Week 7 Riemann Stieltjes Integration: Lectures 19-21

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "Week 7 Riemann Stieltjes Integration: Lectures 19-21"

Transcription

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 Riemnn-Stieltjes (R-S) 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 R-S 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 Riemnn-Stieltjes 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) Semi-Linerity 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 R-S 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 R-S 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 one-one 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 R-S 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 R-S 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 R-S 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 R-S 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 point-wise 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

MAT612-REAL ANALYSIS RIEMANN STIELTJES INTEGRAL

MAT612-REAL ANALYSIS RIEMANN STIELTJES INTEGRAL MAT612-REAL ANALYSIS RIEMANN STIELTJES INTEGRAL DR. RITU AGARWAL MALVIYA NATIONAL INSTITUTE OF TECHNOLOGY, JAIPUR, INDIA-302017 Tble of Contents Contents Tble of Contents 1 1. Introduction 1 2. Prtition

More information

The Regulated and Riemann Integrals

The Regulated and Riemann Integrals Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue

More information

Math 554 Integration

Math 554 Integration Mth 554 Integrtion Hndout #9 4/12/96 Defn. A collection of n + 1 distinct points of the intervl [, b] P := {x 0 = < x 1 < < x i 1 < x i < < b =: x n } is clled prtition of the intervl. In this cse, we

More information

f(x)dx . Show that there 1, 0 < x 1 does not exist a differentiable function g : [ 1, 1] R such that g (x) = f(x) for all

f(x)dx . Show that there 1, 0 < x 1 does not exist a differentiable function g : [ 1, 1] R such that g (x) = f(x) for all 3 Definite Integrl 3.1 Introduction In school one comes cross the definition of the integrl of rel vlued function defined on closed nd bounded intervl [, b] between the limits nd b, i.e., f(x)dx s the

More information

Review of Riemann Integral

Review of Riemann Integral 1 Review of Riemnn Integrl In this chpter we review the definition of Riemnn integrl of bounded function f : [, b] R, nd point out its limittions so s to be convinced of the necessity of more generl integrl.

More information

Advanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004

Advanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004 Advnced Clculus: MATH 410 Notes on Integrls nd Integrbility Professor Dvid Levermore 17 October 2004 1. Definite Integrls In this section we revisit the definite integrl tht you were introduced to when

More information

7.2 Riemann Integrable Functions

7.2 Riemann Integrable Functions 7.2 Riemnn Integrble Functions Theorem 1. If f : [, b] R is step function, then f R[, b]. Theorem 2. If f : [, b] R is continuous on [, b], then f R[, b]. Theorem 3. If f : [, b] R is bounded nd continuous

More information

1 i n x i x i 1. Note that kqk kp k. In addition, if P and Q are partition of [a, b], P Q is finer than both P and Q.

1 i n x i x i 1. Note that kqk kp k. In addition, if P and Q are partition of [a, b], P Q is finer than both P and Q. Chpter 6 Integrtion In this chpter we define the integrl. Intuitively, it should be the re under curve. Not surprisingly, fter mny exmples, counter exmples, exceptions, generliztions, the concept of the

More information

Principles of Real Analysis I Fall VI. Riemann Integration

Principles of Real Analysis I Fall VI. Riemann Integration 21-355 Principles of Rel Anlysis I Fll 2004 A. Definitions VI. Riemnn Integrtion Let, b R with < b be given. By prtition of [, b] we men finite set P [, b] with, b P. The set of ll prtitions of [, b] will

More information

Calculus in R. Chapter Di erentiation

Calculus in R. Chapter Di erentiation Chpter 3 Clculus in R 3.1 Di erentition Definition 3.1. Suppose U R is open. A function f : U! R is di erentible t x 2 U if there exists number m such tht lim y!0 pple f(x + y) f(x) my y =0. If f is di

More information

The Banach algebra of functions of bounded variation and the pointwise Helly selection theorem

The Banach algebra of functions of bounded variation and the pointwise Helly selection theorem The Bnch lgebr of functions of bounded vrition nd the pointwise Helly selection theorem Jordn Bell jordn.bell@gmil.com Deprtment of Mthemtics, University of Toronto Jnury, 015 1 BV [, b] Let < b. For f

More information

Presentation Problems 5

Presentation Problems 5 Presenttion Problems 5 21-355 A For these problems, ssume ll sets re subsets of R unless otherwise specified. 1. Let P nd Q be prtitions of [, b] such tht P Q. Then U(f, P ) U(f, Q) nd L(f, P ) L(f, Q).

More information

UNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3

UNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 UNIFORM CONVERGENCE Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 Suppose f n : Ω R or f n : Ω C is sequence of rel or complex functions, nd f n f s n in some sense. Furthermore,

More information

Properties of the Riemann Stieltjes Integral

Properties of the Riemann Stieltjes Integral Properties of the Riemnn Stieltjes Integrl Theorem (Linerity Properties) Let < c < d < b nd A,B IR nd f,g,α,β : [,b] IR. () If f,g R(α) on [,b], then Af +Bg R(α) on [,b] nd [ ] b Af +Bg dα A +B (b) If

More information

Properties of the Riemann Integral

Properties of the Riemann Integral Properties of the Riemnn Integrl Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University Februry 15, 2018 Outline 1 Some Infimum nd Supremum Properties 2

More information

W. We shall do so one by one, starting with I 1, and we shall do it greedily, trying

W. We shall do so one by one, starting with I 1, and we shall do it greedily, trying Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering)

More information

Lecture 3 ( ) (translated and slightly adapted from lecture notes by Martin Klazar)

Lecture 3 ( ) (translated and slightly adapted from lecture notes by Martin Klazar) Lecture 3 (5.3.2018) (trnslted nd slightly dpted from lecture notes by Mrtin Klzr) Riemnn integrl Now we define precisely the concept of the re, in prticulr, the re of figure U(, b, f) under the grph of

More information

FUNDAMENTALS OF REAL ANALYSIS by. III.1. Measurable functions. f 1 (

FUNDAMENTALS OF REAL ANALYSIS by. III.1. Measurable functions. f 1 ( FUNDAMNTALS OF RAL ANALYSIS by Doğn Çömez III. MASURABL FUNCTIONS AND LBSGU INTGRAL III.. Mesurble functions Hving the Lebesgue mesure define, in this chpter, we will identify the collection of functions

More information

Lecture 1. Functional series. Pointwise and uniform convergence.

Lecture 1. Functional series. Pointwise and uniform convergence. 1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is

More information

Appendix to Notes 8 (a)

Appendix to Notes 8 (a) Appendix to Notes 8 () 13 Comprison of the Riemnn nd Lebesgue integrls. Recll Let f : [, b] R be bounded. Let D be prtition of [, b] such tht Let D = { = x 0 < x 1

More information

Riemann Stieltjes Integration - Definition and Existence of Integral

Riemann Stieltjes Integration - Definition and Existence of Integral - Definition nd Existence of Integrl Dr. Adity Kushik Directorte of Distnce Eduction Kurukshetr University, Kurukshetr Hryn 136119 Indi. Prtition Riemnn Stieltjes Sums Refinement Definition Given closed

More information

1. On some properties of definite integrals. We prove

1. On some properties of definite integrals. We prove This short collection of notes is intended to complement the textbook Anlisi Mtemtic 2 by Crl Mdern, published by Città Studi Editore, [M]. We refer to [M] for nottion nd the logicl stremline of the rguments.

More information

Math 61CM - Solutions to homework 9

Math 61CM - Solutions to homework 9 Mth 61CM - Solutions to homework 9 Cédric De Groote November 30 th, 2018 Problem 1: Recll tht the left limit of function f t point c is defined s follows: lim f(x) = l x c if for ny > 0 there exists δ

More information

Lecture 1: Introduction to integration theory and bounded variation

Lecture 1: Introduction to integration theory and bounded variation Lecture 1: Introduction to integrtion theory nd bounded vrition Wht is this course bout? Integrtion theory. The first question you might hve is why there is nything you need to lern bout integrtion. You

More information

THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS.

THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS. THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrd-lindelof-theorem/ This document is proof of the existence-uniqueness theorem

More information

Homework 4. (1) If f R[a, b], show that f 3 R[a, b]. If f + (x) = max{f(x), 0}, is f + R[a, b]? Justify your answer.

Homework 4. (1) If f R[a, b], show that f 3 R[a, b]. If f + (x) = max{f(x), 0}, is f + R[a, b]? Justify your answer. Homework 4 (1) If f R[, b], show tht f 3 R[, b]. If f + (x) = mx{f(x), 0}, is f + R[, b]? Justify your nswer. (2) Let f be continuous function on [, b] tht is strictly positive except finitely mny points

More information

Lecture 3. Limits of Functions and Continuity

Lecture 3. Limits of Functions and Continuity Lecture 3 Limits of Functions nd Continuity Audrey Terrs April 26, 21 1 Limits of Functions Notes I m skipping the lst section of Chpter 6 of Lng; the section bout open nd closed sets We cn probbly live

More information

Advanced Calculus: MATH 410 Uniform Convergence of Functions Professor David Levermore 11 December 2015

Advanced Calculus: MATH 410 Uniform Convergence of Functions Professor David Levermore 11 December 2015 Advnced Clculus: MATH 410 Uniform Convergence of Functions Professor Dvid Levermore 11 December 2015 12. Sequences of Functions We now explore two notions of wht it mens for sequence of functions {f n

More information

The Henstock-Kurzweil integral

The Henstock-Kurzweil integral fculteit Wiskunde en Ntuurwetenschppen The Henstock-Kurzweil integrl Bchelorthesis Mthemtics June 2014 Student: E. vn Dijk First supervisor: Dr. A.E. Sterk Second supervisor: Prof. dr. A. vn der Schft

More information

The Riemann-Stieltjes Integral

The Riemann-Stieltjes Integral Chpter 6 The Riemnn-Stieltjes Integrl 6.1. Definition nd Eistene of the Integrl Definition 6.1. Let, b R nd < b. ( A prtition P of intervl [, b] is finite set of points P = { 0, 1,..., n } suh tht = 0

More information

MATH 409 Advanced Calculus I Lecture 19: Riemann sums. Properties of integrals.

MATH 409 Advanced Calculus I Lecture 19: Riemann sums. Properties of integrals. MATH 409 Advnced Clculus I Lecture 19: Riemnn sums. Properties of integrls. Drboux sums Let P = {x 0,x 1,...,x n } be prtition of n intervl [,b], where x 0 = < x 1 < < x n = b. Let f : [,b] R be bounded

More information

UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE

UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE 1. Pointwise Convergence of Sequence Let E be set nd Y be metric spce. Consider functions f n : E Y for n = 1, 2,.... We sy tht the sequence

More information

Riemann is the Mann! (But Lebesgue may besgue to differ.)

Riemann is the Mann! (But Lebesgue may besgue to differ.) Riemnn is the Mnn! (But Lebesgue my besgue to differ.) Leo Livshits My 2, 2008 1 For finite intervls in R We hve seen in clss tht every continuous function f : [, b] R hs the property tht for every ɛ >

More information

MAA 4212 Improper Integrals

MAA 4212 Improper Integrals Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly well-defined, is too restrictive for mny purposes; there re functions which

More information

Homework 11. Andrew Ma November 30, sin x (1+x) (1+x)

Homework 11. Andrew Ma November 30, sin x (1+x) (1+x) Homewor Andrew M November 3, 4 Problem 9 Clim: Pf: + + d = d = sin b +b + sin (+) d sin (+) d using integrtion by prts. By pplying + d = lim b sin b +b + sin (+) d. Since limits to both sides, lim b sin

More information

Chapter 0. What is the Lebesgue integral about?

Chapter 0. What is the Lebesgue integral about? Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous

More information

The final exam will take place on Friday May 11th from 8am 11am in Evans room 60.

The final exam will take place on Friday May 11th from 8am 11am in Evans room 60. Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23

More information

Regulated functions and the regulated integral

Regulated functions and the regulated integral Regulted functions nd the regulted integrl Jordn Bell jordn.bell@gmil.com Deprtment of Mthemtics University of Toronto April 3 2014 1 Regulted functions nd step functions Let = [ b] nd let X be normed

More information

Problem Set 4: Solutions Math 201A: Fall 2016

Problem Set 4: Solutions Math 201A: Fall 2016 Problem Set 4: s Mth 20A: Fll 206 Problem. Let f : X Y be one-to-one, onto mp between metric spces X, Y. () If f is continuous nd X is compct, prove tht f is homeomorphism. Does this result remin true

More information

1 The Riemann Integral

1 The Riemann Integral The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re

More information

Chapter 6. Riemann Integral

Chapter 6. Riemann Integral Introduction to Riemnn integrl Chpter 6. Riemnn Integrl Won-Kwng Prk Deprtment of Mthemtics, The College of Nturl Sciences Kookmin University Second semester, 2015 1 / 41 Introduction to Riemnn integrl

More information

Exam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH-1105 Instructor: Attila Máté 1

Exam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH-1105 Instructor: Attila Máté 1 Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH-115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixed-point itertion to converge when solving the eqution

More information

7.2 The Definite Integral

7.2 The Definite Integral 7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where

More information

Math 324 Course Notes: Brief description

Math 324 Course Notes: Brief description 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

More information

Math 360: A primitive integral and elementary functions

Math 360: A primitive integral and elementary functions Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:

More information

Integration Techniques

Integration Techniques Integrtion Techniques. Integrtion of Trigonometric Functions Exmple. Evlute cos x. Recll tht cos x = cos x. Hence, cos x Exmple. Evlute = ( + cos x) = (x + sin x) + C = x + 4 sin x + C. cos 3 x. Let u

More information

Week 10: Riemann integral and its properties

Week 10: Riemann integral and its properties Clculus nd Liner Algebr for Biomedicl Engineering Week 10: Riemnn integrl nd its properties H. Führ, Lehrstuhl A für Mthemtik, RWTH Achen, WS 07 Motivtion: Computing flow from flow rtes 1 We observe the

More information

STUDY GUIDE FOR BASIC EXAM

STUDY GUIDE FOR BASIC EXAM STUDY GUIDE FOR BASIC EXAM BRYON ARAGAM This is prtil list of theorems tht frequently show up on the bsic exm. In mny cses, you my be sked to directly prove one of these theorems or these vrints. There

More information

11 An introduction to Riemann Integration

11 An introduction to Riemann Integration 11 An introduction to Riemnn Integrtion The PROOFS of the stndrd lemms nd theorems concerning the Riemnn Integrl re NEB, nd you will not be sked to reproduce proofs of these in full in the exmintion in

More information

Definite integral. Mathematics FRDIS MENDELU

Definite integral. Mathematics FRDIS MENDELU Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the

More information

Riemann Sums and Riemann Integrals

Riemann Sums and Riemann Integrals Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties

More information

A product convergence theorem for Henstock Kurzweil integrals

A product convergence theorem for Henstock Kurzweil integrals A product convergence theorem for Henstock Kurzweil integrls Prsr Mohnty Erik Tlvil 1 Deprtment of Mthemticl nd Sttisticl Sciences University of Albert Edmonton AB Cnd T6G 2G1 pmohnty@mth.ulbert.c etlvil@mth.ulbert.c

More information

Definite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30

Definite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30 Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function

More information

Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives

Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums - 1 Riemnn

More information

For a continuous function f : [a; b]! R we wish to define the Riemann integral

For a continuous function f : [a; b]! R we wish to define the Riemann integral Supplementry Notes for MM509 Topology II 2. The Riemnn Integrl Andrew Swnn For continuous function f : [; b]! R we wish to define the Riemnn integrl R b f (x) dx nd estblish some of its properties. This

More information

arxiv: v1 [math.ca] 11 Jul 2011

arxiv: v1 [math.ca] 11 Jul 2011 rxiv:1107.1996v1 [mth.ca] 11 Jul 2011 Existence nd computtion of Riemnn Stieltjes integrls through Riemnn integrls July, 2011 Rodrigo López Pouso Deprtmento de Análise Mtemátic Fcultde de Mtemátics, Universidde

More information

Riemann Sums and Riemann Integrals

Riemann Sums and Riemann Integrals Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct

More information

Calculus I-II Review Sheet

Calculus I-II Review Sheet Clculus I-II Review Sheet 1 Definitions 1.1 Functions A function is f is incresing on n intervl if x y implies f(x) f(y), nd decresing if x y implies f(x) f(y). It is clled monotonic if it is either incresing

More information

Review of Calculus, cont d

Review of Calculus, cont d Jim Lmbers MAT 460 Fll Semester 2009-10 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some

More information

The Riemann Integral

The Riemann Integral Deprtment of Mthemtics King Sud University 2017-2018 Tble of contents 1 Anti-derivtive Function nd Indefinite Integrls 2 3 4 5 Indefinite Integrls & Anti-derivtive Function Definition Let f : I R be function

More information

Math 118: Honours Calculus II Winter, 2003 List of Theorems. Lemma 5.1 (Partition Refinement) If P and Q are partitions of [a, b] such that Q P, then

Math 118: Honours Calculus II Winter, 2003 List of Theorems. Lemma 5.1 (Partition Refinement) If P and Q are partitions of [a, b] such that Q P, then Mth 118: Honours Clculus II Winter, 2003 List of Theorems Lemm 5.1 (Prtition Refinement) If P nd Q re prtitions of [, b] such tht Q P, then L(P, f) L(Q, f) U(Q, f) U(P, f). Lemm 5.2 (Upper Sums Bound Lower

More information

MATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1

MATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1 MATH34032: Green s Functions, Integrl Equtions nd the Clculus of Vritions 1 Section 1 Function spces nd opertors Here we gives some brief detils nd definitions, prticulrly relting to opertors. For further

More information

A HELLY THEOREM FOR FUNCTIONS WITH VALUES IN METRIC SPACES. 1. Introduction

A HELLY THEOREM FOR FUNCTIONS WITH VALUES IN METRIC SPACES. 1. Introduction Ttr Mt. Mth. Publ. 44 (29), 159 168 DOI: 1.2478/v1127-9-56-z t m Mthemticl Publictions A HELLY THEOREM FOR FUNCTIONS WITH VALUES IN METRIC SPACES Miloslv Duchoň Peter Mličký ABSTRACT. We present Helly

More information

Math 8 Winter 2015 Applications of Integration

Math 8 Winter 2015 Applications of Integration Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl

More information

Math 118: Honours Calculus II Winter, 2005 List of Theorems. L(P, f) U(Q, f). f exists for each ǫ > 0 there exists a partition P of [a, b] such that

Math 118: Honours Calculus II Winter, 2005 List of Theorems. L(P, f) U(Q, f). f exists for each ǫ > 0 there exists a partition P of [a, b] such that Mth 118: Honours Clculus II Winter, 2005 List of Theorems Lemm 5.1 (Prtition Refinement): If P nd Q re prtitions of [, b] such tht Q P, then L(P, f) L(Q, f) U(Q, f) U(P, f). Lemm 5.2 (Upper Sums Bound

More information

Unit #9 : Definite Integral Properties; Fundamental Theorem of Calculus

Unit #9 : Definite Integral Properties; Fundamental Theorem of Calculus Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl

More information

Entrance Exam, Real Analysis September 1, 2009 Solve exactly 6 out of the 8 problems. Compute the following and justify your computation: lim

Entrance Exam, Real Analysis September 1, 2009 Solve exactly 6 out of the 8 problems. Compute the following and justify your computation: lim 1. Let n be positive integers. ntrnce xm, Rel Anlysis September 1, 29 Solve exctly 6 out of the 8 problems. Sketch the grph of the function f(x): f(x) = lim e x2n. Compute the following nd justify your

More information

IMPORTANT THEOREMS CHEAT SHEET

IMPORTANT THEOREMS CHEAT SHEET IMPORTANT THEOREMS CHEAT SHEET BY DOUGLAS DANE Howdy, I m Bronson s dog Dougls. Bronson is still complining bout the textbook so I thought if I kept list of the importnt results for you, he might stop.

More information

38 Riemann sums and existence of the definite integral.

38 Riemann sums and existence of the definite integral. 38 Riemnn sums nd existence of the definite integrl. In the clcultion of the re of the region X bounded by the grph of g(x) = x 2, the x-xis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These

More information

SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL (1 + µ(f n )) f(x) =. But we don t need the exact bound.) Set

SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL (1 + µ(f n )) f(x) =. But we don t need the exact bound.) Set SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL 28 Nottion: N {, 2, 3,...}. (Tht is, N.. Let (X, M be mesurble spce with σ-finite positive mesure µ. Prove tht there is finite positive mesure ν on (X, M such

More information

Review. April 12, Definition 1.2 (Closed Set). A set S is closed if it contains all of its limit points. S := S S

Review. April 12, Definition 1.2 (Closed Set). A set S is closed if it contains all of its limit points. S := S S Review April 12, 2017 1 Definitions nd Some Theorems 1.1 Topology Definition 1.1 (Limit Point). Let S R nd x R. Then x is limit point of S if, for ll ɛ > 0, V ɛ (x) = (x ɛ, x + ɛ) contins n element s S

More information

The First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).

The First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a). The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples

More information

II. Integration and Cauchy s Theorem

II. Integration and Cauchy s Theorem MTH6111 Complex Anlysis 2009-10 Lecture Notes c Shun Bullett QMUL 2009 II. Integrtion nd Cuchy s Theorem 1. Pths nd integrtion Wrning Different uthors hve different definitions for terms like pth nd curve.

More information

MATH 409 Advanced Calculus I Lecture 18: Darboux sums. The Riemann integral.

MATH 409 Advanced Calculus I Lecture 18: Darboux sums. The Riemann integral. MATH 409 Advnced Clculus I Lecture 18: Drboux sums. The Riemnn integrl. Prtitions of n intervl Definition. A prtition of closed bounded intervl [, b] is finite subset P [,b] tht includes the endpoints

More information

NOTES AND PROBLEMS: INTEGRATION THEORY

NOTES AND PROBLEMS: INTEGRATION THEORY NOTES AND PROBLEMS: INTEGRATION THEORY SAMEER CHAVAN Abstrct. These re the lecture notes prepred for prticipnts of AFS-I to be conducted t Kumun University, Almor from 1st to 27th December, 2014. Contents

More information

Main topics for the First Midterm

Main topics for the First Midterm Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 2-3, Sections 4.1-4.8, nd Sections 5.1-5.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the

More information

LECTURE. INTEGRATION AND ANTIDERIVATIVE.

LECTURE. INTEGRATION AND ANTIDERIVATIVE. ANALYSIS FOR HIGH SCHOOL TEACHERS LECTURE. INTEGRATION AND ANTIDERIVATIVE. ROTHSCHILD CAESARIA COURSE, 2015/6 1. Integrtion Historiclly, it ws the problem of computing res nd volumes, tht triggered development

More information

Mapping the delta function and other Radon measures

Mapping the delta function and other Radon measures Mpping the delt function nd other Rdon mesures Notes for Mth583A, Fll 2008 November 25, 2008 Rdon mesures Consider continuous function f on the rel line with sclr vlues. It is sid to hve bounded support

More information

ON THE C-INTEGRAL BENEDETTO BONGIORNO

ON THE C-INTEGRAL BENEDETTO BONGIORNO ON THE C-INTEGRAL BENEDETTO BONGIORNO Let F : [, b] R be differentible function nd let f be its derivtive. The problem of recovering F from f is clled problem of primitives. In 1912, the problem of primitives

More information

Theoretical foundations of Gaussian quadrature

Theoretical foundations of Gaussian quadrature Theoreticl foundtions of Gussin qudrture 1 Inner product vector spce Definition 1. A vector spce (or liner spce) is set V = {u, v, w,...} in which the following two opertions re defined: (A) Addition of

More information

Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as

Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph

More information

A PROOF OF THE FUNDAMENTAL THEOREM OF CALCULUS USING HAUSDORFF MEASURES

A PROOF OF THE FUNDAMENTAL THEOREM OF CALCULUS USING HAUSDORFF MEASURES INROADS Rel Anlysis Exchnge Vol. 26(1), 2000/2001, pp. 381 390 Constntin Volintiru, Deprtment of Mthemtics, University of Buchrest, Buchrest, Romni. e-mil: cosv@mt.cs.unibuc.ro A PROOF OF THE FUNDAMENTAL

More information

Math Solutions to homework 1

Math Solutions to homework 1 Mth 75 - Solutions to homework Cédric De Groote October 5, 07 Problem, prt : This problem explores the reltionship between norms nd inner products Let X be rel vector spce ) Suppose tht is norm on X tht

More information

a n = 1 58 a n+1 1 = 57a n + 1 a n = 56(a n 1) 57 so 0 a n+1 1, and the required result is true, by induction.

a n = 1 58 a n+1 1 = 57a n + 1 a n = 56(a n 1) 57 so 0 a n+1 1, and the required result is true, by induction. MAS221(216-17) Exm Solutions 1. (i) A is () bounded bove if there exists K R so tht K for ll A ; (b) it is bounded below if there exists L R so tht L for ll A. e.g. the set { n; n N} is bounded bove (by

More information

Review on Integration (Secs ) Review: Sec Origins of Calculus. Riemann Sums. New functions from old ones.

Review on Integration (Secs ) Review: Sec Origins of Calculus. Riemann Sums. New functions from old ones. Mth 20B Integrl Clculus Lecture Review on Integrtion (Secs. 5. - 5.3) Remrks on the course. Slide Review: Sec. 5.-5.3 Origins of Clculus. Riemnn Sums. New functions from old ones. A mthemticl description

More information

ODE: Existence and Uniqueness of a Solution

ODE: Existence and Uniqueness of a Solution Mth 22 Fll 213 Jerry Kzdn ODE: Existence nd Uniqueness of Solution The Fundmentl Theorem of Clculus tells us how to solve the ordinry differentil eqution (ODE) du = f(t) dt with initil condition u() =

More information

Abstract inner product spaces

Abstract inner product spaces WEEK 4 Abstrct inner product spces Definition An inner product spce is vector spce V over the rel field R equipped with rule for multiplying vectors, such tht the product of two vectors is sclr, nd the

More information

g i fφdx dx = x i i=1 is a Hilbert space. We shall, henceforth, abuse notation and write g i f(x) = f

g i fφdx dx = x i i=1 is a Hilbert space. We shall, henceforth, abuse notation and write g i f(x) = f 1. Appliction of functionl nlysis to PEs 1.1. Introduction. In this section we give little introduction to prtil differentil equtions. In prticulr we consider the problem u(x) = f(x) x, u(x) = x (1) where

More information

Line Integrals. Partitioning the Curve. Estimating the Mass

Line Integrals. Partitioning the Curve. Estimating the Mass Line Integrls Suppose we hve curve in the xy plne nd ssocite density δ(p ) = δ(x, y) t ech point on the curve. urves, of course, do not hve density or mss, but it my sometimes be convenient or useful to

More information

Notes on length and conformal metrics

Notes on length and conformal metrics Notes on length nd conforml metrics We recll how to mesure the Eucliden distnce of n rc in the plne. Let α : [, b] R 2 be smooth (C ) rc. Tht is α(t) (x(t), y(t)) where x(t) nd y(t) re smooth rel vlued

More information

Chapter 4. Lebesgue Integration

Chapter 4. Lebesgue Integration 4.2. Lebesgue Integrtion 1 Chpter 4. Lebesgue Integrtion Section 4.2. Lebesgue Integrtion Note. Simple functions ply the sme role to Lebesgue integrls s step functions ply to Riemnn integrtion. Definition.

More information

Numerical Integration

Numerical Integration Chpter 5 Numericl Integrtion Numericl integrtion is the study of how the numericl vlue of n integrl cn be found. Methods of function pproximtion discussed in Chpter??, i.e., function pproximtion vi the

More information

f(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral

f(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one

More information

a n+2 a n+1 M n a 2 a 1. (2)

a n+2 a n+1 M n a 2 a 1. (2) Rel Anlysis Fll 004 Tke Home Finl Key 1. Suppose tht f is uniformly continuous on set S R nd {x n } is Cuchy sequence in S. Prove tht {f(x n )} is Cuchy sequence. (f is not ssumed to be continuous outside

More information

Analysis III. Ben Green. Mathematical Institute, Oxford address:

Analysis III. Ben Green. Mathematical Institute, Oxford  address: Anlysis III Ben Green Mthemticl Institute, Oxford E-mil ddress: ben.green@mths.ox.c.uk 2000 Mthemtics Subject Clssifiction. Primry Contents Prefce 1 Chpter 1. Step functions nd the Riemnn integrl 3 1.1.

More information

MATH 174A: PROBLEM SET 5. Suggested Solution

MATH 174A: PROBLEM SET 5. Suggested Solution MATH 174A: PROBLEM SET 5 Suggested Solution Problem 1. Suppose tht I [, b] is n intervl. Let f 1 b f() d for f C(I; R) (i.e. f is continuous rel-vlued function on I), nd let L 1 (I) denote the completion

More information

SYDE 112, LECTURES 3 & 4: The Fundamental Theorem of Calculus

SYDE 112, LECTURES 3 & 4: The Fundamental Theorem of Calculus SYDE 112, LECTURES & 4: The Fundmentl Theorem of Clculus So fr we hve introduced two new concepts in this course: ntidifferentition nd Riemnn sums. It turns out tht these quntities re relted, but it is

More information

The Bochner Integral and the Weak Property (N)

The Bochner Integral and the Weak Property (N) Int. Journl of Mth. Anlysis, Vol. 8, 2014, no. 19, 901-906 HIKARI Ltd, www.m-hikri.com http://dx.doi.org/10.12988/ijm.2014.4367 The Bochner Integrl nd the Wek Property (N) Besnik Bush Memetj University

More information

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007 A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus

More information

Functions of bounded variation

Functions of bounded variation Division for Mthemtics Mrtin Lind Functions of bounded vrition Mthemtics C-level thesis Dte: 2006-01-30 Supervisor: Viktor Kold Exminer: Thoms Mrtinsson Krlstds universitet 651 88 Krlstd Tfn 054-700 10

More information