For a continuous function f : [a; b]! R we wish to define the Riemann integral


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1 Supplementry Notes for MM509 Topology II 2. The Riemnn Integrl Andrew Swnn For continuous function f : [; b]! R we wish to define the Riemnn integrl R b f (x) dx nd estblish some of its properties. This discussion is bsed on the corresponding section in [1]. 2.1 Step functions A function : [; b]! R is step function if there is prtition P = fx 0 ; x 1 ; : : : ; x n g, = x 0 < x 1 < x 2 < < x n = b; of [; b] so tht is constnt on ech open subintervl (x j 1 ; x j ), j = 1; 2; : : : ; n, i.e., (x) = j ; for x j 1 < x < x j, for some constnts 1 ; : : : ; n. If ech j is positive, then the re under the grph of is the sum of the widths (x j x j 1 ) of the intervls times their heights j. This motivtes us to define I() = I P () = j (x j x j 1 ); which is the vlue we would like to ssign to the integrl of over [; b]. If nother prtition P 0 stisfies P P 0, then we sy tht P 0 refines P. We see tht is constnt on ech open subintervl defined by P 0 nd tht I P 0() = I P (), since we hve done nothing but subdivide the intervls (x j 1 ; x j ) where is constnt. Lemm 2.1. Suppose nd re step functions on [; b]. Then (i) for ny ; 2 R, + is gin step function nd I( + ) = I() + I( ); (2.1) 2.1
2 (ii) 6 implies I() 6 I( ). Proof. Let P nd P be prtitions corresponding to the step functions nd. Put P = P [ P = fx 0 ; x 1 ; : : : ; x N g. Then P refines both P nd P, nd re both constnt on ech open intervl (x j 1 ; x j ), j = 1; 2; : : : ; N. Write (x) = j, (x) = j for x j 1 < x < x j. For the first prt of the Lemm, we hve (+ )(x) = j + j for x j 1 < x < x j, so + is indeed step function. We compute I( + ) = I P ( + ) = = NX NX j (x j x j 1 ) + X ( j + j )(x j x j 1 ) j (x j x j 1 ) = I P () + I P ( ) = I() + I( ); s required. For the second prt, we see tht is step function nd tht > 0. From the definition of I( ), we get immeditely tht 0 6 I( ) = I( ) I(), nd hence I( ) > I(). 2.2 Upper nd lower sums Let f : [; b]! R be bounded function. Suppose nd re step functions such tht 6 f 6 n [; b]. Then we cll I( ) n upper sum for f nd I() lower sum for f. The ssumption tht f is bounded gurntees the existence of such nd. We expect the integrl of f to lie between these two numbers. Note tht Lemm 2.1 shows I() 6 I( ): (2.2) Consider the sets U(f ) = f I( ) : is step function with > f g; L(f ) = f I() : is step function with 6 f g: These sets re nonempty. Furthermore, ech upper sum I( ) for f is n upper bound for L(f ) by (2.2), so sup L(f ) exists nd stisfies sup L(f ) 6 I( ). Now this sys tht sup L(f ) is lower bound for U(f ), so we conclude tht inf U(f ) exists nd stisfies sup L(f ) 6 inf U(f ): (2.3) 2.2
3 Definition 2.2. A bounded function f : [; b]! R is sid to be Riemnn integrble if sup L(f ) = inf U(f ). In this cse we define 2.3 Integrls of continuous functions f (x) dx = sup L(f ) = inf U(f ): Theorem 2.3. Suppose f : [; b]! R is continuous. Then f is Riemnn integrble. Proof. By (2.3) we need only show inf U(f ) 6 sup L(f ). Given " > 0, we prove tht inf U(f ) 6 sup L(f ) + ". Since [; b] is compct nd f is continuous we hve tht f is uniformly continuous by [2, Proposition 5.8.4]. So there is > 0 such tht jf (x) f (y)j < " ; for ll jx yj < : (2.4) b Let P = fx 0 ; x 1 ; : : : ; x n g be prtition of [; b] with ech subintervl (x j 1 ; x j ) of length <. (E.g. choose x j = + j=2 for j = 0; : : : ; n 1, n > 2(b )=.) Put j = inf f (x) nd j = sup f (x) x2[x j 1 ;x j ] x2[x j 1 ;x j ] for j = 1; : : : ; n, so j j < "=(b ) by (2.4), nd define two step functions f; (x) = ( j ; for x j 1 6 x < x j, n ; for x = b nd f; (x) = ( j ; for x j 1 6 x < x j, n ; for x = b. Then f; 6 f 6 f;, so I() 6 sup L(f ) nd I( ) > inf U(f ). We now hve s required. inf U(f ) sup L(f ) 6 I( ) I() = 6 " b (x j x j 1 ) = "; ( j j )(x j x j 1 ) In the proof we defined step functions by tking the mximum nd minimum vlues of f on [x j 1 ; x j ]. However, the proof shows tht ny vlue of f from this intervl will led to t lest s good n pproximtion of the integrl. Corollry 2.4. Suppose f, ",, P, f; nd f; re s in the proof of Theorem 2.3. Suppose tht f; is step function such tht f; (x) = f ( j ) for x j 1 < x < x j nd nd some j 2 [x j 1 ; x j ], j = 1; 2; : : : ; n. Then f; 6 f; 6 f; I( f;) f (x) dx 6 ": 2.3
4 One simple consequence of the setup for the Riemnn integrl is: Theorem 2.5 (Men Vlue Theorem for Integrls). Suppose f : [; b]! R is continuous. Then there is 2 (; b) such tht f (x) dx = f ()(b ): Proof. Since f ([; b]) is compct intervl, there exist 1 ; 2 2 [; b] such tht f ( 1 ) nd f ( 2 ) re the minimum nd mximum of f on [; b]. Put 1 (x) = f ( 1 ) nd 2 (x) = f ( 2 ), for ll x 2 [; b]. Then 1 nd 2 re step functions with 1 6 f 6 2, so f ( 1 )(b ) = I( 1 ) 6 f (x) dx 6 I( 2 ) = f ( 2 )(b ): Thus R b f (x) dx = (b ) for some f ( 1) 6 6 f ( 2 ). But f ([; b]) = [f ( 1 ); f ( 2 )] so = f () for some 2 (; b). 2.4 Integrls nd differentible functions The next two results constitute the Fundmentl Theorem of Clculus relting derivtives nd integrls. Theorem 2.6. Suppose f : [; b]! R is continuous. Let g(x) = Z x f (t) dt: Then g : [; b]! R is differentible function with g 0 (x) = f (x); for ll x 2 [; b]: Proof. Consider x < b nd 0 < h < b we hve for some x < h < x + h. So g(x + h) g(x) = g(x + h) g(x) h x. By the Men Vlue Theorem for Integrls Z x+h x f (t) dt = f ( h )h = f ( h )! f (x) for h! 0 since h! x nd f is continuous t x. Thus g is differentible on the right t x with g 0 + (x) = f (x). Similrly, for x >, g is differentible on the left with g0 (x) = f (x) nd the result follows. 2.4
5 Theorem 2.7. Suppose F : [; b]! R is differentible with f = F 0 continuous. Then F 0 (x) dx = F (b) F (): Proof. As we hve tht f = F 0 is uniformly continuous, so given " > 0 there is > 0 such tht (2.4) holds. Choose P s in tht proof. The Men Vlue Theorem for F on [x j 1 ; x j ] gives the existence of j 2 (x j 1 ; x j ) such tht F (x j 1 ) F (x j ) = f (xi j )(x j x j 1 ). We now hve where jj 6 ". F (b) F () = f ( j )(x j x j 1 ) = I( f; ) = f (x) dx + ; References [1] Supplerende noter MM04, Lecture Notes, Institute for Mthemtics nd Computer Science, University of Southern Denmrk, [2] W. Sutherlnd, Introduction to metric nd topologicl spces, Clrendon Press, Oxford, Lst revised: 6th Februry
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