NOTES AND PROBLEMS: INTEGRATION THEORY


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1 NOTES AND PROBLEMS: INTEGRATION THEORY SAMEER CHAVAN Abstrct. These re the lecture notes prepred for prticipnts of AFSI to be conducted t Kumun University, Almor from 1st to 27th December, Contents 1. Spce of Riemnn Integrble Functions 1 2. Spce of Lebesgue Integrble Functions 2 3. Some Dense subspces of L Spce of Functions of Bounded Vrition 5 5. Riesz Representtion Theorems 7 References Spce of Riemnn Integrble Functions In these notes, we will be concerned bout the spce L 1 (X) of integrble functions on X nd two of its subspces: the subspce C c (X) of continuous functions with compct support nd the subspce R[, b] of Riemnn integrble functions in cse X = [, b]. If not specified, then X = R d or more generlly mesurble subset of R d of positive mesure. It turns out tht the subspce C[, b] := C c [, b] of L 1 [, b] is properly contined in the spce R[, b] of Riemnn integrble functions on [, b]. Indeed, by result of Lebesgue, f : [, b] R is Riemnn integrble if nd only if f is continuous.e. (for proof, see [4, Problem 4, Chpter 1]). Moreover, R[, b] is subspce of L 1 [, b] s shown below. Theorem 1.1. Every Riemnn integrble function on [, b] is Lebesgue integrble. Moreover, the Riemnn integrl of f is sme s the Lebesgue integrl of f. Proof. We give outline of the proof. Suppose f(x) M for ll x [, b] nd some M R. Use the definition of Riemnn integrbility to find sequences {φ k } nd {ψ k } of step functions bounded by M such tht φ k φ nd ψ k ψ for some mesurble functions φ nd ψ such tht φ f ψ. Also, the limits of Riemnn integrls of φ k nd ψ k gree with tht of f. Since Riemnn integrl nd Lebesgue integrl gree for step functions, by bounded convergence theorem, Lebesgue integrls of φ nd ψ re sme. Since φ ψ, we must hve φ = f = ψ.e. This shows tht f is mesurble. It is now esy to see tht the Riemnn integrl of f is sme s its Lebesgue integrl. 1
2 2 SAMEER CHAVAN Remrk 1.2 : The set of Riemnn integrble functions forms subspce of L 1 [, b]. In generl, it is hrd to compute Lebesgue integrl right from the definition. The preceding result, in prticulr, shows tht Lebesgue integrl of continuous functions my be clculted using the methods from Riemnn integrtion theory. The pointwise limit of Riemnn integrble functions need not be Riemnn integrble s shown below. Problem 1.3. Consider the function f m (x) = lim n (cos(m!πx)) n for x R. Find the set of discontinuities of f m. Further, verify the following: (1) {f m } converges pointwise to f, where f(x) = 0 if x R \ Q, nd f(x) = 1 for x Q. (2) f is discontinuous everywhere. Remrk 1.4 : If {r n } n=1 is enumertion of rtionls then the sequence {χ {r1,,r n}} of Riemnn integrble functions converges pointwise to the chrcteristic function of rtionls, which is not Riemnn integrble. Since R[, b] is subspce of L 1 [, b], it is nturl to sk whether it is closed in L 1 [, b]. The nswer is No. Problem 1.5. Let Ĉ denote the Cntorlike set obtined by removing 2k 1 centrlly situted open subintervls I 1k,, I 2 k 1 k of I := [0, 1] ech of length 1/4 k t the kth stge, where k = 1, 2,. Let F k : I I be continuous function such tht F 1 = 1 on I \ 2k 1 i=1 I ik nd F 1 = 0 t the midpoints of I 1k,, I 2 k. For f k 1 n := n i=1 F i for n 1, verify the following: (1) The sequence {f n } of continuous functions decreses to some f : I I pointwise. (2) The limit function f is discontinuous t every point of Ĉ. Conclude tht the set R[0, 1] of Riemnn integrble functions s subspce of L 1 [0, 1] is not closed. Remrk 1.6 : Consider the vector spce R[, b] of Riemnn integrble functions. Then R[0, 1] is subspce of L 1 [, b] with norm f g 1 := b f(t) g(t) dt, which is not complete. The preceding remrk rises the following question: Wht is the closure of R[, b] in L 1 [, b]? 2. Spce of Lebesgue Integrble Functions The mesurble functions, in generl, could be extended relvlued. The following simple observtion shows tht for members of L 1, WLOG, we my confine ourselves to relvlued functions. Problem 2.1. Let f : R d R { } be n extended relvlued function. If f L 1 then the set {x R d : f(x) = } is of mesure zero.
3 NOTES AND PROBLEMS: INTEGRATION THEORY 3 Problem 2.2. Consider the intervls I 1 = [0, 1], I 2 = [0, 1/2], I 3 = [1/2, 1], I 4 = [0, 1/4], I 5 = [1/4, 1/2], I 6 = [1/2, 3/4], I 7 = [3/4, 1] nd so on. For f n = χ In nd f = 0, show tht f n f 1 0, but f n (x) f(x) for ny x [0, 1]. Hint. Any x [0, 1] belongs to infinitely mny I n s nd complements of infinitely mny I n s. Although, the convergence in L 1 need not imply the pointwise convergence, the sitution is not very bd. Lemm 2.3. If {f n } is Cuchy sequence in L 1, then there exists subsequence {f nk } of {f n } such tht f nk +1 f nk 2 k for ll nonnegtive integers k nd f nk (x) f(x).e. where f(x) = f n1 (x) + (f nk +1(x) f nk (x)). k=1 A normed liner spce is complete if every Cuchy sequence is convergent. Theorem 2.4. The normed liner spce L 1 is complete. Proof. Assume tht {f n } is Cuchy sequence in L 1. By the preceding lemm, there exists subsequence {f nk } stisfying the conditions given there. Check tht f f nk g, where g(x) = f n1 (x) + f nk +1(x) f nk (x). k=1 k=1 Since f nk +1 f nk 2 k, f nk +1 f nk < 1. By the monotone convergence theorem, g L 1. On the other hnd, since f f nk g, by the dominted convergence theorem, f nk f 1 0 s k. By generl theory, Cuchy sequence with convergent subsequence is necessrily convergent. The RieszFischer Theorem mkes possible the ppliction of bstrct theory of Bnch spces to L 1. For instnce, one my conclude tht every bsolutely convergent series in L 1 is convergent. An stute reder my notice, however, tht this observtion is implicitly there in the bove proof of RieszFischer theorem. Problem 2.5. Verify tht closed subspce of complete normed liner spce is complete. Conclude tht C[, b] is not closed in L 1 [, b]. Hint. Suppose tht C[, b] is closed in L 1 [, b]. By RieszFischer theorem, (C[, b], ) is complete normed liner spce. We pply the bounded inverse theorem to the mpping f f from (C[, b], ) onto (C[, b], 1 ) to get C > 0 such tht f C f 1 for every f C[, b]. The preceding exercise rises the question: wht is the closure of C[0, 1] in L 1 [0, 1]? We will see in Section 3 tht it is the whole of L 1 [0, 1].
4 4 SAMEER CHAVAN 3. Some Dense subspces of L 1 We sy tht subset Y of X is dense in X if for every f X nd for every ɛ > 0, there exists g in Y such tht f g < ɛ. Mny times it is prcticl to know good dense subsets of given normed liner spce. For instnce, n inequlity for ll members of X could be deduced from the sme inequlity for members of Y (see Exercise 3.9 below for nother ppliction). Problem 3.1. Show tht the set of simple functions is dense in L 1. Hint. Since f = f + f, WLOG, we ssume tht f 0.e. Then there exists sequence of simple functions incresing to f pointwise. Now pply monotone convergence theorem. Problem 3.2. Show tht for ny step function s : R R nd ɛ > 0, there exists continuous function f such tht f s < ɛ. Hint. First try chrcteristic function of n intervl. Problem 3.3. Show tht for ny step function s : R d R in L 1 nd ɛ > 0, there exists continuous function f with compct support such tht f s < ɛ. Hint. Chrcteristic function of cube is product of chrcteristic functions of intervls. Now pply the lst exercise. Problem 3.4. For mesurble subset E of R d of finite mesure nd ɛ > 0, show tht there exists n lmost disjoint fmily of rectngles {R 1,, R M } with m(e M j=1 R j) ɛ. Hint. Cover E by j=1 Q j such tht j=1 Q j m(e) + ɛ/2. There exists N 1 such tht j=n+1 Q j ɛ/2. Verify tht m(e N j=1 Q j) ɛ. Wrning: Q 1,, Q N my not be lmost disjoint. Problem 3.5. Let E be mesurble set of finite mesure nd let ɛ > 0 be given. Show tht there exists step function g such tht χ E g 1 < ɛ. Hint. By the lst exercise, there exists n lmost disjoint fmily of rectngles {R 1,, R M } with m(e M j=1 R j) ɛ/2. Check tht χ E M j=1 χ R j 1 < ɛ. Theorem 3.6. The vector spce of continuous functions with compct support is dense in L 1. Proof. Let f L 1 nd ɛ > 0 be given. By Problem 3.1, there exists simple function s such tht f s < ɛ/3. Since s is finite liner combintion of chrcteristic functions, by the preceding problem, there exists step function g such tht s g 1 < ɛ/3. Now by Problem 3.3, there exists continuous function h with compct support such tht g h 1 < ɛ/3. Finlly, by tringle inequlity, we obtin f h 1 < ɛ. Remrk 3.7 : R[, b] is dense in L 1 [, b].
5 NOTES AND PROBLEMS: INTEGRATION THEORY 5 Corollry 3.8. The spce of polynomils on [, b] re dense in L 1 [, b]. In prticulr, L 1 [, b] is seprble. Let us discuss one ppliction of the density theorem. Problem 3.9. For f : R R nd y R, let f y (x) = f(x y) denote the trnslte of f. If f L 1 then show tht ψ : R L 1 given by ψ(y) = f y is uniformly continuous. Hint. Let g be continuous function in L 1 with support in [ A, A] such tht f g 1 < ɛ. Since g is uniformly continuous on [ A, A], there exists δ < A such tht g(s) g(t) < ɛ/3a whenever s t < δ. Check tht g s g t 1 < (2A+δ)(ɛ/3A) < ɛ, nd hence f s f t 1 < 3ɛ whenever s t < δ. The following is known s RiemnnLebesgue lemm: Corollry If f L 1 then ˆf(ζ) := R d f(x)e 2πixζ dx tends to 0 s ζ. Proof. By the preceding exercise, for f L 1, f f h 0 s h. The desired conclusion follows from ˆf(ζ) = 1 (f(x) f h (x))e 2πixζ dx, 2 R d ζ where h = 1 2 ζ 0 s ζ Spce of Functions of Bounded Vrition For function F : [, b] C nd prtition P : {t 0 < t 1 < < t n } of [, b], let V (F, P ) := n i=1 F (t i) F (t i 1 ). Remrk 4.1 : If c (, b) nd P c denote the refinement of P obtined by djoining c to P, then V (F, P ) V (F, P c ). The totl vrition T F of function F : [0, 1] C is defined s T F := sup V (F, P ), P where sup is tken over ll prtitions P : {t 0 < t 1 < < t n } of [0, 1]. We sy tht F is of bounded vrition if its totl vrition T F is finite. Remrk 4.2 : A function of bounded vrition is necessrily bounded. Just consider the prtition P : {, x, b}, nd note tht V (P, F ) is finite. Let us see some exmples of functions of bounded vrition. Problem 4.3. Show tht every bounded monotonic relvlued function is of bounded vrition. Hint. V (F, P ) = F () F (b) for ny prtition P. Problem 4.4. Show tht every differentible function with bounded derivtive is of bounded vrition.
6 6 SAMEER CHAVAN Hint. Apply men vlue theorem to rel nd imginry prts of F. A differentible function of bounded vrition need not hve bounded derivtive. Problem 4.5. Show tht the function F : [0, 1] R given by is of bounded vrition. F (x) = x 3/2 sin(1/x), for 0 < x 1, = 0, if x = 0 Here is n exmple of continuous monotone function of bounded vrition whose derivtive vnishes lmost everywhere. Exmple 4.6 : Let C be the Cntor set obtined by removing 2 k 1 centrlly situted open subintervls I 1k,, I 2 k 1 k of I := [0, 1] ech of length 1/3 k t the kth stge, where k = 1, 2,. Let F 1 : I I be continuous incresing function such tht F 1 (0) = 0, F 1 = 1/2 on I 11, F 1 (1) = 1, nd liner on remining prt. Similrly, let F 2 : I I be continuous incresing function such tht F 2 (0) = 0, F 2 = 1/4 on I 12, F 2 = 1/2 on I 11, F 2 = 3/4 on I 22, F 1 (1) = 1, nd liner on remining prt. One cn inductively define the sequence {F k } of continuous incresing functions so tht F k+1 (0) = 0, F k+1 = F j on I 1j,, I 2 j 1 j for 1 j k, the vlues of F k+1 on consecutive I ij differ by 1/2 k+1, nd F k+1 (1) = 1. Note tht F k+1 (x) F k (x) 1 for every x [0, 1]. 2k+1 It is esy to see tht F m F n 0, nd hence {F k } converges uniformly to continuous function, sy, F. The continuous incresing function F is known s CntorLebesgue function. By Problem 4.3, F is of bounded vrition. Note further tht F is constnt on ech intervl of the complement of C. Since C hs mesure 0, F (x) = 0 lmost everywhere. Remrk 4.7 : Note tht [0,1] F (x)dx F (1) F (0). Let F : [, b] C nd prtition P : {t 0 < t 1 < < t n } of [, b] be given. If c (, b), then we hve prtitions P 1 nd P 2 of [, c] nd [c, b] respectively such tht P c = P 1 P 2. Moreover, V (F, P c ) = V (F [,c], P 1 ) + V (F [c,b], P 2 ). By Remrk 4.1, we hve T F [,b] = T F [,c] + T F [c,b]. Replcing b by v nd c by u, for u < v b, we obtin (4.1) T F [,v] T F [,u] = T F [u,v]. Remrk 4.8 : If F is of bounded vrition then the mpping x T F[,x] incresing. The following decomposition of functions of bounded vrition is due to Jordn. Theorem 4.9. Any relvlued function of bounded vrition on [, b] is difference of two bounded, incresing functions.
7 NOTES AND PROBLEMS: INTEGRATION THEORY 7 Proof. We decompose F (x) s [F (x) + T F[,x] ] T F[,x] for x [, b]. In view of the lst remrk, it suffices to check tht x F (x) + T F[,x] is incresing. For u < v b, by (4.1), we obtin T F [,v] T F [,u] = T F [u,v] V (F [u,v], {u, v}) = F (v) F (u) F (u) F (v), nd hence the desired conclusion. Problem Show tht the set of discontinuities of monotone function on [, b] is countble, nd hence of mesure 0. Hint. Suppose F is bounded nd incresing. If x is discontinuity of F, then there exists rtionl number r x such tht lim F (t) < r x < lim F (t). Note t x,t<x t x,t>x tht if x < y then r x < r y, nd hence x r x is onetoone. In view of the lst exercise, it is interesting to know the set of points t which bounded, monotone function is nondifferentible. A deep result of Lebesgue sys tht every bounded, monotone function on [, b] is differentible lmost everywhere with integrble derivtive (refer to [4, Chpter 3] for proof). As consequence of this nd Jordn s theorem, we obtin the following: Corollry If F : [, b] C is of bounded vrition then F is differentible lmost everywhere. Moreover, F belongs to L 1 [, b]. 5. Riesz Representtion Theorems In this section, we will see the Riesz representtion theorem for C[, b] nd L 1 [, b]. We strt reclling some preliminries from functionl nlysis. Let X be normed liner spce. The dul spce X of X is defined s the normed liner spce of ll bounded liner functionls f : X C. Problem 5.1. Show tht X is Bnch spce with norm f := sup{ f(x) : x 1}. Hint. Note tht f n (x) f m (x) f n f m x, nd if {f n } is Cuchy then so is {f n (x)}. In this cse, there exists liner functionl f : X C such tht f n (x) f(x) for every x. Show tht for ɛ > 0, there exists N 1 such tht f n (x) f(x) ɛ x for ll n N. Problem 5.2. X {0} if nd only if X {0}. Hint. Let x X be nonzero. Then f(αx) = α x is bounded liner functionl on Cx. Now pply HnhBnch extension theorem. The Riesz representtion theorem for C[, b] identifies the dul spce of C[, b] with subspce of functions of bounded vrition. To estblish it, we need some definitions nd observtions. Consider the vector spce BV [, b] of functions F : [, b] of bounded vrition. Note tht if F BV [, b] then its totl vrition T F is 0 if nd only if F is constnt. This llows us to mke BV [, b] into normed liner spce with norm F := T F + F ().
8 8 SAMEER CHAVAN Problem 5.3. If F BV [, b] then ψ F : C[, b] C given by ψ F (f) := b f(t)df (t) is bounded liner functionl with norm ψ F t most the totl vrition T F of F, where b f(t)df (t) is the limit of the RiemnnStieljes sum n f(s j )[F (t j ) F (t j 1 ], j=1 s the mesh of the prtition { = t 0 < t 1 < < t n = b] tends to 0 nd s j in [t j 1, t j ] for j = 1,, n. Remrk 5.4 : Note tht the bove limit exists in view of Jordn decomposition theorem. The mpping F ψ F is not injective. Indeed, ψ 1 = 0 = ψ 2. Let B[, b] denote the normed liner spce of bounded functions on [, b] with norm. Remrk 5.5 : If φ (C[, b]), then there exists bounded liner functionl ψ (B[0, 1]) such tht ψ(f) = φ(f) for ll f C[, b] nd ψ = φ. This is consequence of HnhBnch extension theorem. Problem 5.6. Let ψ (B[, b]). Define F : [0, 1] K by F (0) = 0 nd F (t) = ψ(χ (0,t] ). Show tht T F ψ. Hint. There exists θ R such tht F (t i ) F (t i 1 ) = e iθ (F (t i ) G(t i 1 )) = ψ(e iθ χ (ti 1,t i]). A function F BV [, b] is sid to be normlized if F () = 0 nd F is right continuous on (, b). Let NBV [, b] denote the spce of normlized functions of bounded vrition. Then NBV [, b] is normed liner spce with norm s the totl vrition. Lemm 5.7. If F BV [, b] then there exists G NBV [, b] such tht b f(t)df (t) = for every f C[, b]. Moreover, T G T F. Proof. Define G by setting G() = 0 nd b f(t)dg(t) G(t) = F (t + ) F () if t (, b) = F (b) F () if t = b. Clerly, G is right continuous. We clim tht T G T F + ɛ for ny ɛ > 0. Given prtition P : {t 0 =,, t n = b} of [, b], find prtition Q : {s 0 =, s 1,, s n 1, s n = b} such tht t j < s j nd F (t + j ) F (s j) < ɛ/2n for j = 1,, n 1. It is esy to see tht V (P, G) V (Q, F ) + ɛ, nd hence the clim stnds verified. However, since ɛ > 0 is rbitrry, we hve T G T F. This shows tht G NBV [, b].
9 NOTES AND PROBLEMS: INTEGRATION THEORY 9 Since F G is constnt lmost everywhere, both the integrls given in the sttement gree. Problem 5.8. The G in the preceding lemm is unique. Hint. Suppose tht b f(t)dg(t) for every f C[, b]. For ny c (, b), show tht G(c) T G [c,c+h] for sufficiently smll h. Now note tht the function T G [,t] is right continuous if so is G. Theorem 5.9. If φ (C[, b]), then φ = φ F for some function F of bounded vrition. Moreover, the mpping F φ F from NBV [, b] onto (C[, b]) is n isometric isomorphism. Proof. For simplicity, we ssume tht = 0 nd b = 1. For f C[0, 1], consider s n := n r=1 f(r/n)χ ((r 1)/n,r/n]. Given ɛ > 0, choose n 1 such tht f(s) f(t) < ɛ for ll x, y such tht x y < 1/n. Note tht s n (t) f(t) = n (f(r/n) f(t))χ ((r 1)/n,r/n] (t). r=1 It is now esy to see tht s n converges uniformly to f. Let ψ (B[0, 1]) be the extension of φ n ensured by Remrk 5.5. We clim tht φ = φ F, where F is the function of bounded vrition given by Problem 5.6. By the continuity of ψ, ψ(s n ) converges to ψ(f) = φ(f), nd hence it suffices to check tht ψ(s n ) = 1 0 s n(t)df (t). This follows from ψ(χ ((r 1)/n,r/n] ) = 1 0 χ ((r 1)/n,r/n](t)dF (t). The equlity T F = φ follows from Problems 5.3 nd 5.6. The remining prt follows from Problems 5.3 nd 5.8 nd Lemm 5.7. Remrk 5.10 : The normed liner spce NBV [, b] with norm s totl vrition is complete. In the remining prt of this section, we present proof of Riesz representtion theorem for L 1 [, b]. To do tht, we stte ( specil cse of) the RdonNikodym theorem (without proof). Theorem Let µ be positive σfinite mesure defined on the σlgebr Σ nd let λ be complex mesure on Σ with the property tht λ(e) = 0 for every E Σ for which µ(e) = 0. Then there exists unique f L 1 (µ) such tht λ(e) = f(t)dµ(t) for every E Σ. E Remrk 5.12 : The RdonNikodym theorem my be viewed s generliztion of the fundmentl theorem of clculus: F (b) F () = F (t)dt. [,b]
10 10 SAMEER CHAVAN For instnce, in view of Corollry 4.11, if F is function of bounded vrition, then we hve complex mesure λ obtined by setting λ(e) = F (t)dt E for every Lebesgue mesurble set E of [, b]. In this cse, F (b) F () = λ([, b]). We wish to recll here tht the conclusion of Fundmentl theorem of clculus my fil even for function of bounded vrition (see Remrk 4.7). A complexvlued function f is sid to be essentilly bounded if φ m, is finite, where φ m, inf{m R + : φ(z) M outside set of Lebesgue mesure 0}. Remrk 5.13 : If f g then f my be not be equl to g. However, we lwys hve f m, = g m,. Problem For mesurble set X, let L (X) denote the set of ll (equivlence clsses of) mesurble functions f for which f m, <. Show tht L (X) is normed liner spce with norm f m,. Problem Let g L [, b]. Define the liner functionl φ g : L 1 [, b] C by φ g (f) := [,b] f(t)g(t)dt. Show tht φ g (L 1 [, b]). Theorem For φ (L 1 [, b]), there exists unique g L [, b] such tht φ = φ g, where φ g is s defined in the preceding problem. Proof. For Lebesgue mesurble subset of [, b], define λ( ) = φ(χ ). Then λ is countbly dditive mesure. If m( ) = 0 then χ = 0 lmost everywhere, nd hence by linerity of φ, λ( ) = φ(0) = 0. By RdonNikodym theorem, there exists g L 1 [, b] such tht λ( ) = g(t)dt for every Lebesgue mesurble subset of [, b]. For ɛ > 0, let A := {x [, b] : g(x) > φ + ɛ} nd let f = χ A (ḡ/g). Clculte f 1 nd exmine φ g (f) to see tht φ g = g. This lso shows tht g L [, b]. To see tht φ = φ g, we check tht φ(s) = φ g (s) for ny simple mesurble function s. By simple ppliction of dominted convergence theorem, we get φ(f) = φ g (f) for ny f L 1 [, b]. We leve the uniqueness of g to the reder. References [1] S. Kumresn, A Problem Course in Functionl Anlysis, privte communiction. [2] B. Limye, Functionl Anlysis, New Age Interntionl Limited, New Delhi, [3] H. Royden nd P. Fitzptrick, Rel Anlysis, Person Prentice Hll, U.S.A [4] E. Stein nd R. Shkrchi, Rel Anlysis: Mesure Theory, Integrtion, nd Hilbert Spces, Princeton University Press, Priceton nd Oxford, 2005.
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