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


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1 MATH 409 Advnced Clculus I Lecture 18: Drboux sums. The Riemnn integrl.
2 Prtitions of n intervl Definition. A prtition of closed bounded intervl [, b] is finite subset P [,b] tht includes the endpoints nd b. Let x 0,x 1,...,x n be the list of ll elements of P ordered so tht x 0 < x 1 < < x n (note tht x 0 = nd x n = b). These points split the intervl [, b] into finitely mny subintervls [x 0,x 1 ], [x 1,x 2 ],..., [x n 1,x n ]. The norm of the prtition P, denoted P, is the mximum of lengths of those subintervls: P = mx 1 j n x j x j 1. Given two prtitions P nd Q of the sme intervl, we sy tht Q is refinement of P (or tht Q is finer thn P) if P Q. Observe tht P Q implies Q P. For ny two prtitions P nd Q of the intervl [,b], the union P Q is lso prtition tht refines both P nd Q.
3 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 function. Definition. The upper Drboux sum (or the upper Riemnn sum) of the function f over the prtition P is the number n U(f,P) = M j (f) j, where j = x j x j 1 nd M j (f) = supf([x j 1,x j ]) for j = 1,2,...,n. Likewise, the lower Drboux sum (or the lower Riemnn sum) of f over P is the number n L(f,P) = m j (f) j, j=1 j=1 where m j (f) = inff([x j 1,x j ]) for j = 1,2,...,n.
4 Properties of the Drboux sums L(f,P) U(f,P). Indeed, inff(j) supf(j) for ny subintervl J [,b]. U(f,P) supf([,b]) (b ). We hve supf(j) supf([,b]) for ny subintervl J [,b]. Then supf(j) J supf([,b]) J, where J is the length of J. Summing up over ll subintervls J creted by the prtition P, we obtin U(f,P) supf([,b]) (b ). inff([,b]) (b ) L(f,P). The proof is nlogous to the previous one. Remrk. Observe tht supf([,b]) (b ) = U(f,P 0 ) nd inff([,b]) (b ) = L(f,P 0 ), where P 0 is the trivil prtition: P 0 = {,b}.
5 Properties of the Drboux sums L(f,P) L(f,Q) U(f,Q) U(f,P) for ny prtition Q tht refines P. Every subintervl J creted by the prtition P is the union of one or more subintervls J 1,J 2,...,J k creted by Q. Since supf(j i ) supf(j) for 1 i k, it follows tht k i=1 supf(j i) J i supf(j) k i=1 J i = supf(j) J. Summing up this inequlity over ll subintervls J, we obtin U(f,Q) U(f,P). The inequlity L(f,P) L(f,Q) is proved in similr wy. L(f,P) U(f,Q) for ny prtitions P nd Q. Since the prtition P Q refines both P nd Q, it follows from the bove tht L(f,P) L(f,P Q) nd U(f,P Q) U(f,Q). Besides, L(f,P Q) U(f,P Q).
6 Upper nd lower integrls Suppose f : [,b] R is bounded function. Definition. The upper integrl of f on [, b], denoted f(x)dx or (U) f(x)dx, is the number inf{u(f,p) P is prtition of [,b]}. Similrly, the lower integrl of f on [, b], denoted f(x)dx or (L) f(x)dx, is the number sup{l(f,p) P is prtition of [,b]}. Remrk. Since < L(f,P) U(f,Q) < + for ll prtitions P nd Q, it follows tht < (L) f(x)dx (U) f(x)dx < +.
7 Integrbility Definition. A bounded function f : [, b] R is clled integrble (or Riemnn integrble) on the intervl [, b] if the upper nd lower integrls of f on [, b] coincide. The common vlue is clled the integrl of f on [, b] (or over [, b]) nd denoted f(x)dx. Theorem A bounded function f : [,b] R is integrble on [,b] if nd only if for every ε > 0 there is prtition P ε of [,b] such tht U(f,P ε ) L(f,P ε ) < ε.
8 Proof of the theorem: The if prt of the theorem follows since 0 (U) f(x)dx (L) f(x)dx U(f,P) L(f,P) for ny prtition P. Conversely, ssume tht f is integrble on [,b]. Given ε > 0, there exists prtition P of [,b] such tht U(f,P) < f(x)dx + ε 2. Also, there exists prtition Q of [, b] such tht L(f,Q) > f(x)dx ε 2. Then U(f,P) L(f,Q) < ε. Now P Q is prtition of [,b] tht refines both P nd Q. It follows tht U(f,P Q) U(f,P) nd L(f,P Q) L(f,Q). Hence U(f,P Q) L(f,P Q) U(f,P) L(f,Q) < ε.
9 Exmples Constnt function f(x) = c is integrble on ny intervl [,b] nd f(x)dx = c(b ). Indeed, for the trivil prtition P 0 = {,b} we obtin U(f,P 0 ) = c(b ) = L(f,P 0 ). { 1 if x > 0, Step function f(x) = 0 if x 0 integrble on [ 1,1] nd 1 1 f(x)dx = 1. For ny ε (0,1) consider prtition P ε = { 1, ε,ε,1}. Then U(f,P ε ) = 1+ε nd L(f,P ε ) = 1 ε. is
10 Exmples { 1 if x Q, Dirichlet function f(x) = 0 if x R\Q is not integrble on ny intervl [, b]. Indeed, ny subintervl of [, b] contins both rtionl nd irrtionl points. Therefore U(f,P) = b nd L(f,P) = 0 for ll prtitions of [,b]. Riemnn function f(x) = is integrble on ny intervl [, b]. { 1/q if x = p/q, 0 if x R\Q For ny δ > 0 the intervl [, b] contins only finitely mny points y 1,y 2,...,y k such tht f(y i ) δ. Let P δ be prtition of [,b] tht includes points y i ±δ/k. Then L(f,P δ ) = 0 nd U(f,P δ ) 2δ +δ(b ).
11 Continuity = integrbility Theorem If function f : [,b] R is continuous on the intervl [, b], then it is integrble on [, b]. Proof: Since the function f is continuous, it is bounded on [, b]. Furthermore, f is uniformly continuous on [, b]. Therefore for every ε > 0 there exists δ > 0 such tht x y < δ implies f(x) f(y) < ε/(b ) for ll x,y [,b]. Obviously, there exists prtition P = {x 0,x 1,...,x n } of [,b] tht stisfies P < δ. Let J = [x j 1,x j ] be n rbitrry subintervl of [,b] creted by P. By the Extreme Vlue Theorem, there re points x,x + J such tht f(x + ) = supf(j) nd f(x ) = inff(j). Since P < δ, the length of J stisfies J < δ. Then x + x J < δ so tht f(x + ) f(x ) < ε/(b ). It follows tht supf(j) J inff(j) J < ε J /(b ). Summing up the ltter inequlity over ll subintervls J, we obtin tht U(f,P) L(f,P) < ε. Thus f is integrble.
12 Riemnn sums Definition. A Riemnn sum of function f : [,b] R with respect to prtition P = {x 0,x 1,...,x n } of [,b] generted by smples t j [x j 1,x j ] is sum S(f,P,t j ) = n j=1 f(t j)(x j x j 1 ). Remrk. Note tht the function f need not be bounded. If f is bounded, then L(f,P) S(f,P,t j ) U(f,P) for ny choice of smples t j. Definition. The Riemnn sums S(f,P,t j ) converge to limit I(f) s P 0 if for every ε > 0 there exists δ > 0 such tht P < δ implies S(f,P,t j ) I(f) < ε for ny prtition P nd choice of smples t j. Theorem The Riemnn sums S(f,P,t j ) converge to limit I(f) s P 0 if nd only if the function f is integrble on [,b] nd I(f) = f(x)dx.
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