Math 324 Course Notes: Brief description


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1 Brief description These re notes for Mth 324, n introductory course in Mesure nd Integrtion. Students re dvised to go through ll sections in detil nd ttempt ll problems. These notes will be modified nd expnded s the course proceeds. i
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3 Contents Mth 324 Course Notes: Brief description i 1 Lebesgue mesurble functions Wht is mesurble function? Properties of mesurble functions Uniform convergence of sequences of functions Approximtion of mesurble functions by simple functions Almost uniform convergence of sequences of functions Mesurble functions versus continuous functions Lebesgue Integrtion Review of the Riemnn integrl Lebesgue integrl for bounded functions on sets of finite mesure Lebesgue integrls of nonnegtive mesurble functions iii
4 Chpter 1 Lebesgue mesurble functions 1.1 Wht is mesurble function? Let f be n extended rel vlues function defined on domin nd let α R. We introduce the following nottion: (f > α) = {x : f(x) > α} (f α) = {x : f(x) α} (f < α) = {x : f(x) < α} (f α) = {x : f(x) α} (f = α) = {x : f(x) = α} Proposition Suppose f is n extended rel vlued function whose domin is mesurble. The following sttements re equivlent: () For ech α R, the set (f > α) is mesurble. (b) For ech α R, the set (f α) is mesurble. (c) For ech α R, the set (f < α) is mesurble. (d) For ech α R, the set (f α) is mesurble. 1
5 2 CHAPTR 1. LBSGU MASURABL FUNCTIONS Proof. If set is mesurble, so is its complement. This shows tht () = (b). To show tht (b) = (c), we note tht (f < α) = n=1 ( f α 1 n nd countble union of mesurble sets is mesurble. (c) = (d) since the complement of mesurble set is mesurble. To show tht (d) = (), we note tht (f > α) = n=1 ( f α + 1 n ), ), nd countble union of mesurble sets is mesurble. Definition An extended rel vlued function f defined on mesurble subset of R is clled Lebesgue mesurble if it stisfies one of the four sttements of the bove proposition. Remrk Unless otherwise specified, the domin of ny function tht we consider will be subset of R nd the rnge will be relvlued or extended relvlued. Moreover, mesurble function will men Lebesgue mesurble function. Henceforth, for nottionl convenience, we sy tht f M() if f : R is mesurble function. Lemm If f : R is mesurble function, then for ech extended rel number α, the set (f = α) is mesurble. Proof. If α R, then (f = α) = (f α) (f α) nd intersection of two mesurble sets is lso mesurble. Moreover, (f = ) = n=1 (f > n) nd therefore mesurble. xercise Let P be nonmesurble set in (0, 1). Define f : (0, 1) R s follows: { x 2, if x P, f(x) = x 2, if x (0, 1) P. Show tht (f = α) is mesurble for ech extended rel number α, but f is not mesurble function. Thus, the converse of Lemm is flse.
6 1.2. PROPRTIS OF MASURABL FUNCTIONS Properties of mesurble functions xercise Prove tht ny constnt function on mesurble set is mesurble. xercise If f M() nd 1 is mesurble set, then f M( 1 ). Theorem Let f be function defined on 1 2, where 1 nd 2 re mesurble sets. A function f : 1 2 R is mesurble if nd only if f 1 nd f 2, tht is, the restrictions of f to 1 nd 2 re mesurble. Proof. This theorem follows from xercise nd the fct tht for ny α R, ( 1 2 )(f < α) = 1 (f < α) 2 (f < α). xercise If f, g M(), then prove tht the set A(f, g) = {x : f(x) < g(x)} is mesurble. [Hint: For ech rtionl number r, the set {x : f(x) < r < g(x)} is mesurble.] xercise Let f be function defined on mesurble domin. Prove tht f is mesurble if nd only if for ny open subset G of R, f 1 (G) is mesurble. Theorem Suppose f nd g re relvlued mesurble functions on mesurble set nd c is rel number. Then ech of the following functions is mesurble on. () f ± c (b) f + g (c) cf (d) f g
7 4 CHAPTR 1. LBSGU MASURABL FUNCTIONS (e) f (f) f 2 (g) fg (h) f/g, provided g(x) 0 on. Proof. Prts () nd (b) re left s esy exercises. To prove (c), for ny α R, (f + g > α) = {x : f(x) > α g(x)}. Since g is mesurble, so is α g. So, {x : α g(x) < f(x)} = r Q((α g < r) (f > r)), which, being countble union of mesurble sets, is mesurble. Prt (d) follows immeditely from (b) nd (c). To prove (e), observe tht { if α < 0, ( f > α) = (f > α) (f < α) if α 0. Thus, ( f > α) is mesurble for ny rel number α. To prove (f), observe tht { (f 2 if α < 0, > α) = ( f > α) if α > 0. Thus, (f 2 > α) is mesurble for every rel number α. Prt (g) immeditely follows by combining prts (b), (c), (d) nd (f), since fg = 1 4 [(f + g)2 (f g) 2 ]. Prt (h) is left s n exercise. Theorem If f 1 nd f 2 re mesurble functions on, so re the functions f = mx{f 1, f 2 } nd f = min{f 1, f 2 }.
8 1.2. PROPRTIS OF MASURABL FUNCTIONS 5 Proof. We observe tht for ny α R, (f > α) = (f 1 > α) (f 2 > α) M nd (f > α) = (f 1 > α) (f 2 > α) M. xercise Given function f : R, define () Show tht f = f + f. (b) Show tht f = f + + f. f + = mx{f, 0}, nd f = mx{ f, 0}. (c) Show tht f is mesurble on if nd only if f + nd f re mesurble on. Definition For sequence of functions {f n } defined on, we define sup n f n to be the function whose vlue t x is the supremum of {f 1 (x), f 2 (x), }. lim sup f n := inf (sup f k ), n n 1 lim inf n inf n k n f n := sup( f n ), n f n := lim sup( f n ) = sup(inf f k). n n k n Theorem Let f n be mesurble function on for ech n 1. The functions sup n f n, inf n f n, lim sup n f n nd lim inf n f n re lso mesurble on. Proof. For ny α R, (sup f n α) = n (f n > α) M. n=1 The theorem follows immeditely from this.
9 6 CHAPTR 1. LBSGU MASURABL FUNCTIONS Corollry If {f n } is sequence of mesurble functions converging to function f on set, then f is lso mesurble on. Proof. Since f n f, we hve lim sup f n = lim inf f n = f M(). n n A property is sid to hold lmost everywhere on set if the subset of those points in where the property does not hold is of mesure zero. Proposition Let f nd g be functions on mesurble set. If f is mesurble nd f = g lmost everywhere on, then g is lso mesurble on. Proof. Let F = {x : f(x) g(x)}. For ny α R, (g > α) = F (g > α) F C (g > α) = F (g > α) F C (f > α). F (g > α), being subset of set with mesure zero, is mesurble. Since F is mesurble, so is F C. Thus, F C (f > α) = (f > α) F C M. Thus, (g > α), being union of two mesurble sets, is mesurble. This proves tht g is mesurble. 1.3 Uniform convergence of sequences of functions It would be worthwhile for us to review the fundmentl notions of convergence nd uniform convergence of sequences of functions. Definition A sequence of functions {f n } defined on set is sid to converge to function f defined on the sme set if for ech x nd for ech ɛ > 0, there exists positive integer N = N(x, ɛ) such tht f n (x) f(x) < ɛ for ll n N.
10 1.3. UNIFORM CONVRGNC OF SQUNCS OF FUNCTIONS 7 The choice of N(x, ɛ) cn depend on both ɛ s well s x. Definition A sequence of functions {f n } is sid to converge uniformly to function f on set if for every ɛ > 0, there exists positive integer N = N(ɛ) such tht for ll n N, f n (x) f(x) < ɛ for ll x. The choice of N(ɛ) depends only on ɛ nd works for ll points x. Thus, the uniform convergence of {f n } to f is equivlent to the condition tht lim sup f n (x) f(x) = 0. n x xmple Let f(x) = 0 for x [0, 1]. Define, for ech n 1, { x n if 0 x < 1, f n (x) = 0 if x = 1. We cn check tht f n (x) f(x) for ech x [0, 1]. However, for ech n 1, sup f n (x) f(x) = 1. x [0,1] Thus, lim sup f n (x) f(x) 0. n x [0,1] This implies tht the sequence {f n } does not converge uniformly to f on [0, 1]. On the other hnd, it is interesting to observe tht if we chose n ɛ with 0 < ɛ < 1 nd consider the set = [1 ɛ, 1], then, for every n 1, Thus, sup f n (x) f(x) = (1 ɛ) n. x [0,1 ɛ] lim sup f n (x) f(x) = 0. n x Thus, {f n } converges uniformly to f on [1 ɛ, 1]. Remrk We will see in the next section tht the bove exmple is specil cse of the generl phenomenon tht if sequence of mesurble functions converges to mesurble function on set, then this convergence is uniform on lrge subset of. xercise Let {f n } be sequence of rel vlued functions on, ech of which is continuous t point c. Suppose {f n } converges uniformly to function f on. Show tht f is continuous t c.
11 8 CHAPTR 1. LBSGU MASURABL FUNCTIONS 1.4 Approximtion of mesurble functions by simple functions Definition [Simple function] A mesurble function f : R is sid to be simple function if is mesurble nd the imge of f() is finite. Tht is, there exists finite set of rel numbers { 1, 2,, n } such tht for every x, f(x) = j for some 1 j n. Definition [Chrcteristic function] Let be subset of R. The chrcteristic function χ : R of is defined s follows: { 1 if x, χ (x) = 0 otherwise. xercise Prove tht χ is mesurble function if nd only if is mesurble subset of R. xercise A function f : R is simple function if nd only if there is finite number of disjoint mesurble sets 1, 2,, n such tht = n i=1 i nd finite set { 1, 2,, n } of rel numbers such tht f(x) = n i χ i (x) for ech x. i=1 [The bove representtion is often referred to s the cnonicl representtion of simple function f.] xercise Let f, g M() be simple functions nd c be rel number. Show tht f + g nd cf re simple functions. Theorem Let f be nonnegtive mesurble function on set. Then there exists sequence {φ k } of simple functions, which re nonnegtive nd monotoniclly incresing nd converge to f t every point x. Proof. For ech k 1, we prtition the intervl [0, ] into disjoint intervl s follows: [0, ] = [0, 1) [1, 2) [k 1, k) [k, ]
12 1.4. APPROXIMATION OF MASURABL FUNCTIONS BY SIMPL FUNCTIONS9 k2 k = i=1 [ ) i 1 2, i [k, ]. k 2 k Define, for ech 1 i k2 k, ([ )) i 1 ki = f 1 2, i k 2 k nd k = f 1 ([k, ]). Define the function φ k (x) s follows: k2 k i=1 i 1 2 k χ k,i + kχ k. If f(x) <, then there exists positive integer k such tht f(x) < k. Since x k,i for some 1 i k2 k, we hve 0 f(x) φ k (x) i 2 k i 1 2 k = 1 2 k. If f(x) =, then φ k (x) = k for every k. Thus, for ech x, lim φ k(x) = f(x). k This gives us sequence φ k (x) of simple functions on converging to f(x). We now show tht φ k (x) φ k+1 (x) for every integer x. To do so, notice tht for ech 1 i k2 k, ([ 2i 2 k+1,2i 1 k+1,2i = f 1 2i 1, 2k+1 2 k+1 )) = f 1 ([ i 1 2 k, i 2 k If x k,i for some 1 i k2 k, then )) ([ f 1 2i 1 2, 2i k+1 2 k+1 = ki. )) On the other hnd, φ k (x) = i 1 2 k. φ k+1 (x) = 2i 2 2 k+1 or 2i 1 2 k+1,
13 10 CHAPTR 1. LBSGU MASURABL FUNCTIONS both of which re i 1 2. k If x k then φ k (x) = k. But, If x f 1 ([k, k + 1]), then f 1 ([k, ]) = f 1 ([k, k + 1]) f 1 ([k + 1, ]). φ k+1 (x) = (k+1)2 k+1 i=k.2 k+1 +1 i 1 2 k+1 χ k,i k2k+1 2 k+1 = φ k(x). If x f 1 ([k + 1, ]), then φ k+1 (x) = k + 1 > k = φ k (x). thus, we hve shown tht for every x, φ k+1 (x) φ k (x). xercise Let f be mesurble function defined on mesurble set. Then there exists sequence of simple functions on which converge to f. If f is bounded on, then this convergence is uniform on. 1.5 Almost uniform convergence of sequences of functions In this section, we prove theorem of goroff, which explins the generl phenomenon tht if sequence of mesurble functions converges to mesurble function on set, then this convergence is uniform on lrge subset of. We strt with the following proposition: Proposition Let be mesurble set with finite mesure nd {f n } be sequence of mesurble functions defined on. Let f be rel vlued mesurble function such tht f n (x) f(x) for ech x. Given ɛ > 0 nd δ > 0, we cn find mesurble subset A of with m(a) < δ nd n integer N such tht f n (x) f(x) < ɛ for ll x A nd ll n N.
14 1.5. ALMOST UNIFORM CONVRGNC OF SQUNCS OF FUNCTIONS11 Proof. For ech n 1, let G n = {x : f n (x) f(x) ɛ}. Since f n nd f re mesurble functions, so re the functions f n f. Thus, G n is mesurble. For ech k 1, define k = G n. Tht is, n=k k = {x : f n (x) f(x) ɛ for some n k}. Clerly, k+1 k. Moreover, since {f n } converges to f for ech x, there must be some positive integer k such tht x k. Thus, nd k k+1 k = ϕ. n=1 Thus, lim m( n) = 0. n [Note tht it is here tht we use the fct tht m( k ) < m() <.] Tht is, given δ > 0, there exists nturl number N such tht m( k ) < δ for ll k N. In prticulr, m( N ) < δ. Moreover, for ech x N, we hve f n (x) f(x) < ɛ for ll n N. xercise Prove the following modifiction of the bove theorem: Let be mesurble set with finite mesure nd {f n } be sequence of mesurble functions defined on. Let f be rel vlued mesurble function such tht f n (x) f(x) lmost everywhere in. Given ɛ > 0 nd δ > 0, we cn find mesurble subset A of with m(a) < δ nd n integer N such tht f n (x) f(x) < ɛ for ll x A nd ll n N. Note tht the condition m() < cnnot be relxed. n 1, { 0 if 0 x n f n (x) = 1 if x > n. Define, for ech One cn check tht the sequence {f n } converges to f on, but the theorem does not hold for ɛ = 1 nd some 0 < δ < 1.
15 12 CHAPTR 1. LBSGU MASURABL FUNCTIONS Definition A sequence {f n } of mesurble functions on set is sid to converge lmost uniformly to mesurble function f on if for ech ɛ > 0, there exists mesurble set A with m(a) < ɛ such tht {f n } converges to f on A. xercise Suppose {f n } converges to f lmost uniformly on. Show tht {f n } converges to f lmost everywhere on. The converse of the sttement in the bove exercise forms the content of very importnt theorem due to goroff. Theorem [goroff s theorem] Let be mesurble set with m() < nd let {f n } be sequence of functions which converge to f lmost everywhere on. Then, the sequence {f n } converges lmost uniformly to f on. Proof. Let η > 0 nd for ech k 1, choose ɛ(k) = 1 k nd δ k = η 2 k. Tking = 0 nd repetedly pplying xercise 1.5.2, we cn find mesurble set A k k 1 nd positive integer N k such tht m(a k ) < δ k nd f n (x) f(x) < 1 k for ll n N k nd for ll x k = k 1 A k. Set Note tht m(a) One cn esily check tht A = A k. k=1 m(a k ) < k=1 A = k=1 k. k=0 η 2 k = η. If x A, then x k for every k 1. For ny ɛ > 0, there exists k 1 such tht ɛ < 1 k. We know from our construction tht there exists N k 1 such tht f n (x) f(x) < 1 k < ɛ for ll n N k. Thu, {f n } f uniformly on A.
16 1.6. MASURABL FUNCTIONS VRSUS CONTINUOUS FUNCTIONS Mesurble functions versus continuous functions We lernt in nlysis tht to sy tht function is continuous is to sy something bout the topologicl properties of inverse imges of open sets under f. More precisely, continuous function pulls bck n open set to n open set. Anlogously, mesurble functions tell us bout the mesurbility of inverse imges. How is the notion of continuity relted to tht of mesurbility? To nswer this question, we observe the following: Theorem Continuous functions defined on mesurble sets re mesurble functions. Proof. Suppose f M(). Since f is continuous, for ny α R, f 1 ((, α)) being the inverse imge of the open set (, α) is n open set nd therefore mesurble. Thus (f < α) = f 1 ((, α)) is the intersection of two mesurble sets nd therefore mesurble. This proves the theorem. Is the converse of the bove theorem true? If not, how close is it to being true? xercise Show tht the following functions re mesurble on their respective domins. () (b) f(x) = x 2 if x < 1, f(x) = 2 if x = 1, 2 x if x > 1. { 1, if x is rtionl number in [0, 1] 0, if x is n irrtionl number in[0, 1].
17 14 CHAPTR 1. LBSGU MASURABL FUNCTIONS Notice tht in the bove exercise, the function defined in () is continuous ll except finitely mny points, where s the function defined in (b) is continuous nowhere. Nonetheless, both re mesurble. In the next exercise, we prove tht every mesurble function is lmost continuous. xercise Let f be mesurble function defined on. Then, for ech ɛ > 0, there exists closed set F with m( F ) < ɛ such tht f is continuous on F. [Hint: Use Problem 2 from Clss Test 1, Theorem nd Theorem ]
18 Chpter 2 Lebesgue Integrtion 2.1 Review of the Riemnn integrl We strt by reviewing the Riemnn integrl vi step functions. Definition A function φ : [, b] R is clled step function if there is prtition = x 0 < x 1 < x n = b of the intervl [, b] such tht phi(x) is constnt function on ech subintervl (x k 1, x k ), 1 k n. Let c 1, c 2,, c n be rel numbers such tht φ(x) = c k for x (x k 1, x k ). The Riemnn integrl of φ on [, b] is defined s φ(x)dx = n c k (x k x k 1 ). k=1 Remrk The vlues tken by φ(x) on the points x 0, x 1, x n hve no effect on vlue of the Riemnn integrl of φ on [, b]. Remrk The vlue of the Riemnn integrl of step function is independent of the choice of the prtition [, b] s long s the step function is constnt on the open subintervls of the prtition. The, the Riemnn integrl of step function is well defined. The following properties of Riemnn integrls of step functions cn be esily verified: 15
19 16 CHAPTR 2. LBSGU INTGRATION Lemm Let φ, ψ be step functions on [, b] nd c R. Then () cφ is step function on [, b] nd cφ(x)dx = c φ(x)dx. (b) φ + ψ is step function on [, b] nd (φ + ψ)(x)dx = φ(x)dx + ψ(x)dx. (c) If φ ψ on [, b], then φ(x)dx ψ(x)dx. (d) If < c < b, then φ(x)dx = c φ(x)dx + c φ(x)dx. Definition Let f(x) be bounded function on [, b], tht is, let α, β be rel numbers such tht α f(x) β for ll x [, b]. The lower Riemnn integrl of f on [, b] is defined s { } f(x)dx := sup φ(x)dx : φ(x) f(x), φ is step function nd the upper Riemnn integrl of f on [, b] is defined s (2.1) { f(x)dx := inf } ψ(x)dx : f(x) ψ(x), ψ is step function (2.2) Remrk Since α f, the set { } φ(x)dx : φ(x) f(x), φ is step function
20 2.1. RVIW OF TH RIMANN INTGRAL 17 is not empty since it contins the function φ(x) = α. Moreover, since φ f β, φ(x)dx βdx = β(b ). Thus, the set is bounded bove. Hence the lower Riemnn integrl in eqution (2.1) is well defined. By nlogous resoning, the upper Riemnn integrl in eqution (2.2) is lso well defined. xercise If f is bounded function on [, b] nd φ, ψ re step functions on [, b] such tht φ f ψ, then φ(x)dx f(x)dx f(x)dx ψ(x)dx. Definition A bounded function f on [, b] is sid to be Riemnn integrble on [, b] if f(x)dx = f(x)dx. The Riemnn integrl of f on [, b] is denoted t f(x)dx nd equls the common vlue bove, tht is f(x)dx = f(x)dx f(x)dx. xercise Check tht the definition for Riemnn integrl for step function in Definition coincides grees with Definition It is not lwys fesible to check the Riemnn integrbility of function by checking the defining property. It is much esier to pply n equivlent condition which essentilly sys tht bounded function f on [, b] is Riemnn integrble if nd only if we cn pproximte f from bove nd below by step functions whose integrls cn be mde rbitrrily close to ech other. Proposition A bounded function f on [, b] is Riemnn integrble if nd only if for ech ɛ > 0, we hve step functions φ nd ψ on [, b] such tht φ f ψ nd 0 [ψ(x) φ(x)]dx < ɛ.
21 18 CHAPTR 2. LBSGU INTGRATION Proof. Assume tht the bounded function f is Riemnn integrble on [, b] nd let ɛ > 0. We cn find step function φ such tht φ f on [, b] nd f(x)dx ɛ 2 < φ(x)dx. Similrly, we cn find step function ψ on [, b] such tht f ψ nd f(x)dx + ɛ 2 > ψ(x)dx. Combining the bove inequlities with the Riemnn integrbility of f, we hve f(x)dx ɛ 2 = f(x)dx ɛ 2 < φ(x)dx f(x)dx ψ(x)dx < Thus, since φ f ψ, we hve 0 f(x)dx + ɛ 2 = f(x)dx + ɛ 2. [ψ(x) φ(x)]dx < ɛ. Conversely, suppose ɛ > 0 nd there re step functions φ f ψ such tht 0 By xercise 2.1.7, we hve Thus, φ(x)dx 0 Since this is true for ny ɛ > 0, [ψ(x) φ(x)]dx < ɛ. f(x)dx f(x)dx f(x)dx = Thus, f is Riemnn integrble. f(x)dx f(x)dx < ɛ. f(x)dx. ψ(x)dx.
22 2.1. RVIW OF TH RIMANN INTGRAL 19 xercise The im of the following exercises is to prove tht every continuous function on [, b] is Riemnn integrble. () Prove tht f(x) is bounded on [, b]. (b) Prove tht f is uniformly continuous on [, b]. Thus, given ɛ > 0, there exists δ > 0 such tht f(x) f(y) < for ll x, y [, b] for which x y < δ. (c) Tke prtition ɛ b = x 0 < x 1 < x 2 < x n = b such tht x k x k 1 < δ for ll 1 k n. Show tht for ech such k, sup f inf f ɛ (x k 1,x k ) (x k 1,x k ) b. (d) Define step functions φ(x) nd ψ(x) such tht φ(x) = inf f(x), x (x k 1, x k ) (x k 1,x k ) nd Show tht ψ(x) = 0 sup f(x), x (x k 1, x k ). (x k 1,x k ) [ψ(x) φ(x)]dx ɛ. (e) Deduce tht f is Riemnn integrble on [, b]. Cn discontinuous functions be Riemnnintegrble? It is not too difficult to see tht function on [, b] with finite number of discontinuities is lso Riemnn integrble. But, how much cn we weken continuity for function to be Riemnn integrble? By Proposition , we observe tht the difference ψ φ is n indiction of how much f cn vry on subintervl.
23 20 CHAPTR 2. LBSGU INTGRATION Definition Let f be bounded function on n intervl [, b]. For ny intervl J, the oscilltion of f on J is defined to be the number ω f (J) := sup{f(x) : x J [, b]} inf{f(x) : x J [, b]}. For x 0 [, b], the oscilltion of f t x 0 is defined s ω f (x 0 ) := inf{ω f (I) : I is n open intervl contining x 0 }. Lemm Let f be s bove. Then f is continuous t x 0 [, b] if nd only if ω f (x 0 ) = 0. Proof. Suppose f is continuous t x 0 nd let ɛ > 0. We cn find δ > 0 such tht for x (x 0 δ, x 0 + δ) [, b], we hve Tht is, f(x) < f(x 0 ) + ɛ. Thus, ɛf(x) f(x 0 ) < ɛ. sup{f(x) : x (x 0 δ, x 0 + δ) [, b]} f(x 0 ) + ɛ. Similrly, since f(x) > f(x 0 ) ɛ, Thus, inf{f(x) : x (x 0 δ, x 0 + δ) [, b]} f(x 0 ) ɛ. We hve shown tht for ny ɛ > 0, ω f ((x 0 δ, x 0 + δ) [, b]) 2ɛ. ω f (x 0 ) ω f ((x 0 δ, x 0 + δ) [, b]) 2ɛ. Thus, ω f (x 0 ) = 0. Conversely, suppose ω f (x 0 ) = 0 nd let ɛ > 0. Then here exists n open intervl I contining x 0 such tht 0 ω f (I) < ɛ. Since I is open, choose δ > 0 such tht (x 0 δ, x 0 + δ) I. Thus, there exists δ > 0 such tht if x (x 0 δ, x 0 + δ) [, b], then ɛ < f(x) f(x 0 ) < ɛ.
24 2.1. RVIW OF TH RIMANN INTGRAL 21 Lemm For n 1, the set D n = {x [, b] : ω f (x) 1 n } is closed set. Proof. We will show tht Dn C is open. Let x Dn C. Tht is, ω f (x) < 1. Thus, there exists n open intervl I n contining x so tht ω f (I) < 1. Since I is open, there exists δ > 0 such tht n (x δ, x + δ) [, b] I [, b]. Let We hve z (x δ, x + δ) [, b]. ω f (z) ω f ((x δ, x + δ)) ω f (I) < 1 n. Tht is, z D C n. This proves tht D C n is open. Theorem [Lebesgue s crieterion for Riemnn integrbility] Let f be bounded function on [, b] nd let D denote the set of discontinuities of f. Then f is Riemnn integrble on [, b] if nd only if D hs mesure zero. Proof. We first prove sufficiency. Suppose m(d) = 0. Tht is, m(d n ) = 0 for every n 1. Let f < B on [, b]. Let ɛ > 0. Since m(d n ) = 0 nd D n is compct subset of [, b], we hve finite collection of open sets {I ki } such tht Thus, the set D n m i=1 I ki, i [, b] l(i ki ) < m i=1 I ki ɛ 4B. is finite union of closed intervls J 1, J 2,, J L. tht is, [, b] = J 1 J 2 J L I k1 I km.
25 22 CHAPTR 2. LBSGU INTGRATION Observe tht since ll points of D n re in I k1 I km, we hve ω f (x) < 1 n, x J 1 J 2 J L. Thus, there exists n open intervl I x contining x such tht ω f (I x ) < 1. So, n {I x } x Ji is n open cover of the closed nd bounded nd therefore compct set J i. Thus, finite subcover must cover J i. Thus, we hve prtition of J i nd sup f inf f < 1 I x I x n on ny of the subintervls of this prtition. We do this for ech J i. Thus, we hve finite collection of subintervls {J i} L i=1 whose union is J 1 J 2 J L such tht on ny of these subintervls, sy J i, ω f (J i) < 1 n. Define Then, φ = inf f on I k1, I k2, I km, J 1, J 2, J L, ψ = sup on I k1, I k2, I km, J 1, J 2, J L. f [ψ(x) φ(x)]dx 1 l(j n i) + 2B i i 1 ɛ (b ) + 2B n 4B < ɛ, l(i ki ) for n sufficiently lrge. So, f is Riemnn integrble on [, b]. Conversely, suppose f is Riemnn integrble on [, b]. We would like to show tht m(d n ) = 0 for ech n 1. Consider D N nd let ɛ > 0. It will be sifficient to show tht m(d N ) < ɛ.
26 2.1. RVIW OF TH RIMANN INTGRAL 23 with Let Since f is Riemnn integrble on [, b], we hve step functions φ f ψ [ψ(x) φ(x)]dx < ɛ N. = x 0 < x 1 < < x n = b be prtition ssocited with φ nd ψ nd split the collection {(x 0, x 1 ), (x 1, x 2 ),, (x n 1, x n )} into two collections D 1 nd D 2 s follows: If (x k 1, x k ) D N φ, then put (x k 1, x k ) in D 1. Otherwise, put this subintervl in D 2. very point of D N is either in D 1 or member of the set So, {, x 1,, x n 1, b}. [ψ(x) φ(x)]dx = D 1 (ψ φ)(x)(x k x k 1 )+ D 2 (ψ φ)(x)(x k x k 1 ) < ɛ N. For D 1, some point of D N is in ech subintervl. Tht is, ech subintervl in D 1 contins point x with ω f (x) 1. Thus, in these subintervls N So, Tht is, ψ φ sup inf 1 f f N. ɛ N > (x k x k 1 ). D 1 (x k 1,x k ) D 1 (x k 1,x k ) D 1 (x k x k 1 ) < ɛ. These intervls of D 1 long with {, x 1,, x n 1, b} contin ll points of D N. Thus, m(d N ) < ɛ.
27 24 CHAPTR 2. LBSGU INTGRATION xercise Let f nd g be bounded, Riemnnintegrble functions on [, b] nd k R. Prove the following sttements: () (kf)(x)dx = k f(x)dx. (b) (f + g)(x)dx = f(x)dx + g(x)dx. (c) If f(x) g(x), then f(x)dx g(x)dx. (d) If < c < b, then f is Riemnn integrble on [, c] nd [c, b] nd f(x)dx = (e) If α f(x) β on [, b], then α(b ) c f(x)dx + c f(x)dx. f(x)dx β(b ). xercise For 0 x 1, let { 1, if x Q, f(x) = 0, otherwise. Let r 1, r 2,, r k, be n enumertion of rtionl numbers. 0 x 1, { 1, if x = r 1, r 2,, r k f k (x) = 0, otherwise. Define, for () Prove tht f k (x) converges to f(x) on [0, 1] s k, but not uniformly. (b) vlute 1 0 f k(x)dx for ech k 1. (c) Is it true tht 1 0 f(x)dx = lim k 1 0 f k (x)dx?
28 2.1. RVIW OF TH RIMANN INTGRAL 25 xercise Suppose {f k (x)} is sequence of Riemnn integrble functions on [, b]. If f k converges to function f on [, b], prove tht () f is Riemnn integrble on [, b]. (b) Prove tht f(x)dx = lim k f k (x)dx. Theorem If f is continuous on [, b], differentible on (, b) nd if the derivtive f of f is Riemnn integrble on [, b], then f (x)dx = f(b) f() Proof. Since f is Riemnn integrble on [, b], we hve, for every ɛ > 0, step functions φ, ψ such tht nd φ(x)dx f (x)dx (ψ(x) φ(x))dx < ɛ. ψ(x)dx Tke the common prtition ssocited with φ nd ψ, sy, { = x 0, x 1, x n = b}. Suppose n φ(x) = c i χ (xi 1,x i ) nd ψ(x) = i=1 n d i χ (xi 1,x i ). i=1 We know, by the men vlue theorem, tht f(b) f() = n f(x i ) f(x i 1 ) = i=1 n f (θ i ) i=1 for some x i 1 θ i x i.
29 26 CHAPTR 2. LBSGU INTGRATION Since c i f d i on (x i 1, x i ), we hve c i (x i x i 1 ) f(b) f() i i d i (x i x i 1 ), tht is, We lredy know tht φ(x)dx f(b) f() Thus, for ny ɛ > 0, f(b) f() φ(x)dx f (x)dx f (x)dx ψ(x)dx. ψ(x)dx. φ(x) ψ(x)dx < ɛ. This implies, f (x)dx = f(b) f(). 2.2 Lebesgue integrl for bounded functions on sets of finite mesure Recll from Section 1.4 tht function φ whose domin is mesurble set is clled simple function provided there exist rel numbers c 1, c 2, c n nd disjoint mesurble subsets { k } of such tht = n k=1 k nd φ(x) = n c k χ k (x). k=1
30 2.2. LBSGU INTGRAL FOR BOUNDD FUNCTIONS ON STS OF FINIT MASUR27 xercise Prove tht every step function is simple function. Is it true tht every simple function is step function? Definition Suppose φ is simple function on mesurble set with m() <, tht is, n φ(x) = c k χ k (x), k=1 where c k s nd k s re s bove. The Lebesgue integrl of φ on is defined s n φ = c k m( k ). xercise Suppose = k=1 n i = i=1 where i s re mesurble nd mutully disjoint nd so re F j s. Suppose φ(x) is simple function such tht φ(x) = c i for every x i nd φ(x) = d j for every x F j. Prove tht m j=1 F j n c i m( i ) = i=1 m d j m(f j ). j=1 This shows tht the Lebesgue integrl of simple function is well defined. xercise Let f(x) be s in xercise vlute [0,1] f. xercise Let φ nd ψ be simple functions on nd m() <. Let k R. Prove the following sttements: () kφ is simple function on nd k φ = (b) φ + ψ is simple function on nd (φ + ψ) = kφ. φ + ψ.
31 28 CHAPTR 2. LBSGU INTGRATION (c) If φ ψ on, then φ (d) Let = 1 2, 1 nd 2 being disjoint nd mesurble. Then φ = φ + 1 φ. 2 Definition Let f be bounded function on mesurble set, m() <. The lower Lebesgue integrl of f on is defined s f = sup{ φ : φ f, φ simple function} nd the upper Lebesgue integrl of f on is defined by f = inf{ ψ : f ψ, ψ simple function}. xercise Show tht the bove notions re well defined. Show tht if φ nd ψ re simple functions on, then φ f f ψ. Definition A bounded function f defined on mesurble set with finite mesure is sid to be Lebesgue integrble on if f = f. The common vlue is clled the Lebesgue integrl of f on nd is denoted by f. Theorem Let f be bounded function on [, b]. If f is Riemnn integrble on [, b], then f is Lebesgue integrble on [, b] nd f(x)dx = f. [,b] ψ.
32 2.2. LBSGU INTGRAL FOR BOUNDD FUNCTIONS ON STS OF FINIT MASUR29 Proof. f(x)dx = sup{ = sup{ sup{ = [,b] = inf{ inf{ = inf{ = [,b] [,b] f [,b] [,b] f(x)dx. Since, f is Riemnn integrble, φ(x)dx : φ f, φ step function} φ : φ f, φ step function} φ : φ f, φ simple function} [,b] f ψ : f ψ, ψ simple function} ψ : f ψ, ψ step function} ψ(x)dx : f ψ, ψ simple function} f(x)dx = f(x)dx. Thus, f = f. [,b] [,b] xercise Show tht bounded function f defined on mesurble set with finite mesure is Lebesgue integrble if nd only if for every ɛ > 0, we hve simple functions φ f ψ on such tht 0 ψ φ < ɛ.
33 30 CHAPTR 2. LBSGU INTGRATION Theorem Let f be bounded function on set with finite mesure. Then f is Lebesgue integrble on if nd only if f is mesurble on. Proof. Let f be bounded, mesurble function on. Let m f(x) < M on nd n 1. For ech 1 k n, define { k = x : m + k 1 n (M m) f(x) < m + k } (M m). n Since f is mesurble, ech k is mesurble. Moreover, { k } n k=1 re disjoint subsets of nd = k k. We now define step functions φ n nd ψ n on s follows: φ n (x) = m + k 1 n (M m), x k, Clerly, φ n (x) f(x) ψ n (x) nd (ψ n (x) φ n (x)) = ψ n (x) = m + k n (M m), x k. n k=1 M m n m( k ) = M m (b ). n Thus, by xercise , f is Lebesgue integrble on. Conversely, suppose f is bounded nd Lebesgue integrble on. Let n 1. By xercise , we hve simple functions φ n nd ψ n on such tht φ n f ψ n nd (ψ n φ n ) < 1 n. Define φ = sup{φ 1, φ 2, }, ψ = sup{ψ 1, ψ 2, }. We observe tht φ nd ψ re mesurble functions on nd for ech n 1, φ n φ f ψ ψ n. (2.3) If we cn show tht f = φ lmost everywhere on, then by Proposition , we deduce tht f is mesurble on. By virtue of the inequlity in
34 2.2. LBSGU INTGRAL FOR BOUNDD FUNCTIONS ON STS OF FINIT MASUR31 eqution (2.3), it is sufficient to show tht φ = ψ lmost everywhere on. For positive integer l, define Clerly, for every n 1, 1 n > (ψ n φ n ) > Thus, 1 (l) = {x : ψ n φ n > 1 l }. 1 (l) (ψ n φ n ) > 1 l m( 1(l)). m( 1 (l)) < l for ll n 1. n Thus, m( 1 (l)) = 0. By the inequlity in eqution (2.3), we hve {x : ψ(x) φ(x) > 0} = {x : ψ(x) φ(x) > 1 l } l=1 l 1 (l), n 1. Since m( 1 (l)) = 0, we hve This proves the theorem. m ({x : ψ(x) φ(x) > 0}) = 0. xercise Let f be bounded, Lebesgue integrble function on mesurble set with finite mesure. Let g be bounded function on nd g = f lmost everywhere on. Show tht g is Lebesgue integrble on. The following theorem encpsultes some importnt properties of Lebesgue integrls of bounded functions on sets of finite mesure. Theorem Let f nd g be bounded nd mesurble functions defined on set of finite mesure nd let k be rel number. Then () kf is Lebesgue integrble on nd kf = k f.
35 32 CHAPTR 2. LBSGU INTGRATION (b) f + g is Lebesgue integrble on nd (f + g) = f + (c) If f = g lmost everywhere on, then f = g. (d) If f g lmost everywhere on, then f g. Thus, (e) If α f(x) β, then αm() f f. g. f βm(). (f) If 1, 2 re disjoint, mesurble subsets of, then f is Lebesgue integrble on 1 nd 2 nd f = 1 2 f + 1 Proof. Prts (), (b), (d), (e), (f) re left s exercises. To prove (c), let ψ be simple function on such tht ψ f g. Since f g = 0 lmost everywhere on, we get tht ψ 0 lmost everywhere on. Let 1 = {x : ψ(x) 0}. 2 f By xercise (b), we hve ψ ψ = 0, since m(1 C ) = 0. C 1
36 2.2. LBSGU INTGRAL FOR BOUNDD FUNCTIONS ON STS OF FINIT MASUR33 Thus, (f g) = inf{ ψ : ψ simple ndψ f g} 0. Similrly, (g f) 0, tht is, by (), (f g) 0. We conclude tht (f g) = 0. xercise With f nd s bove, if m() = 0, then show tht f = 0. xercise If f(x) = r lmost everywhere on, show tht f = km(). Thus, if f(x) = 0 lmost everywhere on, then f = 0. Remrk Converse of the sttement in the bove theorem is not true. There exist functions which re not zero nywhere on, but f = 0. For exmple, for 1 x 1, let f(x) = 1 if x 0 nd 1 otherwise. Then f = 0. [ 1,1] Theorem If f = 0 nd f(x) 0 on, then f = 0 lmost everywhere on. Proof. Suppose hs subset A where f(x) > 0. Tht is, A = {x : f(x) > 1 n }. n=1 Let 1 (n) = { x : f(x) > 1 }. n If possible, suppose there is positive integer N such tht m( 1 (N)) > 0. Then f f 1 N m( 1(N)) > 0, 1 (N) which contrdicts the fct tht f = 0. Thus,m( 1(n)) = 0 for ll n 1. This proves the theorem. We now ddress the issue of interchnging limit nd integrl opertions with respect to the Lebesgue integrl.
37 34 CHAPTR 2. LBSGU INTGRATION Theorem [Bounded Convergence Theorem] Let {f n } be sequence of mesurble functions defined on mesurble set of finite mesure. Suppose there exists nonnegtive rel number M such tht f n (x) M for ll x nd ll n. If f(x) = lim n f n (x) for ech x, then f is Lebesgue integrble on nd f = lim f. n n Proof. Since f is limit of mesurble functions, f is mesurble. Being bounded nd mesurble function on, it is Lebesgue integrble. By Proposition 1.5.1, we know tht given ɛ > 0, there exists mesurble subset A of with m(a) < ɛ 4M nd positive integer N 1 such tht f n (x) f(x) < ɛ 2m() + 1 Since f n (x) M for ll n N nd x, Thus, f n = < for ll n N. Thus, f n (x) f(x) 2M,. A for ll n N, x A. f = f n f f n f f n f + f n f ɛ m( A) + 2M ɛ 2m() + 1 A lim f n = f. n 4M < ɛ
38 2.3. LBSGU INTGRALS OF NONNGATIV MASURABL FUNCTIONS35 xercise Let f n nd f stisfy the sme conditions s in the theorem bove nd f n f lmost everywhere on. Show tht f is Lebesgue integrble on nd f = lim f. n n Remrk The limit function of convergent sequence of Riemnn integrble functions my not be Riemnn integrble, but the Bounded Convergence Theorem tells us tht this limittion does not hold in the cse of bounded, Lebesgue mesurble (k integrble) functions on sets of finite mesure. xercise Let {f n } be sequence of Riemnn integrble functions defined on [, b] such tht f n (x) M for ll n nd x [, b]. Show tht if {f n } converges lmost everywhere to Riemnn integrble function f on [, b], then f(x)dx = lim n f n (x)dx. 2.3 Lebesgue integrls of nonnegtive mesurble functions Definition Let φ be nonnegtive simple function on mesurble set. Tht is, there exist mutully disjoint mesurble sets { i } n i=1 such tht = n i=1 i nd nonnegtive rel numbers c i, 1 i n such tht φ(x) = n c i χ i (x). i=1 The Lebesgue integrl of φ on is defined s n φ = c i m( i ). Observe tht 0 φ. i=1 Remrk One cn redily verify tht the Lebesgue integrl for simple functions stisfies ll properties listed out in xercises nd lso hold when m() =.
39 36 CHAPTR 2. LBSGU INTGRATION Definition Let f(x) be nonnegtive, mesurble function on mesurble set. Recll, from Theorem 1.4.6, tht there exists sequence of nonnegtive, monotoniclly incresing simple functions 0 φ 1 (x) φ 2 (x) φ n (x) φ n+1 (x) + f(x) such tht lim φ n(x) = f(x), x. n The Lebesgue integrl of f on is defined s f = lim φ n (x). n f is nonnegtive member of the extended rels. xercise Suppose φ n nd ψ n re two sequences of nonnegtive, monotoniclly incresing simple functions converging to f on. Show tht lim φ n = lim ψ m. n m This shows tht the bove definition for the Lebesgue integrl of nonnegtive, mesurble function is independent of the choice of the sequence of simple functions tht converge to the function. Solution. Let us choose n 1 nd observe tht 0 φ n f = lim m ψ m. Suppose we hve mutully disjoint mesurble sets { i } N i=1 such tht = N i=1 i nd nonnegtive rel numbers c i, 1 i N such tht φ n = N c i m( i ). i=1 By xercise 2.2.5(d), which shows the dditivity of the integrl over disjoint mesurble components of the domin, ψ m = N i=1 i ψ m.
40 2.3. LBSGU INTGRALS OF NONNGATIV MASURABL FUNCTIONS37 We first show tht lim m ψ m c i m( i ). i (2.4) If c i = 0, the bove follows immeditely. if c i > 0, let us choose α R such tht 0 < α < 1. For m 1, let B m = {x i : ψ m (x) αc i }. Clerly B m is mesurble for every m 1. Since ψ m ψ m+1, we hve Moreover, for ny x i, Thus, nd B m B m+1. lim ψ m φ n (x) = c i. m i = m=1 B m lim m(b m) = m( i ). m But, since B m i, we hve ψ m ψ m α i m(b m ). k B m Thus, lim ψ m lim αc im(b m ) = αc i m( i ). m m i This holds for ny 0 < α < 1. We get qution (2.4). From this, we deduce lim ψ m lim ( N (c i m( i )) = m m As the bove is true for every n 1, lim m i=1 ψ m lim φ n. n φ n.
41 38 CHAPTR 2. LBSGU INTGRATION By similr resoning, we cn show lim n φ n lim m xercise Let f be nonnegtive, mesurble function on mesurble set. Show tht f = sup{ φ : φ nonnegtive simple function, φ f}. ψ m. [Hint: Use the rgument to prove qution (2.4).] xercise Let f nd g be nonnegtive mesurble functions on mesurble set nd c be nonnegtive rel number. Show tht () cf = c f. (b) If 0 f g, then (c) (f + g) = f + g. 0 f g. (d) If 1 nd 2 re two disjoint, mesurble subsets of, then f = 1 2 f + 1 f. 2 Definition is techniclly cumbersome wy to clculte Lebesgue integrls. We look for esier wys to clculte Lebesgue integrls. For exmple, suppose we wnt to evlute 1 x dx. 2 [1, ) Insted of pproximting by monotone sequence of simple functions, let us pproximte f(x) = 1/x 2, x 1 by monotone sequence of nonnegtive mesurble functions { 1, 1 x n, x f n (x) = 2 0, x > n.
42 2.3. LBSGU INTGRALS OF NONNGATIV MASURABL FUNCTIONS39 Then lim f n(x) = f(x). n If we re llowed to interchnge limit nd integrl, then ( f(x) = lim f n = lim 1 [1, ) n [1, ) 1 ) = 1. n n We ddress the issue of interchngebility of integrl nd limit opertions. Theorem [Lebesgue Monotone Convergence Theorem] Let {f k } be monotoniclly incresing sequence of nonnegtive functions on mesurble set nd let lim k f k = f. Then, f = lim f k = lim f k. k k Proof. Recll, from xercise tht f = sup{ φ : φ nonnegtive simple function, φ f}. We first observe tht f is nonnegtive, mesurble function on. Since, {f k } is monotoniclly incresing sequence of nonnegtive functions, by xercise 2.3.6, we hve 0 f k f k+1 f. Thus, lim f k f. k We now show tht f lim f k. k let φ be ny simple function such tht 0 φ f. If we re ble to show tht φ lim f k,
43 40 CHAPTR 2. LBSGU INTGRATION this would imply tht lim f k is n upper bound of the set { φ : φ nonnegtive simple function, φ f}. Thus, In order to show tht we observe tht φ = f lim f k. k φ lim f k, N c i m( i ), i =, c i 0. i=1 Since f k is dditive on, it is sufficient to show tht lim k N i=1 i f k N c i m( i ). i=1 To do so, we show tht for ech 1 i N, lim k i f k c i m( i ). This cn be done by repeting the sme rgument used to prove eqution (2.4), by simply replcing φ m by f k. This proves the theorem.
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