21-355 Principles of Rel Anlysis I Fll 2004 A. Definitions VI. Riemnn Integrtion Let, b R with < b be given. By prtition of [, b] we men finite set P [, b] with, b P. The set of ll prtitions of [, b] will be denoted by P[, b]. The set of ll bounded functions f : [, b] R will be denoted by B[, b]. Given P P[, b] nd f B[, b] we write P = {x 0, x 1, x 2,... x n } where = x 0 < x 1 < x 2 <... < x n = b, nd put x i = x i x i 1, m i (f) = inf{f(x) : x i 1 x x i } nd M i (f) = sup{f(x) : x i 1 x x i }. We define the lower nd upper sums of f for the prtition P by L(f, P ) = n m i (f) x i i=1 U(f, P ) = Notice tht for every P P[, b] we hve n M i (f) x i. i=1 nd m(f)(b ) L(f, P ) U(f, P ) M(f)(b ), where m(f) = inf{f(x) : x [, b]} nd M(f) = sup{f(x) : x [, b]}. Definition 1: Let f [, b] be given. The lower integrl of f is defined by The upper integrl of f is defined by f = sup{l(f, P ) : P P[, b]}. f = inf{u(f, P ) : P P[, b]}. Definition 2. Let f B[, b] be given. We sy tht f is Reimnn integrble if in this cse we write Sometimes we write f = f = f(t)dt in plce of functions f : [, b] R will be denoted by R[, b]. f; f. f. The set of ll Riemnn integrble 1
Definition 3. Let f R[, b] be given. Then we define f = 0 nd b f = Definition 4: Let P, Q P[, b] be given. If P Q we sy tht Q is refinement of P. Definition 5: Let P 1, P 2 P[, b] be given. The prtition P = P 1 P 2 is clled the common refinement of P 1 P 2. B. Some Key Results VI.1 Proposition: Let f B[, b] nd P, Q P[, b] with P Q be given. Then L(f, P ) L(f, Q) nd U(f, P ) U(f, Q). VI.2 Proposition: Let f B[, b] be given. Then VI.3 Theorem: Let f B[, b] be given. Then f R[, b] if nd only ɛ > 0, P P[, b] such tht U(f, P ) L(f, P ) < ɛ. f VI.4 Theorem: Assume tht f[, b] R is monotonic. Then f R[, b]. VI.5 Theorem: Assume tht f : [, b] R is continuous. Then f R[, b]. VI.6 Theorem: Let f R[, b] be given nd choose c, d, R such tht c < d nd c f(x) d for ll x [, b]. Let ϕ : [c, d] R be given nd ssume tht ϕ is continuous. Then ϕ f R[, b]. VI.7 Theorem: Let f, g R[, b] nd α R be given. Then f. f. i. f + g R[, b] nd ii. αf R[, b] nd iii. fg R[, b]. (f + g) = αf = α f + f. g; iv. If f(x) g(x) x [, b] then f g. v. f R[, b] nd f f. VI.8 Theorem: Let f R[, b] nd c, d R with c < d b be given. Then the restriction of f to [c, d] is integrble on [c, d]. 2
VI.9 Theorem: Let f B[, b] nd c (, b) be given. If f is integrble on [, c] nd on [c, b] then f R[, b] nd f = c f + c f. VI.10 Fundmentl Lemm of Clculus: Let f R[, b] nd c, x 0 (, b) be given. Define F : [, b] R by F (x) := x c f(t)dt x [, b]. Then F is uniformly continuous on [, b]. Moreover if f is continuous t x 0 then F is differentible t x 0 nd F (x 0 ) = f(x 0 ). VI.11 Fundmentl Theorem of Clculus: Let f R[, b] be given nd ssume tht f is continuous on (, b). Let F : [, b] R be ny function tht is continuous on [, b], differentible on (, b) nd such tht F (x) = f(x) for ll x [, b]. Then f = F (b) F (). VI.12 Men Vlue Theorem for Integrls. Let f R[, b] be given nd ssume tht f is continuous on (, b). Then there exists c (, b) such tht C. Some Remrks. f(c) = 1 b VI.13 Remrk: It is strightforwrd to verify tht f. VI.14 Remrk: Define f : [, b] R by 0 x [, b]\q f(x) = 1 x [, b] Q. It is strightforwrd to verify tht nd consequently f / R[, b]. f = 0 nd f = b 1 = b. VI.15 Remrk: Let f, g R[, b] be given nd ssume tht f(x) < g(x) for ll x (, b). Then we hve 3
f < lthough this seems much more difficult to prove thn Theorem VI.7 (iv). D. Some Proofs. Proof of VI.3: Assume first tht f R[, b]. Let ɛ > 0 be given. Choose P 1, P 2 P[, b] such tht g (1) U(f, P 1 ) f < ɛ 2 (2) f L(f, P 2 ) < ɛ 2 nd put P = P 1 P 2. By Proposition VI.1 we hve (3) U(f, P ) f < ɛ 2 (4) f L(f, P ) < ɛ 2. Since f = f we my dd (3) nd (4) to obtin (5) U(f, P ) L(f, P ) < ɛ To prove the converse impliction let ɛ > 0 be given nd choose P such tht (5) holds. Then, by Proposition VI.2 we hve (6) L(f, P ) Combining (5) nd (6) we get f U(f, P ) (7) 0 b f < ɛ. 4
Since ɛ > 0 ws rbitrry we conclude tht (8) b f f = 0 nd f R[, b]. Proof of VI.4: We tret the cse when f is incresing. [The cse when f is decresing very similr.] We use Theorem VI.3. Let ɛ > 0 be given. Choose n N such tht (9) n > (f(b) f())(b ). ɛ Let P be the uniform prtition of [, b] with n sub-intervls, i.e. the prtition chrcterized by ( ) b (10) x i = + i, i = 0, 1,..., n. n Let (11) x = nd notice tht (b ) n (12) x i x i 1 = x, i = 1, 2,..., n. Since f is incresing we hve (13) m i (f) = f(x i 1 ), M i (f) = f(x i ) i = 1, 2,..., n. It follows tht (14) U(f, P ) L(f, P ) = n [f(x i ) f(x i 1 )] x = i=1 (b ) [f(b) f()]. n Combining (9) nd (14) we get (15) U(f, P ) L(f, P ) < ɛ. 5
Proof of VI.5. Once gin, we pply Theorem VI.3. Le ɛ > 0 be given. Since f is continuous on [, b] nd [, b] is compct, we know tht f is uniformly continuous. Therefore we my choose δ > 0 so tht (16) f(t) f(s) < ɛ b s, t [, b], t s < δ. Let P be ny prtition of [, b] such tht (17) x i < δ, i = 1, 2,..., n. Since f is continuous, for ech i {1, 2,..., n} we my choose x i, x i [x i 1, x i ] such tht (18) f( x i ) f(x) f(x i ) x [x i 1, x i ]. It follows tht (19) U(f, P ) L(f, P ) = n [f(x i ) f( x i )] x i i=1 Since x i x i < δ for ll i {1, 2,..., n} we know tht (20) f(x i ) f( x i ) < ɛ b i {1, 2,..., n}. It follows from (19) nd (20) tht (21) U(f, P ) L(f, P ) < n i=1 ɛ (b ) x i = ɛ. Proof of VI.6: Once gin, we use Theorem IV.3. Let ɛ > 0 be given. Since ϕ is continuous on the compct set [c, d] it is uniformly continuous nd it is bounded. Choose δ > 0 such tht (22) ϕ(t) ϕ(s) < ɛ 2(b ) s, t [c, d], t s > δ nd choose K > 0 such tht (23) ϕ(s) K s [c, d]. 6
Since f R[, b] we my choose P P[, b] such tht (24) U(f, P ) L(f, P ) < δɛ 4K. Split the index {1, 2,..., n} set into two pieces A, B s follows: (25) i A M i (f) m i (f) < δ, (26) i B M i (f) m i (f) δ. Notice tht U(ϕ f, P ) L(ϕ f, P ) (27) = i A [M i(ϕ f) m i (ϕ f)] x i It follows from (22) nd (25) tht + i B [M i(ϕ f) m i (ϕ f] x i. (28) M i (ϕ f) m i (ϕ f) ɛ 2(b ) i A. Consequently (29) [M i (ϕ f) m i (ϕ f)] x i i A i A ɛ x i n 2(b ) i=1 ɛ x i 2(b ) = ɛ 2(b ). Notice tht (30) δ i B x i i B It follows from (30) tht [M i (f) m i (f)] x i U(f, P ) L(f, P ) < δɛ 4k. (31) x i < i B ɛ 4K 7
For i B, we hve (32) [M k (ϕ f) m i (ϕ f)] M i (ϕ f) + m i (ϕ f) 2K nd consequently (33) i B[M i (ϕ f) m i (ϕ f)] x i i B 2K x i < 2K ( ɛ ) = ɛ 4K 2 by virture of (31) nd (32). Combining (27), (29), nd (33), we rrive t (34) U(ϕ f, P ) L(ϕ f, P ) < ɛ. Proof of VI.7(i). Let ɛ > 0 be given. Choose P 1, P 2 P[, b] such tht (35) U(f, P 1 ) L(f, P 1 ) < ɛ/2 (36) U(g, P 2 ) L(g, P 2 ) < ɛ/2. Let P = P 1 P 2 nd observe tht (37) U(f, P ) L(f, P ) < ɛ/2 (38) U(g, P ) L(g, P ) < ɛ 2. Notice tht for ech i {1, 2,..., n} we hve (39) m i (f) f(x) M i (f) x [x i 1, x i ] (40) m i (g) g(x) M i (g) x [x i 1, x i ], nd consequently (41) m i (f) + m i (g) f(x) + g(x) M i (f) + M i (g) x [x i 1, x i ]. It follows tht 8
(42) m i (f) + m i (g) m i (f + g) M i (f + g) M i (f) + M i (g) i {1, 2,..., n}. Multiplying (42) by x i nd summing over i we get (43) L(f, P ) + L(g, P ) L(f + g, P ) U(f + g, P ) U(f, P ) + U(g, P ). It follows from (37), (38), nd (43) tht (44) U(f + g, P ) L(f + g, P ) < ɛ. We conclude tht f + g R[, b]. Notice tht (45) L(f, P ) f U(f, P ), (46) L(g, P ) g U(g, P ), (47) L(f + g, P ) (f + g) U(f + g, P ). Combining (37), (38), (45), (46), nd (47) in strightforwrd (but perhps tedious) fshion we rrive t (48) ɛ + f + g < (f + g) < f + Since ɛ > 0 ws rbitrry we conclude tht g + ɛ. (49) (f + g) = f + g. The proofs of VI.7 (ii) nd VI.7 (iv) re left s exercises. Proof of VI.7(iii). The function t t 2 is continuous on R. Therefore, by Theorem VI.6, F 2 R[, b] for every F R[, b]. We conclude tht (f + g) 2 R[, b] nd (f g) 2 R[, b] by virtue of Theorem VI.7 (i), (ii) nd the observtion bove. The fct tht f, g R[, b] now follows from the eqution 9
(50) fg = 1 4 [(f + g)2 (f g) 2 ] nd nother ppliction of Theorem VI.7(i), (ii). Proof of VI.7(v): The fct tht f R[, b] follows from Theorem VI.6 nd continuity of the function t t on R. The desired inequlity follows form Theorem VI.7(ii), (iv) nd the observtion (51) f(x) f(x) x [, b] (52) f(x) f(x) x [, b]. Proof of VI.10: The uniform continuity of F is homework problem. For h 0 nd h smll enough so tht x 0 + h [, b] we hve F (x 0 + h) = x 0 +h f(t)dt c (53) nd consequently = x 0 f(t)dt + x 0 +h f(t)dt c x 0 = F (x 0 ) + x 0 +h x 0 f(t)dt (52) F (x 0 + h) F (x 0 ) h = 1 h x0 +h x 0 f(t)dt. Let ɛ > 0 be given. Since f is continuous t x 0 we my choose δ > 0 such tht (53) f(t) f(x 0 ) < ɛ 2 t B δ (x 0 ) [, b]. Observe tht for h 0 we hve (56) f(x 0 ) = 1 h x0 +h x 0 f(x 0 )dt. Let h B δ (0) be given such tht x 0 + h [, b]. Then we hve (57) F (x 0 + h) F (x 0 ) h f(x 0 ) = 1 h x0 +h x 0 (f(t) f(x 0 ))dt 10
by virtue of (54) nd (56). It follows from (55) nd (57) tht (58) F (x 0 + h) F (x 0 ) f(x 0 ) h 1 h x0 +h x 0 f(t) f(x 0 ) dt 1 h ɛ 2 h < ɛ. We conclude tht F is differentible t x 0 nd F (x 0 ) = f(x 0 ). Proof of VI.II: Define F, G : [, b] R by x (59) F (x) = f(t)dt x [, b] (60) G(x) = F (x) F (x) x [, b]. Notice tht F, G re continuous on [, b], differentible on (, b) nd (61) G (x) = F (x) F (x) = f(x) f(x) = 0 x (, b). We conclude tht G is constnt on [, b], i.e. (62) G(x) = G() x [, b]; in prticulr (63) G(b) = G(). Notice tht (64) G() = F () F () = F () Combining (63) nd (64) yields (65) G(b) = F () Observe tht (66) f(t)dt = F (b) = F (b) G(b) = F () () by virtue of (59), (60), nd (65). 11
Proof of VI.12: Define F : [, b] R by (67) F (x) = x ft)dt x [, b] Then F is continuous on [, b], differentilbe on (, b) nd F (x) = f(x) for ll x (, b). By the Men Vlue Theorem for derivtives we my choose c (, b) such tht (68) f(c) = F (c) = = 1 b = 1 b F (b) F () b [ f(t)dt f(t)dt. b ] f(t)dt 12