MEASURE THEORY NOTES (2010)

Transcription

1 MEASURE THEORY NOTES (2010) Mr. Andrew L Pinchuck eprtment of Mthemtics (Pure & Applied) Rhodes University

2 Contents Introduction 1 1 Riemnn integrtion Introduction Bsic definitions nd theorems Fundmentl theorem of clculus Mesure spces Introduction Nottion nd preliminries Mesure on set Outer mesure Lebesgue mesure on R The Lebesgue integrl Introduction Mesurble functions Bsic notions Convergence of sequences of mesurble functions efinition of the integrl Integrl of nonnegtive simple function Integrl of nonnegtive mesurble function Integrble functions Lebesgue nd Riemnn integrls L p spces Hölder nd Minkowski inequlities ecomposition of mesures Signed mesures

3 Introduction In mthemtics, more specificlly in mesure theory, mesure on set is systemtic wy to ssign to ech suitble subset number, intuitively interpreted s the size of the subset. In this sense, mesure is generliztion of the concepts of length, re, volume, etc. A prticulrly importnt exmple is the Lebesgue mesure on Eucliden spce, which ssigns the conventionl length, re nd volume of Eucliden geometry to suitble subsets of R n. Most mesures met in prctice in nlysis (nd in mny cses lso in probbility theory) re Rdon mesures. Rdon mesures hve n lterntive definition in terms of liner functionls on the loclly convex spce of continuous functions with compct support. In this course, however, we will minly concentrte on the Lebesgue mesure. Mesure theory ws developed in successive stges during the lte 19th nd erly 20th century by Emile Borel, Henri Lebesgue, Johnn Rdon nd Murice Frchet, mong others. The min pplictions of mesures re in the foundtions of the Lebesgue integrl, in Andrey Kolmogorov s xiomtistion of probbility theory nd in ergodic theory. In integrtion theory, specifying mesure llows one to define integrls on spces more generl thn subsets of Eucliden spce; moreover, the integrl with respect to the Lebesgue mesure on Eucliden spces is more generl nd hs richer theory thn its predecessor, the Riemnn integrl. Probbility theory considers mesures tht ssign to the whole set, the size 1, nd considers mesurble subsets to be events whose probbility is given by the mesure. Ergodic theory considers mesures tht re invrint under, or rise nturlly from, dynmicl system. The importnce of integrtion nd how mesure theory puts integrtion nd probbility theory on n xiomtic foundtion is principle motivtion for the development of this theory. Mesure theory is useful in functionl nlysis in defining L p function spces, which cnnot otherwise be defined properly. These notes re intended to be n introduction to mesure theory nd integrtion. I mke no clims of originlity with regrds to this mteril, nd I hve used number of different sources s references in the compiltion of these notes. Chpter 1 dels with the theory of Riemnn integrtion nd highlights some of its shortcomings. In Chpter 2, we define the fundmentl concepts of mesure theory nd present some well-known, significnt results. In Chpter 3, we define the Lebesgue integrl nd consider some results tht follow from this nd lstly, in Chpter 4, we provide brief introduction to the study of signed mesures. Throughout the course notes you will find exercises tht need to be ttempted for complete understnding of the text. Once gin, I should mention tht I hve found wikipedi to be useful resource nd recommend its use to students. All the course informtion nd mteril cn be found on the Rhodes mthemtics web site: by following the pproprite links. 1

4 Chpter 1 Riemnn integrtion 1.1 Introduction In the brnch of mthemtics known s rel nlysis, the Riemnn integrl, creted by Bernhrd Riemnn ( ), ws the first rigorous definition of the integrl of function on n intervl. The Riemnn integrl ws clerly n improvement upon the wy in which integrtion ws initilly conceived by Leibnitz nd Newton, i.e., simply s the nti-derivtive. While the Riemnn integrl is one of the esiest integrls to define, nd hs obvious uses in numericl nlysis, it is unsuitble for mny theoreticl purposes. Some of these technicl deficiencies cn be remedied by the RiemnnStieltjes integrl, nd most of them dispper in the Lebesgue integrl. The ide behind the Riemnn integrl is to use very simple pproximtions for the re of S. By tking better nd better pproximtions, we cn sy tht in the limit, we get exctly the re of S under the curve. 1.2 Bsic definitions nd theorems In this section we discuss the construction of the Riemnn integrl. This discussion revels some of the shortcomings of the Riemnn integrl nd we present these s motivtion for the construction of the Lebesgue integrl efinition Let Œ; b be closed intervl in R. A prtition of Œ; b is set P fx 0 ; x 1 ; x 2 ; : : : ; x n g of points in R such tht x 0 < x 1 < x 2 < < x n b: Let f be rel-vlued function which is bounded on Œ; b nd let P fx 0 ; x 1 ; x 2 ; : : : ; x n g be prtition of Œ; b. enote by M sup f.x/ nd m inf f.x/: xb xb Since f is bounded on Œ; b, it is bounded on ech subintervl Œx i 1 ; x i, for ech i 1; 2; : : :; n. Let M i supff.x/ W x i 1 x x i g nd m i infff.x/ W x i 1 x x i g; for ech i 1; 2; : : :; n. Clerly, We cn now form the sums U.f; P/ m m i M i M; for ech i 1; 2; : : :; n: n M i.x i x i 1 / nd L.f; P/ n m i.x i x i 1 /: 2

5 1.2.2 efinition The sums U.f; P/ nd L.f; P/ re clled, respectively, the upper nd the lower sum of f reltive to the prtition P Remrk [1] It is importnt to note tht U.f; P/ nd L.f; P/ depend on the prtition P. If f is nonnegtive on Œ; b, then the upper sum U.f; P/ is the sum of the res of the rectngles whose heights re M i nd whose bses re Œx i 1 ; x i. Similrly, L.f; P/ is the sum of the res of the rectngles whose heights re m i, nd whose bses re Œx i 1 ; x i. [2] It lso follows tht U.f; P/ L.f; P/ Theorem Let f be rel-vlued function which is bounded on Œ; b nd let P fx 0 ; x 1 ; x 2 ; : : : ; x n g be prtition of Œ; b. Then m.b / L.f; P/ U.f; P/ M.b /: Since M i M, for ech i 1; 2; : : :; n, it follows tht U.f; P/ n M i.x i x i 1 / n M.x i x i 1 / M Similrly, since m m i, for ech i 1; 2; : : :; n, it follows tht L.f; P/ n m i.x i x i 1 / n m.x i x i 1 / m n.x i x i 1 / M.b /: n.x i x i 1 / m.b /: This theorem sys tht the set A fu.f; P/ W P is prtition of Œ; b g is bounded below by m.b Hence, A hs n infimum,.f /, sy. Tht is, /..f / inf U.f; P/; P where the infimum is tken over ll possible prtitions P of Œ; b. This theorem lso shows tht the set B fl.f; P/ W P is prtition of Œ; b g is bounded bove by M.b / nd hence B hs supremum,.f /, sy. Tht is,.f / sup L.f; P/; P where the supremum is tken over ll possible prtitions P of Œ; b. It follows tht m.b /.f / M.b /; nd m.b /.f / M.b /: efinition Let f be rel-vlued function which is bounded on Œ; b. The upper integrl of f on Œ; b is defined by b f.x/dx inf U.f; P/; P 3

6 nd the lower integrl of f on Œ; b is defined by b f.x/dx sup L.f; P/; P where, once gin, the infimum nd supremum re tken over ll possible prtitions P of Œ; b. It is intuitively cler tht R b f.x/dx R b f.x/dx. Of course, this is not enough nd we will therefore prove this fct shortly efinition Let P fx 0 ; x 1 ; x 2 ; : : : ; x n g be prtition of Œ; b. A prtition P of Œ; b is clled refinement of P, denoted by P P, if x i 2 P, for ech i 0; 1; 2; : : :; n. A prtition P is clled common refinement of the prtitions P 1 nd P 2 of Œ; b if P is refinement of both P 1 nd P 2. The next theorem sttes tht refining prtition decreses the upper sum nd increses the lower sum Lemm (Refinement Lemm) Let f be rel-vlued function which is bounded on Œ; b. If P is refinement of prtition P fx 0 ; x 1 ; x 2 ; : : : ; x n g of Œ; b, then L.f; P/ L.f; P / U.f; P / U.f; P/: Suppose tht P hs one more point tht P, sy point x which lies in the subintervl Œx r 1 ; x r. Let L 1 supff.x/ W x r 1 x x g; L 2 supff.x/ W x x x r g nd l 1 infff.x/ W x r 1 x x g; l 2 infff.x/ W x x x r g: Reclling tht M r supff.x/ W x r 1 x x r g; nd m r infff.x/ W x r 1 x x r g; we observe tht It follows tht m r l 1 ; m r l 2 ; L 1 M r ; nd L 2 M r : m r.x r x r 1 / m r.x r x / C m r.x x r 1 / l 2.x r x / C l 1.x x r 1 /: Hence, r 1 n L.f; P / m j.x j x j 1 / C l 1.x x r 1 / C l 2.x r x / C m j.x j x j 1 / jrc1 r 1 n m j.x j x j 1 / C m r.x r x r 1 / C m j.x j x j 1 / n m j.x j x j 1 / L.f; P/: jrc1 Similrly, M r.x r x r 1 / M r.x r x / C M r.x x r 1 / L 2.x r x / C L 1.x x r 1 /; 4

7 nd so r 1 U.f; P/ M j.x j x j 1 / C L 1.x x r 1 / C L 2.x r x / C r 1 M j.x j x j 1 / C M r.x r x r 1 / C n M j.x j x j 1 / U.f; P/: n jrc1 n jrc1 M j.x j x j 1 / M j.x j x j 1 / the cse where P contins k 2 more points thn P cn be proved by simply repeting the bove rgument k times Theorem Let f be rel-vlued function which is bounded on Œ; b. Then, b f.x/dx b f.x/dx: Let P 1 nd P 2 be ny two prtitions of Œ; b nd let P be their common refinement. Then by Lemm 1.2.7, L.f; P 1 / L.f; P / U.f; P / U.f; P 2 /: Since P 1 is ny prtition of Œ; b, it follows tht nd since P 2 is ny prtition of Œ; b, we hve tht sup L.f; P/ U.f; P 2 /; P sup P L.f; P/ inf U.f; P/; P where the infimum nd the supremum re tken over ll possible prtitions P of Œ; b. Thus, b f.x/dx b f.x/dx: Remrk Implicit in the proof of Theorem 1.2.8, is the fct tht no lower sum cn exceed n upper sum. Tht is, every lower sum is less thn or equl to every upper sum efinition Let f be rel-vlued function on Œ; b. We sy tht f is Riemnn-integrble on Œ; b if f is bounded on Œ; b nd b f.x/dx b f.x/dx: 5

8 If f is Riemnn-integrble on Œ; b, we define the integrl of f on Œ; b to be the common vlue of the upper nd lower integrls, i.e, b f.x/dx b f.x/dx b f.x/dx: We shll denote by RŒ; b the collection of ll functions tht re Riemnn-integrble on Œ; b Remrk In the definition of the integrl of f on Œ; b, we hve tcitly ssumed tht < b. If b, then R b f.x/dx R f.x/dx 0. Also if b <, then define b f.x/dx b f.x/dx: Exmples [1] Show tht if f is constnt function on Œ; b, then f 2 RŒ; b nd find it s integrl. Solution: Let f.x/ k, for ll x 2 Œ; b nd let P fx 0 ; x 1 ; x 2 ; : : : ; x n g by ny prtition of Œ; b. Then M i supff.x/ W x i 1 x x i g k nd m i infff.x/ W x i 1 x x i g k; for ech i 1; 2; : : :; n. Therefore, U.f; P/ L.f; P/ n M i.x i x i 1 / k n m i.x i x i 1 / k n.x i x i 1 / k.b / nd n.x i x i 1 / k.b / Since P is ny prtition of Œ; b, it follws tht U.f; P/ L.f; P/ k.b of Œ; b. Therefore, Tht is, f is integrble on Œ; b nd b b f.x/dx k.b / f.x/dx: /, for ll prtitions P b f.x/dx k.b /: [2] Let f be function defined by Show tht f is not Riemnn-integrble on Œ0; 1. [3] 1 if x 2 Q \ Œ0; 1 f.x/ 0 if x 62 Q \ Œ0; 1 : 6

9 Solution: Firstly we observe tht f is bounded on Œ0; 1. Let P fx 0 ; x 1 ; x 2 ; : : : ; x n g be ny prtition of Œ0; 1. Since for ech i 1; 2; : : :; n, the subintervl Œx i 1 ; x i contins both rtionl nd irrtionl numbers, we hve tht M i supff.x/ W x i 1 x x i g supf0; 1g 1; nd m i infff.x/ W x i 1 x x i g inff0; 1g 0; for ech i 1; 2; : : :; n. Thus, n U.f; P/ M i.x i x i n 1 /.1/.x i x i 1 / 1 0 1, nd L.f; P/ n m i.x i x i 1 /.0/ n.x i x i 1 / 0: Since P is ny prtition of Œ0; 1, it follows tht U.f; P/ nd L.f; P/ 0, for ll prtitions of P of Œ0; 1. Therefore, 1 1 f.x/dx 0 nd f.x/dx 1; 0 0 nd hence f is not Riemnn-integrble on Œ0; 1. Notice tht in the lst exmple, the given function f is discontinuous t every point Theorem (rboux s integrbility condition) Let f be rel-vlued function which is bounded on Œ; b. Then f is integrble on Œ; b if nd only if, for ny > 0, there exists prtition P of Œ; b such tht U.f; P / L.f; P / < : Assume tht f is integrble on Œ; b nd let > 0. Since b there is prtition P 1 of Œ; b such tht Agin, since b f.x/dx b b f.x/dx f.x/dx there exists prtition P 2 of Œ; b such tht b U.f; P 2 / < 7 f.x/dx sup L.f; P/; P 2 < L.f; P 1/: f.x/dx inf U.f; P/; P b f.x/dx C 2 :

10 Let P be common refinement of P 1 nd P 2. Then b f.x/dx It now follows tht Then, Thus, b 2 < L.f; P 1/ L.f; P / U.f; P / U.f; P 2 / < U.f; P / L.f; P / < : For the converse, ssume tht given > 0, there is prtition P of Œ; b such tht b U.f; P / L.f; P / < : f.x/dx inf P U.f; P/ U.f; P /; nd 0 b f.x/dx Since > 0 is rbitrry, we hve tht b b b f.x/dx C 2 : f.x/dx sup L.f; P/ L.f; P /: P f.x/dx U.f; P / L.f; P / < : f.x/dx b f.x/dx: Tht is, f 2 RŒ; b. We now highlight the following importnt fct which is contined in the first prt of the proof of rboux s integrbility condition: Theorem If f is integrble on Œ; b, then for ech > 0, there exists prtition P of Œ; b such tht b f.x/dx < L.f; P/ U.f; P/ < Theorem If f is continuous on Œ; b, then it is integrble there. b f.x/dx C : Since f is continuous on Œ; b, we hve tht f is bounded on Œ; b. Furthermore, f is uniformly continuous on Œ; b. Hence, given > 0, there is ı > 0 such tht jf.x/ f.y/j <, whenever x; y 2 Œ; b nd jx yj < ı: b Let P fx 0 ; x 1 ; x 2 ; : : : ; x n g be ny prtition of Œ; b such tht x i x i 1 < ı, for ech i 1; 2; : : :; n. By the Extreme-Vlue Theorem (pplied to f on Œx i 1 ; x i, for ech i 1; 2; : : :; n), there exist points t i nd s i in Œx i 1 ; x i, for ech i 1; 2; : : :; n such tht f.t i / supff.x/ W x i 1 x x i g M i nd f.s i / infff.x/ W x i 1 x x i g m i : 8

11 Since x i x i 1 < ı, it follows tht jt i s i j < ı, nd so Thus, U.f; P/ M i m i f.t i / f.s i / jf.t i / f.s i /j < ; for ll i 1; 2; : : :; n: b L.f; P/ n M i.x i x i 1 / n m i.x i x i 1 / n jf.t i / f.s i /j.x i x i 1 / <.b / : b It now follows from Theorem , tht f is integrble on Œ; b Theorem If f is monotone on Œ; b, then f is integrble there. n.m i m i /.x i x i 1 / n b.x i x i 1 / Assume tht f is monotone incresing on Œ; b nd f./ < f.b/. Since f./ f.x/ f.b/, for ll x 2 Œ; b, f is clerly bounded on Œ; b. We wnt to show tht, given ny > 0, there is prtition P of Œ; b such tht U.f; P/ L.f; P/ <. Let > 0 be given n let P fx 0 ; x 1 ; x 2 ; : : : ; x n g be ny prtition of Œ; b such tht x i x i 1 <, for ech i 1; 2; : : :; n. Since f is incresing on Œ; b, f.b/ f./ we hve tht M i supff.x/ W x i 1 x x i g f.x i / nd m i infff.x/ W x i 1 x x i g f.x i 1 /; for ech i 1; 2; : : :; n. Hence, U.f; P/ L.f; P/ n M i.x i x i 1 / n m i.x i x i 1 / n.m i m i /.x i x i 1 / n Œf.x i / f.x i 1 /.x i x i 1 / < f.b/ f./ Œf.b/ f./ : f.b/ f./ n Œf.x i / f.x i 1 / It then follows from Theorem , tht f is Riemnn-integrble on Œ; b. The cse where f is monotone decresing cn be proved in exctly the sme wy Theorem If f is integrble on Œ; b nd c < d b, then f is integrble on Œc; d. Since f is integrble on Œ; b, it is bounded there. Hence f is bounded on Œc; d. Furthermore, given ny > 0, there is prtition P of Œ; b, such tht U.f; P/ L.f; P/ < : Let P P [ fc; dg. Then P is refinement of P, nd hence U.f; P / L.f; P / U.f; P/ L.f; P/ < : 9

12 Let Q 1 P \ Œ; c ; Q 2 P \ Œc; d ; Q 3 P \ Œd; b. Then P Q 1 [ Q 2 [ Q 3, nd so, U.f; P / U.f; Q 1 / C U.f; Q 2 / C U.f; Q 3 /; nd L.f; P / L.f; Q 1 / C L.f; Q 2 / C L.f; Q 3 /: Hence, ŒU.f; Q 1 / L.f; Q 1 / C ŒU.f; Q 2 / L.f; Q 2 / C ŒU.f; Q 3 / L.f; Q 3 / U.f; P / L.f; P / < : Note tht ll terms on the left re nonnegtive. Therefore Q 2 is prtition of Œc; d with the property tht U.f; Q 2 / L.f; Q 2 / < : This implies tht f is integrble on Œc; d Corollry If f is integrble on Œ; b nd < c < b, then f is integrble on both Œ; c nd Œc; b. The following theorem sttes tht the converse of Corollry lso holds Theorem If < c < b nd f is integrble on both Œ; c nd Œc; b, then f is integrble on Œ; b nd b f.x/dx c b f.x/dx C f.x/dx: c Let > 0 be given. Since f is integrble on Œ; c nd on Œc; b, there re prtitions P 1 nd P 2 of Œ; c nd Œc; b respectively, such tht U.f; P 1 / Let P P 1 [ P 2. Then P is prtition of Œ; b nd L.f; P 1 / < 2 ; nd U.f; P 2 / L.f; P 2 / < 2 : U.f; P/ L.f; P/ U.f; P 1 / C U.f; P 2 / L.f; P 1 / L.f; P 2 / Therefore f is integrble on Œ; b. Furthermore, U.f; P 1 / L.f; P 1 / C U.f; P 2 / L.f; P 2 / < 2 C 2 : b f.x/dx U.f; P/ U.f; P 1 / C U.f; P 2 / < L.f; P 1 / C C L.f; P 2 / C 2 2 L.f; P 1 / C L.f; P 2 / C c b f.x/dx C f.x/dx C : c 10

13 Since is rbitrry, it follows tht b f.x/dx c b f.x/dx C c f.x/dx (1.1) Also, b Since is rbitrry, it follows tht f.x/dx L.f; P/ L.f; P 1 / C L.f; P 2 / > U.f; P 1 / U.f; P 1 / C U.f; P 2 / C U.f; P 2 / 2 2 b Combining (1.1) nd (1.2), we hve tht b f.x/dx f.x/dx c c b f.x/dx C c b f.x/dx C c f.x/dx: (1.2) f.x/dx: Corollry Let f be defined on Œ; b nd suppose tht c 0 < c 1 < < c n 1 < c n b. Then f is integrble on Œ; b if nd only if f is integrble on Œc k 1 ; c k, for ech k 1; 2; : : :; n. In this cse, b f.x/dx n k1c k c k 1 f.x/dx: Corollry If f is continuous t ll but finite set of points in Œ; b, then f is Riemnn-integrble on Œ; b. 1.3 Fundmentl theorem of clculus Theorem (Fundmentl theorem of clculus) Let f be bounded rel-vlued integrble function on Œ; b. Suppose tht F W Œ; b! R is continuous on Œ; b nd differentible on.; b/ nd tht F 0.x/ f.x/; for ech x 2.; b/: Then, b f.x/dx F.b/ F./: 11

14 Let P fx 0 ; x 1 ; x 2 ; : : : ; x n g be prtition of Œ; b. M i supff.x/ W x i 1 x x i g nd m i infff.x/ W x i 1 x x i g; for ech i 1; 2; : : :; n. For ech i 1; 2; : : :; n, F is continuous on Œx i.x i 1 ; x i /. By the Men Vlue Theorem, there is i 2.x i 1 ; x i / such tht 1 ; x i nd differentible on F.x i / F.x i 1 / F 0. i /.x i 1 x i / f. i /.x i 1 x i /: Since, for ech i 1; 2; : : :; n m i f. i / M i, it follows tht m i.x i x i 1 / f. i /.x i 1 x i / M i.x i x i 1 /, m i.x i x i 1 / F.x i / F.x i 1 / M i.x i x i 1 / n n n ) m i.x i x i 1 / ŒF.x i / F.x i 1 / M i.x i x i 1 /, L.f; P/ F.b/ F./ U.f; P/: Since P is n rbitrry prtition of Œ; b, it follows tht F.b/ F./ is n upper bound for the set fl.f; P/ W P is prtition of Œ; b g nd lower bound for the set fu.f; P/ W P is prtition of Œ; b g. Therefore, b f.x/dx sup L.f; P/ F.b/ P F./ inf U.f; P/ P b f.x/dx: Since f is integrble on Œ; b, we hve tht b b b f.x/dx f.x/dx f.x/dx; b nd consequently f.x/dx F.b/ F./ Exercise [1] Let f be the function on Œ0; 1 given by Show tht f is not Riemnn-integrble on Œ0; 1. [2] Let f be the function on Œ0; 1 given by x if x is rtionl f.x/ 0 if x is irrtionl: f.x/ 1 if 0 x < 1 2 x 1 2 if 1 2 x < 1: () Show, from first principles, tht f is Riemnn-integrble on Œ0; 1. 12

15 (b) Quote result tht ssures us tht f is Riemnn-integrble. 1 (c) Find f.x/dx: 0 (d) Let fr 1 ; r 2 ; r 3 ; : : : g be n enumertion of rtionls in the intervl Œ0; 1. For ech n 2 N, define 1 if x 2 fr1 ; r f n.x/ 2 ; r 3 ; : : : ; r n g 0 otherwise: Show tht.f n / is nondecresing sequence of functions tht re Riemnn-integrble on Œ0; 1. Show lso tht the sequence.f n / converges pointwise to the function nd tht f is not Riemnn-integrble. 1 if x 2 Q \ Œ0; 1 f.x/ 0 otherwise; 13

16 Chpter 2 Mesure spces 2.1 Introduction We now concern ourselves with the ide of defining mesure on the rel numbers, which puts the intuitive ide of length on n xiomtic foundtion. This turns out to be more difficult thn you might expect nd we begin with the motivtion for developing mesure theory. If we re to properly define the definite integrl of rel-vlued function, we need meningful notion of the length of set A R. If we consider the set of rel numbers, we would like to be ble to define function tht mesures the length of set. Tht is function m W P.R/! Œ0; 1 such tht: [1].Œ; b / b, for b. [2] is trnsltion invrint, i.e.,.a/ m.x C A/, for A R nd x 2 R. [3].;/ 0. [4] If.A i / is sequence of mutully disjoint sets in.r/ (i.e., A i \ A j ;, for every i 6 j ), then [ 1 1 A i A i : 14

17 This turns out to be impossible. Tht is, you cnnot define such function whose domin consists of the whole of P.R/. To show this we present the following exmple by Giuseppe Vitli (1905): Exmple A Vitli set V, is subset of Œ0; 1, which for ech rel number r, V contins exctly one number v such tht v r is rtionl (This implies tht V is uncountble, nd lso, tht v u is irrtionl, for ny u; v 2 V, u 6 v). Such sets cn be shown to exist provided we ssume the xiom of choice. To construct the Vitli set V, consider the dditive quotient group R=Q. Ech element of this group is shifted copy of the rtionl numbers. Tht is, set of the form Q C r, for some r 2 R. Therefore, the elements of this group re subsets of R nd prtition R. There re uncountbly mny elements. Since ech element intersects Œ0; 1, we cn use the xiom of choice to choose set V Œ0; 1 contining exctly one representtive out of ech element R=Q. A Vitli set is non-mesurble. To show this, we rgue by contrdiction nd ssume tht V is mesurble. Let q 1 ; q 2 ; : : : be n enumertion of the rtionl numbers in Œ 1; 1. From the construction of V, note tht the trnslted sets V k V C q k fv C q k W v 2 V g, k 1; 2; : : : re pirwise disjoint, nd further note tht Œ0; 1 S k V k Œ 1; 2 (to see the first inclusion, consider ny rel number r in Œ0; 1 nd let v be the representtive in V for the equivlence clss Œr, the r v q for some rtionl number q in Œ 1; 1 ). Now ssume tht there exists tht stisfies the four conditions listed bove. Apply to these inclusions using condition [4]: 1 1.V k / 3: k1 Since is trnsltion invrint,.v k /.V / nd 1 1.V / 3: k1 This is impossible. Summing infinitely mny copies of the constnt.v / yields either zero or infinity, ccording to whether the constnt is zero or positive. In neither cse is the sum in Œ1; 3. So, V cnnot hve been mesurble fter ll, i.e., must not define ny vlue for.v /. It is importnt to stress tht this whole rgument depends on the ssumption of the xiom of choice. 2.2 Nottion nd preliminries Let be set. We will use the stndrd nottion for the compliment of A: A c fx 2 W x 62 Ag We denote the complement of B with respect to A by AnB A \ B c. We denote by P. /, the powerset of, i.e., the set of ll subsets of. Let A nd B be in P. /. The symmetric difference of A nd B, denoted by A4B, is given by A4B.AnB/ [.BnA/: A subset C of P. / is sid to be: 15

18 [1] closed under finite intersections if, whenever fa 1 ; A 2 ; : : : ; A n g is finite collection of elements n\ of C, then A j 2 C. [2] closed under finite unions if, whenever fa 1 ; A 2 ; : : : ; A n g is finite collection of elements of C, n[ then A j 2 C. [3] closed under countble intersections if, whenever.a n / is sequence of elements of C, then 1\ A j 2 C. [4] closed under countble unions if, whenever.a n / is sequence of elements of C, then C. Let.A n / be sequence in P. /. We denote by lim sup A n n!1 1\ 1[ mn A m, nd lim inf n!1 A n 1[ mn 1\ A m : 1[ A j Exercise Let be nonempty set. Prove tht if collection C of P. / is closed under countble intersections (resp., unions), then it is closed under finite intersections (resp., unions) efinition An lgebr (in ) is collection of subsets of such tht: () ;; 2. (b) If A 2, then A c 2. (c) If A 1 ; : : : ; A n 2, then [ n A i nd \ n A i re in. is -lgebr (or -field) if in ddition: (d) If A 1 ; A 2 ; : : : re in, then [ 1 A i nd \ 1 A i re in Exmples [1] Let be ny nonempty set. Then there re two specil -lgebrs on : 1 f;; g nd 2 P. / (the power set of ). [2] Let Œ0; 1. Then f;; ; Œ0; 1 2 ;. 1 ; 1 g is -lgebr on. 2 [3] Let R nd let fa R W A is countble or A c is countbleg: is -lgebr on. Prts () nd (b) of the definition re esy to show. Suppose A 1 ; A 2 ; : : : re ll in. If ech of the A i s re countble, then [ i A i is countble,nd so in. If A c i 0 is countble for some i 0, then.[ i A i / c \ i A c i Ac i 0 is countble, nd gin [ i A i is in. Since \ i A i.[ i A c i /c, then the countble intersection of sets in is gin in Remrk Let R nd let be -lgebr on. If.A n / is sequence of sets in, then lim inf n!1 A n 2. 16

19 If S is ny collection of subsets of set, then by tking the intersection of ll -lgebrs contining S, we obtin the smllest -lgebr contining S. Such -lgebr, denoted by.s/ is sid to be generted by S. It is cler tht if S is -lgebr, then.s/ S efinition The smllest -lgebr contining ll open subsets of R is clled the Borel -lgebr of R nd is denoted by B.R/. Elements of this -lgebr re clled Borel sets Proposition Let I fœ; b/ W ; b 2 R; bg. Then.I/ B.R/. If ; b 2 R nd b, then since Œ; b/ 1\ 1 n ; b ; it follows tht Œ; b/ 2 B.R/ nd consequently, I B.R/. It now follows tht.i/ B.R/. Conversely, let A be n open subset of R. Then A is countble union of open intervls. Since ny open intervl.; b/ cn be expressed in the form where 1 n 0 < b.; b/ 1[ nn 0 h C 1 n ; b ;, it follows tht A 2.I/. Thus.I/ B.R/ Exercise Let be nonempty set nd S the smllest -lgebr in which contins ll singletons fxg. Show tht S fa 2 P. / W either A or A c is countbleg: 2.3 Mesure on set efinition A mesurble spce is pir.; /, where is set nd is -lgebr on. The elements of the -lgebr re clled mesurble sets efinition Let.; / be mesurble spce. A mesure on.; / is function W! Œ0; 1 such tht: ().;/ 0, nd (b) if.a i / is sequence of mutully disjoint sets in (i.e., A i 2, for ech i 1; 2; 3; : : :, nd A i \ A j ; for every i 6 j ), then [ 1 1 A i A i : Property (b) in the bove definition is known s the countble dditivity property of the mesure. If (b) is replced by the condition tht. S n A i/ P n A i for every finite sequence A 1 ; A 2 ; : : : ; A n of pirwise disjoint sets in, then we sy tht is finitely dditive. It follows immeditely, tht if mesure 17

20 is countbly dditive, then it must lso be finitely dditive. If you hve finite sequence A 1 ; A 2 ; : : : ; A n in, then you cn consider it to be n infinite sequence.a i / with A nc1 A nc2 : : : ;. Then [ n [ 1 1 A i A i A i n A i : efinition A mesure spce is triple.; ; /, where is set, -lgebr in, nd mesure on Remrk Let.; ; / be mesure spce. If. / 1, then.; ; / is clled probbility spce nd is clled probbility mesure Exmples [1] Let be nonempty set nd P. /. efine W! Œ0; 1 by n if A hs n elements.a/ 1 otherwise: Then is mesure on clled the counting mesure. [2] Let.; / be mesurble spce. Choose nd fix x 2. efine x W! Œ0; 1 by 1 if x 2 A x.a/ 0 otherwise: Then x is mesure on clled the unit point mss t x Theorem Let.; ; / be mesure spce. Then the following sttements hold: [1] If A; B 2 nd A B, then.a/.b/. [2] If.A n / is sequence in, then [ 1 A n 1.A n /: [3] If.A n / is sequence in such tht A n A nc1, for ech n 2 N, then [ 1 A n lim n!1.a n/: [4] If.A n / is sequence in such tht A n A nc1, for ech n 2 N nd.a 1 / < 1, then \ 1 A n lim n!1.a n/: [1] Write B A [.BnA/, disjoint union. Since is finitely dditive, we hve tht.b/.a/ C.BnA/.A/: 18

21 [2] Set B 1 A 1, nd for ech nturl number n > 1, let B n A n n S n 1 A i. Then n\ 1 n\ 1 B n.a n na i / A n A c i : Therefore, for ech n 1; 2; 3; : : :, we hve tht B n 2. Clerly, B n A n, for ech n 1; 2; 3; : : :. Let l nd k be integers such tht l < k. Then B l A l, nd so Tht is, B i Ti6j B j ;: k\ 1 B l \ B k A l \ B k A l \ A k A c i A l \ A k \ \ A c l \ A l \ A c l \ A k \ \ ; \ A k \ \ ;: Clim: S 1 A n S 1 B n: Since B n A n, for ech n 1; 2; 3; : : :; we hve tht S 1 B n S 1 A n. Let x 2 S 1 A n: Then x 2 A n for some n. Let k be the smllest index such tht x 2 A n. Then x 2 A k nd x 62 A j, for ech j 1; 2; : : :; k 1. Therefore, x 2 Aj c, for ech j 1; 2; : : :; k 1. T Tht is, x 2 B k A k 1 k Ac j, whence x 2 S 1 B n, which proves the clim. Therefore, by prt [1]. [ 1 [ 1 A n 1 1 B n.b n /.A n /; T [3] Let B 1 A 1 nd B n A n na n 1 for ll n > 1. Then B i i6j B j ;; A k S k B n nd S 1 A n S 1 B n. Therefore, [ 1 [4] Let A T 1 A n. Then [ 1 A n B n 1.B n / lim k k!1 1\ 1[ A 1 na A 1 n A n.a 1 na n /; nd A 1 na n A 1 na nc1, for ech nturl number n. Therefore, by [3] bove,.b n / lim k!1.a k/: [ 1.A 1 na/.a 1 na n / lim.a 1nA n /: (2.1) n!1 Since A A 1 nd.a 1 / < 1, we hve tht.a/ < 1. Also, since A n A 1, for ech nturl number n, it follows tht.a/ < 1, for ech n. Writing A 1 s disjoint union A 1.A 1 na/ [ A, we hve tht.a 1 /.A 1 na/ C.A/. It now follows tht.a 1 na/.a 1 /.A/: (2.2) 19

22 Similrly, by expressing A 1 s A 1.A 1 na n / [ A n, we get tht From equtions 2.1, 2.2 nd 2.3, we hve tht.a 1 na n /.A 1 /.A n /: (2.3).A 1 /.A/ lim n!1 Œ.A 1/.A n /.A 1 / lim n!1.a n/: Since.A 1 / < 1, we hve tht.a/ lim.a \ 1 n/; tht is n!1 A n lim n!1.a n/: Theorem (Borel-Cntelli Lemm) Let.; ; / be mesure spce. If.A n / is sequence of mesurble sets such tht P 1.A n/ < 1, then lim sup n!1 A n 0. Since lim sup n!1 A n T 1 S 1 kn A k, it follows tht for ech n 2 N, s n! 1. [ 1 lim sup A n A k n! efinition Let.; ; / be mesure spce. The mesure is sid to be: [1] finite if. / < 1, nd 1.A k /! 0 [2] -finite if there is sequence. n / of sets in with. n / < 1, for ech n such tht S 1 n. 2.4 Outer mesure Constructing mesure is nontrivil exercise s illustrted in the introduction to this chpter. The bsic ide tht we will use is s follows: We will begin by defining n outer mesure on ll subsets of set. Then we will restrict the outer mesure to resonble clss of subsets nd in so doing we will define mesure. This is indeed how Lebesgue solved the problems ssocited with mesure nd integrtion. Both were published s prt of his disserttion in efinition Let be set. An outer mesure on is set function W P. /! Œ0:1 stisfying the following properties: ().;/ 0: (b) If A B, then.a/.b/. (c) For ny sequence.a n / of subsets of, S1 A n P 1.A n /: Property (b) bove is referred to s monotonicity, nd property (c) is countble subdditivity 20

23 2.4.2 Exmples [1] Let be set. efine on P. / by: Then is n outer mesure on..a/ 1 if A 6 ; 0 if A ;: [2] Let be n uncountble set. efine on P. / by: 0 if A is countble.a/ 1 if A is uncountble: Then is n outer mesure on. We cn now use the concept of outer mesure to construct mesure efinition Let be set nd n outer mesure on. A subset E of is sid to be -mesurble if, for every subset A of, we hve.a/.a \ E/ C.A \ E c /: Remrk [1] Since A.A \ E/ [.A \ E c / nd is subdditive, we lwys hve tht.a/.a \ E/ C.A \ E c /: Therefore, subset E of is -mesurble if, for every subset A of,.a/.a \ E/ C.A \ E c /: [2] Replcing E by E c in the definition of -mesurbility, we hve tht.a/.a \ E c / C ŒA \.E c / c.a \ E c / C.A \ E/: Thus, E is -mesurble if nd only if E c is -mesurble Lemm Let be set, n outer mesure on nd E subset of. If.E/ 0, then E is -mesurble. Let A be ny subset of. Then A \ E E. Since is monotone,.a \ E/.E/ 0. Also, since A \ E c A, it follows tht.a \ E c /.A/. Therefore,.A/.A \ E/ C.A \ E c /; nd therefore, E is -mesurble Corollry The empty set ; is -mesurble Lemm Let be set, n outer mesure on, nd E nd F -mesurble subsets of. Then E [ F is -mesurble. 21

24 Let A be ny susbset of. Then since E is -mesurble, Also, since F is -mesurble, From the two equtions bove we get: Therefore, whence E [ F is mesurble..a/.a \ E/ C.A \ E c /:.A \ E c / Œ.A \ E c / \ F C Œ.A \ E c / \ F c :.A/.A \ E/ C Œ.A \ E c / \ F C Œ.A \ E c / \ F c :.A/.A \ E/ C Œ.A \ E c / \ F C Œ.A \ E c / \ F c Œ.A \ E/ [..A \ E c / \ F/ C Œ.A \ E c / \ F c ŒA \.E [ F/ C ŒA \.E [ F/ c ; Theorem Let be set nd n outer mesure on. enote by M, the collection of ll -mesurble subsets of. Then M is n lgebr on. This follows from Remrk [2], Corollry nd Lemm Proposition Let be set nd n outer mesure on. If E 1 ; E 2 2 M nd E 1 \ E 2 ;, then ŒA \.E 1 [ E 2 /.A \ E 1 / C.A \ E 2 /: Since E 1 is -mesurble,.a/.a \ E 1 / C.A \ E c 1 /; for ech subset A of. Replcing A by A \.E 1 [ E 2 /, we hve ŒA \.E 1 [ E 2 / Œ.A \.E 1 [ E 2 // \ E 1 C Œ.A \.E 1 [ E 2 // \ E c 1 Œ..A \ E 1 / [.A \ E 2 // \ E 1 C Œ..A \ E 1 / [.A \ E 2 // \ E c 1 Œ.A \ E 1 \ E 1 / [.A \ E 2 \ E 1 / C Œ.A \ E 1 \ E c 1 / [.A \ E 2 \ E c 1 /.A \ E 1 / C.A \ E 2 /: Theorem Let be set nd n outer mesure on. Suppose tht.e n / be sequence of mutully disjoint sets in M. Then for ny n 1 nd ny subset A of, A \ n[ E j n.a \ E j /: (2.4) 22

25 We prove this theorem by induction. The cse n 1 is obviously true. The cse n 2 hs been proved in Proposition Assume tht Firstly, we note tht Since E n is -mesurble, [ n A \ E j n[ 1 A \ E j [ n A \ n 1.A \ E j /: (2.5) E j \ E n A \ E n, nd [ n n[ 1 A \ E j \ En c A \ E j : [ n A \.A \ E n / C A \ [ n E j \ E n C A \ n 1 [ E j n 1.A \ E n / C.A \ E j / (by 2.5) n.a \ E j /: E j \ E c n By tking A in (2.4), we obtin [ n E j n.e j /: i.e., is finitely dditive on M. The following theorem is the key to using n outer mesure to construct mesure Theorem Let be set nd n outer mesure on. enote by M, the collection of ll -mesurble subsets of. Then [1] M is -lgebr, nd [2] the restriction of j M of to M is mesure on.; M/. [1] We hve lredy shown in Theorem tht M is n lgebr. It remins to show tht M is closed under countble unions. We will show tht M is in fct closed under pirwise disjoint countble 23

26 unions. Let.E n / be pirwise disjoint sequence of sets in M, S n S n E j nd S \ 1 E j. Then, since S n 2 M, we hve tht for ny A,.A/.A \ S n / C.A \ S c n /.A \ S n / C.A \ S c / (since Sn c Sc nd is monotone) n.a \ E j / C.A \ S c / (by Theorem ). Since this is true for every nturl number n, it follows tht.a/ 1.A \ E j / C.A \ S c / Therefore, S 1 E j 2 M..A \ S/ C.A \ S c / (since is countbly subdditive). [2] By definition of the outer mesure,.;/ 0. We hve to show tht is countbly dditive. We do this s follows, let.e n / be pirwise disjoint sequence of sets in M. As shown in [1], for ny A, 1 [ 1 c.a/.a \ E j / C A \ E j : If A is replced by [ 1 E j, we hve tht [ 1 1 E j.e j /: (2.6) Since is, by definition, countbly subdditive, we hve tht From (2.6) nd (2.7), we conclude tht Thus, is mesure on M. [ 1 1 E j.e j /: (2.7) [ 1 1 E j.e j /: The mesure j M on is often referred to s the mesure on induced by the outer mesure Exercise [1] In proving Theorem [1], we showed tht M is closed under pirwise disjoint countble unions. educe from this tht M is closed under countble unions. [2] Let be n outer mesure on set such tht.a/ 0, for some subset A of. Show tht if B is subset of, then.a [ B/.B/. 24

27 2.5 Lebesgue mesure on R We now use the pproch discussed in the previous section to construct the Lebesgue mesure on R. The Lebesgue mesure will led to the definition of the Lebesgue integrl in the next chpter. Let I be collection of open intervls in R. For I 2 I, let l.i/ denote the length of I Theorem For ny subset A of R, define m W P.R/! Œ0; 1 by m.a/ inf l.i n / W I n 2 I, for ech n nd A [ n n I n : Then m is n outer mesure on R. (i) Since ;.; /, for ny rel number, it follows tht 0 m.;/ 0. Thus, m.;/ 0. (ii) Let A nd B be subsets of R such tht A B. If B S n I n, where.i n / is sequence of open intervls in R, then A S n I n. Therefore, m.a/ P 1 l.i n/. It follows tht m.a/ m.b/. (iii) Let.A n / be sequence of subsets of R. If m.a n / 1 for some n, then [ 1 1 m A n m.a n /: Suppose tht m.a n / < 1, for ech n. Then given > 0, there is sequence.i nk / k of open intervls in R such tht A n S 1 k1 I nk, nd 1 k1 l.i nk / < m.a n / C 2 n : Now, since S 1 A n S 1 S 1 k1 I nk, it follows tht [ m A n l.i nk / < m.a n / C 1 m.a 2 n n / C : k1 Since is rbitrry, we hve tht m S1 A n P 1 m.a n /. The outer mesure defined in Theorem is clled the Lebesgue outer mesure efinition A set E R is sid to be Lebesgue-mesurble (or simply mesurble) if m.a/ m.a \ E/ C m.a \ E c /; for every set A R. enote by M, the set of ll Lebesgue-mesurble subsets of R. We hve lredy shown in Theorem , tht M is -lgebr nd tht m m j M is mesure on M. This mesure is clled the Lebesgue mesure on R. The Lebesgue outer mesure hs other importnt properties tht re listed in the following propositions. 25

28 2.5.3 Proposition The outer mesure of n intervl I is its length. Tht is, m.i/ l.i/: Cse 1: Let I be closed nd bounded intervl, sy I Œ; b. Then Œ; b. > 0. Thus, m.œ; b / l. ; b C / b C 2: ; b C /, for ech Since is rbitrry, m.i/ m.œ; b / b. It remins to show tht m.i/ b. Suppose tht I S 1 k1 I k, where I k is open for ech k 2 N. Since I is compct, there is finite subcollection fi k g m k1 of intervls such tht I S m k1. Since 2 I, there is n intervl. 1; b 1 / in the collection fi k g m k1 such tht 1 < < b 1. If b b 1, then we re done. Otherwise, < b 1 < b nd so b 1 2 I. Therefore, there is n intervl. 2 ; b 2 / in the collection fi k g m k1 such tht 2 < b 1 < b 2. If b < b 2, then we re done. Otherweise, < b 2 < b. Tht is, b 2 2 I. So there is n intervl. 3 ; b 3 / in the collection fi k g m k1 such tht 3 < b 2 < b 3. Continuing in this fshion, we obtin sequence of intervls. 1 ; b 1 /;. 2 ; b 2 /; : : : ;. r ; b r / from the collection fi k g m k1 such tht i < b i 1 < b i for i 2; 3; : : :; r. Since fi k g m is finite k1 collection, this process must terminte fter s m steps with b < b s. Now, 1 l.i j / m l.i j / s l. j ; b j /.b s s / C.b s 1 s 1 / C.b s 2 s 2 / C C.b 1 1 / b s. s b s 1 /. s 1 b s 2 /. 2 b 1 / 1 > b s 1 > b : It follows from this tht m.i/ b l.i/. Hence, Cse 2: Let I be n intervl of the form.; b/,.; b or Œ; b/. Then for ech 0 < <.b Œ C 2 ; b m Œ C 2 ; b Since is rbitrry, it follows tht m.i/ b 2 I Œ 2 ; b C 2 : 2 m.i/ m Œ 2 ; b C 2,l Œ C 2 ; b 2 m.i/ Œ 2 ; b C 2,b m.i/ b C ( by cse 1). l.i/: / with Cse 3: Suppose tht I is n unbounded intervl. Then for ny rel number k, there is closed intervl J I such tht l.j/ k. Hence, m.i/ m.j/ l.j/ k: Thus, m.i/ Proposition For ech x 2 R, m.fxg/ 0: 26

29 Given ny > 0, fxg.x 2 ; x C 2 /. Thus 0 m.fxg/ m.x 2 ; x C 2 / l.x 2 ; x C 2 / : Since is rbitrry, we hve tht m.fxg/ Corollry Every countble subset of R hs outer mesure zero. Let A be countble subset of R. Then A S n f ng. Therefore, [ 0 m.a/ m f n g m.f n g/ 0: n n Thus, m.a/ Corollry The intervl Œ0; 1 is uncountble Proposition The Lebesgue outer mesure is trnsltion-invrint. Tht is, m.a C x/ m.a/ for ll A R nd x 2 R. Let.I n / be sequence of open intervls such tht A S n I n. Then A C x S n.i n C x/. Thus, m.a C x/ n l.i n C x/ n l.i n /: Tht is, m.a C x/ is lower bound for the sums P n l.i n/, with A S n I n. Therefore, m.a C x/ m.a/: (2.8) To prove the reverse inequlity, we only need to notice tht m.a/ m.a C x C. x// m.a C x/; where the inequlity follows from (4.1). Thus, m.a C x/ m.a/ Proposition For ny set A R nd ny > 0, there is n open set V such tht A V nd m.v / < m.a/ C. By definition of m.a/, there is sequence.i n / of open intervls such tht A S n I n nd l.i n / < m.a/ C 2 : n If I n. n ; b n /, let J n. n ; b 2 nc1 n /. Then A S n J n. Let V S n J n. Then V is n open set in R nd m.v / l.j n / l.i n / C 2 < m.a/ C : n n 27

30 2.5.9 Proposition For ny 2 R, the set.; 1/ is mesurble. Let A be ny subset of R. We must show tht m.a/ m.a \.; 1// C m.a \.; 1/ c / m.a \.; 1// C m.a \. 1; /: Let A 1 A \.; 1/ nd A 2 A \. 1;. Then we need to show tht m.a/ m.a 1 / C m.a 2 /: If m.a/ 1, then there is nothing to prove. Assume tht m.a/ < 1. From the definition of m, given > 0, there is sequence.i n / of open intervls such tht A S n I n nd m.a/ C > n l.i n /: Let J n I n \.; 1/ nd Jn 0 I n \. 1;. Then J n nd Jn 0 re disjoint (possibly empty) intervls nd I n J n [ J 0 n. Therefore, l.i n / l.j n / C l.j 0 n / m.j n / C m.j 0 n /: Since A 1 S n J n nd A 2 S n J n 0, it follows tht [ m.a 1 / m J n n n [ m.a 2 / m Jn 0 n n m.j n / n m.j 0 n / n l.j n /; l.j 0 n /: nd Therefore, m.a 1 / C m.a 2 / n Œl.J n / C l.j 0 n / n l.i n / < m.a/ C : Since is rbitrry, m.a 1 / C m.a 2 / m.a/ Corollry [1] Every intervl in R is mesurble. [2] Every open set in R is mesurble. [3] Every closed set in R is mesurble. [4] Every set tht is countble intersection of open sets in R is mesurble. [5] Every set tht is countble union of closed sets in R is mesurble. [1] We hve lredy shown tht for ech 2 R, the intervl.; 1/ is mesurble. It follows from the fct tht M is n lgebr tht. 1;.; 1/ c is mesurble for ech 2 R. Since Œ; 1/ fg [.; 1/ nd m.fg/ 0, it follows from Lemm tht fg is mesurble. Therefore, Œ; 1/ is mesurble. Since M is n lgebr, we hve tht. 1; / Œ; 1/ c is mesurble for ech 2 R. If nd b re rel numbers such tht < b, then.; b/. 1; b/ \.; 1/. Since. 1; b/ nd.; 1/ re both mesurble, it follows tht.; b/ is lso mesurble. If nd b re rel numbers such tht < b, then Œ; b/ fg [.; b/,.; b.; b/ [ fbg, nd Œ; b fg [.; b/ [ fbg. Therefore, the sets Œ; b/,.; b, nd Œ; b re ll mesurble. 28

31 [2] Every open set in R is countble union of open intervls. Since ech open intervl is mesurble nd M is -lgebr, it follows tht the union of open intervls is lso mesurble. [3] A closed set F in R is complement of n open set V in R. Tht is F V c for some open set V in R. Since V is mesurble, V c F is lso mesurble. [4] Let U T 1 V i, where V i is open in R for ech i 2 N. Then by [2], V i is mesurble, for ech i 2 N, nd so is Vi c for ech i 2 N. Since M is -lgebr, we hve tht 1\ [ 1 c U V i Vi c is mesurble. [5] Let F S 1 F i, where F i is closed in R, for ech i 2 N. Then Fi c is open in R, for ech i 2 N. By [4], we hve tht T 1 F i c is mesurble. Since M is -lgebr, it follows tht c T1 S 1 F i F is mesurble. F c i efinition () A set tht is union of countble collection of closed sets is clled F -set. (b) A set tht is n intersection of countble collection of open sets is clled G ı -set Remrk [1] The previous corollry sys, mongst other things, tht every F -set nd every G ı -set is mesurble. [2] The previous corollry lso sttes tht ll nice sets in R re mesurble. There re however sets tht re not mesurble. The point being tht these sets re certinly not nturl but somewht strnge. [3] The previous corollry shows tht every open set in R is mesurble. The smllest -lgebr contining ll the open sets is clled the Borel -lgebr. The following result sttes tht ny mesurble subset of R cn be pproximted by the nice ones Proposition Let E R nd m the Lebesgue outer mesure on R. The following sttements re equivlent: [1] E is mesurble. [2] For ech > 0, there is n open set V such tht E V nd m.v ne/ <. [3] There is G ı -set G such tht E G nd m.gne/ 0. [4] For ech > 0, there is closed set F such tht F E nd m.enf/ <. [5] There is n F -set H such tht H E nd m.enh/ 0. [1] ) [2]: Suppose tht E is mesurble nd let > 0. If m.e/ < 1, then by Proposition 2.5.8, there is n open set V such tht E V nd m.v / < m.e/ C : Since V nd E re mesurble, we cn rewrite the eqution bove s m.v / < m.e/ C : 29

32 Hence, m.v ne/ m.v / m.e/ <. Suppose tht m.e/ 1. Write R S n2n I n, union of disjoint finite intervls. For ech n 2 N, let E n E \ I n. Then E S n2n E n, nd for ech n 2 N, m.e n / < 1, Therefore, by the first prt, there is n open set V n such tht E n V n nd m.v n ne n / <. Write V S 2 n n2n V n. Then V is n open set nd V ne [ n2n V n n [ n2n E n [ n2n.v n ne n /: Hence m.v ne/ n2n m.v n ne n / < n2n 2 n : [2] ) [3]: For ech n 2 N, let V n be n open set such tht E V n nd m.v n ne/ < 1 n. Let G T n2n V n. Then G is G ı -set, E G, nd for ech n 2 N, \ m.gne/ m n ne/ m n2n.v.v n ne/ 1 n : Thus, m.gne/ 0: [3] ) [1]: Since G is G ı -set, it is mesurble. Also GnE is mesurble since m.gne/ 0. Therefore, the set E Gn.GnE/ is lso mesurble. [1] ) [4]: Suppose tht E is mesurble nd let > 0. Then E c is lso mesurble. Since [1] implies [2], we hve, by [2], tht there is n open set V such tht E c V nd m.v ne c / <. Note tht V ne c E \ V EnV c. Tke F V c. Then F is closed set, F E, nd m.enf/ <. [4] ) [5]: For ech n 2 N, let F n be closed set such tht F n E nd m.enf n / < 1 n. Let H S n2n F n. Then H is n F -set, H E, nd for ech n 2 N, m.enh/ m En [ F n \ n / m n2n n2n.enf.enf n / < 1 n : Thus, m.enh/ 0. [5] ) [1]: Since H is n F -set, it is mesurble. Also EnH is mesurble since m.enh/ 0. Therefore, the set E.EnH/ [ H is lso mesurble. 30

33 Chpter 3 The Lebesgue integrl 3.1 Introduction We now turn our ttention to the construction of the Lebesgue integrl of generl functions, which, s lredy discussed, is necessry to void the technicl deficiencies ssocited with the Riemnn integrl. 3.2 Mesurble functions Bsic notions The extended rel number system, R, is the set of rel numbers together with two symbols 1 nd C1. Tht is, R R [ f 1; C1g Π1; 1. The lgebric opertions for these two infinities re: [1] For r 2 R, 1 C r 1. [2] For r 2 R, r. 1/ 1, if r > 0 nd r. 1/ 1 if r < 0. [3] C1 C.C1/ C1 nd 1 C. 1/ 1. [4] 1 C. 1/ is undefined. [5] 0. 1/ efinition Let.; / be mesurble spce nd E 2. A function f W E! R is sid to be mesurble if for ech 2 R, the set fx 2 E W f.x/ > g is mesurble Proposition Let.; / be mesurble spce nd E 2, nd f! R. Then the following re equivlent: [1] f is mesurble. [2] For ech 2 R, the set fx 2 E W f.x/ g is mesurble. [3] For ech 2 R, the set fx 2 E W f.x/ < g is mesurble. [4] For ech 2 R, the set fx 2 E W f.x/ g is mesurble. 31

34 [1] ) [2]: If f is mesurble, then the set fx 2 E W f.x/ g \ n2n x 2 E W f.x/ > 1 ; n which is n intersection of mesurble sets, is mesurble. [2] ) [3]: If the set fx 2 E W f.x/ g is mesurble, then so is the set Enfx 2 E W f.x/ g fx 2 E W f.x/ < g: [3] ) [4]: If the set fx 2 E W f.x/ < g is mesurble, then so is the set fx 2 E W f.x/ g \ x 2 E W f.x/ < C 1 n n2n since it is n intersection of mesurble sets. [4] ) [1]: If the set fx 2 E W f.x/ g, is mesurble, so is its complement in E. Hence, fx 2 E W f.x/ > g Enfx 2 E W f.x/ g is mesurble. It follows from this tht f is mesurble Exmples [1] The constnt function is mesurble. Tht is, if.; / is mesurble spce, c 2 R, nd E 2, then the function f W E! R given by f.x/ c, for ech x 2 E, is mesurble. Let 2 R. If c, then the set fx 2 E W f.x/ > g ;, nd is thus mesurble. If < c, then the set fx 2 E W f.x/ > g E, nd is therefore mesurble. Hence, f is mesurble. [2] Let.; / be mesurble spce nd let A 2. The chrcteristic function, A, is defined by 1 if x 2 A A.x/ 0 if x 62 A: The chrcteristic function A is mesurble. This follows immeditely from the observtion tht, for ech 2 R nd E 2, the set fx 2 E W A > g is either E, A or ;. [3] Let B be the Borel -lgebr in R nd E 2 B. Then, ny continuous function f W E! R is mesurble. This is n immedite consequence of the fct tht, if f W E! R is continuous nd 2 R, then the set fx 2 E W f.x/ > g is open nd hence belongs to B Proposition [1] If f nd g re mesurble rel-vlued functions defined on common domin E 2 nd c 2 R, then the functions () f C c, (b) cf, (c) f g, (d) f 2, 32

35 (e) f g, (f) jf j, (g) f _ g, (h) f ^ g re lso mesurble. [2] If.f n / is sequence of mesurble functions defined on common domin E 2, then the functions () sup n f n, (b) inf n f n, (c) lim sup n f n, (d) lim inf n f n re lso mesurble. [1] () For ny rel number, fx 2 E W f.x/ C c > g fx 2 E W f.x/ > cg: Since the set on the right-hnd side is mesurble, we hve tht f C c is mesurble. (b) If c 0, then cf is obviously mesurble. Assume tht c < 0. Then, for ech rel number, fx 2 E W cf.x/ > g x 2 E W f.x/ < : c Since the set fx 2 E W f.x/ < g is mesurble, it follows tht cf is lso mesurble. c (c) Let be rel number. Since the rtionls re dense in the rels, there is rtionl number r such tht f.x/ < r < g.x/; whenever f.x/ C g.x/ <. Therefore, fx 2 E W f.x/ C g.x/ < g [ r2q.fx 2 E W f.x/ < rg \ fx 2 E W g.x/ < rg/: Since the sets fx 2 E W f.x/ < rg nd fx 2 E W g.x/ < rg re mesurble, so is the set fx 2 E W f.x/ < rg \ fx W g.x/ < rg, nd consequently, the set fx 2 E W f.x/ C g.x/ < g, being countble union of mesurble sets, is lso mesurble. If g is mesurble, it follows from (b) tht. 1/g is lso mesurble. Hence, so is f C. 1/g f g. (d) Let 2 R nd E 2. If < 0, then fx 2 E W f 2.x/ > g E, which is mesurble. If 0, then fx 2 E W f 2.x/ > g fx 2 E W f.x/ > p g [ fx 2 E W f.x/ < p g: Since the two sets on the right hnd side re mesurble, it follows tht the set fx 2 E W f 2.x/ > g is lso mesurble. Hence f 2 is mesurble. (e) Since f g 1 4 Œ.f C g/2.f g/ 2, it follows from (b), (c), nd (d) tht f g is mesurble. (f) Let 2 R nd E 2. If < 0, then fx 2 E W jf.x/j > g E, which is mesurble. If 0, then fx 2 E W jf.x/j > g fx 2 E W f.x/ > g [ fx 2 E W f.x/ < g: Since the two sets on the right hnd side re mesurble, it follows tht the set fx 2 E W f.x/ 2 > g is lso mesurble. Thus, jf j is mesurble. 33

36 (g) It is sufficient to observe tht f _ g 1 2ff C g C jf (d) tht f _ g is mesurble. gjg. It now follows from (b), (c), nd (h) It is sufficient to observe tht f ^ g 1 ff C g jf gjg. It now follows from (b), (c), nd 2 (d) tht f ^ g is mesurble. [2] () Let 2 R. Then fx 2 E W sup f n > g n 1[ fx 2 E W f n.x/ > g: Since for ech n 2 N, f n is mesurble, it follows tht the set fx 2 E W f n.x/ > g is mesurble for ech n 2 N. Therefore, the set fx 2 E W sup n f n.x/ > g is mesurble s it is countble union of mesurble sets. (b) It is sufficient to note tht inf n f n sup n. f n /. (c) Notice tht lim sup n f n inf n1 sup kn f k nd use () nd (b) bove. (d) Notice tht lim inf n f n sup n1 inf kn f k nd use () nd (b) bove Corollry Let.; / be mesurble spce. If f is pointwise limit of sequence.f n / of mesurble functions defined on common domin E 2, then f is mesurble. If the sequence.f n / converges pointwise to f, then, for ech x 2 E, Now, by Proposition [2], f is mesurble. f.x/ lim f n.x/ lim sup f n.x/: n! efinition Let f be rel-vlued function defined on set. The positive prt of f, denoted by f C, is the function f C mxff; 0g f _0 nd the negtive prt of f, denoted by f, is the function f mxf f; 0g. f / _ 0. Immeditely we hve tht if f is rel-vlued function defined on, then f f C f nd jf j f C C f : n Note tht f C 1 2 Œjf j C f nd f 1 Œjf j f : 2 Let.; / be mesurble spce nd f W! R. It is trivil to deduce from Proposition tht f is mesurble if nd only if f C nd f re mesurble efinition Let.; / be mesurble spce. A simple function on is function of the form P n c j Ej, where, for ech j 1; 2; : : :n, c j is n extended rel number nd E j 2. Since Ej mesurble. is mesurble for ech j 1; 2; : : :; n, it follows from Proposition tht is lso 34