- 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 intervl I = [, b], prtition of I is ny finite strictly incresing sequence of points P = {x 0, x 1,...x n 1, x n } such tht = x 0 nd b = x n. The mesh of the prtition is defined by meshp = mx 1 j n (x j x j 1 ). Ech prtition P = {x 0, x 1,...x n 1, x n } of I decomposes I into n subintervls I j = [x j 1, j ], j=1,2,...,n, such tht I j I k = { xj, if k = j + 1 φ, if k j or k j + 1 Ech such decomposition of I into subintervls is clled subdivision of I.
Prtition Riemnn Stieltjes Sums Refinement Definition Given function f tht is bounded nd defined on the closed intervl I = [, b], function α tht is defined nd monotoniclly incresing on I, nd prtition P = {x 0, x 1,...x n 1, x n } of I. Let M j = sup x Ij f (x); m j = inf x Ij f (x), for I j = [x j 1, x j ]. Then, upper nd lower Riemnn Stieltjes sum of f over α with respect to the prtition P is defined by U(P, f, α) = n M j α j, L(P, f, α) = j=1 n m j α j j=1 where α j = (α(x j ) α(x j 1 )).
Prtition Riemnn Stieltjes Sums Refinement Definition For prtition P k = {x 0, x 1,...x k 1, x k } of I = [, b]. If P n nd P m re prtitions of [,b] hving n + 1 nd m + 1 points, respectivly, nd P n P m, then P m is sid to be refinement of P n. If the prtitions P n nd P m re chosen independently, then the prtition P n P m is clled common refinement of P n nd P m.
Boundedness of Riemnn Stieltjes Sums Remrk Definition Bounds on Riemnn Stieltjes Integrls Lemm Our next result reltes the Riemnn sums tken over vrious prtitions of n intervl. Suppose f is rel vlued bounded function defined on I=[,b], nd prtition P = {x 0, x 1,...x n 1, x n } of I. Then m(α(b) α()) L(P, f, α) U(P, f, α) M(α(b) α()) nd L(P, f, α) L(P, f, α) U(P, f, α) U(P, f, α) for ny refinement P of P.
Boundedness of Riemnn Stieltjes Sums Remrk Definition Bounds on Riemnn Stieltjes Integrls The lemm ssures tht lower nd upper Riemnn Stieltjes sums will remin bounded bove by l(i)sup x I f (x) nd bounded below by l(i)inf x I f (x). sup {L(P, f, α); P P} nd inf {U(P, f, α); P P} exists. with the refinement of prtition lower sum increses while upper sum decreses.
Boundedness of Riemnn Stieltjes Sums Remrk Definition Bounds on Riemnn Stieltjes Integrls Definition Suppose tht f is rel vlued bounded function defined on I = [, b], P = P[, b] be the set of ll prtitions of [, b] nd α monotoniclly incresing function defined on I. Then the upper nd lower Riemnn Sieltjes integrls re defined by f (x)dα(x) = inf P U(P, f, α); f (x)dα(x) = sup P L(P, f, α), respectively. If f (x)dα(x) = f (x)dα(x) then f is sid to be Riemnn Stieltjes integrble.
Boundedness of Riemnn Stieltjes Sums Remrk Definition Bounds on Riemnn Stieltjes Integrls It is rther short jump from previous Lemm to upper nd lower bounds on the Riemnn integrls. They re given by: Theorem Suppose tht f is bounded rel vlued function defined on I = [, b], α monotoniclly incresing function on I, nd m f (x) M for ll x I. Then m(α(b) α()) f (x)dα(x) f (x)dα(x) M(α(b) α()). Furthermore, if f is Riemnn Stieltjes integrble on I, then m(α(b) α()) f (x)dα(x) M(α(b) α()).
Necessry nd Sufficient Condition Clss of Riemnn Stieltjes Integrble Functions Remrk It is not worth our while to grind out some tedious processes in order to show tht specil functions re integrble. Towrds this end, we wnt to seek some properties of functions tht would gurntee integrbility. Theorem Suppose tht f is function tht is bounded on n intervl I = [, b] nd α is monotoniclly incresing on I. Then f R(α) on I if nd only if for every ǫ > 0 there exists prtition P of I such tht U(P, f, α) L(P, f, α) < ǫ. (1)
Necessry nd Sufficient Condition Clss of Riemnn Stieltjes Integrble Functions Remrk Proof : Prt. Let f be function tht is bounded on n intervl I = [, b] nd α be monotoniclly incresing on I. Suppose tht for every ǫ > 0 there exists prtition P of I such tht U(P, f, α) L(P, f, α) < ǫ. From the definition of the Riemnn Stieltjes integrl nd Lemm 4, 0 f (x)dα(x) f (x)dα(x) U(P, f, α) L(P, f, α) < ǫ. Since ǫ > 0 ws chosen rbitrrily, it follows tht f (x)dα(x) f (x)dα(x) = 0 f R(α).
Necessry nd Sufficient Condition Clss of Riemnn Stieltjes Integrble Functions Remrk Proof: Prt b. Conversely suppose tht f R(α) nd let ǫ > 0 is geven. For ǫ 2 > 0 definition of supremum nd infimum suggests there exists prtitions P 1, P 2 P[, b] such tht U(P 1, f, α) < f (x)dα(x)+ ǫ 2 & L(P 2, f, α) > Let P be the common refinement of P 1 nd P 2, then U(P, f, α) U(P 1, f, α) < L(P, f, α) L(P 2, f, α) > f (x)dα(x) + ǫ 2, nd f (x)dα(x) ǫ 2. f (x)dα(x) ǫ 2.
Necessry nd Sufficient Condition Clss of Riemnn Stieltjes Integrble Functions Remrk Proof: Prt b continues. Moreover, U(P, f, α) < f (x)dα(x) + ǫ 2, nd L(P, f, α) < f (x)dα(x) + ǫ 2. Combining bove inequlities U(P, f, α) L(P, f, α) < = ǫ. ( f R(α) which implies f (x)dα(x) f (x)dα(x) + ǫ f (x)dα(x) = ) f (x)dα(x).
Necessry nd Sufficient Condition Clss of Riemnn Stieltjes Integrble Functions Remrk As firly immedite consequence of preceding results, we hve Corollry Suppose tht f is bounded on [, b] nd α is monotoniclly incresing on [, b]. 1 If (1) holds for some prtition P P[, b] nd ǫ > 0, then (1) holds for every refinement P of P. 2 If (1) holds for some prtition P P[, b] nd s j, t j re rbitrry points in I j = [x j 1, x j ], then n j=1 f (s j) f (t j ) α j < ǫ. 3 If f R(α), eqution (1) holds for the prtition P P[, b] nd t j is n rbitrry point in I j = [x j 1, x j ], then n j=1 f (t j ) α j fdα < ǫ.
Necessry nd Sufficient Condition Clss of Riemnn Stieltjes Integrble Functions Remrk Proof- Prt 1. For ny refinement P of P, Lemm 4 gives L(P, f, α) L(P, f, α) U(P, f, α) U(P, f, α). From this it is esy to observe tht U(P, f, α) L(P, f, α) U(P, f, α) L(P, f, α) < ǫ, from (1).
Necessry nd Sufficient Condition Clss of Riemnn Stieltjes Integrble Functions Remrk Proof- Prt 2. Suppose tht M j = sup x Ij f (x), m j = inf x Ij f (x) nd s j, t j re rbitrry points in I j, j = 1, 2,...,n. Then f (s j ), f (t j ) [m j, M j ] nd hence f (s j ) f (t j ) M j m j, i.e. n n n f (s j ) f (t j ) α j M j α j m j α j, j=1 j=1 j=1 = U(P, f, α) L(P, f, α), < ǫ, from (1).
Necessry nd Sufficient Condition Clss of Riemnn Stieltjes Integrble Functions Remrk Proof- Prt 3. From the definition of Riemnn Stieltjes integrl nd Lemm 4, L(P, f, α) f (x)dα(x) U(P, f, α). (2) Moreover, for m j, M j re s defined erlier nd j = 1, 2,...,n, t j [x j 1, x j ] therefore f (t j ) [m j, M j ]. From this it is esy to construct the inequlity L(P, f, α) n f (t j ) α j U(P, f, α). (3) j=1 From inequlities (2) nd (3) it cn be concluded tht n j=1 f (t j) α j fdα U(P, f, α) L(P, f, α) < ǫ.
Necessry nd Sufficient Condition Clss of Riemnn Stieltjes Integrble Functions Remrk So fr we hve gone through results which will be useful to us whenever we hve wy of closing the gp between functionl vlues on the sme intervls. Next two results give us two big clsses of integrble functions in the sense of Riemnn Stieltjes integrtion. Theorem If f is function tht is continuous on the intervl I = [, b], then f is Riemnn Stieltjes integrble on [,b].
Necessry nd Sufficient Condition Clss of Riemnn Stieltjes Integrble Functions Remrk Proof. Let α be monotoniclly incresing on I nd f be continuous on I. Suppose tht ǫ > 0 is given. Then there exists n η > 0 such tht [α(b) α()]η < ǫ. Clerly I = [, b] is compct nd therefore f is uniformly continuous in [, b]. Hence, there exists δ > 0 such tht x, t I, x t < δ f (x) f (t) < η.
Necessry nd Sufficient Condition Clss of Riemnn Stieltjes Integrble Functions Remrk Proof Continues. Let P = {x 0, x 1,...x n 1, x n } be the prtition of I for which mesh P < δ i.e., x j = (x j x j 1 ) < δ for ny j. Since f is uniformly continuous this implies tht M j m j < η for ny i. Consider U(P, f, α) L(P, f, α) = < η < ǫ. n (M j m j ) α j j=1 n α j j=1
Necessry nd Sufficient Condition Clss of Riemnn Stieltjes Integrble Functions Remrk Proof Continues. In view of the Integrbility Criterion, f R(α). Becuse α ws rbitrry, we conclude tht f is Riemnn Stieltjes Integrble (with respect to ny monotoniclly incresing function on [,b].
Necessry nd Sufficient Condition Clss of Riemnn Stieltjes Integrble Functions Remrk As n immedite consequence of the bove theorem, we hve Corollry If f is function tht is monotonic on the intervl I = [, b] nd α is continuous nd monotoniclly incresing on I, then f R(α).
Necessry nd Sufficient Condition Clss of Riemnn Stieltjes Integrble Functions Remrk So we cn summrize the results s follows; 1 Bounded nd continous function f cn be integrted with respect to ny monotonic incresing function α. 2 Bounded nd monotonic function f cn be integrted with repsect to ny monotonic incresing nd continous function α.
1 Wlter Rudin : Principles of Mthemticl Anlysis, McGrw Hill Pulishers. 2 T. Apostol, Mthemticl Anlysis, Nros Publiction. 3 A. Kushik, Lecture Notes, Directorte of Distnce Eduction, Kurukshetr University Kurukshetr.
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