Chapter 7 Riemann-Stieltjes Integration

1 Chapter 7 Riemann-Stieltjes IntegrationCalculus provides us with tools to study nicely behaved phenomena using smalldiscrete increments for information collection. The general idea is to (intelligently)connect information obtained from examination of a phenomenon over a lot of tinydiscrete increments of some related quantity to close in on or approximate some-thing that behaves in a controlled ( , bounded, continuous, etc.) way. The clos-ing in on approach is useful only if we can get back to information concerning thephenomena that was originally under study. The bene t of this approach is mostbeautifully illustrated with the elementary theory of integral calculus us to adapt some limiting formulas that relate quantities of physical interestto study more realistic situations involving the three formulas that are encountered frequently in most standard phys-ical science and physics classes at the pre-college level:A l *d r tm d the space that is provided to indicate what you know about these use of these formulas is limited to situations where the quantities on theright are constant.

2 The minute that we are given a shape that is not rectangular,a velocity that varies as a function of time, or a density that is determined by ourposition in (or on) an object, at rst, we appear to be out of luck. However, whenthe quantities given are well enough behaved, we can obtain bounds on what we275276 Chapter 7. Riemann-Stieltjes Integration wish to study, by making certain assumptions and applying the known that except for the units, the formulas are indistinguishable. Consequently,illustrating the closing in on" or approximating process with any one of them car-ries over to the others, though the physical interpretation (of course) s get this more down to earth! Suppose that you build a rocket launcher aspart of a physics project. Your launcher res rockets with an initial velocity of 25ft/min, and, due to various forces, travels at a rate) t given by) t 25 t2ft/minwheretis the time given in minutes.

3 We want to know how far the rocket travels inthe rst three minutes after launch. The only formula that we have isd r t,butto use it, we need a constant rate of can make use of the formula to obtainbounds or estimates on the distance do this, we can take increments inthe time from 0 minutes to 3 minutes and pick a relevant rate to compute a boundon the distance travelled in each section of time. For example, over the entire threeminutes, the velocity of the rocket is never more that 25ft does this tell us about the product 25 ft/min 3mincompared to the distance that we seek?How does the product 16 ft/min 3min relate to the distance that we seek?We can improve the estimates by taking smaller increments (subintervals of 0minutes to 3 minutes) and choosing a different estimating velocity on each subin-terval. For example, using increments of 1 5 minutes and the maximum velocitythat is achieved in each subinterval as the estimate for a constant rate through riemann SUMS AND INTEGRABILITY277subinterval, yields an estimate of 25 ft/min 1 5min tt25 94uft/minu 1 5 min the estimate for the distance travelled taking increments ofone minute (which is not small for the purposes of calculus) and using the minimumvelocity achieved in each subinterval as the estimating velocity.

4 **Hopefully, you obtained 61 feet.**Notice that none of the work done actually gave us the answer to the originalproblem. Using Calculus, we can develop the appropriate tools to solve the problemas an appropriate limit. This motivates the development of the very important anduseful theory of Integration . We start with some formal de nitions that enable us tocarry the closing in on process to its logical riemann Sums and IntegrabilityDe nition a closed interval I [a b],apartitionof I is any nitestrictly increasing sequence of pointsS x0 x1 xn 1 xn such that a x0and b of the partition x0 x1 xn 1 xn is de ned bymeshS max1njnnbxj xj 1c Each partition of I , x0 x1 xn 1 xn , decomposes I into n subintervals Ij dxj 1 xje,j 1 2 n, such that IjDIk xjif and only if k j 1and isempty for k/ jork/ j 1 . Each such decomposition of I into subintervals iscalled asubdivision a partitionS x0 x1 xn 1 xn of an interval I [a b], the two notations xjand(bIjcwill be used forbxj xj 1c, the length ofthe jthsubinterval in the partition.)

5 The symbol or I will be used to denotean arbitrary subdivision of an interval I .278 Chapter 7. Riemann-Stieltjes INTEGRATIONI ffis a function whose domain contains the closed intervalIandfis boundedon the intervalI, we know thatfhas both a least upper bound and a greatest lowerbound onIas well as on each interval of any subdivision nition a function f that is bounded and de nedontheintervalI and a partitionS x0 x1 xn 1 xn of I , let Ij dxj 1 xje,Mj supx+Ijf x and mj infx+Ijf x for j 1 2 n. Then theupper riemann sum offwith respect to the partitionS, denoted by U S f ,isde ned byU S f n;j 1Mj xjand thelower riemann sum offwith respect to the partitionS, denoted byL S f ,isde ned byL S f n;j 1mj xjwhere xj bxj xj the subdivision notation the upper and lower riemann sumsfor f are denoted by U f and L f , r f x 2x 1in I [0 1]andS |0 14 12 34 1},U S f 14t32 2 52 3u 94and L S f 14t1 32 2 52u r g x 0, forx+TD[0 2]1, forx +TD[0 2]U I g 2and L I g 0for any subdivision of[0 2].

6 To build on the motivation that constructed some riemann sums to estimate adistance travelled, we want to introduce the idea of re ning or adding points topartitions in an attempt to obtain better riemann SUMS AND INTEGRABILITY279De nition a partitionSk x0 x1 xk 1 xk of an interval I [a b],let kdenote to corresponding subdivision of[a b].IfSnandSmarepartitions of[a b]having n 1and m 1points, respectively, andSntSm, thenSmis are nementofSnor mis are nementof n. If the partitionsSnandSmare independently chosen, then the partitionSnCSmis acommon re nement ofSnandSmand the resulting SnCSm is called acommon re nement of nand |0 12 34 1}andS` |0 14 13 12 58 34 1}.(a) If and `are the subdivisions of I [0 1]that correspondSandS`,respectively, then |v0 12w v12 34w v34 1w}.Find `.(b) Set I1 v0 12w,I2 v12 34w,andI3 v34 1 2 3,let k bethe subdivision of Ikthat consists of all the elements of `that are containedin k for k 1 2 and3.

7 (c) For f x x2and the notation established in parts (a) and (b), nd each ofthe following.(i) m infx+If x 280 Chapter 7. Riemann-Stieltjes Integration (ii) mj infx+Ijf x for j 1 2 3(iii) m`j inf|infx+Jf x :J+ j }(iv) M supx+If x (v) Mj supx+Ijf x for j 1 2 3(vi) M`j sup|supx+Jf x :J+ j }(d) Note how the values m, mj,m`j,M,Mj, and M`jcompare. What you ob-served is a special case of the general situation. LetS x0 a x1 xn 1 xn b be a partition of an interval I [a b], be the corresponding subdivisionof[a b]andS`denote a re nement ofSwith corresponding subdivision de-noted by `.Fork 1 2 n, let k be the subdivision of Ikconsistingof the elements of `that are contained in Ik. Justify each of the followingclaims for any function that is de ned and bounded on I .(i) If m infx+If x and mj infx+Ijf x , then, for j 1 2 n, mnmjand mjninfx+Jf x for J+ j . riemann SUMS AND INTEGRABILITY281(ii) If M supx+If x and Mj supx+Ijf x ,then,for j 1 2 n, MjnM and Mjosupx+Jf x for J+ j.

8 Our next result relates the riemann sums taken over various subdivisions of that f is a bounded function with domain I [a b].Let be a subdivision of I , M supx+If x , and m infx+If x . Thenm b a nL f nU f nM b a ( )andL f nLb ` fcnUb ` fcnU f ( )for any re nement `of . Furthermore, if <and Dare any two subdivisionsof I , thenLb < fcnU D f ( )Excursion in what is missing to complete the following thatfis a bounded function with domainI [a b],M supx+If x ,andm infx+If x .For Ik:k 1 2 n an arbitrary subdi-vision ofI,letMj supx+Ijf x andmj infx+Ijf x . ThenIjtIfor eachj 1 2 n, we have thatmnmjn 1 , for eachj 1 2 xj bxj xj 1co0 for eachj 1 2 n, it follows immediatelythat 2 mn;j 1bxj xj 1cnn;j 1mj xj L f 282 Chapter 7. Riemann-Stieltjes Integration andn;j 1mj xjnn;j 1Mj xj U f n 3 M b a .Therefore,m b a nL f nU f nM b a as claimed in equation( ).Let `be a re nement of and, for eachk 1 2 n,let k be thesubdivision ofIkthat consists of all the elements of `that are contained view of the established conventions for the notation being used, we know that 1J J+ `" 2!

9 K k+ 1 2 n FJ+ k also, for eachJ+ k ,JtIk"mk infx+Ikf x ninfx+Jf x andMk supx+Ikf x osupx+Jf x . Thus,mk( Ik nL k f andMk( Ik oU k f from which it follows thatL f n;j 1mj(bIjcnn;j 1L j f Lb ` fcandU f 4 on;j 1U j f 5 .From equation ( ),L ` f nU ` f . Finally, combining the inequalitiesyields thatL f nLb ` fcnUb ` fcnU f which completes the proof of equation ( ).Suppose that <and Dare two subdivisions ofI. Then <C Dis 6 <and D. Because is a re nement of <,bythecomparison of lower sums given in equation ( ),Lb < fcnL f .Ontheother hand, from being a re nement of D,itfollowsthat 7 .Combining the inequalities with equation ( ) leads to equation ( ). riemann SUMS AND INTEGRABILITY283**Acceptable responses are: (1)MjnM,(2)m b a ,(3)M3nj 1bxj xj 1c,(4)3nj 1Mj(bIjc,(5)U ` f , (6) the common re nement of, and(7)U f nU f D .**Iffis a bounded function with domainI [a b]and, ,[a b]istheset of all partitions of [a b], then the Lemma assures us that L f : +, isbounded above by b a supx+If x and U f : +, is bounded below by b a infx+If x.))))

10 Hence, by the least upper bound and greatest lower bound prop-erties of the reals both sup L f : +, and inf U f : +, exist tosee that they need not be equal, note that for the bounded functionggiveninExam-ple we have that sup L g : +, 0 while inf U g : +, nition that f is a function onUthat is de ned and bounded onthe interval I [a b]and, ,[a b]is the set of all partitions of[a b]. Thentheupper riemann integraland thelower riemann integralare de ned by=baf x dx inf +,U S f and=baf x dx sup +,L S f ,respectively. If5baf x dx 5baf x dx, then f isRiemann integrable, or justintegrable, on I , and the common value of the integral is denoted by=baf x f x 5x 3, forx +T0, forx+ r e a c h n+M,let ndenote the subdivision of the interval[1 2]that con-sists of n segments of equal length. Use n:n+M to nd an upper bound for284 Chapter 7. Riemann-Stieltjes INTEGRATION521f x d x.