Week 7 Riemann Stieltjes Integration: Lectures 1921

 Noel Rodgers
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1 Week 7 Riemnn Stieltjes Integrtion: Lectures Lecture 19 Throughout this section α will denote monotoniclly incresing function on n intervl [, b]. Let f be bounded function on [, b]. Let P = { = 0 < 1,, n = b} be prtition of [, b]. Put α i = α( i ) α( i 1 ). M i = sup{f(x) : i 1 x i }. m i = inf{f(x) : i 1 x i }. U(P, f) = M i α i ; L(P, f) = m i α i. i=1 fdα = inf{u(p, f) : P }; Definition 1 If fdα = i=1 fdα = sup{l(p, f) : P }. fdα, then we sy f is RiemnnStieltjes (RS) integrble w.r.t. to α nd denote this common vlue by b fdα := f(x)dα(x) := fdα = fdα. 0
2 Let R(α) denote the clss of ll RS integrble functions on [, b]. Definition 2 A prtition P of [, b] is clled refinement of nother prtition P of [, b] if, points of P re ll present in P. We then write P P. Lemm 1 If P P then L(P ) L(P ) nd U(P ) U(P ). Enough to do this under the ssumption tht P hs one extr point thn P. And then it is obvious becuse if < b < c then inf{f(x) : x c} min{inf{f(x) : x b}, inf{f(x) : b x c} etc. Theorem 1 fdα fdα. For first of ll, becuse for every prtition P we hve U(P, f) L(P, f). Let P nd Q be ny two prtitions of [, b]. By tking common refinement T = P Q, nd pplying the bove lemm we get U(P ; f) U(T ; f) L(T ; f) L(Q; f) Now vrying Q over ll possible prtitions nd tking the supremum, we get U(P ) fdα. Now vrying P over ll prtitions of [, b] nd tking the infimum, we get the theorem. Theorem 2 Let f be bounded function nd α be monotoniclly incresing function. Then the following re equivlent. (i) f R(α). (ii) Given ɛ > 0, there exists prtition P of [, b] such tht U(P, f) L(P, f) < ɛ. 1
3 (iii) Given ɛ > 0, there exists prtition P of [, b] such tht for ll refinements of Q of P we hve U(Q, f) L(Q, f) < ɛ. (iv) Given ɛ > 0, there exists prtition P = { 0 < 1 < cdots < n } of [, b] such tht for rbitrry points t i, s i [ i 1, i ] we hve f(s i ) f(t i ) α i < ɛ. i=1 (v) There exists rel number η such tht for every ɛ > 0, there exists prtition P = { 0 < 1, < < n } of [, b] such tht for rbitrry points t i [ i 1, i ], we hve n i=1 f(t i) α i η < ɛ. Proof: (i) = (ii): By definition of the upper nd lower integrls, there exist prtitions Q, T such tht U(Q) fdα < ɛ/2; fdα L(T ) < ɛ/2. Tke common refinement P to Q nd T nd replce Q, T by P in the bove inequlities, nd then dd the two inequlities nd use the hypothesis (i) to conclude (ii). (ii) = (i): Since L(P ) fdα b fdα U(P ) the conclusion follows. (ii) = (iii): This follows from the previous theorem for if P P then (iii) = (ii): Obvious. L(P ) L(P ) U(P ) U(P ). (iii) = (iv): Note tht f(s i ) f(t i ) M i m i. Therefore, f(s i ) f(t i ) α i i i (M i m i ) α i = U(P, f) L(P, f) < ɛ. 2
4 (iv) = (iii): Choose points t i, s i [ i 1, i ] such tht m i f(s i ) < ɛ 2n α i, M i f(t i ) < ɛ 2n α i. Then U(P, f) L(P, f) i (M i m i ) α i i [ M i f(t i ) + m i f(s i ) + f(t i ) f(s i )] α i < 2ɛ. Thus so fr, we hve proved tht (i) to (iv) re ll equivlent to ech other. (i) = (v): We first note tht hving proved tht (i) to (iv) re ll equivlent, we cn use ny one of them. We tke η = fdα. Given ɛ > 0 we choose prtition P such tht L(P ) η < ɛ/3. nd prtition Q such tht (iv) holds with ɛ replced by ɛ/3. We then tke common refinement T of these two prtitions for which gin the sme would hold becuse of (iii). We now choose s i [ i 1, i ] such tht m i f(s i ) < ɛ 3n α i whenever α i is non zero. (If α i = 0 we cn tke s i to be ny point.) Then for rbitrry points t i [ i 1, i ], we hve = i f(t i) α i η [(f(t i ) f(s i ) + (f(s i ) m i ) + m i ] α i η i i f(s i ) f(t i ) α i + f(s i ) m i α i + L(P ) η i ɛ/3 + ɛ/3 + ɛ/3 = ɛ. (v) = (iv): Given ɛ > 0 choose prtition s in (v) with ɛ replced by ɛ/2. 3
5 Lecture 20 Fundmentl Properties of the RiemnnStieltjes Integrl Theorem 3 Let f, g be bounded functions nd α be n incresing function on n intervl [, b]. () Linerity in f : This just mens tht if f, g R(α), λ, µ R then λf + µg R(α). Moreover, (λf + µg) = λ fdα + µ fdα. (b) SemiLinerity in α. This just mens if f R(α j ), j = 1, 2 λ j > 0 then f R(λ 1 α 1 + λ 2 α 2 ) nd moreover, fd(λ 1 α 1 + λ 2 α 2 ) = λ fdα 1 + µ fdα 2. (c) Let < c < b. Then f R(α) on [, b] if f R(α) on [, c] s well s on [c, b]. Moreover we hve fdα = c fdα + c fdα. (d) f 1 f 2 on [, b] nd f i R(α) then f 1dα f 2dα. (e) If f R(α) nd f(x) M then fdα M[α(b) α()]. (f) If f is continuous on [, b] then f R(α). (g) f : [, b] [c, d] is in R(α) nd φ : [c, d] R is continuous then φ f R(α). (h) If f R(α) then f 2 R(α). (i) If, f, g R(α) then fg R(α). (j) If f R(α) then f R(α) nd fdα f dα. 4
6 Proof: () Put h = f + g. Given ɛ > 0, choose prtitions P, Q of [, b] such tht U(P, f) L(P, f) < ɛ/2, P (Q, g) L(Q, g) < ɛ/2 nd replce these prtitions by their common refinement T nd then ppel to L(T, f) + L(T, g) L(T, h) U(T, h) U(T, f) + U(T, g). For constnt λ since U(P, λf) = λu(p, f); L(P, λf) = λl(p.f) it follows tht proof of (). λfdα = λ (b) This is esier: In ny prtition P we hve fdα. Combining these two we get the from which the conclusion follows. (λ 1 α 1 + λ 2 α 2 ) = λ 1 α 1 + λ 2 α 2 (c) All tht we do is to stick to those prtitions of [, b] which contin the point c. (d) This is esy nd (e) is consequence of (d). ɛ (f) Given ɛ > 0, put ɛ 1 =. Then by uniform continuity of f, α(b) α() there exists δ > 0 such tht f(t) f(s) < ɛ 1 whenever t, s [, b] nd t s < δ. Choose prtition P such tht α i < δ for ll i. Then it follows tht M i m i < ɛ 1 nd hence U(P ) L(P ) < ɛ. (g) Given ɛ > 0 by uniform continuity of φ, we get ɛ > δ > 0 such tht φ(t) φ(s) < ɛ for ll t, s [c, d] with t s < δ. There is prtition P of [, b] such tht U(P, f) L(P, f) < δ 2. 5
7 The differences M i m i my behve in two different wys: Accordingly let us define A = {1 i n : M i m i < δ}, B = {1, 2,..., n} \ A. Put h = φ f. It follows tht M i (h) m i (h) < ɛ, i A. Therefore we hve δ( α i ) i m i ) α i < U(P, f) L(P, f) < δ i B i B(M 2. Therefore we hve i B α i < δ. Now let K be bound for φ(t) on [c, d]. Then U(P, h) L(P, h) = i (M i(h) m i (h)) α i = i A (M i(h) m i (h)) α i + i B (M i(h) m i (h)) α i ɛ(α(b) α()) + 2Kδ < ɛ(α(b) α() + 2K). Since ɛ > 0 is rbitrry, we re done. (h) Follows from (g) by tking φ(t) = t 2. (i) Write fg = 1 4 [(f + g)2 (f g) 2 ]. (j) Tke φ(t) = t nd pply (g) to see tht f R(α). Now let λ = ±1 so tht λ fdα 0. Then fdα = λ fdα = λfdα f dα. This completes the proof of the theorem. Theorem 4 Suppose f is monotonic nd α is continuous nd monotoniclly incresing. Then f R(α). Proof: Given ɛ > 0, by uniform continuity of α we cn find prtition P such tht ech α i < ɛ. 6
8 Now if f is incresing, then we hve M i = f( i ), m i = f( i 1 ). Therefore, U(P ) L(P ) = i [f( i ) f( i 1 )] α i < f(b) f())ɛ. Since ɛ > 0 is rbitrry, we re done. Lecture 21 Theorem 5 Let f be bounded function on [, b] with finitely mny discontinuities. Suppose α is continuous t every point where f is discontinuous. Then f R(α). Proof: Becuse of (c) of theorem 3, it is enough to prove this for the cse when c [, b] is the only discontinuity of f. Put K = sup f(t). Given ɛ > 0, we cn find δ 1 > 0 such tht α(c + δ 1 ) α(c δ 1 ) < ɛ. By uniform continuity of f on [, b] \ (c δ, c + δ) we cn find δ 2 > 0 such tht x y < δ 2 implies f(x) f(y) < ɛ. Given ny prtition P of [, b] choose prtition Q which contins the points c nd whose mesh is less thn min{δ 1, δ 2 }. It follows tht U(Q) L(Q) < ɛ(α(b) α()) + 2Kɛ. Since ɛ > 0 is rbitrry this implies f R(α). Remrk 1 The bove result leds one to the following question. Assuming tht α is continuous on the whole of [, b], how lrge cn be the set of discontinuities of function f such tht f R(α)? The nswer is not within RS theory. Lebesgue hs to invent new powerful theory which not only nswers this nd severl such questions rised by Riemnn integrtion theory but lso provides sound foundtion to the theory of probbility. Exmple 1 We shll denote the unit step function t 0 by U which 7
9 is defined s follows: U(x) = { 0, x 0; 1, x > 0. By shifting the origin t other points we cn get other unit step function. For exmple, suppose c [, b]. Consider α(x) = U(x c), x [, b]. For ny bounded function f : [, b] R, let us try to compute fdα. Consider ny prtition P of [, b] in which c = k. The only non zero α i is α k = 1. Therefore U(P ) L(P ) = M k (f) m k (f). Now ssume tht f is continuous t c. Then by choosing k+1 close to k = c, we cn mke M k m k 0. This mens tht f R(α). Indeed, it follows tht M k f(c) nd m k f(c). Therefore, fdα = f(c). Now suppose f hs discontinuity t c of the first kind i.e, in prticulr, f(c + ) exists. It then follows tht M k m k f(c) f(c + ). Therefore, f R(α) iff f(c + ) = f(c). Thus, we see tht it is possible to destroy integrbility by just disturbing the vlue of the function t one single point where α itself is discontinuous. In prticulr, tke f = α. It follows tht α R(α) on [, b]. We shll now prove prtil converse to (c) of Theorem 3. Theorem 6 Let f be bounded function nd α n incresing function on [, b]. Let c [, b] t which (t lest) f or α is continuous. If f R(α) on [, b] then f R(α) on both [, c] nd [c, b]; moreover, in tht cse, fdα = c fdα + c fdα. 8
10 Proof: Assume α is continuous t c. If T c is the trnsltion function T c (x) = x c then the functions g 1 = U T nd g 2 = 1 U T re both in R(α) since they re discontinuous only t c. Therefore fg 1, fg 2 R(α). But these respectively imply tht f R(α) on [c, b] nd on [, c]. We now consider the cse when f is continuous t c. We shll prove tht f R(α) on [, c], the proof tht f R(α) on [c, b] being similr. Recll tht the set of discontinuities of monotonic function is countble. Therefore there exist sequence of points c n in [, c] (we re ssuming tht < c) such tht c n c. By the erlier cse f R(α) on ech of the intervls [, c n ]. We clim tht the sequence s n := cn fdα converges to limit which is equl to c fdα. Let K > 0 be bound for α. Given ɛ > 0 we cn choose δ > 0 such tht for x, y [c δ, c + δ], f(x) f(y) < ɛ/2k. If n 0 is big enough then n, m n 0 implies tht s n s m < ɛ. This mens {s n } Cuchy nd hence is convergent with limit equl to sy, s. Now choose n so tht s s n < ɛ. Put = α(c) α(c ). Since c n c, from the left, it follows tht α(c n ) α(c ). Choose n lrge enough so tht where L is bound for f. α(c n ) α(c ) < ɛ/l Now, choose ny prtition Q of [, c n ] so tht U(Q, f) s n < ɛ. This is possible becuse f R(α) on [, c n ]. Put P = Q {c}, M = mx{f(x) : x [c n, c]}. Then s + f(c) U(P, f) s s n + s n U(Q, f) + f(c) (α(c) α(c n ))M ɛ + ɛ + (f(c) M) + (α(c n ) α(c ))M 2ɛ + ɛ 2K + M ɛ K 4ɛ. 9
11 Theorem 7 Let {c n } be sequence of non negtive rel numbers such tht n c n <. Let t n (, b) be sequence of distinct points in the open intervl nd let α = n c nu T tn. Then for ny continuous function f on [, b] we hve fdα = n c n f(t n ). Proof: Observe tht for ny x [, b], 0 n U(x t n) n c n nd hence α(x) mkes sense. Also clerly it is monotoniclly incresing nd α() = 0 nd α(b) = n c n. Given ɛ > 0 choose n 0 such tht n>n 0 c n < ɛ. Tke α 1 = n n 0 U T tn, α 2 = n>n 0 U T tn. By (b) of theorem 3, nd from the exmple bove, we hve fdα 1 = n n 0 c n f(t n ). If K is bound for f on [, b] we lso hve fdα 2 < K(α 2 (b) α 2 ()) = K c n = Mɛ. n>n 0 Therefore, This proves the clim. fdα c n f(t n ) < Kɛ. n n 0 Theorem 8 Let α be n incresing function nd α R(x) on [, b]. Then for ny bounded rel function on [, b], f R(α) iff fα R(x). Furthermore, in this cse, fdα = f(x)α (x)dx. 10
12 Proof: Given ɛ > 0, since α is Riemnn integrble, by (iv) of theorem 2, there exists prtition P = { = 0 < 1 < < n = b} of [, b] such tht for ll s i, t i [ i 1, i ] we hve, α (s i ) α (t i ) x i < ɛ. i=1 Apply MTV to α to obtin t i [ i 1, i ] such tht α i = α (t i ) x i. Put M = sup f(x). Then f(s i ) α i = i=1 f(s i )α (t i ) x i. i=1 Therefore, f(s i ) x i i=1 f(s i )α (s i ) x i < i i=1 f(s i ) α (s i ) α (t i ) x i > Mɛ. Therefore f(s i ) x i i=1 f(s i )α (s i ) x i + Mɛ U(P, fα ) + Mɛ. i=1 Since this is true for rbitrry s i [ i 1, i ], it follows tht U(P, f, α) U(P, fα ) + Mɛ. Likewise, we lso obtin U(P, fα ) U(P, f, α)) + Mɛ. Thus U(P, f, α) U(P, fα ) < Mɛ. Exctly in the sme mnner, we lso get L(P, f, α L(P, fα ) < Mɛ. 11
13 Note tht the bove two inequlities hold for refinements of P s well. Now suppose f R(α). we cn then ssume tht the prtition P is chosen so tht It then follows tht U(P, f, α) L(P, fα) < Mɛ. U(P, fα ) L(P, fα ) < 3Mɛ. Since ɛ > 0 is rbitrry, this implies fα is Riemnn integrble. The other wy impliction is similr. Moreover, the bove inequlities lso estblish the lst prt of the theorem. Remrk 2 The bove theorems illustrte the power of Stieltjes modifiction of Riemnn theory. In the first cse, α ws stircse function (lso clled pure step function). The integrl therein is reduced to finite or infinite sum. In the ltter cse, α is differentible function nd the integrl reduced to the ordinry Riemnn integrl. Thus the RS theory brings brings unifiction of the discrete cse with the continuous cse, so tht we cn tret both of them in one go. As n illustrtive exmple, consider thin stright wire of finite length. The moment of inerti bout n xis perpendiculr to the wire nd through n end point is given by l 0 x 2 dm where m(x) denotes the mss of the segment [0, x] of the wire. If the mss is given by density function ρ, then m(x) = x ρ(t)dt or 0 equivlently, dm = ρ(x)dx nd the moment of inerti tkes form l 0 x 2 ρ(x)dx. On the other hnd if the mss is mde of finitely mny vlues m i 12
14 concentrted t points x i then the inerti tkes the form x 2 i m i. i Theorem 9 Chnge of Vrible formul Let φ : [, b] [c, d] be strictly incresing differentible function such tht φ() = c, φ(b) = d. Let α be n incresing function on [c, d] nd f be bounded function on [c, d] such tht f R(α). Put β = α φ, g = f φ. Then g R(β) nd we hve gdβ = d c fdα. Proof: Since φ is strictly incresing, it defines oneone correspondence of prtitions of [, b] with those of [c, d], given by { = 0 < 1 < < n = b} {c = φ() < φ( 1 ) < < φ( n ) = d}. Under this correspondence observe tht the vlue of the two functions f, g re the sme nd lso the vlue of function α, β re lso the sme. Therefore, the two upper sums lower sums re the sme nd hence the two upper nd lower integrls re the sme. The result follows. 13
15 Week 8 Functions of Bounded Vrition Lectures Lecture 22 : Functions of bounded Vrition Definition 3 Let f : [, b] R be ny function. For ech prtition P = { = 0 < 1 < < n = b} of [, b], consider the vritions Let V (P, f) = f( k ) f( k 1 ). k=1 V f = V f [, b] = sup{v (P, f) : P is prtition of [, b]}. If V f is finite we sy f is of bounded vrition on [, b]. Then V f is clled the totl vrition of f on [, b]. Let us denote the spce of ll functions of bounded vritions on [, b] by BV[, b]. Lemm 2 If Q is refinement of P then V (Q, f) V (P, f). Theorem 10 () f, g BV[, b], α, β R = αf + βg BV[, b]. Indeed, we lso hve V αf+βg α V f + β V g. (b) f BV[, b] = f is bounded on [, b]. 14
16 (c) f, g BV[, b] = fg BV[, b]. Indeed, if f K, g L then V fg LV f + KV g. (d) f BV[, b] nd f is bounded wy from 0 then 1/f BV[, b]. (e) Given c [, b], f BV[, b] iff f BV[, c] nd f BV[c, b]. Moreover, we hve V f [, b] = V f [, c] + V f [c, b]. (f) For ny f BV[, b] the function V f : [, b] R defined by V f () = 0 nd V f (x) = V f [, x], < x b, is n incresing function. (g) For ny f BV[, b], the function D f = V f f is n incresing function on [, b]. (h) Every monotonic function f on [, b] is of bounded vrition on [, b]. (i) Any function f : [, b] R is in BV[, b] iff it is the difference of two monotonic functions. (j) If f is continuous on [, b] nd differentible on (, b) with the derivtive f bounded on (, b), then f BV[, b]. (k) Let f BV[, b] nd continuous t c [, b] iff V f : [, b] R is continuous t c. Proof: () Indeed for every prtition, we hve V (P, αf+βg) = αv (P, f)+ βv (P, g). The result follows upon tking the supremum. (b) Tke M = V f + f(). Then f(x) f(x) f() f() V (P, f) + f() M, where P is ny prtition in which, x re consecutive terms. (c) For ny two points x, y we hve, f(x)g(x) f(y)g(y) f(x) f(x) g(y) + g(y) f(x) f(y) Therefore, it follows tht V (P, f) KV g + LV f. 15
17 (d) Let 0 < m < f(x) for ll x [, b]. Then 1 f(x) 1 f(y) = f(x) f(y) f(x)f(y) It follows tht V (P, 1/f) V f for ll prtitions P nd hence the result. m 2 (e) Given ny prtition P of [, c] we cn extend it to prtition Q of [, b] by including the intervl [c, b]. Then V (Q, f) = V (P, f) + f(b) f(c) nd hence it follows tht if f BV[, b] then f BV[, c]; for similr reson, f BV[c, b] s well. Conversely suppose f BV[, c] BV[c, b]. Given prtion P of [, b] we first refine it to P by dding the point c nd then write Q = Q 1 Q 2 where Q i re the resstrictions of Q to [, c], [c, b] respectively. It follows tht V (P, f) V (Q, f) = V (Q 1, f) + V (Q 2, f) V f ([, c] + V f [c, b]. Therefore f BV[, b]. In either cse, the bove inequlity lso shows tht V f [, b] V f [, c + V f [c, b]. On the other hnd, since V (P, f) V (P, f) for ll P it follows tht V f [, b] = sup{v (P, f) : P is prtition of [, b]}. Since every prtition P is of the form P = Q 1 Q 2 where Q 1, Q 2 re rbitrry prtitions of [, c] nd [c, b] respectively, nd V (P, f) = V (Q 1, f) + V (Q 2, f) it follows tht V f [, b] = V f [, c] + V f [c, b]. (f) Follows from (e) (g) Let x < y b. Proving V f [, x] f(x) V f [, y] f(y) is the 16
18 sme s proving V f, x] + f(y) f(x) V f [, y]. For ny prtition P of [, x] let P = P {y}. Then V (P, f) + f(y) f(x) V (P, f) + f(y) f(x) = V (P, f) V f [, y]. Since this is true for ll prtitions P of [, x] we re through. (h) My ssume f is incresing. But then for every prtition P we hve V (P, f) = f(b) f() nd hence V f = f(b) f(). (i) If f BV[, b], from (f) nd (g), we hve f = V f (V f f) s difference of two incresing functions. The converse follows from () nd (h). (j) This is becuse then f stisfies Lipschitz condition f(x) f(y) M x y for ll x, y [, b]. Therefore for every prtition P we hve V (P, f) M(b ). (k) Observe tht V f is incresing nd hence V f (c ± ) exist. By (h) it follows tht sme is true for f. We shll show tht f(c) = f(c ± ) iff V f (c) = V f (c ± ) which would imply (k). So, ssume tht f(c) = f(c + ). Given ɛ > 0 we cn find δ 1 > 0 such tht f(x) f(c) < ɛ for ll c < x < c + δ 1, x [, b]. We cn lso choose prtition P = {c = x 0 < x 1 < < x n = b} such tht V f [c, b] ɛ < k f k. Put δ = min{δ 1, x 1 c}. Let now c < x < c + δ. Then V f (x) V f (c) = V f [c, x] = V f [c, b] V f [x, b] < ɛ + k f k V f [x, b] ɛ + f(x) f(c) + f(x 1 ) f(x) + k 2 f k V f [x, b] ɛ + ɛ + V f [x, b] V f [x, b] = 2ɛ. This proves tht V f (c + ) = V f (c) s required. 17
19 Conversely, suppose V f (c + ) = V f (c). Then given ɛ > 0 we cn find δ > 0 such tht for ll c < x < c + δ we hve V f (x) V f (c) < ɛ. But then given x, y such tht c < y < x < c + δ it follows tht f(y) f(c) + f(x) f(y) V f ([c, x] = V f (x) V f (c) < ɛ which definitely implies tht f(x) f(y) ɛ. This completes the proof tht V f (c + ) = V f (c) iff f(c + ) = f(c). Similr rguments will prove tht V f (c ) = V f (c) iff f(c ) = f(c). Exmple 2 Not ll continuous functions on closed nd bounded intervl re of bounded vrition. A typicl exmples is f : [0, π] R defined by f(x) = { For ech n consider the prtition P = {0, x cos ( π x), x 0 0, x = n, 1 2n 1,..., 1} Then V (P, f) = n 1 k=1. As n, we know this tends to. k However, the function g(x) = xf(x) is of bounded vrition. To see this observe tht g is differentible in [0, 1] nd the derivtive is bounded (though not continuous) nd so we cn pply (j) of the bove theorem. Also note tht even prtil converse to (j) is not true, i.e., differentible function of bounded vrition need not hve its derivtive bounded. For exmple h(x) = x 1/3, being incresing function, is of bounded vrition on [0, 1] but its derivtive is not bounded. Remrk 3 We re now going extend the RS integrl with integrtors α not necessrily incresing functions. In this connection, it should be noted tht condition (v) of theorem 63 becomes the strongest nd hence we dopt tht s the definition. 18
20 Definition 4 Let f, α : [, b] R be ny two functions. We sy f is RS integrble with respect to to α nd write f R(α) if there exists rel number η such tht for every ɛ > 0 there exists prtition P of [, b] such tht for every refinement Q = { + x 0 < x 1 < < x n = b} of P nd points t i [x i 1, x i ] we hve f(t i ) α i η < ɛ. i=1 We then write η = fdα nd cll it RS integrl of f with respect to α. It should be noted tht, in this generl sitution, severl properties listed in Theorem 64 my not be vlid. However, property (b) of Theorem 64 is vlid nd indeed becomes better. Lemm 3 For ny two functions α, β nd rel numbers λ, µ, if f R(α) R(β), then f R(λα + µβ). Moreover, in this cse we hve fd(λ α + µβ) = λ fdα + µ fdβ. Proof: This is so becuse for ny fixed prtition we hve the linerity property of : (λα + µβ) i = (λα + µβ)(x i x i 1 ) = λ( α) i + µ( β) i And hence the sme is true of the RS sums. Therefore, if η = fdα, γ = fdβ, then it follows tht λη + µγ = fd(λα + µβ). 19
21 Theorem 11 Let α be function of bounded vrition nd let V denote its totl vrition function V : [, b] R defined by V (x) = V α [, x]. Let f be ny bounded function. Then f R(α) iff f R(V α ) nd f R(V α). Proof: The if prt is esy becuse of (). Also, we need only prove tht if f R(α) then f R(V ). Given ɛ > 0 choose prtition P ɛ so tht for ll refinements P of P ɛ, nd for ll choices of t k, s k [ i 1, i ], we hve, k ) f(s k )) α k < ɛ, V f (b) < k=1(f(t k We shll estblish tht U(P, f, V ) L(P, f, V ) < ɛk α k + ɛ. for some constnt K. By dding nd subtrcting, this tsk my be broken up into estblishing two inequlities [M k (f) m k (f)][ V k α k ] < ɛk/2; k [M k (f) m k (f)] α k < ɛk/2. Now observe tht V k α k 0 for ll k. Therefore if M is bound for f, then k [M k(f) m k (f)][ V k α k ] 2M k ( V k α k ) 2M(V f (b) α k ) < 2Mɛ. To prove the second inequlity, let us put A = {k : α k 0}; B = {1, 2,..., n} \ A. For k A choose t k, s k [ k 1, k ] such tht f(t k ) f(s k ) > M k (f) m k (f) ɛ; 20 k
22 nd for k B choose them so tht We then hve f(s k ) f(t k ) > M k (f) m k (f) ɛ. k [M k(f) m k (f)] α k < k A (f(t k) f(s k )) α k + k B (f(s k) f(t k )) k + ɛ k α k = k [f(t k) f(s k )) α k + ɛv (b) = ɛ(1 + V (b)). Putting K = mx{2m, 1 + V (b)} we re done. Corollry 1 Let α : [, b] R be of bounded vrition nd f : [, b] R be ny function. If f R(α) on [, b] then it is so on every subintervl [c, d] of [, b]. Corollry 2 Let f : [, b] R be of bounded vrition nd α : [, b] R be continuous of bounded vrition. Then f R(α). Proof: By (k) of the bove theorem, we see tht V (α) nd V (α) α re both continuous nd incresing. Hence by previous theorem, V (f) nd V (f) f re both integrble with respect to V (α) nd V (α) α. Now we just use the dditive property. Lecture 24 Exmple 3 : 1. Consider the double sequence, s m,n = m, m, n 1. m + n 21
23 Compute the two iterted limits nd record your results. lim m lim n s m,n, lim n lim m s m,n x 2 2. Let f n (x) = (1 + x 2 ), x R, n 1 nd put f(x) = n n f n(x). Check tht f n is continuous. Compute f nd see tht f is not continuous. 3. Define g m (x) = lim n (cos m!πx) 2n nd put g(x) = lim m g m (x). Compute g nd see tht g is discontinuous everywhere. Directly check tht ech g m is Riemnn integrble wheres g is not Riemnn integrble. sin nx 4. Consider the sequence h n (x) = nd put h(x) = lim n f n (x). n Check tht h 0. On the other hnd, compute lim n h n(x). Wht do you conclude? 5. Put λ n (x) = n 2 x(1 x 2 ) n, 0 x 1. Compute the lim n λ n (x). On the other hnd check tht Therefore we hve 1 0 = lim n [ 1 0 λ n (x)dx = ] λ n (x)dx n2 2n [lim n λ n (x)]dx = 0. These re ll exmples wherein certin nice properties of functions fil to be preserved under pointwise limit of functions. And we hve seen enough results to show tht these properties re preserved under uniform convergence. We know tht if sequence of continuous functions converges uniformly to function, then the limit function is continuous. We cn 22
24 now sk for the converse: Suppose sequence of continuous functions f n converges pointwise to function f which is lso continuous. the convergence uniform? The nswer in generl is NO. But there is sitution when we cn sy yes s well. Theorem 12 Let X be compct metric spce f n : X R be sequence of continuous functions converging pointwise to continuous function f. Suppose further tht f n is monotone. Then the f n f uniformly on X. Proof: Recll tht f is monotone incresing mens f n (x) f n+1 (x) for ll n nd for ll x. Likewise f n is monotone decresing mens f n (x) f n+1 (x) for ll n nd for ll x. Therefore, we cn define g n (x) = f(x) f n (x) to obtin sequence of continuous functions which monotoniclly decreses to 0. It suffices to prove tht g n converges uniformly. Given ɛ > 0 we wnt to find n 0 such tht g n (x) < ɛ for ll n n 0 nd for ll x X. Put K n = {x X : g n (x) ɛ}. Then ech K n is closed subset of X. Also g n (x) g n+1 (x) it follows tht K n+1 K n. On the other hnd, sunce g n (x) 0 it follows tht n K n =. Since this is hppening in compct spce X we conclude tht K n0 = for n 0. Remrk 4 The compctness is crucil s illustrted by the exmple: f n (x) = 1 nx + 1, 0 < x < 1. Is 23
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