1 i n x i x i 1. Note that kqk kp k. In addition, if P and Q are partition of [a, b], P Q is finer than both P and Q.


 Carol Cooper
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1 Chpter 6 Integrtion In this chpter we define the integrl. Intuitively, it should be the re under curve. Not surprisingly, fter mny exmples, counter exmples, exceptions, generliztions, the concept of the integrl my seem strnge. And not surprisingly, there is more thn one wy to define the integrl. We strt with the Drboux integrl nd lter show it is equivlent to the Riemnn integrl. 6.. Drboux Integrl Throughout this chpter we ssume [, b] be closed, bounded intervl, nd tht the functions f : [, b] R under considertion re bounded. Definition 6.. Let [, b] be closed, bounded intervl. (i) A Prtition of [, b] is set of points P = {x, x, x 2,..., x n, x n } with = x < x < < x n = b. (ii) The norm of prtition is kp k = mx i n x i x i. (iii) A Refinement of P is prtition Q of [, b] such tht P Q. We sy Q is finer thn P. Note tht kqk kp k. In ddition, if P nd Q re prtition of [, b], P Q is finer thn both P nd Q. Definition 6.2. Let f : [, b] R be bounded. Set M j (f) = () The upperdrboux Sum of f over P is sup f(x) m j (f) = inf f(x). x j x x j x j x x j U(P, f) = M j (f)(x j x j ). j= 65
2 66 6. Integrtion (b) The lowerdrboux Sum of f over P is L(P, f) = m j (f)(x j x j ). j= The following lemm provides reltionship between the upper/lower Drboux sums nd prtitions. Lemm 6.3. Let P be ny prtition of [, b] nd Q be ny refinement of P. Then L(P, f) L(Q, f) U(Q, f) U(P, f). Proof. Let P = {x, x,..., x n }. By ssumption P Q. Let us suppose one point, z, hs been dded to P. Specificlly, suppose x j < z < x j for some j. Cll the new prtition P. Let M j (f) = sup{f(x) : x [x j, x j ]}, m j (f) = inf{f(x) : x [x j, x j ]}, β (f) = inf{f(x) : x [x j, z]}, β 2 (f) = inf{f(x) : x [z, x j ]}. Note m j β nd m j β 2 since m j is the infimum over lrger set. Moreover, we my write x j x j = (x j z) + (z x j ). Thus L(P, f) = m i (f)(x i x i ) = Xj m i (f)(x i x i ) + m j (f)(x j x j ) + i=j+ Xj m i (f)(x i x i ) + β (z x j ) + β 2 (x j z) + i=j+ = L(P, f) m i (f)(x i x i ) m i (f)(x i x i ) A similr rgument shows U(P, f) U(P, f). If Q contins points not in P, we repet the rgument (there cn only be finite number of dditionl points) to obtin L(P, f) L(Q, f) nd U(Q, f) U(P, f). The definition of the upper nd lower sums implies L(Q, f) U(Q, f), nd the inequlity follows. Corollry 6.4. We hve L(P, f) U(Q, f) for ny prtitions P nd Q of [, b]. Proof. Since P Q is refinement of both P nd Q, L(P, f) L(P Q, f) U(P Q, f) U(Q, f). We re redy to define the integrl. Definition 6.5. The function f : [, b] R is sid to be Drboux Integrble on [, b] if nd only if f is bounded nd, for ll ǫ >, there exists prtition, P, of [, b] such tht U(P, f) L(P, f) < ǫ. In this cse we sy f R[, b] (the R stnds for Riemnn).
3 6.. Drboux Integrl 67 is. Note the definition of the existence of the integrl does not sy wht the integrl Definition 6.6. Let f : [, b] R be bounded. Set (i) (ii) f(x) dx = inf{u(p, f) P prtition of [, b]} f(x) dx = sup{l(p, f) P prtition of [, b]} If f(x) dx = f(x) dx, we sy the integrl of f is f(x) dx. You might wonder why the sup nd inf exist in the definition. Lemm 6.7. If f : [, b] R is bounded, then both exist. Moreover, f(x) dx f(x) dx. f(x) dx nd f(x) dx Proof. Recll tht L(P, f) U(Q, f) for ny prtitions P, Q of [, b]. Thus the numbers L(P, f) re bounded bove nd by the completeness xiom sup L(P, f) P exists. Similrly, U(Q, f) is bounded below by ny L(P, f), nd so inf U(P, f) P exist. Tking the supremum of the left side of the inequlity L(P, f) U(Q, f), we find f(x) dx U(Q, f). Tking the infimum of the right side provides the inequlity in the lemm. You might hope too tht f R[, b] implies R b f(x) dx exists. Theorem 6.8. Suppose f R[, b]. Then f(x) dx = f(x) dx. Proof. Since f R[, b], for ll ǫ >, there exists prtition of [, b] such tht U(P, f) L(P, f) < ǫ. Thus Tht is, L(P, f) nd the result follows. f(x) dx f(x) dx f(x) dx U(P, f). f(x) dx U(P, f) L(P, f) < ǫ,
4 68 6. Integrtion Exmple 6.9. Prove the function f(x) = is in R[, ] nd find the integrl. x <, 2 x =. Proof. Let ǫ >. Set P = {, ǫ 2, }. Then M (f) = = m (f), nd M 2 (f) = 2, m 2 (f) = ( sketch of the function nd the prtition might be useful). Thus U(P, f) = ǫ ( ǫ 2 ) = + ǫ 2 L(P, f) = ǫ 2 + ( ǫ 2 ) =, nd hence U(P, f) L(P, f) = ǫ/2 < ǫ nd f R[, ]. or To find the integrl note which implies Z L(P, f) f(x) dx =. Z Z f(x) dx U(P, f) f(x) dx + ǫ 2 Exmple 6.. Show tht Dirichlet s function x Q, f(x) = x Q c. is not in R[, ]. Proof. To show function is not integrble we must show the existence of ǫ > such tht for ll prtitions of [, b], U(P, f) L(P, f) ǫ. Let ǫ = /2. For ny prtition of [, ] we hve M j (f) = nd m j (f) =. Thus U(P, f) L(P, f) = U(P, f) = (x j x j ) = ǫ nd f is not integrble. Integrble functions cn be complicted s the next exmple shows. j= Exmple 6.. Consider the function on [, ] x = n f(x) = x 6= n. Prove f R[, ]; i.e. f is Drboux integrble nd find the integrl.
5 6.. Drboux Integrl 69 Proof. Let ǫ >. By the Archimeden Property n n N exists so tht 2/ǫ < n. Tht is /n < ǫ/2. By the WellOrdering Property, minimum such n (gin clled n ) exists. Tht is, (6.) < ǫ n 2 n. We ssume, without loss of generlity, tht n > 2. Behold: Set P to be the prtition P =, + ǫ2 n 6, n ǫ2 6, n + ǫ2 6, n 2 ǫ2 6, n 2 + ǫ2 6,..., ǫ2 6, = {, x, x 2,..., x 2n 2, }. Before we check tht P is in fct prtition of [, ], we compute the upperdrboux sum. We hve, using (6.) nd /(n ) < 2/n (becuse n > 2), U(P, f) = + ǫ2 ǫ28 ǫ2 + (n 2) + n 6 6 = + (n ) ǫ 2 n 2 2 < ǫ (n ) n = ǫ 2 + 2(n ) < ǫ 2 + < ǫ n 2 + ǫ 2 = ǫ. Moreover, L(P, f) =. Thus, U(P, f) L(P, f) < ǫ nd f is Drboux Integrble provided P is prtition of [, ]. To show P is prtition, we need only check x < x 2 since the gps grow for incresing x i. Tht is, if x < x 2, the rest of the terms in the prtition re ordered. Here, x 2 x = n ǫ2 + ǫ2 n 2 = 6 n 6 2n (n ) 2. This is positive for n > 2, which we hve ssumed without loss of generlity. or To find the integrl note which implies Z L(P, f) f(x) dx =. Z Z f(x) dx U(P, f) f(x) dx ǫ
6 7 6. Integrtion 6.2. Chrcteriztions of The Integrl In this section we give five different equivlent chrcteriztions of the integrl. For given problem, one chrcteriztion my be esier to pply thn nother. Definition 6.2. Let P be ny prtition of [, b] nd f : [, b] R is bounded. The Riemnn Sum with respect to P is S(P, f) = f(t i )(x i x i ), where x i t i x i. Agin throughout f : [, b] R is bounded. We recll the definition of f R[.b] s the first (zeroth) chrcteriztion. The other chrcteriztions re stted s theorems. However, when we refer to the chrcteriztions we men the lterntive description of f R[, b] given in the chrcteriztion. Definition. f R[, b] if nd only if, for ll ǫ >, there exists prtition, P, of [, b] such tht U(P, f) L(P, f) < ǫ. Chrcteriztion I. f R[, b] if nd only if f(x) dx = f(x) dx = f(x) dx. Chrcteriztion II. f R[, b] if nd only if, for ll ǫ >, there exists prtition, P ǫ, of [, b] such tht, for ny refinement P ǫ P, U(P, f) L(P, f) < ǫ. Chrcteriztion III. f R[, b] if nd only if there exists I R such tht for ll ǫ > there exists prtition, P ǫ, of [, b] such tht, for ny refinement P ǫ P nd ny Riemnn sum S(P, f), we hve S(P, f) I < ǫ. In this cse I = Chrcteriztion IV. f R[, b] if nd only if there exists I R such tht I = lim S(P, f) = lim f(t i )(x i x i ). kp k kp k f(x) dx. Tht is, for ll ǫ > there exists δ > such tht, for ll prtitions, P, of [, b], kp k < δ implies S(P, f) I < ǫ. Agin I = f(x) dx. Chrcteriztion IV cn be tken s the definition of Riemnn integrl. Thus Chrcteriztion IV shows the Drboux nd Riemnn integrls re the sme.
7 6.2. Chrcteriztions of The Integrl 7 Theorem 6.3. Chrcteriztion I nd f R[.b] re equivlent. Proof. We showed in Theorem 6.8 f R[.b] implies Chrcteriztion I. Suppose f(x) dx = f(x) dx. By the properties of supremums nd infimums, given ny ǫ >, there exists prtitions P, P 2 of [, b] such tht Let P = P P 2. Then Tht is, f(x) dx ǫ 2 < L(P, f), f(x) dx + ǫ 2 > U(P 2, f). f(x) dx ǫ 2 < L(P, f) L(P, f) U(P, f) U(P 2, f) < U(P, f) L(P, f) < nd thus f R[, b]. = ǫ, f(x) dx + ǫ 2! f(x) dx ǫ 2 f(x) dx + ǫ 2. Theorem 6.4. Both Chrcteriztions III nd IV imply f R[.b]. In prticulr, if bounded function is Riemnn integrble, it is Drboux integrble. Proof. Suppose bounded function f stisfies Chrcteriztion III or IV. Let ǫ >. Then either prtition P ǫ = {x, x,..., x n } of [, b] exists or δ > exists such tht ny prtition of [, b] with kp k < δ implies f(t i )(x i x i ) I < ǫ/3 i=n for ll t i [x i, x i ], i =, n. We need to show U(P, f) L(P, f) < ǫ. By the properties of supremum nd infimum, there exist t i nd s i such tht ǫ M i (f) 6(b ) < f(t ǫ i), m i (f) + 6(b ) > f(s i). This implies Thus U(P ǫ ) L(P ǫ ) = M i (f) m i (f) < f(t i ) f(s i ) + (M i (f) m i (f))(x i x i ) ǫ, ǫ 3(b ). ǫ (f(t i ) f(s i )(x i x i ) + (x i x i ) 3(b ) f(t i )(x i x i ) I + n I X f(s i )(x i x i ) + ǫ 3!
8 72 6. Integrtion nd f R[, b]. Next we show I = f(x) dx. Let ǫ >. Agin by the properties of supremum nd infimum there exists prtitions, P, P ǫ nd P 2 such tht L(P, f) > f(x) dx ǫ 3, S(P ǫ, f) I < ǫ 3, U(P 2, f) L(P 2, f) < ǫ 3. Set P = P P ǫ P 2. Then L(P, f) S(P, f) U(P, f) < L(P, f) + ǫ 3, or S(P, f) L(P, f) < ǫ/3, nd I f(x) dx Since this is true for ll ǫ >, I = < ǫ. I S(P, f) + S(P, f) L(P, f) + L(P, f) f(x) dx. f(x) dx Theorem 6.5. f R[.b] implies Chrcteriztion III. Proof. Let ǫ > be given. By the properties of supremums nd infimums, given ny ǫ, the exists prtitions, P, P 2 of [, b] such tht f(x) dx ǫ 2 < L(P, f), Set P ǫ = P P 2. For ny refinement P ǫ P we hve Similrly Thus Tht is, f(x) dx + ǫ 2 > U(P 2, f). U(P, f) < U(P, f) + ǫ 2 U(P, f) + ǫ 2 < f(x) dx + ǫ. L(P, f) > f(x) dx ǫ. f(x) dx ǫ < L(P, f) S(P, f) U(P, f) < S(P, f) f(x) dx < ǫ f(x) dx + ǫ. for ny prtition P of [, b] with P ǫ P. This is Chrcteriztion III. The proof tht bounded function is Drboux integrble implies it is Riemnn integrble is much hrder. We first must relte the upper nd lower sums of given fixed prtition of [, b] to ny prtition with kp k smll.
9 6.2. Chrcteriztions of The Integrl 73 Lemm 6.6. f : [, b] R is bounded. Let P be ny prtition of [, b]. For ny ǫ > there exists δ >, such tht, for ny prtition P of [, b] with kp k < δ, U(P, f) U(P P, f) < ǫ. Proof. Since f is bounded, there exists M such tht f(x) M for ll x [.b]. Let ǫ >. Suppose the given prtition P = {x, x..., x n }. Set δ = ǫ/4nm. Let P be ny prtition of [, b] with kp k < δ ( sketch of the prtitions P nd P would be beneficil. In sketching the prtitions, think of P s much more refined thn P ). To sy more, suppose P = {y i }, nd P P = {x i } {y i } = {z i } with the {z i } ordered. For exmple, perhps the order might be {z i } = {, y, x, y 2, y 3, y 4, y 5,...}. In the event cell [z i, z i ] does not contin x i, then both U(P, f) nd U(P P, f) contin M i (f)(y i y i ) nd tht term cncels. So in our exmple the first few terms re U(P, f) U(P P, f) = sup f(x)(y ) + sup f(x)(y 2 y ) +... x y y x y 2 sup f(x)(y ) + sup f(x)(x y ) + sup f(x)(y 2 x ) +... x y y x x x x y 2 sup f(x)(y 2 y ) +... M y x y 2 (y 2 y ) + (y 3 y 2 ) +... < M2δ +..., where the like terms cncel, nd the remining terms in U(P P, f) re dropped. Thus, we see every occurrence of x i between two y i does not cncel nd leds to term from U(P, f) in the form sup f(x)(y i y i ) which cn by estimted y i x y i by M2δ. Since there re n terms in P (i.e. totl of n x i s), this cnnot hppen more thn n times, nd we hve U(P, f) U(P P, f) < 2nMδ = ǫ. Theorem 6.7. f R[.b] implies Chrcteriztion IV. Tht is, if bounded function is Drboux integrble, it is Riemnn integrble. Proof. With Lemm 6.6, the proof is lmost identicl to the proof of Theorem 6.5 (tht f R[, b] implies Chrcteriztion III). Let ǫ > be given nd P be chosen so tht U(P, f) < f(x) dx + ǫ 2. Set δ = ǫ/8nm nd let P be ny prtition of [, b] with kp k < δ. By the previous lemm U(P, f) < U(P P, f) + ǫ 2 U(P, f) + ǫ 2 < f(x) dx + ǫ, for ll prtitions of [, b] with kp k < δ.
10 74 6. Integrtion Similrly, there is δ 2 > such tht L(P, f) > f(x) dx ǫ for ll prtitions of [, b] with kp k < δ 2. Let δ = min{δ, δ 2 }. Then Tht is, f(x) dx ǫ < L(P, f) S(P, f) U(P, f) < S(P, f) f(x) dx < ǫ f(x) dx + ǫ. for ny prtition P of [, b] with kp k < δ. This is Chrcteriztion IV. Chrcteriztion II is resttement of the definition. It is designed to pper more like limit Algebr of Integrble Functions In this section we estblish some well known properties of integrls. The proofs in this section could be shortened by using Chrcteriztion IV in the previous section. This would involve using the Riemnn integrl insted of the Drboux integrl. In n effort to keep the exposition self contined s possible, we eschew use of the Riemnn integrl when prcticble. We strt with severl lemms which build towrd the expected lgebr involving integrls. Lemm 6.8. If f R[, b], f R[, b] nd (6.2) f dx = f dx. Proof. We estblished in the proof Theorem 2.4 tht sup I ( f) = inf(f). Moreover, inf I so tht ( f) = sup I (f). Let ǫ >. Since f R[, b], prtition P of [, b] exists U(P, f) L(P, f) = (M i (f) m i (f))(x i x i ) < ǫ. However, m i (f) = M i ( f) nd M i (f) = m i ( f). Thus ( m f ( f) + M i ( f))(x i x i ) < ǫ. This shows U(P, f) L(P, f) < ǫ nd estblishes f R[, b]. To find the integrls, we clculte f dx = f dx U(P, f) = L(P, f) U(P, f) + ǫ f dx + ǫ. I
11 6.3. Algebr of Integrble Functions 75 Similrly, f dx = f dx L(P, f) = U(P, f) L(P, f) ǫ Combining the two we conclude Z! b f dx f dx ǫ. Since ǫ > is rbitrry, (6.2) is estblished. f dx! < ǫ. Lemm 6.9. If f R[, b] nd c R, c f R[, b] nd (6.3) c f dx = c f dx. Proof. We first suppose c >. It is strightforwrd to show sup I (c f(x)) = c sup I (f(x)). Similrly, we see inf I (c f(x)) = c inf I (f(x)). Hence, for ny prtition P of [, b] U(P, c f) = c U(P, f) nd L(P, c f) = c L(P, f). Let ǫ >. Then prtition P of [, b] exists so tht U(P, f) L(P, f) < ǫ/c. Then nd so c f R[, b]. U(P, c f) L(P, c f) = c (U(P, f) L(P, f)) < ǫ To estblish the integrls we rgue s in the previous lemm. We clculte In the sme wy, we find c f dx = c f dx U(P, c f) = c U(P, f) < c (L(P, f) + ǫ/c ) c f dx + ǫ. c f dx L(P, c f) = c L(P, f) c U(P, f) ǫ c f dx ǫ. Combining these we find Tht is, c f dx ǫ Since ǫ > is rbitrry, (6.3) follows. c f dx c f dx + ǫ. c f dx c f dx < ǫ.
12 76 6. Integrtion If c <, then c >. We pply the bove nd the Lemm 6.8 to gin conclude c f R[, b] nd (6.3). Theorem 6.2. Suppose f, f 2 R[, b]. Then (i) for ny c, c 2 R, c f + c 2 f 2 R[.b] nd (c f + c 2 f 2 ) dx = c f dx + c 2 f 2 dx. (ii) If f (x) f 2 (x) for ll x [, b], then f (x) dx (iii) If m f(x) M for ll x [.b], (iv) If c (, b), m(b ) f(x) dx = Z c f 2 (x) dx. f(x) dx M(b ). f(x) dx + c f(x) dx. Proof. To prove (i) we first show f + f 2 R[, b]. Problem 2.5 shows, for ny prtition P of [, b], U(P, f + f 2 ) U(P, f ) + U(P, f 2 ) L(P, f + f 2 ) L(P, f ) + L(P, f 2 ). Let ǫ >. Then prtitions P nd P 2 of [, b] exist so tht U(P, f ) L(P, f ) < ǫ/2 nd U(P 2, f 2 ) L(P 2, f 2 ) < ǫ/2. Set P = P P 2. Then U(P, f + f 2 ) L(P, f + f 2 ) (U(P, f ) L(P, f )) + (U(P, f 2 ) L(P, f 2 )) < ǫ. This proves f + f 2 R[, b]. To find reltionship between the integrls we copy the ides of the previous two lemms. We hve A similr clcultion shows (f + f 2 ) dx U(P, f + f 2 ) U(P, f ) + U(P, f 2 ) < L(P, f ) + ǫ 2 + L(P, f 2) + ǫ 2 (f + f 2 ) dx > As before this shows (f + f 2 ) dx Since ǫ > is rbitrry, (6.4) (f + f 2 ) dx = f dx + f dx + f dx + f 2 dx + ǫ. f dx + f 2 dx ǫ. f 2 dx! f 2 dx. < ǫ.
13 6.3. Algebr of Integrble Functions 77 For rbitrry constnts c, c 2 R it follows from Lemm 6.9 tht c f R[, b] nd c 2 f 2 R[, b]. The bove shows c f + c 2 f 2 R[, b]. To estblish the integrl in (i), we pply (6.4) nd (6.3). To prove (ii) note tht L(P, f ) L(P, f 2 ) for ny prtition P of [, b]. Thus f (x) dx = f (x) dx f 2 (x) dx = f 2 (x) dx. To prove (iii) we ppel to the definition of f R[, b]. We note tht P = {, b} is prtition of [, b], nd L(P, f) = m(b ), U(P, f) = M(b ). Using the properties of prtitions given in Lemm 6.3, we hve for ny prtition, P of [, b] m(b ) L(P, f) f(x) dx U(P, f) M(b ). s required. Prt (iv) is left to the exercises (see Problem 6.9). Theorem 6.2. Suppose f R[, b]. Then f R[, b] nd (6.5) f(x) dx f(x) dx. Proof. We clim (6.6) M i ( f ) m i ( f ) M i (f) m i (f). Indeed, suppose x, y [x i, x i ]. If if f(x), f(y) f(x) f(y) = f(x) f(y) M i (f) m i (f); if f(x) f(y), then m i (f) nd f(x) f(y) = f(x) + f(y) M i (f) + M i (f) m i (f); if f(y) f(x), then, f(x) f(y) f(x) + f(y) M i (f) m i (f). In ll cses we see f(x) M i (f) m i (f)+ f(y). Tking the supremum of the left side M i ( f ) M i (f) m i (f)+ f(y). Then f(y) M i ( f ) M i (f)+m i (f). This implies m i ( f ) M i ( f ) M i (f)+m i (f), nd M i ( f ) m i ( f ) M i (f) m i (f). Let ǫ >. Then there exists prtition of [, b] such tht U(P, f) L(P, f) < ǫ. Inequlity (6.6) implies U(P, f ) L(P, f ) U(P, f) L(P, f) < ǫ, nd f R[, b]. Since f(x) f(x) f(x), Theorem 6.2, (ii) implies (6.5). Theorem Suppose f, g R[, b]. Then fg R[, b].
14 78 6. Integrtion Proof. Suppose f 2 R[, b] when f R[, b]. Then f(x)g(x) = (f(x) + g(x))2 f 2 (x) g 2 (x) 2 is in R[, b] by Theorem 6.2. To show f 2 R[, b], note tht M i (f 2 ) = (M i ( f )) 2 nd m i (f 2 ) = (m i ( f )) 2 (see Problem 2.6). Thus M i (f 2 ) m i (f 2 ) = (M i ( f )) 2 (m i ( f )) 2 = (M i ( f ) + m i ( f ))(M i ( f ) m i ( f )) 2M(M i ( f ) m i ( f )), where M is the bound on f. Since f R[, b], there exists prtition of [, b] such tht U(P, f ) L(P, f ) < ǫ/(2m) for ny ǫ >. Then U(P, f 2 ) L(P, f 2 ) 2M(U(P, f ) L(P, f )) < ǫ. This shows f 2 R[, b] Clsses of Integrble Functions We would like to chrcterize functions which re Riemnn integrble. mke some hedwy in tht direction. We cn Theorem If f : [, b] R is continuous (we sometimes write this f C[, b]), then f R[, b]. Proof. Since f C[, b], f is uniformly continuous (Theorem 4.4). Let ǫ >. Thus, there exists δ > such tht x, y [, b], x y < δ implies f(x) f(y) < ǫ/(b ). Let P be ny prtition of [, b] with kp k < δ. On nd subintervl [x i, x i ] there exists x m, x m [x i, x i ] (Theorem 4.26) such tht f(x M ) = M i (f) nd f(x m ) = m i (f). Since kp k < δ, x M x m < δ nd f(x M ) f(x m ) = M i (f) m i (f) < ǫ/(b ). Thus U(P, f) L(P, f) = < nd f R[, b]. (M i (f) m i (f))(x i x i ) ǫ b (x i x i ) = ǫ, Theorem If f : [, b] R is monotone, then f R[, b]. Proof. Suppose f is monotone incresing. Note f is bounded since f(x) f(b) for ll x [, b]. Let ǫ >. Choose δ > so tht (f(b) f())δ < ǫ. Choose
15 6.4. Clsses of Integrble Functions 79 prtition of [, b] so tht kp k < δ. Since f is incresing, m i (f) = f(x i ) nd M i (f) = f(x i ). Thus U(P, f) L(P, f) = (M i (f) m i (f))(x i x i ) < (M i (f) m i (f))δ = (f(b) f())δ < ǫ, nd f R[, b]. If f is decresing, then f is incresing, nd f R[, b]. By the lgebr of integrtion f R[, b]. Theorem Suppose f R[, b], nd g = f except t finite number of points. Then g R[, b], nd g(x) dx = f(x) dx. Proof. Suppose f differs from g t exctly one point in [, b], sy t z. Given ǫ > there exists prtition P of [, b] such tht U(P, f) L(P, f) < ǫ/2. We my ssume, by refining P if necessry, tht z is in P. Then Xj U(P, g) L(P, g) = (M i (f) m i (f))(x i x i ) +(M j (g) m j (g))(z x j ) + (M j+ (g) m j+ (g))(x j+ z ) + (M i (f) m i (f))(x i x i ). i=j+2 We note tht (though we do not need explicitly) nd M j (g) = mx{m j (f), g(z )}, M j+ (g) = mx{m j+ (f), g(z )}, m j (g) = min{m j (f), g(z )}, m j+ (g) = min{m j+ (f), g(z )}. The differences M j (g) m j (g) nd M j+ (g) m j+ (g) re therefore bounded. We cll the bound M. We my refine the prtition P, if necessry, so tht (M j (g) m j (g))(z x j ) + (M j+ (g) m j+ (g))(x j+ x ) M (z x j ) + (x j+ z ) < ǫ/2. It follows U(P, g) L(P, g) < ǫ nd g R[, b]. Next we show the integrls of f nd g re the sme. Agin by properties of supremum (Theorem 2.) there exist prtitions P nd P 2 of [, b] such tht U(P, f) < f(x) dx + ǫ 3, U(P 2, g) < g(x) dx + ǫ 3. Set P = P P 2. By refining P if necessry, s we did in computing U(P, g) L(P, g) bove, we my require U(P, f) U(P, g) < ǫ 3.
16 8 6. Integrtion Combining these we find f(x) dx g(x) dx < ǫ. f(x) dx U(P, f) + U(P, f) U(P, g) + U(P, g) g(x) dx Since ǫ > is rbitrry, we conclude R b g = R b f. If g differs from f on [, b] t m points, we mke sequence of functions g i, i m with g m = g, g differing from f t one point nd g i+ differing from g i t one point. We then iterte the bove proof m times to conclude R b f = R b g = R b g 2 = = R b g. While we will not hve time to develop the tools necessry to completely chrcterize which functions re Riemnn integrble, we stte the theorem here. Theorem A bounded function f is Riemnn integrble if nd only if it is continuous lmost everywhere. Here lmost everywhere mens in the sense of Lebesgue mesure. We will not discuss this further. However, the following corollry my be bit more pltble. It shows Theorem 6.25 remins vlid even if f differs from g t countble number of points. Corollry A bounded function f is Riemnn integrble if it is continuous everywhere except t countble number of points. We close with one more theorem  whose proof we leve to the exercises (see Problem 6.6). Theorem Suppose f is bounded on [, b] nd f is Riemnn integrble on every closed subintervl of (, b), then f R[, b]. Exmple Consider the function on [, ] x = n f(x) = x 6= n. Prove f R[, ] using three different methods. Proof. In Exmple 6. we proved directly tht f R[, b]. Note tht f is equl to the continuous function zero everywhere except t countble number of points. Corollry 6.27 pplies nd f R[, b]. Finlly, note on every subintervl of (, ) f is equl to zero everywhere except finite number of points. Thus by Theorem 6.25 f is integrble on every subintervl of (, ). Now Theorem 6.28 pplies nd f R[, b].
17 6.5. The Fundmentl of Clculus nd Derivtives of Integrls The Fundmentl of Clculus nd Derivtives of Integrls As you might expect there is reltion between integrls nd derivtives from your elementry clculus dys. In prticulr, one might expect the function defined by F (x) = Z x f(x) dx not only to be differentible but lso F (x) = f(x). In prticulr, given ny function f, we might expect R x f(x) dx to be n ntiderivtive of f. If you think this, the next exmples show you would be wrong. Exmple 6.3. Consider (yet gin) the function x = n f(x) = x 6= n, on [, ]. Set F (x) = Z x f(x) dx. We know from Exmple 6. tht F (x) =. Thus F is differentible, but F 6= f. Exmple 6.3. Consider the function x 2 sin(/x 2 ) < x F (x) = x =. One cn check tht F = f exists on [, ]. However, F is not bounded on [, ] nd R therefore not Riemnn integrble. Thus f hs n ntiderivtive, but F (x) 6= x f(x) dx. Exmple Consider the function x < f(x) = x on [, ]. As we sw Exmple 5.7, there is no function such tht F = f on [, ]. The function f does not hve n ntiderivtive. However, F (x) = R x f(s) ds mkes sense. On my wonder wht is the reltionship between F nd f in (6.7) in generl.we hve Theorem Let f be bounded on [, b] nd Riemnn integrble on [, b]. Set for x [, b]. Then (i) F is continuous on [, b] F (x) = Z x f(s) ds (ii) If f is continuous t x [, b], F (x ) = f(x )
18 82 6. Integrtion Proof. We re given M > exists such tht f(x) M for ll x [, b]. We pply the definition of continuity. Let ǫ > be given. Set δ = ǫ/m. If x, y [, b], x y nd x y < δ, then, using Theorem 6.2 (iii) nd (iv), Z y F (x) F (y) = f(s) ds < ǫ. x Z y x f(s) ds M x y Thus F is in fct Lipschitz on [, b] nd uniformly continuous there. Next, suppose f is continuous t x [, b]. Given ǫ >, choose δ > so tht x y < δ nd y [, b] implies f(y) f(x ) < ǫ/2. Then F (x ) F (y) R x f(x ) x y = f(s) ds R y f(s) ds f(x ) x y Z = x f(s) ds Z x f(x ) ds x y x y This implies F (x ) = f(x ). x y < ǫ. y Z x y y ǫ f(s) f(x ) ds < x y 2 x y The next version of the fundmentl theorem of clculus completely chrcterizes the nti derivtives of continuous functions. It lso vlidtes the reson so much effort ws mde in finding ntiderivtives in your elementry clculus clss. Theorem (Fundmentl Theorem of Clculus) Let f be continuous on [, b]. A function on [, b] stisfies (6.7) F (x) F () = for ll x [, b] if nd only if F = f. Z x f(s) ds Proof. Suppose F is defined by (6.7). Since f is continuous on [, b], Theorem 6.33 pplies nd we see F = f on [, b]. Conversely suppose F (x) = f(x) on [, b]. Set G(x) = Z x f(s) ds. Agin by Theorem 6.33, G (x) = f(x). Thus G (x) = F (x) on [, b]. Set H(x) = G(x) F (x) on [, b]. Then H is continuous nd differentible on [, b] nd H (x) = there. By the MenVlue Theorem H(x) = C (Theorem 5.5) for some rel constnt C. Thus G(x) = F (x) + C. However, we see G() = nd so C = F (). Tht is, s required. G(x) = Z x f(s) ds = F (x) F ()
19 6.5. The Fundmentl of Clculus nd Derivtives of Integrls 83 There is nother version of the Fundmentl Theorem of Clculus which requires Chrcteriztion IV. Theorem (Fundmentl Theorem of ClculusVersion II) Suppose F : [, b] R is differentible on [, b] nd F = f with f R[, b]. Then f(x) dx = F (b) F (). Proof. We pply Chrcteriztion IV. Let P be ny prtition of [, b]. If we pply the menvlue theorem to ech [x i, x i ], there exists t i [x i, x i ] such tht Thus f(t i )(x i x i ) = F (x i ) F (x i ). f(t i )(x i x i ) = F (b) f(). Let {P n } n N be ny sequence of prtitions of [, b] such tht kp n k s n, nd with the Riemnn sum constructed with t i s bove. Then, by the sequentil chrcteriztion of limits, Theorem 4.2, lim S(P n, f) = n f(x) dx = F (b) F (). We end this chpter with the (fmilir) chnge of vribles. Theorem Suppose g is differentible on [, b] nd g is Riemnn integrble on [, b]. If f is continuous on the rnge of g, then f(g(x))g (x) dx = Z g(b) g() f(t) dt. (Formlly obtined by setting t = g(x) nd dt = g (x)dx.) Exmple Before proceeding to the proof, we mke sure the nottion is understood. If we wnted to integrte Z + x dx, we would set t = + x = g(x), dt = dx, nd f(x) = x. Then Z Z 2 + x dx = t dt. Proof. Define F on the rnge of g to be F (x) = Z x g() Thus F (x) = f(x). By the chin rule, Theorem 5.6, (F g) (x) = F (g(x))g (x) = f(g(x))g (x) f.
20 84 6. Integrtion for ll x [, b]. By the fundmentl theorem of clculus f(g(x))g (x) dx = F g(b) F g() = F g(b) = Z g(b) g() f. Summry of Ides If we wnted conditions tht put us bck in n engineering world where functions re well behved nd the big theorems pply, we might consider the following theorem. Theorem Suppose I is closed, bounded intervl nd f : I R is continuously differentible on I. Then f is uniformly continuous on I. f is uniformly continuous on I. f hs minimum nd mximum. f enjoys the intermeditevlue property. If f injective, f is continuous nd differentible. The integrl of f exists nd my be evluted using the fundmentl theorem of clculus.
21 6.5. The Fundmentl of Clculus nd Derivtives of Integrls 85 The Completeness Axiom MonotoneConvergence Theorem NestedCell Theorem BolznoWeierstrss Theorem Unif.Cont. ExtremeVlue Thm Cuchy Thm Archimeden Principle Density of Q f cont f R(I) Rolle s Theorem MenVlue Theorem Cuchy MVT L Hôsiptl s Rule The big theorems of the course. We note tht the following re equivlent: () The completeness xiom. (2) Every Cuchy sequence of rel numbers converges. (3) Every bounded monotone sequence of rel numbers converges. (4) The BolznoWeierstrss theorem. (5) The nestedcell theorem.
22 86 6. Integrtion 6.6. Homework Exercise 6.. Let f(x) = c on [, b]. Show f R[, b] nd R b f(x) dx = c(b ). Exercise 6.2. Let f(x) = x = x >. Show tht f is Drboux integrble on [, ] nd find R f(x)dx. Exercise 6.3. Let x x Q f(x) = x Q c Show tht f is not Drboux integrble on [, ]. Exercise 6.4. x < f(x) = 2 x = < x 2 Show tht f is Drboux integrble on [, 2] nd find R 2 f dx. Exercise 6.5. (Thome s Function) Let x Q c f(x) = n x Q, x = m n x = Show tht f is Drboux integrble on [, ]. Exercise 6.6. Show the functions in Problems (6.), (6.2), nd (6.4) re integrble by pplying theorems. Exercise 6.7. Give exmples of functions f nd g with fg nd f integrble on some bounded closed intervl, but with neither f or g Drboux integrble on tht intervl. Exercise 6.8. Give exmples of function f : [, ] R nd g : [, ] R both Riemnn integrble on [, ] such tht f g is not Riemnn integrble on [, ]. So f, g R[, b] does not imply f g R[, b]. Hint: look t Problem 4.7. Exercise 6.9. Let < c < b, nd let f be defined on [, b]. Show tht f R[, b] if nd only if f R[, c] nd f R[c, b]. Moreover, R b f = R c f + R b c f. Exercise 6.. Suppose f R[, b]. Show tht f = lim c + Exercise 6.. Suppose lim S(f, P ) = I for some I R. Let {P n} n N be ny kp k sequence of prtitions such tht lim kp nk =. Show lim S(f, P n) = I. n n Exercise 6.2. Suppose f C[, b] nd f(x) for ll x [, b] with f not identiclly zero. Show tht R b f >. Exercise 6.3. Suppose f, g C[, b] nd R b f = R b g. Show tht f(c) = g(c) for some c [, b]. (For fun cn you supply physicl interprettion of this?) c f.
23 6.6. Homework 87 Exercise 6.4. Suppose f C[, b]. Show there is c [, b] such tht f(c) = b Exercise 6.5. (MenVlue Theorem for Integrls) Generlize the previous problem: Suppose f : [, b] R is continuous nd g : [, b] R is integrble on [, b] with R b g 6=. Moreover, suppose g(x) for ll x [, b]. Then, there exists c [, b] such tht f(x)g(x) dx = f(c) f. g(x) dx. Exercise 6.6. Prove Theorem Tht is, suppose f is bounded on [, b] nd f is Riemnn integrble on every closed subintervl of (, b), then f R[, b]. Use this to show the function sin f(x) = x < x x = is Riemnn integrble on [, ]. Exercise 6.7. Find with proof the following limits () i lim n n n. (2) lim n i + 2n. (3) lim n n iπ sin. 2n Exercise 6.8. We proved in Exmple 3.28 tht the sequence S n = n + + n n is monotone incresing nd bounded bove (you do not need to show this). We were not ble to find the limit then  now we cn. Find the limit (Hint: it is Riemnn sum). Exercise 6.9. Define Z x ln x = t dt for x >. Prove tht ln x is continuously differentible nd injective on (, ). We define the number e so tht ln e =. Prove tht e exists. Also show tht f(x) = e x is the continuous inverse of ln x. You need only pply pproprite theorems for this problem. Exercise 6.2. (integrtion by prts). Suppose u nd v re differentible functions on [, b] nd tht both u nd v re integrble on [, b]. Show tht both uv nd u v re Riemnn integrble on [, b]. Then prove tht Z uv = uv b b u v.
24 88 6. Integrtion Exercise 6.2. By considering the upper nd lower sums for R n ln x dx show lim n n! (n!) /n = nd lim nn n n = e. Exercise Suppose f C[, b] nd R x f = for ll x [, b]. Show f = on [, b]. Exercise Suppose f C[, b] nd R x f = R b f for ll x [, b]. Show f = x on [, b]. Exercise Suppose f C[, b] nd strictly positive on [, b]. F (x) = R x f is strictly incresing on [, b]. Exercise Compute d Z x 3 cos(t 2 ) dt. dx x 2 Exercise If f is Riemnn integrble on [, b], show R b f dx = lim Suppose f is given by x f(x) = < x Show tht c + R b c f dx. Compute F (x) = R x f dx nd F () using the fundmentl theorem of clculus. Recll f does not hve n ntiderivtive. Exercise Suppose f is continuous on [, ]. Define g n (x) = f(x n ) for n N. Prove tht lim n Z g n (x) dx = f(). Exercise We sy continuous function on R is periodic with period p > if f(x + p) = f(x) for ll x R. Show by exmple tht ntiderivtives of periodic functions need not be periodic while derivtives of continuously differentible periodic function re lwys periodic. Prove tht necessry nd sufficient condition for periodic function to hve periodic ntiderivtive is for f to hve zero verge over one period. Tht is, R p f(t) dt = is required. Exercise Let f be Riemnn integrble on [, ], where >. Prove the following. () If f is even, then R f dx = 2 R f dx. (b) If f is odd, then R f dx =. Exercise 6.3. Let f be continuous nd nonnegtive on [, ]. Define M = mx{f(x) x [, ]}. Show tht Z /n M = lim f dx n. n
25 6.6. Homework 89 Exercise 6.3. Suppose f is continuous on [, b] nd tht f(x)ϕ(x) dx = for ll continuous ϕ on [, b] with ϕ() = = ϕ(b). Then f(x) = on [, b]. Exercise Suppose f is continuous on [, b] nd tht f(x)ϕ (x) dx = for ll continuously differentible functions ϕ on [, b] with ϕ() = = ϕ(b). Then f(x) is constnt on [, b]. Exercise Suppose f : [, b] R is Riemnn integrble on [, b]. Show F (x) = R x f(t) dt stisfies the inequlity F (x) F (y) C x y for ll x [, b]. Then show F is uniformly continuous on [, b]. Clculte, justifying your steps, F (x) = R x f(t) dt, where t f(t) = < t 2 Exercise Show x f(x) = < x 2 is Riemnn integrble on [, 2], nd find F (x) = R x f(t) dt. Wht properties does F hve? Exercise Cler up the devstting Exmple.2 in Chpter one. A function f : [, ) R is clled uniformly Lipschitz  there exists M such tht f(x) f(y) M x y for ll x, y [, ). Now prove the following theorem. Theorem Suppose f : [, ) R is uniformly Lipschitz nd Z f(x) dx <. Then lim f(x) =. x Prove the theorem by proving the contrpositive. Tht is, negte lim f(x) = x, use the properties of f to show R f(x) dx =. Exercise Repet Problem 6.35 only ssume f is differentible on [, ) nd Z (f (x)) 2 dx <.
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