Properties of the Riemann Integral


 Egbert Ramsey
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1 Properties of the Riemnn Integrl Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University Februry 15, 2018
2 Outline 1 Some Infimum nd Supremum Properties 2 Riemnn Integrl Properties 3 Wht Functions Are Riemnn Integrble? 4 Homework
3 Some Infimum nd Supremum Properties Theorem If f is bounded function on the finite intervl [, b], then 1 sup x b f (x) = inf x b ( f (x)) nd sup x b ( f (x)) = inf x b (f (x)) 2 sup (f (x) f (y)) = sup sup (f (x) f (y)) x,y [,b] y [,b] x [,b] = sup sup (f (x) f (y)) = M m x [,b] y [,b] where M = sup x b f (x) nd m = inf x b f (x). 3 sup x,y [,b] f (x) f (y) = M m First let Q = sup x b ( f ) nd q = inf x b ( f ).
4 Some Infimum nd Supremum Properties (1): Let (f (x n ) be sequence which converges to M. Then since f (x n ) q for ll n, letting n, we find M q. Now let ( f (z n )) be sequence which converges to q. Then, we hve f (z n ) M for ll n nd letting n ]infty, we hve q M or q M. Combining, we see q = M which is the first prt of the sttement; i.e. sup x b f (x) = inf x b ( f (x)). Now just replce ll the f s by f s in this to get sup x b ( f (x)) = inf x b ( f (x)) or sup x b ( f (x)) = inf x b (f (x)) which is the other identity. (2): We know f (x) f (y) ( ) sup x b f (x) f (y) = M f (y) But f (y) inf y [,b] f (y) = m, so f (x) f (y) M f (y) M m
5 Some Infimum nd Supremum Properties Thus, sup (f (x) f (y)) M m x,y [,b] So one side of the inequlity is cler. Now let f (x n ) be sequence converging to M nd f (y n ) be sequence converging to m. Then, we hve Letting n, we see f (x n ) f (y n ) sup (f (x) f (y)) x,y [,b] M m sup (f (x) f (y)) x,y [,b] This is the other side of the inequlity. We hve thus shown tht the equlity is vlid.
6 Some Infimum nd Supremum Properties (3): Note f (x) f (y) = { f (x) f (y), f (x) f (y) f (y) f (x), f (x) < f (y) In either cse, we hve f (x) f (y) M m for ll x, y using Prt (2) implying tht sup x,y f (x) f (y) M m. To see the reverse inequlity holds, we first note tht if M = m, we see the reverse inequlity holds trivilly s sup x,y f (x) f (y) 0 = M m. Hence, we my ssume without loss of generlity tht the gp M m is positive. Then, using the STL nd ITL, given 0 < 1/j < 1/2(M m), there exist, s j, t j [, b] such tht M 1/(2j) < f (s j ) nd m + 1/(2j) > f (t j ), so tht f (s j ) f (t j ) > M m 1/j. By our choice of j, these terms re positive nd so we lso hve f (s j ) f (t j ) > M m 1/j.
7 Some Infimum nd Supremum Properties It follows tht sup f (x) f (y) f (s j ) f (t j ) > M m 1/j. x,y [,b] Since we cn mke 1/j rbitrrily smll, this implies tht sup f (x) f (y) M m. x,y [x j 1,x j ] This estblishes the reverse inequlity nd proves the clim. We re now redy to look t some of the properties of the Riemnn Integrl.
8 Riemnn Integrl Properties Theorem Let f, g RI [, b]. Then (1) f RI [, b]; (2) b f (x)dx b f dx; (3) f + = mx{f, 0} RI [, b]; (4) f = mx{ f, 0} RI [, b]; (5) b f (x)dx = b [f + (x) f (x)]dx = b f + (x)dx b f (x)dx b f (x) dx = b [f + (x) + f (x)]dx = b f + (x)dx + b f (x)dx; (6) f 2 RI [, b]; (7) fg RI [, b]; (8) If there exists m, M such tht 0 < m f M, then 1/f RI [, b].
9 Riemnn Integrl Properties (1) f RI [, b]: Note given prtition π = {x 0 =, x 1,..., x p = b}, for ech j = 1,..., p, using the result bove, we know sup (f (x) f (y)) = M j m j x,y [x j 1,x j ] Now, let m j nd M j be defined by m j = inf f (x), [x j 1,x j ] M j = sup f (x). [x j 1,x j ] Then, pplying the first Theorem to f, we hve ( ) M j m j = sup x,y [x j 1,x j ] f (x) f (y).
10 Riemnn Integrl Properties For ech j = 1,..., p, we hve M j m j = sup f (x) f (y). x,y [x j 1,x j ] So, since f (x) f (y) f (x) f (y) for ll x, y, it follows tht M j m j M j m j. This implies π (M j m j ) x j π (M j m j ) x j. This mens U( f, π) L( f, π) U(f, π) L(f, π) for the chosen π. Since f is integrble by hypothesis, we know the Riemnn criterion must lso hold for f. Thus, given ɛ > 0, there is prtition π 0 so tht U(f, π) L(f, π) < ɛ for ny refinement π of π 0. Therefore f lso stisfies the Riemnn Criterion nd so f is Riemnn integrble.
11 Riemnn Integrl Properties (2) b f (x)dx b f dx: We hve f f nd f f, so tht b b from which it follows tht f (x)dx b f (x)dx b f (x) dx f (x) dx, nd so b f (x) dx b f (x)dx b b f b f f (x) dx
12 Riemnn Integrl Properties (3) f + = mx{f, 0} RI [, b] nd (4) f = mx{ f, 0} RI [, b]: This follows from the fcts tht f + = 1 2 ( f +f ) nd f = 1 2 ( f f ) nd the Riemnn integrl is liner mpping. (5) b f (x)dx = b [f + (x) f (x)]dx = b f + (x)dx b f (x)dx b f (x) dx = b [f + (x) + f (x)]dx = b f + (x)dx + b f (x)dx: This follows from the fcts tht f = f + f nd f = f + + f nd the linerity of the integrl. (6): f 2 RI [, b] Note tht, since f is bounded, there exists K > 0 such tht f (x) K for ll x [, b]. Applying our infimum/ supremum properties theorem to f 2, we hve sup (f 2 (x) f 2 (y)) = M j (f 2 ) m j (f 2 ) x,y [x j 1,x j ]
13 Riemnn Integrl Properties where [x j 1, x j ] is subintervl of given prtition π nd M j (f 2 ) = sup x [xj 1,x j ] f 2 (x) nd m j (f 2 ) = inf x [xj 1,x j ] f 2 (x). Thus, for this prtition, we hve U(f 2, π) L(f 2, π) = π (M j (f 2 ) m j (f 2 )) x j But we lso know sup (f 2 (x) f 2 (y)) = sup (f (x) + f (y))(f (x) f (y)) x,y [x j 1,x j ] x,y [x j 1,x j ] 2K sup ((f (x) f (y)) x,y [x j 1,x j ] = 2K (M j m j ).
14 Riemnn Integrl Properties Thus, U(f 2, π) L(f 2, π) = π (M j (f 2 ) m j (f 2 )) x j 2K π (M j m j ) x j = 2K (U(f, π) L(f π)). Now since f is Riemnn Integrble, it stisfies the Riemnn Criterion nd so given ɛ > 0, there is prtition π 0 so tht U(f, π) L(f π) < ɛ/(2k) for ny refinement π of π 0. Thus, f 2 stisfies the Riemnn Criterion too nd so it is integrble.
15 Riemnn Integrl Properties (7): fg RI [, b] To prove tht fg is integrble when f nd g re, simply note tht ) fg = (1/2) ((f + g) 2 f 2 g 2. Property (6) nd the linerity of the integrl then imply fg is integrble. (8) If there exists m, M such tht 0 < m f M, then 1/f RI [, b]: Suppose f RI [, b] nd there exist M, m > 0 such tht m f (x) M for ll x [, b]. Note tht 1 f (x) 1 f (y) f (x) = f (y) f (x)f (y).
16 Riemnn Integrl Properties Let π = {x 0 =, x 1,..., x p = b} be prtition of [, b], nd define Then we hve M j = sup [x j 1,x j ] 1 f (x), m j = inf [x j 1,x j ] M j m j f (y) f (x) = sup x,y [x j 1,x j ] f (x)f (y) f (y) f (x) sup x,y [x j 1,x j ] f (x) f (y) 1 f (x). 1 m 2 sup f (y) f (x) x,y [x j 1,x j ] M j m j m 2.
17 Riemnn Integrl Properties Since f RI [, b], given ɛ > 0 there is prtition π 0 such tht U(f, π) L(f, π) < m 2 ɛ for ny refinement, π, of π 0. Hence, the previous inequlity implies tht, for ny such refinement, we hve ( 1 ) ( 1 ) U f, π L f, π = π (M j m j) x j 1 m 2 (M j m j ) x j π 1 ) (U(f m 2, π) L(f, π) < m2 ɛ m 2 = ɛ. Thus 1/f stisfies the Riemnn Criterion nd hence it is integrble.
18 Wht Functions Are Riemnn Integrble? Now we need to show tht the set RI [, b] is nonempty. We begin by showing tht ll continuous functions on [, b] will be Riemnn Integrble. Theorem If f C[, b], then f RI [, b]. Since f is continuous on compct set, it is uniformly continuous. Hence, given ɛ > 0, there is δ > 0 such tht x, y [, b], x y < δ f (x) f (y) < ɛ/(b ). Let π 0 be prtition such tht π 0 < δ, nd let π = {x 0 =, x 1,..., x p = b} be ny refinement of π 0. Then π lso stisfies π < δ. Since f is continuous on ech subintervl [x j 1, x j ], f ttins its supremum, M j, nd infimum, m j, t points s j nd t j, respectively. Tht is, f (s j ) = M j nd f (t j ) = m j for ech j = 1,..., p.
19 Wht Functions Are Riemnn Integrble? Thus, the uniform continuity of f on ech subintervl implies tht, for ech j, M j m j = f (s j ) f (t j ) < ɛ b. Thus, we hve U(f, π) L(f, π) = π (M j m j ) x j < ɛ b x j = ɛ. Since π ws n rbitrry refinement of π 0, it follows tht f stisfies Riemnn s criterion. Hence, f RI [, b]. π Theorem If f : [, b] R is constnt function, f (t) = c for ll t in [, b], then f is Riemnn Integrble on [, b] nd b f (t)dt = c(b ).
20 Wht Functions Are Riemnn Integrble? For ny prtition π of [, b], since f is constnt, ll the individul m j s nd M j s ssocited with π tke on the vlue c. Hence, U(f, π) U(f, π) = 0 lwys. It follows immeditely tht f stisfies the Riemnn Criterion nd hence is Riemnn Integrble. Finlly, since f is integrble, by our fundmentl integrl estimtes, we hve Thus, b f (t)dt = c(b ). c(b ) RI (f ;, b) c(b ). Theorem If f is monotone on [, b], then f RI [, b].
21 Wht Functions Are Riemnn Integrble? As usul, for concreteness, we ssume tht f is monotone incresing. We lso ssume f (b) > f (), for if not, then f is constnt nd must be integrble by the previous theorem. Let ɛ > 0 be given, nd let π 0 be prtition of [, b] such tht π 0 < ɛ/(f (b) f ()). Let π = {x 0 =, x 1,..., x p = b} be ny refinement of π 0. Then π lso stisfies π < ɛ/(f (b) f ()). Thus, for ech j = 1,..., p, we hve x j < ɛ f (b) f (). Since f is incresing, we lso know tht M j = f (x j ) nd m j = f (x j 1 ) for ech j. Hence, U(f, π) L(f, π) = (M j m j ) x j = [f (x j ) f (x j 1 )] x j π π ɛ < [f (x j ) f (x j 1 )]. f (b) f () π
22 Wht Functions Are Riemnn Integrble? But this lst sum is telescoping nd sums to f (b) f (). So, we hve U(f, π) L(f, π) < Thus, f stisfies Riemnn s criterion. ɛ (f (b) f ()) = ɛ. f (b) f ()
23 Wht Functions Are Riemnn Integrble? Let f n, for n 2 be defined by f n (x) = { 1, x = 1 1/(k + 1), 1/(k + 1) x < 1/k, 1 k < n 0, 0 x < 1/n We know f n is RI [0, 1] becuse it is monotonic lthough we do not know wht the vlue of the integrl is. Define f by f (x) = lim n f n (x). Then given x, we cn find n integer N so tht 1/(N + 1) x < 1/N telling us f (x) = 1/(N + 1). Moreover f (x) = f N+1 (x). So if x < y, y is either in the intervl [1/(N + 1), 1/N) or y [1/N, 1] implying f (x) f (y). Hence f is monotonic. At ech 1/N, the right nd left hnd limits do not mtch nd so f is not continuous t countble number of points yet it is still Riemnn Integrble.
24 Homework 12.1 If you didn t know f (x) = x ws continuous, why would you know f is RI [, b] for ny [, b]? 12.2 Use induction to prove f (x) = x n is RI [, b] for ny [, b] without ssuming continuity Use induction to prove f (x) = 1/x n is RI [, b] on ny [, b] tht does not contin 0 without ssuming continuity For f (x) = sin(2x) on [ 2π, 2π], drw f + nd f Prove f is RI [0, 1] where f is defined by { x sin(1/x), x (0, 1] f (x) = 0, x = Let f n (x) = { 1, x = 1 1/(k + 1), 1/(k + 1) x < 1/k, 1 k < n 0, 0 x < 1/n Grph f 5 nd f 8 nd determine where the set of cluster points S(p) contin two vlues.
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