0 was arbitrary we have 11 AnE 11 AnEc S 11 A Thus E is 11 7 measurable by the Note after Definition 24 Because rlkC R is a o iring that is now shown to contain R but EUR by definition is the smallest 0 iring that contains R we must have 6R g rlkC R QED Dr Neal Spring 2009 Synopsis l X is a noniempty set The power set TX is the set of all subsets of X Z R is a ring of subsets of X ie R is a noniempty collection that is closed under finite unions and set differences 3 1 R gt 0 00 is a finiteivalued countably additive measure 4 C R is the o ring of all subsets of X that have countable coverings of sets from R and C91 contains ZR 5 1 C R gt 0 00 is the outer measure and 1 1 on R 6 rlkC R is a smaller class of subsets within C R that consists of 1 imeasurable sets rlkC R is also a o ring that contains ZR 7 1 MC R gt 0 00 is a countably additive measure and still 1 1 on R 8 EUR is the smallest 0 ring containing R R g 6R g rMC R g CiR g TX Exercises Exercise 21 Let S be a airing of sets and let 1 S gt 0 00 be a measure Let R A ES 1A lt oo Prove that R is a ring Exercise 22 Let R be a Boolean ring of sets and let 1 R gt 0 00 be a measure Suppose E e R El 6 R for all z39 with all EZ mutually disjoint and U EZ g E Prove iil that 21w s ME 1 Exercise 23 Let R be a Boolean ring of sets and let 1 R gt 0 00 be a measure For AB e R say that A NB ifand only ifl1AAB 0 a Prove that is an equivalence relation ie that it is re exive symmetric and transitive b Prove If A NB then 1A 13 1AnB c Define the distance between equivalence classes of sets in R by dA B 1A AB Prove that d satisfies the axioms ofa metric d Suppose Al N A2 and 31 N 32 Prove that dA1 31 dA2 32 Dr Neal Spring 2009 MATH 532 Axioms of Measure Theory I The Space 7 Throughout the course we shall let X denote a generic noneempty set In general we shall not assume that any algebraic structure exists on X so that X is merely a collection of objects under consideration However at times we may let X be a metric space with distance function d or let X be a normed linear space with norm II and distance function dxy II x y A special case of both is when X is the real line with the absolute value being the norm When we consider probability applications the set of all possible objects under consideration is called the sample space and is denoted by Q In this case Q also could be the set of all possible outcomes in a random procedure or sequence of random procedures A single element from Q is called an outcome and is denoted by a After determining Q we should determine its cardinality denoted by QI The cardinality may be finite denumerable countably infinite or uncountable Example 11 Describe the sample space Q in the following scenarios Then give I QI a Pick one student at random from a class of 40 b Roll two dice c Roll a single sixesided die 10 times in a row d Play the lottery until you win e Pick a real number at random from the interval 72 2 f Pick a rational number at random from the interval 0 1 Solution a Q is the set of all students in the class I QI 40 b Q l l 1 Z 6 6 xy xy el 6 IQI 36 c Q u102 w10mie16Q 61060466176 d Q W LW LLW LLLW LLLLW LLLLLW QI N0 denumerable e Q 72 Z Q c the uncountable cardinality of SR f Q Q n 0 1 QI No because the rationals are denumerable II The 0 algebra 7 Given a set X we shall assume the existence of a certain collection of subsets of X denoted by F In order to work with these subsets we must assume that we can perform various set operations such as unions and complements The following definition lays out the minimum requirements necessary from which other operations will follow as consequences Definition 11 Let F be a collection of subsets of a set X Then F is called a a algebra and its elements are called measurable sets or even ts provided 1 F contains at least one set ii F is closed under complements iii F is closed under denumerable unions Dr Neal Spring 2009 Note If F is merely closed under finite unions then F is called an algebra We shall show below that every 0 algebra is an algebra Proposition 11 Let F be a a ralgebra ofa set X a XeFandQ eF b F is closed under denumerable intersections c F is closed under finite unions d F is closed under finite intersections Proof a By Axiom i F contains at least one set A1 Then A2 A10 6 F because F is closed under complements Now let AZ A1 for all 139 Z 3 Because F is closed under 00 denumerable unions we have X A1 U A10 U AZ 6 F Then Q X6 e F as well 139l b Let AZ 6 F for 139 Z 1 Because F is closed under complements and denumerable 00 unions A e F for all z39 and thus UAZc e F Thus by DeMorgan s Law and closure i l 00 00 under complements we have H Al U AZc 0 e F il il c d Let AZ 6 F for 1 Si S n Then let AZ Q for 139 Z n1 Because F is closed under n 00 denumerable unions we have UAZ U AZ 6 F Thus F is also closed under finite 139l iil n unions Now because F is closed under complements and finite unions we have H Al 139 l n U A 0 e F thus F is also closed under finite intersections QED il Note a F is also closed under set difference which is defined as A B A nBc b Given any set X the power set of X which is the collection of all subsets of X denoted by PX is clearly a a falgebra The Generated o falgebra Let X be a set and let A be a noneempty collection of subsets of X Then we can generate a oealgebra from A denoted by 0A as follows i PX is a oealgebra that contains the smaller collection of subsets A ii Let F 0LgtCl be the collection of all a falgebras that contain A iii the o falgebra generated by A is given by 0A Fu39 a Dr Neal Spring 2009 So if Fa is any cealgebra that contains A then 0A g Fa That is 0A is the smallest possible a falgebra that contains A Of course it remains to be shown that 0A actually is a o falgebra but we leave this as an exercise Exercise 11 Let A be a noneempty collection of subsets of a set X i Prove that 0A contains A ii Prove that 0A is a o falgebra The Borel o falgebra Let X SR the real line and let A be the collection of all intervals of the form a b where a b 6 ER with a S b Then 0A is the Borel o falgebra which we shall denote as B Then B contains SR B is closed under complements and B is closed under both finite and denumerable unions and intersections The following sequence of exercises shows that the Borel 0 falgebra also contains various other types of intervals Exercise 12 Let B be the Borel oealgebra of the real line Let a b c 6 SR Use the properties of a 0 algebra to prove the following results a B contains all intervals of the form a 00 b B contains all intervals of the form 00 a c B contains all intervals of the form 00 b d B contains all intervals of the form I 00 e B contains all singleton sets c f B contains all intervals of the form a b a b and a b for a lt b The Borel o falgebra can be generated by other types of intervals other than a b For example we could let A1 be the collection of all closed intervals of the form a b where a b 6 ER with a lt b and let B1 0A1 Then the Borel oealgebra B contains A1 by Exercise 12 f But because B1 is the smallest o falgebra that contains A1 we must have B1 g B However as an exercise we can show that B1 contains all intervals of the form a b where a b 6 ER with a S b But because B is the smallest oealgebra that contains these halfeopen intervals we must have B Q B1 Hence B B1 Exercise 13 Prove that B1 contains all intervals 61 b where a b 6 ER with a S b Dr Neal Spring 2009 We also can weaken the condition of the generating class of sets and still generate B Specifically let Aq be the collection of all intervals of the form a b with a b e Q and a lt b and let Bq 0Aq Then the Borel 0 ialgebra B contains A Bq is the smallest o ialgebra that contains Aq we must have Bq g B q But because As another exercise we can show that Bq contains all intervals of the form a b where a b 6 ER with a S b But because B is the smallest o ialgebra that contains these open intervals we must have B Q Bq Hence B Bq Exercise 14 Prove that Bq contains all intervals 61 b where a b 6 ER with a S b III The measure 7 Given a set X with a o ialgebra F a measure u is an extended real valued function defined on F that satisfies the following properties i 0 SuAS 00 for all events A eF with u 0 00 00 ii If A11 is a denumerable sequence of disjoint events then U A1 2 uAZ il 11 1e 11 is countably additive Collectively the triple X F p is called a measure space If uA 0 then A is said to have measure 0 and A is said to be a negligible or null set If uA lt 00 for all events A then u is called a nite measure A special case of a finite measure is a probability measure P defined on a sample space Q But in this case we need the additional condition that PQ 1 Some basic properties of a measure 11 and of a probability measure P are given next Any property of 1 also holds for P but there are some additional properties that hold only for a probability measure P Proposition 12 Let X F u be a measure space and let Q FP be a probability space n n a Let AZH11 be a finite collection of disjoint events Then p991 AZ 1 AZ ie p S I is finitely additive In particular A U B uA pB for disjoint events A and B b For all events A eF pA MAC pLX and PAC1 PA c If A B are events witthA then pB S pA If in addition pB lt 00 then A B uA pB when BgA d For all events A 6F 0 SpAS uX and 0 SPAS1 Dr Neal Spring 2009 e Let M be a finite measure For all A B e F A U B pA uB pAn B and A A B pAuB 2 A nB f Let AZ1 be a collection of events Then U Ai S 2 A0 i1 il g Let A11 be a collection of events that each have measure 0 Then U A 0 T il h Let u be a finite measure and let B be a collection of events such that 1 B1 pX for all 139 Then 0 B1 pX il Proof a Let AZ Q for z39gt n The infinite sequence of sets A1 is still disjoint and because u 0 we have ult1Aoult31Aigt 21uAi Elm10 2 ylt gt Elm1i I I i i in i b Because events A and Ac are disjoint and union to X we have uX uAuAc ultAgt MAC For a probability space Q F P we have in particular that 1P 2PAUAcPAPAc c If B gA then A B U A B is a disjoint union Thus uA B U A B pB A B Z uB because pA B Z 0 From uA uB A B we obtain uA pB A B provided uB lt 00 d By axiom and Part c we have 0 S uA S uX because any event A is contained in X For a probability measure we have 0 S PA S PQ 1 e The event A UB can be written as the disjoint union A U B A and the event B can be written as the disjoint union A n B U B A Thus pB pAmBpB A so that pB A pB pAnB We then have A U B uA pB A pAuB A MB Dr Neal Spring 2009 The event A AB is the disjoint union of A B and B A hence MAM MABuBA MAuAoB MBuAoB MAuB2XuAoB f We define a collection of disjointevents B as follows Let B1 A1 B2 A2 A1 and in general let Bl AZ UA for 1392 2 Then B Q Al for all 139 thus B1 S A0 lt1 00 00 00 for all 139 by Part c Moreover UBZ g UAi But if u e UAi then we can choose the il il il 00 00 smallest indeXz39 such that a 6 Ai Then 03 6 AZ U A B Q UBi Hence UAZ g jlti il il U B which gives U Al U Bi Therefore by countable additivity we have il il il V 14139 MU Bi 2 M303 2 Ai39 21 21 11 11 g Because A0 0 for all 139 we have by f that 0 S uU Ai S 2 A0 0 il 11 h For each 139 uBlc uX B1 0 Thus by DeMorgan s Law Part c and Part g we have MEIRF uXuElBZc uXu 13ic uX0 MX Note Whenever we are subtracting a measure value then we should assume that it is finite For example let A be the lrrationals on the real line and let B be the Positive lrrationals Then B Q A We shall see that both of these events have infinite measure uA MB 00 But A B is the set of negative irrationals which also has infinite measure So although we can write uB pA B pA it does not follow that u A B uA p B because all three values are infinite Nested Events 00 Definition 12 a A sequence of events AZL1 for all 2 1 A1 g A2 gA3 b The sequence is called nested decreasing if A QAJH foralljZl 1412142 2A3 is called nested increasing if A j gA 1H Dr Neal Spring 2009 Proposition 13 a Let 141 be a nested increasing sequence of events Then MU A um Win 11 neoo b Let A 11 be a nested decreasing sequence of events with pAllt 00 Then 00 MO A1 lim NW 1 1 n em 00 n n Proof a We again create a disjoint sequence of events Blh1 such that U Bl U A 7 11 1 1 for 1S ns 00 To do so let 31 A1 and let Bl AZ UA for 139 Z 2 Because 141 is jltl n n nested increasing then An U Al U Bl for all n lt 00 Thus we have 11 11 n n 00 00 00 lim uAn lim uVBz lim 2 uBz Z uBi UB139 MU A1 new new 11 neooi1 11 11 11 b Because pAllt 00 then all measurable subsets of A1 have finite measure by Proposition 12 c For all n 2 1 we have An g A1 thus uA1 An 11 Any But A1 Ai1 is a nested increasing sequence with U Al Al A1 H Al Thus by 11 11 Part a and Proposition 12 c we have m 1411 MAI lim 141 An A1 HU 141 141 new new 11 uA1 Ai 5A1 ultA1 uA1 u An a A1 11 11 11 Some Examples of Probability Measures 1 EquieProbable Outcomes e Let Q be a finite set with IQI n and let F be the power set of Q which consists of all 2quot possible subsets of Q For every 1 e Q we can let Pu 1 n to denote that each individual outcome is equally likely Then for any event A e F we have PA IAI n In particular PQ Q I n 1 n U141 11 follows that P is finitely additive In this case we only need P to be finitely additive since there is no need to have infinite unions of sets Because for a finite collection of disjoint finite sets we have n Ell11quot it 11 Dr Neal Spring 2009 2 Binomial Probability 7 Let Q ml can u e W L which represents the set of outcomes in a sequence of n independent attempts On the ith attempt ml is either W for a win or L for a loss We assume that each win occurs with probability p and each loss occurs with probability q 1 p Then IQI 2quot We again let F be the power set of Q which consists of the 22quot subsets of Q With n 3 then Q has 23 8 elements and F has 28 256 elements In this case Q WW W WWL W LW LW W W LL L WL LL W LLL Various events in F can be described in words For example A1 a win on the first try A2 at least one win A3 never the same outcome back to back Then A1 W W W W WL WL W WLL A2 W W W W WL WL W LWW WLL L WL LL W A3 WLW LWL In all there are 256 possible events We define a probability measure P on the set of events F as follows For a single outcome 0 m1 an we let Pw1 mn pk qnik where k is the number ofwins in the sequence 11 can Then for A e F we let PA Z Pm Then 0 EA P is finitely additive as with equiiprobable outcomes Moreover the individual outcomes are equally likely if and only if p q 1 2 Now suppose we let Wk be the event that there are exactly k wins in a sequence for k 0 1 n Then the events W0 W1 Wn are disjoint and union to Q Because there distinct ways to have exactly k wins in a sequence the probability of Wk is pk qnik Then by the binomial expansion theorem we have n n n k 7k 139 ZPWk Z kjp qquot pqquot 1 1 k0 k0 3 Geometric Probability 7 Again let p be the probability of a win on any independent attempt and let q 1 p be the probability of a loss But now assume that 0 lt p lt 1 Let Q W LW LL W LLL W which represents the possible outcomes in a sequence of attempts where we stop play upon the first win Then I QI N0 Let F be the power set of Q which is uncountable with cardinality 2N0 c For each n e Q define Pw qnil p where n is the total number of attempts made in sequence n Then for A e F let PA Z Pw u EA Dr Neal Spring 2009 In this scenario 1 represents the sequence of outcomes up to the first win So if the first win is on the nth attempt then there must have been n 1 losses followed by a nil single win hence Pw q p Then the probability of the entire space is 00 n7 00 n l PQ ZPw Zq1pp2q pX1 1 neg nl n0 q Because 2 Pw converges absolutely to 1 then 0 S 2 Pm S l for any 069 DEA event A g Q Moreover no rearrangement of the terms in A will affect the sum of the 00 series So if 1331 is a collection of disjoint events with A U AZ then 139l 00 00 00 PU 14139 ZPw Z ZPm Z PAi 11 DEA il DEAi il Hence P is countably additive Although there are an uncountable number of events in F the events of most interest usually can be described easily in words For example compute the probability of the following events A the first win is within k attempts for k 2 1 B it takes at least k attempts for the first win for k 2 1 C it takes an odd number of attempts for the first win Here A is the complement of the event of there being k initial losses in a row hence PA 1 qk Event B is equivalent to the event of there being k 1 initial losses in a row hence PB qkil Finally 00 Zn 00 2n P 1 PC Zq p p2q z n0 n0 1 9 1q 4 Uniform Probability 7 Let Q a b where 61 6 ER with a lt b Let F a b n B B e B where B be the Borel o algebra of the real line For a subeinterval d c d of 9 let Pc 51 b c a point lie from c to d when picking a point at random from a b The probability measure P is initially only defined on subeintervals c d g a b 1 glee oi b 61139 Later we shall show that we can use this basic defining of P to extend P to be defined for all events in F In this case P measures the probability of having a 00 If C U cl all is a union of disjoint subeintervals then we let PC 139l Dr Neal Spring 2009 Exercise 15 Let A1 be a sequence of events in a oialgebra F Define the liminf and limsup of this sequence by 00 oo oo oo U 0 Ai n U Ai nl in nl in a Explain why 1imsupAZ and lim inf A1 are in F b Show that liminf AZDc 1imsupAf and limsupAZc lim inf AlC c Prove that an element x is in 1imsupAZ if and only if x belongs to an infinite number of the sets AZ d Prove that an element x is in lim inf A if and only if x belongs to all but a finite number of the sets AZ e Show that lim inf 141 g 1imsupAZ 00 f If A11 is nested increasing prove that lim inf Ai 1imsupAZ U An 7 nl 00 g If A11 is nested decreasing prove that lim inf Ai 1imsupAZ n An 7 nl Exercise 16 Let A B be events such that A A B 0 Prove that uA uB Exercise 17 Let A B be events with B Q A such that pA lt uB a for all a gt 0 Prove that uA uB Exercise 18 Let p be a finite measure and let A1 be nested increasing with 00 00 UAiX Provethatu Alc0 lim 4 i1 i1 new Dr Neal Spring 2009 MATH 532 Integration of Simple Functions Throughout let X T u be a measure space We now shall develop the theory of integration with respect to the measure M We begin with the integration of simple functions Simple Functions Let fX gt SR be asimple function That is f has the form n f 2 611114 a11A1anlAn il where all an are real numbers andAZ1 is a collection of disjoint events from T n De nition 61 a The integral of a simple function f Z ail1i with respect to the il n measure it is given by I f du 2 cy AZy b A simple function f is called integrable il if I f du is finite c For E e T the de nite integral of f over E is given by n lfdu If1E du Z ellMA 0E E i1 In particular I f du I f dp If I f dp is finite then we say that f is integrable over E X E Notes a A simple function f is integrable over E provided Al n E lt 00 for all z39 for which al at 0 and f is integrable provided AZlt oo whenever al at 0 Every simple function f is integrable over a null set and I f cl 0 whenever uE 0 E b If f is integrable then f is integrable over E for all E e T c On a probability space Q RP a simple function X Q gt SR is called a discrete random variable Because P is always finite every discrete random variable is integrable and the integral is called the mean average or expected value of X and is denoted by X EX IXdP a1PA1 anPAn d Given a discrete random variable X Q gt SR and a nonenegligible event E e T the conditional average of X over E is given by EX E gagHA ME Dr Neal Spring 2009 Example 61 Let Q be a set of college students with 9quot consisting of all subsets of Q Assume Pw 1 X for all m e X Let A be the subset of all undergraduate males which is 50 of the population let B be the subset of all undergraduate females which is 30 of the population and let C be the subset of all other students which is 20 of the population Let X Q gt SR be given by X 51A 1013 151C a Compute EX b Compute the conditional average of X among just undergraduates Solution a EX 5PA10PB15PC 50510031502 85 b EXIAU3 5gtltPAnAUB10gtltPBnAUB15gtltPCnAUB PAUB 5 gtltPA10 gtltPB 5051003 6 875 PAUB 08 39 39 Theorem 61 Let f and g be simple functions that are integrable over E e T a For all a 6 SR the function af is integrable over E and Iafdu a I fdu E E b The function f g is integrable over E and f g du Ifdu Igdu E E E c lffZO aeover Ethen Ifdp 20 E d lffZg ae over E then Ifdu Z Igdu E E e lfmeSM aeover EthenmpES IfduSMpE E f The function fl is integrable over E and I f du E sllfldu E g Forallevents EAwithEgA Ilfldusjlfldp Sjlfldu E A h lfglduSlfdullgldu E E E Dr Neal Spring 2009 n m Proof Letf 2 6111141 andg Z bJIBj il jl n a We have af 2aai1Ai If aal 0 then al at 0 which means AZ nE lt oo il because f is integrable over E Thus a f is integrable over E and MM 21 aaiuAinEazlcyuAinE alfdu E 1 1 E n m b Let AnH 1811406 Bm1JEJlBjC and rewrite f all11 anlAn 0114 1 n1 ml and g b11131 meBm OIBmH We then have fg 21 210139 b1Ai Bj J Now if 611 b i 0 then one of 611 or b must not equal 0 So one of AZ nE or Bf nE is finite which makes uAZ n B n E finite Thus f g is integrable over E and n1 ml Ifgdu Z Z 69 bjuA1ij oE E il jl n1 ml ml n1 Z 0i Z MAi Bj39 E Z bf Z MAz39 Bj E 139 jl jl il ml m1 n1 ai Ai E lej Z 1 J jl I n1 ml 2 al AZME 2 b BJ WE il jl IfduIgdw E E n c Suppose f 2 0 ae over E Then consider flE Z ailAi E On the set AZ nE f il equals ai So if 611 lt 0 then we must have uAZ n E 0 because f 2 0 ae over E So n ifpAZnEgt0thenwemusthave 61120 Hence Ifdp Z cyuAlnEZO E il d We have f g Z 0 ae over E thus Ifdp Igdu If gdu Z 0 which gives E E E IfduZIgdu E E e From Part dwe have muEIm1Edp S Ifdu S IMlEdu SMME E E E Dr Neal Spring 2009 n f We have If ZIcyIlAi and ifIcyIi 0 then a 0 which means uAZrEltoo 11 Thus I f I is integrable over E Moreover Ifdu zlaimAmEszllaimAmEnazllailultAmEIIfIdu E 1 1 1 E n n g Ilfldu EloyIMAmE zlceIuAnAIIfIdu Now applying this E 11 11 A resulttoAgXwehave IIfIdusIIfIduIIfdu A X h Because If g S I fl I gI everywhere on E we have IfgdMS I fgdu I I flduIgdu E E E E Note A special case of all preceding results in when E X In this case we are assuming that the functions are integrable and simply write I f du rather than I f du X The next result also applies in particular when E X Theorem 62 Let f be an integrable simple function Then I I f Idp 0 if and only if E f0 ae on the set E 11 Proof Suppose f 0 ae over E Then consider flE Z ailAmE On the set AZ nE 139l f equals ai For each 139 either all 0 or all 0 But if all 0 then we must have n AZnE 0 because f0 ae over E Hence IIfIdu Z lollIMAZ nE0 E 11 71 Next suppose Z lollIMAZ nE I Ifldu 0 Then for each 139 lollIMAZ nE 0 So 11 E if al at 0 then pAZ n E 0 Let I be the set of indices such that al at 0 Then we have x eE fx 0 UAZ n E which is a set ofmeasure 0 So f 0 ae on the set E i6 QED Dr Neal Spring 2009 The next result shows that the definite integral creates a new measure on T Theorem 63 Let f be an integrable simple function The set function a T gt oo 00 defined by xE Ifdp is countably additive If f 2 0 ae over the space X then a is E a measure n Proof Let f 2 6111141 and let E32071 be a sequence of disjoint measurable sets with 139l T 00 E U E Then because u is countably additive we have J391 E Ifdu Z aiuAioE Z anvilo U Ej E il i1 J391 2 am UA nE Z a Z MAmEj il jl il jl Z Z CtMAi EjF Z Z aiuAinEJ 1 11 il jl J Z Ifdu 2 E1 J391EJ J391 Thus a is countably additive n Because p 0 we have u Ifdu Z aiuAl n Q 0 And if f2 0 ae il then aE Ifdu Z 0 by Theorem 61 c thus a is a measure QED E Corollary 61 Let f be an integrable simple function The set function aE I f do is E additive In particular i If A B are disjoint events then I f dp I f du I f dp AUB A B n lfBgA then IN Ifdu Ifdu AiB A B De nition 62 Let f be an integrable simple function The set function given by aE I f du is called the de nite integral measure The function f is called the derivative E ofa with respect to u and can be denoted by f do d Dr Neal Spring 2009 Theorem 64 Let f be a simple function that is integrable with respect to u and such that f 2 0 ae over the space X Let a be the definite integral measure Any other simple function g is integrable with respect to a and for all measurable sets E we du d have gala g l I an MIgfdw E n m m n Proof Let f 2 ail1i andg Z bJlBj Then gf Z Z bJaZlBj Ai Thus il jl jl il m m m n jgdaZbJQBjMEij jfduZbJchuAlnBjME E F1 F1 3 mE j1 il Z Z bJceMBJoAi oE I gfdu E J 1 l l Example 62 Let f SR gt SR be a continuous function Then f is Borel measurable by Exercise 47 For any interval a b we can estimate the standard calculus integral 1 I f xdx with the integral of a simple function with respect to Lebesgue measure in a several ways First partition 61 b by a b 611 b1 U 02 b2 U U aw bn where aa1ltb1a2 ltb2 anltbnb I Monte Carlo Integration For each 139 pick any point cl 6 ab bi and let fn f611a1b1 f021a2b2 fCn1ab b n Then Ifxdx m If a7 21f0ibi ai a 11 I II Upper Riemann Sum For each 139 let M1 be the maximum value of fx on the closed interval ah bi and let un M11a1b1 M21a2b2 Mn1anbn Then I n mxmx s Jundx 21Mib cy a 11 1 III Lower Riemann Sum For each 139 let m be the minimum value of fx on the closed interval ah bi and leth m11a1b1 m21a2 b2 mn1an bn Then n b 2 mlbl a j 1 am 3 jfxdx 151 11 11 Dr Neal Spring 2009 De nition 63 Let a and u be set functions defined on T Then a is said to be absolutely continuous with respect to 11 provided lim 1 E 0 That is aE tends uEa0 to 0 whenever pE tends to 0 Theorem 65 Let f be a simple function that is integrable with respect to u The definite integral set function 1E I f dp is absolutely continuous with respect to u E n Proof Let f 2 ail1i If all 611 0 then aE Ifdu 0 for all E and a is il E absolutely continuous with respect to any measure Otherwise let a gt 0 be given Let 6 4 gt 0 and assume uE lt5 Then nmaXa1 Ianl n n 1de s llfldu ZlailmAmE zlailmE E E il il n n 8 n a ltZMWZMI SZ i1 1391 quotmaX ilalLu Ianl i1quot Hence lim uE0 QED pEgt0 Finally we can use the integral operator to define a distance between simple integrable functions Definition 64 For simple integrable functions f g we define the distance between f and g by pmg I I f gldu We note that p acts like a metric i pfg Z 0 by Theorem 61 d ii pfg pgf iii For any other simple integrable function h pfgI I fgldu llfhhgdu SllfhIHhgldu 1 l f hl du IIhgl du pfhphg However pfg 0 if and only if f g ae Rather than pfg 0 if and only if fgJ Sequences of Simple Functions Our main goal is to see how to extend all the results about integrable simple functions to a larger class of integrable functions To do so we want to use functions that are in some way limits of integrable simple functions Again we let X T p be a measure space throughout Dr Neal Spring 2009 De nition 65 Let fn be a sequence of integrable simple functions Then fn is called Cauchy in mean if lim I Ifn fmI cl 0 That is for every 8 gt 0 there exists an nmgtoo integer N 21 such that I Ifn fmI dp lt 5 whenever mm 2 N Lemma 61 Let fn be a sequence of integrable simple functions that is Cauchy in mean Then I fn I is also Cauchy in mean Proof By Theorem 61 f we already know that each Ifn I is integrable For all n m we have and thus San fml 61W Hence Iim IIInt Ifmlldu0 QED nmgtoo By Lemma 61 the following results also apply to I fn I when fn is Cauchy in mean Theorem 66 Let fn be a sequence of integrable simple functions If fn is Cauchy in mean then fn is Cauchy in measure Proof Let a gt 0 and e39 gt 0 be given For all nm let En x fnx fmx Z 2 Then 21Equot m SIfnx fmx1En m hence by Thoerem 61 e mam Ilfnx fmxldu s I Ifnx fmxldu Because fn is Cauchy in mean there exists an integer N Z 1 such that I Ifnx fmx Id lt 539s whenever mm 2 N So for mm 2 N we have 1 ultfn fmI2e ultEnmsJIfnx fmxldulta39 Hence fn is Cauchy in measure QED Theorem 67 Let fn be a sequence of integrable simple functions that is Cauchy in mean The definite integral set functions are uniformly absolutely continuous with respect to u Proof Let a gt 0 be given There exists an integer N Z 1 such that I I fn fm Idp lt s 2 whenever mm 2 N Because each I fn I is also integrable each set function I I fn Idu is E absolutely continuous for 1 S n S N So there exists a single 6 gt 0 such that IIntdu lt s 2 for 1 S nS N whenever pE lt6 Butfor n gtN wehave E Dr Neal Spring 2009 Ifndl i E s Ilfnldu Ilfn fNledu SIIfnledu IIfNIdu E E E E San ledHIIfNdH lt8 E Thus there is a single 6 gt 0 such that lt e for all n whenever pE lt 5 QED Ifndl E Theorem 68 Let fn be a sequence of integrable simple functions that is Cauchy in mean For every measurable set E the value aE lim I fn dp exists as a finite n 00 E number Moreover a is a countably addititive set function on T Proof For all nm IfndIL Ifmdu Ifnfmdu Sjlfnfmdugt0 Thus E E E E quotJquot the sequence fn nip is Cauchy on the real line which is complete Thus E 1E lim I fn dp exists as a real number for all measurable sets E n 00 E 00 Next let E32071 be a sequence of disjoint measurable sets with E U E Now let F1 an be the definite integral set function of fn which we know is countably additive by Theorem 63 We now may write 1E lim an E for all E e 9quot Using the fact that n 00 the limit of a finite sum is the sum of the limits we have for all k k I k I k k I k a U Ej 11m an U Ej 11m 2 anEJ 2 11m anEJ Z xEJ j1 new j1 ngt00391 11 ngtoo F1 That is a is finitely additive Now for each n and k S0t1 30ln1 3 k 0E 2 ME F1 k anEan U Ej J391 k k M U E Z aE J391 F1 GE0lnE k k k anE an UEJJ anUEJ uUEJ J391 J391 J391 ltXEanE 00 k k anE U Ej an U Eja U Ej jkl jl jl Dr Neal Spring 2009 k Now let a gt 0 be given Because E U E is nested decreasing with an empty 11 k k intersection E UEJ gt 0 as k increases Then because an is uniformly J391 absolutely continuous with respect to u we can make lt e 3 for all n 00 an E U E jkl whenever k 2 K for some integer K 2 1 Now fix k 2 K By the definition of a we then have gt 0 k k A U E u U E J391 J391 and aE an E gt 0 as n increases So we can choose n large enough so that each of these terms is less than B 3 We therefore can make lt a for all k aw 2 E19 F1 k 00 kZK so that lim 2 aEJ aE That is aE Z uEJ and a is countably k quotjl jl additive QED Theorem 69 Let fn and gn be a sequences of mean Cauchy integrable simple functions that both converge in measure to a measurable function f Let aE lim jfn d and WE lim jgn d Then aE 3E forallE e T n 900 E ngtooE Proof Let E e T If uE 0 then If du 0 gm dp for all n thus aE E E E So assume uE gt 0 and let a gt 0 be given For each n let En x ilfnx gnx2E3HE Soon E5 fnx gnx lt e 3 uE Moroever because fn gt f and gn gt f then u M fn gn gt 0 hence pEn gt 0 as n increases Because Ifnl and Ign I are Cauchy in mean Lemma 61 the collections of measures I I fn Idp and I I gn Idu are both uniformly absolutely continuous with E E respect to 11 Theorem 67 So there exists a 6 gt0 such that if uElt6 then anldult and gnldult foralln E 3 E 3 Dr Neal Spring 2009 Now for all n we have Ifndugndu E E Ifn gndl E S I Ifn gn Id E angndl d Ilfngndlquot EiE EOEquot uE za II Adu IIEEm EOEquot HEquot S 3mm uE Ilfnldu Ilgn Idu En En S 3MB 8 3lfndulgndw En n We now choose N large enough so that pEn lt 6 for all n 2 N Then for all n 2 N wehave fndu Igndp lts Hence lim aw gnaw 0 and uEBE E quot9 C E E E for all E e T QED Exercises Exercise 61 Let p be as in Definiton 64 For measurable sets EF prove that MhJ u A Exercise 62 Prove Corollary 61 Exercise 63 Let f be an integrable simple function such that f gt0 ae over a measurable set E Suppose I f du 0 Prove that uE 0 E Exercise 64 Let f be an integrable simple function such that I f dp 0 for all E e T E Prove that f 0 ae

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All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.