MATH 409 Advanced Calculus I Lecture 19: Riemann sums. Properties of integrals.
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1 MATH 409 Advnced Clculus I Lecture 19: Riemnn sums. Properties of integrls.
2 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.
3 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. For ny prtitions P nd Q of the intervl [,b], L(f,P) (L) f(x)dx (U) f(x)dx U(f,Q).
4 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]). 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. Theorem If function is continuous on the intervl [, b], then it is integrble on [, b].
5 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 the norm 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.
6 Drboux sums nd Riemnn sum x 0 x 1 x 2 x 3 x 4 x 0 t 1 x 1 t 2 x 2 t 3 x 3 t 4 x 4
7 Proof of the theorem ( only if ): Assume tht the Riemnn sums S(f,P,t j ) converge to limit I(f) s P 0. Given ε > 0, we choose δ > 0 so tht for every prtition P with P < δ, we hve S(f,P,t j ) I(f) < ε for ny choice of smples t j. Let t j be different set of smples for the sme prtition P. Then S(f,P, t j ) I(f) < ε. We cn choose the smples t j, t j so tht f(t j ) is rbitrrily close to supf([x j 1,x j ]) while f( t j ) is rbitrrily close to inff([x j 1,x j ]). Tht wy S(f,P,t j ) gets rbitrrily close to U(f,P) while S(f,P, t j ) gets rbitrrily close to L(f,P). Hence it follows from the bove inequlities tht U(f,P) I(f) ε nd L(f,P) I(f) ε. As consequence, U(f,P) L(f,P) 2ε. In prticulr, the function f is bounded. We conclude tht f is integrble. Let I = f(x)dx. The number I lies between L(f,P) nd U(f,P). The inequlities U(f,P) L(f,P) 2ε nd U(f,P) I(f) ε imply tht I I(f) 3ε. As ε cn be rbitrrily smll, I = I(f).
8 Integrtion s liner opertion Theorem 1 If functions f,g re integrble on n intervl [,b], then the sum f +g is lso integrble on [,b] nd ( ) b f(x)+g(x) dx = f(x)dx + g(x)dx. Theorem 2 If function f is integrble on [,b], then for ech α R the sclr multiple αf is lso integrble on [,b] nd αf(x)dx = α f(x)dx.
9 Proof of Theorems 1 nd 2: Let I(f) denote the integrl of f nd I(g) denote the integrl of g over [,b]. The key observtion is tht the Riemnn sums depend linerly on function. Nmely, S(f +g,p,t j ) = S(f,P,t j )+S(g,P,t j ) nd S(αf,P,t j ) = α S(f,P,t j ) for ny prtition P of [,b] nd choice of smples t j. It follows tht S(f +g,p,t j ) I(f) I(g) S(f,P,t j ) I(f) + S(g,P,t j ) I(g), S(αf,P,t j ) αi(f) = α S(f,P,t j ) I(f). As P 0, the Riemnn sums S(f,P,t j ) nd S(g,P,t j ) get rbirrily close to I(f) nd I(g), respectively. Then S(f +g,p,t j ) will be getting rbitrrily close to I(f)+I(g) while S(αf,P,t j ) will be getting rbitrrily close to αi(f). Thus I(f)+I(g) is the integrl of f +g nd αi(f) is the integrl of αf over [, b].
10 Theorem If function f is integrble on [,b], then it is integrble on ech subintervl [c,d] [,b]. Proof: Since f is integrble on the intervl [, b], for ny ε > 0 there is prtition P ε of [,b] such tht U(f,P ε ) L(f,P ε ) < ε. Given subintervl [c,d] [,b], let P ε = P ε {c,d} nd Q ε = P ε [c,d]. Then P ε is prtition of [,b] tht refines P ε. Hence U(f,P ε ) L(f,P ε ) U(f,P ε) L(f,P ε ) < ε. Since Q ε is prtition of [c,d] contined in P ε, it follows tht U(f,Q ε ) L(f,Q ε ) U(f,P ε) L(f,P ε) < ε. We conclude tht f is integrble on [c,d].
11 Theorem If function f is integrble on [,b] then for ny c (,b), f(x)dx = c f(x)dx + c f(x)dx. Proof: Since f is integrble on the intervl [, b], it is lso integrble on subintervls [, c] nd [c, b]. Let P be prtition of [,c] nd {t j } be some smples for tht prtition. Further, let Q be prtition of [c,b] nd {τ i } be some smples for tht prtition. Then P Q is prtition of [,b] nd {t j } {τ i } re smples for it. The key observtion is tht S(f,P Q,{t j } {τ i }) = S(f,P,t j )+S(f,Q,τ i ). If P 0 nd Q 0, then P Q = mx( P, Q ) tends to 0 s well. Therefore the Riemnn sums in the ltter equlity will converge to the integrls f(x)dx, c f(x)dx, nd f(x)dx, respectively. c
12 Theorem If function f is integrble on [,b] nd f([,b]) [A,B], then for ech continuous function g : [A,B] R the composition g f is lso integrble on [, b]. Corollry If functions f nd g re integrble on [,b], then so is fg. Proof: We hve (f +g) 2 = f 2 +g 2 +2fg. Since f nd g re integrble on [,b], so is f +g. Since h(x) = x 2 is continuous function on R, the compositions h f = f 2, h g = g 2, nd h (f +g) = (f +g) 2 re integrble on [,b]. Then fg = 1 2 (f +g)2 1 2 f g2 is integrble on [, b] s liner combintion of integrble functions.
13 Comprison Theorem for integrls Theorem If functions f,g re integrble on [,b] nd f(x) g(x) for ll x [,b], then f(x)dx g(x)dx. Proof: Since f g on the intervl [, b], it follows tht S(f,P,t j ) S(g,P,t j ) for ny prtition P of [,b] nd choice of smples t j. As P 0, the sum S(f,P,t j ) gets rbitrrily close to the integrl of f while S(g,P,t j ) gets rbitrrily close to the integrl of g. The theorem follows. Corollry 1 If f is integrble on [,b] nd f(x) 0 for x [,b], then f(x)dx 0.
14 Corollry 2 If f is integrble on [,b] nd m f(x) M for x [,b], then m(b ) f(x)dx M(b ). Corollry 3 If f is integrble on [,b], then the function f is lso integrble on [, b] nd f(x)dx f(x) dx. Proof: The function f is the composition of f with continuous function g(x) = x. Therefore f is integrble on [,b]. Since f(x) f(x) f(x) for x [,b], the Comprison Theorem for integrls implies tht f(x) dx f(x)dx f(x) dx.
15 Integrl with vrible limit Suppose f : [,b] R is n integrble function. For ny x [,b] let F(x) = x f(t)dt (we ssume tht F() = 0). Theorem The function F is well defined nd continuous on [,b]. Proof: Since the function f is integrble on [, b], it is lso integrble on ech subintervl of [, b]. Hence the function F is well defined on [,b]. Besides, f is bounded: f(t) M for some M > 0 nd ll t [,b]. For ny x,y [,b], x y, we hve y f(t)dt = x f(t)dt + y f(t)dt. It x follows tht y y F(y) F(x) = f(t)dt f(t) dt M y x. x Thus F is Lipschitz function on [,b], which implies tht F is uniformly continuous on [, b]. x
16 Sets of mesure zero Definition. A subset E of the rel line R is sid to hve mesure zero if for ny ε > 0 the set E cn be covered by countbly mny open intervls J 1,J 2,... such tht n=1 J n < ε. Exmples. Any countble set hs mesure zero. Indeed, suppose E is countble set nd let x 1,x 2,... be list of ll elements of E. Given ε > 0, let ( J n = x n ε 2 n+1, x n + ε ), n = 1,2,... 2 n+1 Then E J 1 J 2... nd J n = ε/2 n for ll n N so tht n=1 J n = ε. A nondegenerte intervl [, b] is not set of mesure zero. There exist sets of mesure zero tht re of the sme crdinlity s R.
17 Lebesgue s criterion for Riemnn integrbility Definition. Suppose P(x) is property depending on x S, where S R. We sy tht P(x) holds for lmost ll x S (or lmost everywhere on S) if the set {x S P(x) does not hold} hs mesure zero. Theorem A function f : [,b] R is Riemnn integrble on the intervl [,b] if nd only if f is bounded on [, b] nd continuous lmost everywhere on [, b].
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