MAT612REAL ANALYSIS RIEMANN STIELTJES INTEGRAL


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1 MAT612REAL ANALYSIS RIEMANN STIELTJES INTEGRAL DR. RITU AGARWAL MALVIYA NATIONAL INSTITUTE OF TECHNOLOGY, JAIPUR, INDIA Tble of Contents Contents Tble of Contents 1 1. Introduction 1 2. Prtition 2 3. Riemnn Stieltjes Sum 2 4. Riemnn Stieltjes Integrl 2 5. Properties 3 6. Step function s integrtor Step function s integrtor Reduction of RS integrl to finite sum Reduction of RS integrl to finite sum 7 7. Upper nd lower Stieltjes sums Upper nd lower Stieltjes sums Upper nd Lower integrls Riemnn Condition Comprison Theorem Men vlue theorems Second men vlue theorem for RS integrls 12 References Introduction In mthemtics, the RiemnnStieltjes integrl is generliztion of the Riemnn integrl, nmed fter Bernhrd Riemnn nd Thoms Jonnes Stieltjes, first published in Riemnn Stieltjes integrl (RSI) is given by f(x)d(α(x)) If α(x) = x, RSI reduces to RI. RSI mkes senses when α is not differentible or discontinuous. By suitble choice of discontinuous α, ny finite or infinite sum cn be expressed s Stieltjes integrl nd summtion. Applictions 1
2 2 DR. RITU AGARWAL It ppers in the originl formultion of F. Riesz s theorem which represents the dul spce of the Bnch spce C[, b] of continuous functions in n intervl [, b] s Riemnn Stieltjes integrls ginst functions of bounded vrition. It lso ppers in the formultion of the spectrl theorem for (noncompct) selfdjoint opertors in Hilbert spce. In this theorem, the integrl is considered with respect to spectrl fmily of projections. Applictions in physics in mss distributions which re prtly discrete, prtly continuous. Mthemticl theory of probbility 2. Prtition Prtition of set If [, b] is compct intervl, set of points P = {x 0, x 1,...x n } stisfying the inequlities = x 0 < x 1 <... < x n 1 < x n = b is clled prtition of [, b]. The intervl [x k 1, x k ] is clled the k th subintervlof P nd we write x k = x k x k 1, so tht n x k = b The set of ll prtitions of [, b] is denoted by P[, b] Refinement nd Norm A prtition P is clled refinement of P or finer thn P if P P. A refinement of the prtition P is nother prtition P tht contins ll the points from P nd some dditionl points. Norm of prtition P is the length of lrgest subintervl of P. It is denoted by P. P = mx{x j x j 1, j = 1,..., n} P P P P. If the prtitions Pn nd Pm re independently chosen, then the prtition P n P m is common refinement of Pn nd Pm. 3. Riemnn Stieltjes Sum Let P = {x 0, x 1,...x n } be prtition of [, b] nd let t k be point in the subintervl [x k 1, x k ]. A sum of the form S(P, f, α) = f(t k ) α k is clled Riemnn Stieltjes sum of f with respect to α. Riemnn Stieltjes Integrl 4. Riemnn Stieltjes Integrl f is sid to be Riemnn integrble w.r.to α on [, b] nd write f R(α) on [, b] if there exists number A hving the following property: For every ɛ > 0, prtition P ɛ of [, b] such tht for every prtition P finer thn P ɛ nd for the choice of points t k in [x k 1, x k ] S(P, f, α) A < ɛ
3 MAT612REAL ANALYSIS:RIEMANN STIELTJES INTEGRAL 3 When such number exists, it is uniquely determined nd is denoted by f(x)d(α(x)) or b fdα. When α(x) = x, S(P, f, α) is written S(P, f) nd f R(α) is written f R, clss of Riemnn integrble functions. 5. Properties If f R(α), g R(α) then c 1 f + c 2 g R(α) on [, b] nd we hve (c 1 f + c 2 g)dα = c 1 fdα + c 2 gdα Proof: Let c 1 f + c 2 g = h. Given prtition P of [, b], we cn write S(P, h, α) = h(t k ) α k = c 1 f(t k ) α k + c 2 = c 1 S(P, f, α) + c 2 S(P, g, α) f(t k ) α k (5.1) Given ɛ > 0, choose P ɛ P, then S(P, f, α) fdα < ɛ. Similrly, choose P ɛ P, then S(P, g, α) gdα < ɛ. Tke P ɛ = P ɛ b P ɛ P (P is finer thn P ɛ ) S(P, h, α) c 1 fdα c 2 c 1 ɛ + c 2 ɛ. Hence proved. If f R(α), f R(β) on [, b], then f R(c 1 α + c 2 β) nd we hve fd(c 1 α + c 2 β) = c 1 fdα + c 2 gdβ Theorem 5.1 (Intervl of integrtion is dditive). Assume c [, b]. integrls in fdα = c exists, then third lso exists nd stisfy (5.2). fdα + c gdα < If two of the three fdα (5.2) Proof: LetP is prtition of [, b] such tht c P. Let P = P [, c] nd P = P [c, b] denote the corresponding prtitions of [, c] nd [c, b] respectively. Clerly, P = P P. The RiemnnStieltjes sum for these prtitions re connected by the eqution S(P, f, α) = S(P, f, α) + S(P, f, α) (5.3) Assume c fdα nd fdα exist. Then given ɛ > 0, prtitions P c ɛ of [, c] nd P ɛ of [c, b] such tht for prtition P finer thn P ɛ nd P finer thn P ɛ, we hve S(P, f, α) c fdα < ɛ 2 (5.4)
4 4 DR. RITU AGARWAL nd S(P, f, α) c fdα < ɛ 2, (5.5) respectively. P ɛ = P ɛ P ɛ is prtition of [, b] nd P = P P is finer thn P ɛ. Combining (5.4) nd (5.5), we get S(P, f, α) c fdα Hence, fdα exists nd is equl to c fdα + c fdα. c fdα < ɛ (5.6) Theorem 5.2 (Integrtion by Prts). If f R(α) on [, b] then α R(f) on [, b] nd we hve f(x)dα(x) + α(x)df(x) = f(b)α(b) f()α() (5.7) Proof: Let ɛ > 0 be given. Since fdα exists, there is prtition P ɛ of [, b] such tht for every prtition P finer thn P ɛ we hve S(P, f, α) fdα < ɛ (5.8) Consider the rbitrry RiemnnStieltjes sum for the integrl α(x)df(x) for the prtition P = { = x 0, x 1,...x n = b} finer then P ɛ S(P, α, f) = α(t k ) f k = α(t k )f(x k ) α(t k )f(x k 1 ) (5.9) Writing A = f(b)α(b) f()α(), we hve A = α(x k )f(x k ) Subtrcting 5.9 from 5.10, we get A S(P, α, f) = f(x k ){α(x k ) α(t k )} = S(P, f, α) α(x k 1 )f(x k 1 ) (5.10) f(x k 1 ){α(t k 1 ) α(x k 1 )} (5.11) where P is prtition of [, b] obtined by tking the points x k nd t k together i.e. P = { = x 0, t 1, x 1, t 2,..., t n, x n = b} finer thn P ɛ. Hence, A S(P, f, α) Therefore, α(x)df(x) exists nd is given by Hence proved. α(x)df(x) = f(b)α(b) f()α() fdα < ɛ (5.12) fdα (5.13)
5 MAT612REAL ANALYSIS:RIEMANN STIELTJES INTEGRAL 5 Theorem 5.3 (Chnge of Vrible). If f R(α) on [, b] nd g be strictly monotonic continuous function defined on intervl S hving the end points c nd d. Assume tht = g(c) nd b = g(d). Let h nd β be the composite functions defined s h(x) = f[g(x)] nd β(x) = α[g(x)], x [c, d]. Then h R(β) on S = [c, d] nd we hve f(x)dα(x) = d c h(x)dβ(x) = d c f[g(x)]dα[g(x)] (5.14) Proof: Assume for definiteness c < d nd g is onetoone nd hs strictly incresing nd continuous inverse g 1 defined on [, b]. For every prtition P = {c = y 0, x 1,...y n = d} of [c, d], there corresponds one nd only one prtition P = { = x 0, x 1,...x n = b} of [, b] with x k = g(y k ). In fct, we cn write, P = g(p ) nd P = g 1 (P ). Also, refinement of P produces refinement of P nd vicevers. If ɛ > 0 be given, there is prtition P ɛ of [, b] such tht for every prtition P finer thn P ɛ we hve S(P, f, α) fdα < ɛ (5.15) Let P ɛ = g 1 (P ɛ) be the corresponding prtition of [c, d] nd P be finer thn P ɛ. Consider the RS sum S(P, h, β) = h(u k ) β k, u k [y k 1, y k ] (5.16) Put t k = g(u k ) nd x k = g(y k ), then P = { = x 0, x 1,...x n = b} is finer then P ɛ nd S(P, h, β) = f(g(u k )){α[g(y k )] α[g(y k 1 )]} Since t k [x k 1, x k ] = f(t k ){α(x k ) α(x k 1 )} = S(P, f, α) S(P, h, β) (5.17) fdα < ɛ (5.18) Theorem 5.4. If f R(α) on [, b] nd α hs continuous derivtive α on [, b]. Then the Riemnn integrl f(x)α (x)dx exists nd we hve f(x)dα(x) = f(x)α (x)dx (5.19) Proof: Let g(x) = f(x)α (x). Consider the Riemnn Sum S(P, g) = g(t k ) x k = f(t k )α (t k ) x k (5.20) For the sme choice of t k nd prtition P S(P, f, α) = f(t k ) α k (5.21)
6 6 DR. RITU AGARWAL () Step Function (b) Floor Function Applying the Men Vlue Theorem, we cn write nd hence α k = α (v k ) x k, v k (x k 1, x k ) S(P, f, α) S(P, g) = f(t k )[α (v k ) α (t k )] x k (5.22) SInce f is bounded, we hve f(x) M, M > 0 x [, b]. Continuity of α on [, b] implies uniform continuity on [, b]. Hence if ɛ > 0 is given, δ > 0 (depending on ɛ) such tht 0 x y < δ α (x) α ɛ (y) < 2M(b ) Tke prtition P ɛ with norm P ɛ < δ then for ny finer prtition P, we hve α (v k ) α ɛ (t k ) < 2M(b ) For such P, therefore, we hve S(P, f, α) S(P, g) < ɛ/2 (5.23) On the other hnd, since f R(α) on [, b], there exists prtition P ɛ it, implies S(P, f, α) such tht P finer thn f(x)dα(x) < ɛ/2 (5.24) Combining these two, we get S(P, g) f(x)dα(x) < ɛ for prtition P finer thn P ɛ = P ɛ P ɛ 6. Step function s integrtor Definition 6.1 (Step Function). A function α defined on [, b] is clled step function if there is prtition P = { = x 0, x 1,...x n = b} such tht α is constnt in ech of its subintervls (x k 1, x k ). The number α(x k +) α(x k ) is clled the jump t x k if 1 < k < n. The jump t x 0 is α(x 0 +) α(x 0 ) nd t x n is α(x n ) α(x n ). A piecewise function contining ll constnt pieces. Exmple: Heviside function, Stircse function etc Step function s integrtor. Theorem 6.2. Given < c < b. Define α on [, b] s follows: { α(), x < c α(x) = α(b), c < x b (6.1)
7 MAT612REAL ANALYSIS:RIEMANN STIELTJES INTEGRAL 7 If f be defined on [, b] in such wy tht t lest one of the functions f or α is continuous from the left t c nd t lest one is continuous from the right t c. Then f R(α) on [, b] nd we hve f(x)dα(x) = f(c)[α(c+) α(c )] (6.2) Proof: If c P, every term in the sum S(P, f, α) is zero except the two terms rising from the subintervl seprted by c, for t k 1 c t k Let S(P, f, α) = f(t k 1 )[α(c) α(c )] + f(t k )[α(c+) α(c)]. (6.3) =S(P, f, α) f(c)[α(c+) α(c )] =[f(t k ) f(c)][α(c+) α(c)] + [f(t k 1 ) f(c)][α(c) α(c )] f(t k ) f(c) α(c+) α(c) + f(t k 1 ) f(c) α(c) α(c ) If f is continuous t c for every ɛ > 0, there is δ > 0 such tht P < δ f(t k 1 ) f(c) < ɛ nd f(t k ) f(c) < ɛ ɛ[ α(c+) α(c) + α(c) α(c ) ] (6.4) This inequlity holds whether or not, f is continuous t c, since If f is continuous from both sides: α(c) = α(c+) = α(c ) = 0 If f is continuous from right nd discontinuous from left: α(c) = α(c+) ɛ α(c) α(c ) If f is continuous from left nd discontinuous from right: α(c) = α(c ) ɛ α(c+) α(c) Hence proved Reduction of RS integrl to finite sum. Theorem 6.3. Let α be step function defined on [, b] with jump α k t x k, where P = { = x 0, x 1,...x n = b} is prtition of [, b]. Let f be defined on [, b] in such wy tht t lest one of the functions f or α is continuous from the left nd t lest one is continuous from the right t ech of x k. Then f(x)dα(x) exists nd is given by f(x)dα(x) = f(x k )α k. (6.5) 6.3. Reduction of RS integrl to finite sum. Theorem 6.4. Every finite sum cn be written s RS integrl. In fct, given sum n k, define f on [0, n] s follows: f(x) = k, k 1 < x k, k = 1, 2,..., n, f(0) = 0. Then n k = n f(k) = n f(x)d[x] 0 [x] is the gretest integer x.
8 8 DR. RITU AGARWAL Proof: Gretest integer function is step function continuous from right nd hving jump 1 t ech integer. The function f is continuous from left t 1, 2,..., n. Applying previous theorem, we get the desired result. Assignment Complete the proof of the theorem. 7. Upper nd lower Stieltjes sums Upper nd lower Stieltjes sums Let P = {x 0, x 1,...x n } be prtition of [, b] nd M k (f) = sup{f(x) : x [x k 1, x k ]} m k (f) = inf{f(x) : x [x k 1, x k ]} The numbers U(P, f, α) = L(P, f, α) = M k (f) α k (7.1) m k (f) α k re clled the upper nd lower Stieltjes sums of f w.r. to α for the prtition P. Remrk m k (f) M k (f) s Infimum is lwys less thn Supremum. If α increses on [, b], α k 0 m k (f) α k M k (f) α k L(P, f, α) U(P, f, α). Also t k [x k 1, x k ] m k (f) f(t k ) M k (f) L(P, f, α) S(P, f, α) U(P, f, α) Upper nd lower Stieltjes sums. Theorem 7.1. Assume α is incresing on [, b]. Then (1) If P is finer thn P U(P, f, α) U(P, f, α) nd L(P, f, α) L(P, f, α). (2) For ny two prtitions P 1 nd P 2 L(P 1, f, α) U(P 2, f, α). Proof: To prove (i), let P contins exctly one more point c(sy), (c [x i 1, x i ]) thn P. U(P, f, α) = M k (f) α k + M [α(c) α(x i 1 )] (7.2),k i + M [α(x i ) α(c)] where M nd M denote the supremums of f in [x i 1, c] nd [c, x i ], respectively. But since M M i (f) nd M M i (f) U(P, f, α) U(P, f, α). Similrly, m m i (f) nd m m i (f)
9 L(P, f, α) L(P, f, α). MAT612REAL ANALYSIS:RIEMANN STIELTJES INTEGRAL 9 (ii) Let P = P 1 P 2. Then P is common refinement of P 1 nd P 2 L(P 1, f, α) L(P, f, α) U(P, f, α) U(P 2, f, α). Remrk For incresing α m[α(b) α()] L(P, f, α) U(P, f, α) M[α(b) α()] where m=infimum nd M=Supremum of f on [, b] Upper nd Lower integrls. Assume tht α is incresing on [, b]. The upper Stieltjes Integrl of f with respect to α is defined s follows Ī(f, α) = b fdα = inf {U(P, f, α) : P P[, b]} (7.3) The lower Stieltjes Integrl of f with respect to α is defined s follows I(f, α) = fdα = sup {L(P, f, α) : P P[, b]} (7.4) For α(x) = x, we get Riemnn integrls first introduced by Drboux J.G. (1875). Theorem 7.2. Assume tht α is incresing on [, b]. Then I(f, α) Ī(f, α). Proof: If ɛ > 0, prtition P such tht U(P, f, α) Ī(f, α) + ɛ Ī(f, α) + ɛ is n upper bound to ll lower sums L(P, f, α). I(f, α) < Ī(f, α) + ɛ. Since ɛ is rbitrry, I(f, α) Ī(f, α). Exmple 7.3. Let f(x) = 1, if x Q nd f(x) = 0 if x / Q in [0, 1]. In ech subintervl [x k 1, x k ], M k (f) = 1 nd m k (f) = 0 for ll P P[, b]. I(f, α) = 0 nd Ī(f, α) = Riemnn Condition. A function f is sid to stisfy Riemnn Condition with respect to α on [, b] if for every ɛ > 0, prtition P ɛ such tht for prtition P finer thn P ɛ. 0 U(P, f, α) L(P, f, α) < ɛ Theorem 7.4. Assume tht α is incresing on [, b]. Then following sttements re equivlent: (1) f R(α) on [, b]. (2) f stisfies Riemnn condition with respect to α on [, b]. (3) I(f, α) = Ī(f, α). Proof: (i) (ii) If α(b) = α(), then the result is trivil. (??)
10 10 DR. RITU AGARWAL Let α(b) > α(). Given ɛ > 0, choose P ɛ so tht for ny prtition P finer thn it nd for ll choices of t k nd t k in [x k 1, x k ], we hve f(t k ) α k A < ɛ 3 nd f(t k) α k A < ɛ 3 where A = fdα. Combining these inequlities, [f(t k ) f(t k)] α k < 2ɛ 3 Since M k (f) m k (f) = sup{f(x) f(x ) : x, x [x k 1, x k ]}, for every h > 0, choose t k nd t k such tht f(t k ) f(t k ) > M k(f) m k (f) h. Mking choice corresponding to h =, we cn write U(P, f, α) L(P, f, α) = ɛ 3[α(b) α()] < < ɛ [M k (f) m k (f)] α k [f(t k ) f(t k)] α k + h α k (ii) (iii) If ɛ > 0 is given, there exists prtition P ɛ such tht P finer thn P ɛ implies U(P, f, α) < L(P, f, α) + ɛ. Hence for such P, we hve Ī(f, α) U(P, f, α) < L(P, f, α) + ɛ I(f, α) + ɛ Ī(f, α) < I(f, α) + ɛ for every ɛ > 0. Ī(f, α) I(f, α) But I(f, α) Ī(f, α). Combining, we get I(f, α) = Ī(f, α). (iii) (i) Let I(f, α) = Ī(f, α) = A (sy). Given ɛ > 0, choose prtition P ɛ such tht P finer thn P ɛ implies U(P, f, α) < Ī(f, α) + ɛ. Choose prtition P ɛ such tht P finer thn P ɛ implies If P ɛ = P ɛ P ɛ, we cn write L(P, f, α) > I(f, α) ɛ. I(f, α) ɛ < L(P, f, α) S(P, f, α) U(P, f, α) < Ī(f, α) + ɛ, (7.5)
11 MAT612REAL ANALYSIS:RIEMANN STIELTJES INTEGRAL 11 for every P finer thn P ɛ. Since I(f, α) = Ī(f, α) = A, S(P, f, α) A < ɛ, fdα = A. Corollries (Assignment) (1) If f is function tht is monotonic on the intervl I = [, b] nd α is continuous nd monotoniclly incresing on, then f R(α). (2) Suppose tht f is bounded on [, b], f hs only finitely mny points of discontinuity in I = [, b], nd tht the monotoniclly incresing function α is continuous t ech point of discontinuity of f. Then f R(α). (3)If f is function tht is continuous on the intervl I = [, b], then f is RiemnnStieltjes integrble on [, b] Comprison Theorem. Theorem 7.5 (Comprison Theorem). Assume tht α is incresing on [, b]. If f R(α) nd g R(α)on [, b] nd f(x) g(x) x [, b], then we hve f(x)dα(x) g(x)dα(x). For every prtition P, corresponding RiemnnStieltjes sum stisfy S(P, f, α) = f(t k ) α k g(t k ) α k = S(P, g, α) (7.6) Since α is incresing on [, b] fdα Theorem 7.6. Assume tht α is incresing on [, b]. If f R(α)on [, b] then f R(α) on [, b] nd we hve f(x)dα(x) f(x) dα(x) (7.7) gdα Proof:M k (f) m k (f) = sup{f(x) f(y) : x, y [x k 1, x k ]}. Since f(x) f(y) f(x) f(y) M k ( f ) m k ( f ) M k (f) m k (f). Multiplying by α k nd summing on k, we hve U(P, f, α) L(P, f, α) U(P, f, α) L(P, f, α) for every prtition P of [, b]. By pplying Riemnn condition, we find tht f R(α) on [, b]. Further, f(x) f(x) f(x) f(x) dα(x) f(x)dα(x) f(x)dα(x) f(x) dα(x) f(x) dα(x)
12 12 DR. RITU AGARWAL Assignement: Assume tht α is incresing on [, b]. If f R(α)on [, b] then f 2 R(α) on [, b] 8. Men vlue theorems Theorem 8.1. Assume tht α is incresing on [, b]. If f R(α)on [, b]. Let M nd m denote respectively the supremum nd infimum of the set {f(x) : x [, b]}. Then there exists rel number c stisfying m c M such tht f(x)dα(x) = c dα(x) = c[α(b) α()] (8.1) In prticulr, if f is continuous on [, b] then c = f(x 0 ) for some x 0 [, b]. Proof: If α(b) = α(), the result is obvious. Let α(b) > α(). Since ll upper nd lower sums stisfy the integrl lies between m nd M. m[α(b) α()] L(P, f, α) U(P, fα) M[α(b) α()], f(x)dα(x) must lie between the sme bounds. Therefore, the quotient c = fdα b dα When f is continuous on [, b], then Intermedite vlue Theorem yields, c = f(x 0 ) for some x 0 in [, b] Second men vlue theorem for RS integrls. Theorem 8.2. Assume tht α is incresing on [, b]. If f R(α)on [, b]. Then there exists point x 0 [, b] such tht x0 f(x)dα(x) = f() dα(x) + f(b) dα(x) x 0 (8.2) Proof: Integrtion by prts gives f(x)dα(x) = f(b)α(b) f()α() Applying first men vlue theorem, we obtin for x 0 [, b] Hence proved. α(x)df(x) f(x)dα(x) = f(b)α(b) f()α() α(x 0 )[f(b) f()] = f(b)[α(b) α(x 0 )] + f()[α(x 0 ) α()] = f(b) x 0 dα(x) + f() References x0 dα(x) [1] Tom M. Apostol, Mthemticl Anlysis, Nros Publishing House, ISBN X [2] Wikipedi, the encyclopedi.
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