Review of the Riemann Integral

Transcription

1 Chpter 1 Review of the Riem Itegrl This chpter provides quick review of the bsic properties of the Riem itegrl. 1.0 Itegrls d Riem Sums Defiitio Let [, b] be fiite, closed itervl. A prtitio P of [, b] is fiite subset of [, b] of the form P {x i : x 0 < x 1 < < x b}. Defiitio For prtitio P of [, b] hvig elemets, 1. defie x i : x i x i 1, d 2. defie the orm of P, deoted s P, by P : mx{ x i : i 1, 2,..., }. Exmple For y N we c costruct the -regulr prtitio P of [, b] by subdividig [, b] ito ideticl prts with the legth of ech subitervl beig b. I.e., x i b b for ech i, d P. Note tht x 1 + b, x b,. x i + i b,. x + b b. Assume tht f(x) is bouded o [, b]. Let P be prtitio o [, b]. Let M i sup{f(x) : x [x i 1, x i ]}, d let m i if{f(x) : x [x i 1, x i ]}. The m i f(x) M i for ll x [x i 1, x i ]. 2

2 Defiitio Give f(x), bouded o [, b], d prtitio P of [, b], we defie the upper Riem sum of f(x) with respect to P by U(f, P) : U b (f, P) d the lower Riem sum of f(x) with respect to P by L(f, P) : L b (f, P) M i x i, m i x i, where M i d m i re defied s bove. Filly, for ech i, choose c i [x i 1, x i ]; Riem sum for f(x) o [, b] with respect to P is defied by S b (f, P) f(c i ) x i. The bouds of summtio d b from defiitio re usully omitted. Remrk Sice m i f(c i ) M i d x i > 0, we c coclude tht L b (f, P) S b (f, P) U b (f, P). Defiitio Give prtitio P, we sy tht prtitio Q is refiemet of P if d oly if P Q. We lso sy tht Q refies P or P is refied by Q. Theorem Let f(x) be bouded o [, b]. Let P, Q be prtitios o [, b] with Q refiemet of P. The L(f, P) L(f, Q) U(f, Q) U(f, P). Proof. Assume tht Q P {y 0 } for some y 0 [, b] \ P. Hece Q {x i, y 0 : x 0 < x 1 < < x i 1 < y 0 < x i < < x b} d P {x i : x 0 < x 1 < < x b}. Let Also let M j sup{f(x) : x [x j 1, x j ]}, m i if{f(x) : x [x j 1, x j ]}. M i,1 sup{f(x) : x [x i 1, y 0 ]}, m i,1 if{f(x) : x [x i 1, y 0 ]}, M i,2 sup{f(x) : x [y 0, x i ]}, m i,2 if{f(x) : x [y 0, x i ]}. Note tht sice f([x i 1, x i ]) f([x i 1, y 0 ]) d f([x i 1, x i ]) f([y 0, x i ]), we hve tht M i,1 M i d 3

3 M i,2 M i. Similrly, m i,1 m i d m i,2 m i. Now Similrly, U(f, P) L(f, P) M j x j +M j x j + M i x i j i +M j x j + M i (y 0 x i 1 ) + M i (x i y 0 ) j i +M j x j + M i,1 (y 0 x i 1 ) + M i,2 (x i y 0 ) j i U(f, Q). m j x j +m j x j + m i x i j i +m j x j + m i (y 0 x i 1 ) + m i (x i y 0 ) j i +m j x j + m i,1 (y 0 x i 1 ) + m i,2 (x i y 0 ) j i L(f, Q). Due to remrk 1.0.5, we coclude tht L(f, P) L(f, Q) U(f, Q) U(f, P). We c use iductio o the umber of poits i Q \ P to estblish the theorem. Corollry If P d Q re prtitios of f(x) o [, b], the L(f, P) U(f, Q). Proof. Let T P Q. The T refies both P d Q. Hece L(f, P) L(f, T ) U(f, T ) U(f, Q), s desired. Defiitio Let f(x) be bouded o [, b]. The upper Riem itegrl for f(x) over [, b] is d the lower Riem itegrl for f(x) over [, b] is f(x) dx if{u(f, P) : P prtitio of [, b]}, f(x) dx sup{l(f, P) : P prtitio of [, b]}, 4

4 Observtio. f(x) dx f(x) dx. 1.1 Riem Itegrble Fuctios Defiitio We sy tht f(x) is Riem itegrble o [, b] if i which cse we deote this commo vlue by f(x) dx f(x) dx. f(x) dx, I defiitios d 1.1.1, we ecoutered ew ottios: the is the itegrl sig, d b re edpoits of itegrtio, f(x) is the itegrd, d dx refers to the vrible of itegrtio (lso clled the dummy vrible) i this cse it is x. Exmple Let f(x) If P {x i : 0 x 0 < x 1 < < x 1}, the let We hve tht Hece U(f, P) L(f, P) { 1 if x [0, 1] Q, 1 if x [0, 1] \ Q. M i sup{f(x) : x [x i 1, x i ]}, 1 m i if{f(x) : x [x i 1, x i ]}. 0 M i x i m i x i f(x) dx 1 1 d hece this fuctio is ot Riem itegrble o [0, 1]. x i 1, d x i f(x) dx, Observtio. f(x) i the bove exmple is discotiuous everywhere o [, b]. Theorem Let f(x) be bouded o [, b]. The f(x) is Riem itegrble o [, b] if d oly if for every > 0 there exists prtitio P of [, b] such tht U(f, P) L(f, P) <. Proof. Assume tht f(x) is Riem itegrble o [, b]. I.e., f(x) dx f(x) dx. 5

5 Let > 0. Sice f(x) dx if{u(f, P) : P is prtitio}, there exists prtitio P 1 such tht f(x) dx U(f, P 1 ) < f(x) dx + 2. Similrly, sice f(x) dx sup{l(f, P) : P is prtitio}, there exists prtitio P 2 such tht Let Q P 1 P 2. The This implies tht f(x) dx 2 f(x) dx 2 < L(f, P 2) < L(f, P 2 ) f(x) dx 2 f(x) dx. L(f, Q) (by theorem 1.0.7) U(f, Q) U(f, P 1 ) (by theorem 1.0.7) < f(x) dx + 2 f(x) dx + 2. U(f, Q) L(f, Q) <. To prove the coverse, ssume tht for ech > 0 there exists prtitio P of [, b] such tht However, by defiitio 1.0.9, we hve tht d hece Sice is rbitrry, we obti L(f, P) 0 U(f, P) L(f, P) <. f(x) dx f(x) dx f(x) dx therefore f(x) is Riem itegrble o [, b], s required. f(x) dx U(f, P), f(x) dx <. f(x) dx; Exmple Let f(x) x 2. Let P be the -regulr prtitio of [0, 1]. The U 1 0(f, P ) f(x i ) x i i d L 1 0(f, P ) f(x i 1 ) x i (i 1)

6 The U(f, P ) L(f, P ) ( ) ( ) 1. This shows tht for ll > 0, we c fid prtitio P such tht U(f, P) L(f, P) < (simply select lrge eough by the Archimede Priciple d use the -regulr prtitio P ). Hece by theorem 1.1.4, f(x) is Riem itegrble o [0, 1]. Lter, we will be ble to show tht 1 0 x 2 dx 1 3. Recll the followig defiitio. Defiitio We sy tht f(x) is uiformly cotiuous o itervl I if for every > 0, there exists δ > 0 such tht for ll x, y I, Also recll the followig theorem. x y < δ implies f(x) f(y) <. Theorem [Sequetil Chrcteriztio of Uiform Cotiuity]. A fuctio f(x) is uiformly cotiuous o itervl I if d oly if wheever {x }, {y } re sequeces i I, lim (x y ) 0 implies lim (f(x ) f(y )) 0. Exmple Let I R d f(x) x 2. Tke {x } { + } 1, {y } {}. The lim (x y ) 0, but lim 1 Hece f(x) x 2 is ot uiformly cotiuous o R. lim ) f(y )) [ ( lim + 1 ) ] 2 2 lim (2 + 1 ) 2 0. Oe should lredy be fmilir with the followig theorem. Theorem If f(x) is cotiuous o [, b], the f(x) is uiformly cotiuous o [, b]. Proof. Assume tht f(x) is cotiuous o [, b] but ot uiformly cotiuous o [, b]. The by theorem 1.1.7, there exists sequeces {x } d {y } with lim (x y ) 0 but lim (f(x ) f(y )) 0. (Note tht this limit my ot eve exist.) By choosig subsequece if ecessry, we c ssume without loss of geerlity tht there exists 0 > 0 such tht (1.1) (f(x ) f(y )) 0 f(x ) f(y ) 0 for ll N. 7

7 Sice {x } [, b], by the Bolzo-Weierstrss Theorem {x } hs subsequece {x k } with lim k x k x 0 [, b]. Note tht lim k (x k y k ) 0, hece lim k y k x 0 lso. By the sequetil chrcteriztio of cotiuity, we hve Therefore, we c select K N with lim f(x k ) f(x 0 ), k lim f(y k ) f(x 0 ). k for ll k K. Hece we hve, for ll k K, f(x k ) f(x 0 ) < 0 2 f(y k ) f(x 0 ) < 0 2 d 0 f(x k ) f(y k ) f(x k ) f(x 0 ) + f(y k ) f(x 0 ) < , directly cotrdictig equtio (1.1). Hece f(x) is uiformly cotiuous o [, b]. Theorem [Itegrbility Theorem for Cotiuous Fuctios]. If f(x) is cotiuous o [, b], the f(x) is Riem itegrble o [, b]. Proof. Let > 0. Sice f(x) is cotiuous o [, b], f(x) is lso uiformly cotious o [, b] by theorem bove. We c fid δ > 0 such tht if x y < δ, the f(x) f(y) < b. Let P be prtitio of [, b] with P mx{ x i } < δ. For exmple, we c choose the -regulr prtitio of [, b] with lrge eough (chose by the Archimede Priciple) so tht b P < δ. After we hve chose P, let M i sup{f(x) : x [x i 1, x i ]}, m i if{f(x) : x [x i 1, x i ]}. By the extreme vlue theorem, there exists c i, d i [x i 1, x i ] such tht f(c i ) m i d f(d i ) M i ; ote tht M i m i M i m i f(d i ) f(c i ). Sice c i d i < x i < δ, we hve tht f(d i ) f(c i ) < b. This shows tht U(f, P) L(f, P) < M i x i m i x i (M i m i ) x i b b. Hece f(x) is Riem itegrble by theorem b x i x i (b ) 8

8 Remrk Assume tht f(x) is cotiuous o [, b]. For ech, let P be the -regulr prtitio. The if S(f, P ) is y Riem sum ssocited with P, we hve f(x) dx lim S(f, P ). Proof. Give > 0, we c fid N lrge eough so tht if N, the b bove. Hece U(f, P ) L(f, P ) <, vi similr rgumet mde i theorem f(x) dx L(f, P ). This implies tht d so lim S(f, P ) f(x) dx, s remrked. < δ s defied i theorem But U(f, P ) S(f, P ) L(f, P ) d U(f, P ) f(x) dx S(f, P ) < Note tht the bove remrk shows tht if f(x) is cotiuous, we hve f(x) dx lim [ ( f + i b ) b ]. Defiitio Give prtitio P {x i : x 0 < x 1 < < x b}, defie the right-hd Riem sum, S R, by S R : f(x i ) x i ; defie the left-hd Riem sum, S L, by defie the midpoit Riem sum, S M, by S M : S L : f(x i 1 ) x i ; ( ) xi 1 + x i f x i. 2 Theorem Assume tht f(x) is Riem itegrble o [, b]. Give > 0, there exists δ > 0 such tht if P is prtitio of [, b] with P < δ, the S(f, P) f(x) dx < for y Riem sum S(f, P). Proof. Sice f(x) is Riem itegrble, we c fid prtitio P 1 of [, b] with U(f, P 1 ) L(f, P 1 ) < 2. I prticulr, for y Riem sum S(f, P 1 ), we hve U(f, P 1 ) S(f, P 1 ) L(f, P 1 ) d U(f, P 1 ) f(x) dx L(f, P 1). Suppose P 1 {x i : x 0 < x 1 < < x b}. The let M sup{f(x) : x [, b]}, m if{f(x) : x [, b]}. 9

9 If M m 0, the f(x) is costt o [, b], so sy f(x) c for ll x [, b]. I this cse U(f, P) c(b ) L(f, P) for y prtitio P. Hece y δ > 0 would stisfy the theorem d we re doe. So ssume M > m. Let δ < 2(M m). Let P {y i : y 0 < y 1 < < y j < < y k b} be y prtitio of [, b] with P < δ. Let T {j : j {1, 2,..., k}, d [y j 1, y j ] [x i 1, x i ] for some i}. Also, let The we hve tht U(f, P) L(f, P) M j sup{f(x) : x [y j 1, y j ]}, m j if{f(x) : x [y j 1, y j ]}. k M j y j k m j y j k (M j m j ) y j j T(M j m j ) y j + j {1,2,...,k}\T (M j m j ) y j. Note tht (M j m j ) y j U(f, P 1 P) L(f, P 1 P) j T U(f, P 1 ) L(f, P 1 ) < 2. Observe tht {1, 2,..., k} \ T hs t most elemets (sice for ech j T, we hve uique i such tht y j 1 < x i 1 < y j < x i ; coversely for ech such i there exists uique j T but there re oly poits i P 1 ). Hece for ech j {0, 1,..., k} \ T, we hve This shows tht (M j m j ) y j (M m) P j {1,2,...,k}\T < (M m) 2. (M j m j ) y j < 2. 2(M m) j {1,2,...,k}\T 2 2 Hece, if P < δ, U(f, P) L(f, P) < ; therefore, if S(f, P) is y Riem sum, the S(f, P) f(x) dx <, s required. 10

10 This thoerem is geerliztio of remrk I prticulr, this theorem provides esy, lterte proof of remrk Here we stte the remrk gi s corollry. Corollry If f(x) is Riem itegrble o [, b], the f(x) dx lim S(f, P ) lim f(c i ) b, where P is the -regulr prtitio of [, b] d c i [x i 1, x i ]. ( S(f, P ) is y Riem sum.) Proof. Let > 0. By the bove theorem, we c fid δ > 0 so tht if P 1 < δ, S(f, P ) f(x) dx < for y Riem sum S(f, P ). The result follows by choosig N (usig the Archimede Priciple) so tht 1 1 N < δ for ll N. I geerl, we write f(x) dx lim S(f, P). P 0 Theorem Let f(x) be mootoic o [, b]. The f(x) is Riem itegrble o [, b]. Proof. Let P be the -regulr prtitio. Assume, without loss of geerlity, tht f(x) is odecresig o [, b]. The So U(f, P ) S R (f, P ) L(f, P ) S L (f, P ) U(f, P ) L(f, P ) f(x i ) b, d f(x i 1 ) b. f(x i ) b 1 i0 f(x i ) b b [(f(x 1) + + f(x )) (f(x 0 ) + + f(x 1 ))] b (f(x ) f(x 0 )) b (f(b) f()). b Sice lim (f(b) f()) 0, f(x) is Riem itegrble o [, b] by theorem Questio: Assume tht f(x) is cotiuous o [, b] except t c. (Note tht f(x) is bouded o [, b].) Is f(x) still Riem itegrble o [, b]? Theorem Assume tht f(x) is cotiuous o [, b] except t c [, b]. itegrble o [, b]. The f(x) is Riem 11

11 Proof. Let P {x i : x 0 < x 1 < < x b} be prtitio of [, b] so tht c (x i0 1, x i0 ) for some i 0. Let > 0. Let M sup{f(x) : x [, b]}, m if{f(x) : x [, b]}, M i sup{f(x) : x [x i 1, x i ]}, m i if{f(x) : x [x i 1, x i ]}. We c choose P so tht x i0 < 3(M m). The (M i 0 m i0 ) x i0 (M m) x i0 < (M m) 3(M m) 3. Sice f(x) is uiformly cotiuous o [, x i0 1], by refiig s ecessry, we c ssume tht if 0 i i 0, the M i m i < 3(x i0 ). We ow hve i 0 1 (M i m i ) x i < i 0 1 A similr rgumet shows, by refiig if ecessry, tht ii 0+1 3(x i0 1 ) x i i (M i m i ) x i < 3. x i x i The U(f, P) L(f, P) (M i m i ) x i i 0 1 (M i m i ) x i + (M i0 m i0 ) x i0 + (M i m i ) x i ii 0+1 < This shows tht f(x) is Riem itegrble o [, b] by theorem Observtio. Observe the followig fcts: 1. If f(x) is bouded o [, b] d cotiuous except t possibly t fiitely my poits, the f(x) is Riem itegrble o [, b]. 2. If f(x) is Riem itegrble o [, b] d g(x) f(x) except t fiitely my poits, the g(x) is Riem itegrble d f(x) dx g(x) dx. 1.2 Flws i the Riem Itegrl The followig problem will be key to our motivtio for the costructio of coutbly dditive mesures. 12

12 Problem Assume tht {f } is sequece of Riem itegrble fuctios with {f } covergig poitwise o [, b] to fuctio f 0. Is f 0 Riem itegrble, d if so is f 0 (x) dx lim f (x) dx? Ufortutely, s we will see below, the swer to both questios bove is egtive. Exmple ) Let [0, 1] Q {r 1, r 2,, r, } d { 0 if x [0, 1] \ {r 1, r 2,, r 1 } f (x) 1 if x {r 1, r 2,, r 1 }. The ech f (x) is Riem itegrble. Moreover, f (x) f 0 (x) poitwise, where { 0 if x [0, 1] \ Q f 0 (x) 1 if x [0, 1] Q. d f 0 (x) is ot Riem itegrble. 2) Let 4 2 x if x [0, 1 2 ] f (x) (x 1 2 ) if x ( 1 2, 1 ] 0 if x ( 1, 1] (x) The digrm bove represets the grph of f 5 (x). It is esy to see geometriclly tht 1 0 f 5(t) dt 1 sice the itegrl represets the re of trigle with bse 0.2 d height 10. Geericlly, 1 0 f (t) dt 1 sice the itegrl represets the re of trigle with bse 1 d height 2. From this it follows tht f (x) f 0 (x) 0 for ech x [0, 1], d 1 lim 1 0 f (t) dt 1 0 lim f (t) dt 0. 13

13 1.3 A Vector-Vlued Riem Itegrl I this sectio we will see how we my defie vector-vlued versio of the Riem itegrl. I doig so we will ssume tht B is Bch spce d tht F : [, b] B is cotiuous fuctio. Defiitio Let B be Bch spce. Let F : [, b] B be cotiuous. Give prtitio P { t 0 < t 1 < < t b} of [0, b], by Riem sum we will me sum S(F, P ) of the form where c i [t i 1, t i ]. S(F, P) F (c i )(t i t i 1 ) Lemm Assume tht F : [, b] B is cotiuous. Let > 0. The there exists δ > 0 such tht if P is y prtio of [, b] with P < δ d if P 1 is refiemet of P, the for y Riem sums S(F, P) d S(F, P 1 ) we hve S(F, P) S(F, P 1 ) <. Proof. Give > 0, sice F is uiformly cotiuous, we c fid δ such tht if x y < δ, the F (x) F (y) < (b ). Assume tht P mx{ t i } < δ. Let P 1 be refiemet of P with ech itervl [t i 1, t i ] beig prtitioed ito {t i 1 s i,0 < s i,1 < < s i,ki t i }. Let S(F, P 1 ) be y Riem sum ssocited with P 1. Sice we kow tht k i F (c i )(t i t i 1 ) F (c i )(s i,j s i,j 1 ), we hve S(F, P ) S(F, P 1 ). F (c i )(t i t i 1 ) k i F (c i,j )(s i,j s i,j 1 ) k i F (c i )(s i,j s i,j 1 ) k i F (c i,j )(s i,j s i,j 1 ) k i F (c i ) F (c i,j ) (s i,j s i,j 1 ) b k i (s i,j s i,j 1 ) Theorem Assume tht F : [, b] B is cotiuous. Let > 0. The there exists δ > 0 such tht if P d Q re y two prtitios of [, b] with P < δ d Q < δ, the for y Riem sums s(f, P) d s(f, Q) we hve S(F, P) S(F, Q) <. Proof. From the previous lemm, we kow tht we c choose δ > 0 such tht if P 1 is y prtio of [, b] with P 1 < δ d if R is refiemet of P 1, the for y Riem sums S(F, P 1 ) d S(F, R) we hve S(F, P 1 ) S(F, R) < 2. 14

14 Next ssume tht P d Q re y two prtitios of [, b] with P < δ d P < δ. Let R P Q be refiemet of both P d Q, d let S(F, P) d S(F, Q) be Riem sums for F ssocited with P d Q respectively. Next let S(F, R) be y Riem sums for F ssocited with R respectively. The S(F, P) S(F, Q) S(F, P) S(F, R) + S(F, R) S(F, Q) < Corollry Let F : [, b] B be cotiuous. If {P k } is sequece of prtitios of [, b] such tht lim P k 0, the for y choice of Riem sums {S(F, P k )} is Cuchy i B, d hece coverges. k Moreover, the limit is idepedet of both the choice of prtitios d of the choice of Riem sums. Proof. Sice lim P k 0, the previous theorem llows us to immeditely deduce tht {S(F, P k )} is k Cuchy i B. Let {P k } d {Q k } be two sequece of prtitios of [, b] such tht Let {S(F, P k )} d {S(F, Q k )} be such tht d lim P k lim Q k 0. k k lim {S(F, P k)} x 0 B k lim {S(F, Q k)} y 0 B. k The we crete the sequece {P 1, Q 1, P 2, Q 2, P 3, Q 3, } of prtitios. It follows tht the sequece {S(F, P 1 ), S(F, Q 1 ), S(F, P 2 ), S(F, Q 2 ), } is lso Cuchy i B. From this we c coclude tht x 0 y 0 d hece tht the limit is i fct idepedet of both the choice of prtitios d of the choice of Riem sums. Defiitio [Vector-vlued Riem Itegrl] Let F : [, b] B be cotiuous. We defie the B-vlued Riem itegrl of F (X) o [, b] by F (t) dt lim S(F, P). P 0 where lim P 0 S(F, P) is the uique limit obtied s i the Corollry bove. 15