Chapter 7 Riemann-Stieltjes Integration Calculus provides us with tools to study nicely behaved phenomena using small discrete increments for information collection. The general idea is to (intelligently) connect information obtained from examination of a phenomenon over a lot of tiny discrete increments of some related quantity to “close in on” or approximate some- thing that behaves in a controlled (i.e., bounded, continuous, etc.) way. The “clos- ing in on” approach is useful only if we can get back to information concerning the phenomena that was originally under study. The bene¿t of this approach is most beautifully illustrated with the elementary theory of integral calculus over U.Iten- ables us to adapt some “limiting” formulas that relate quantities of physical interest to study more realistic situations involving the quantities. Consider 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 l. Use the space that is provided to indicate what you “know” about these formulas. Our use of these formulas is limited to situations where the quantities on the right are constant. 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 our position in (or on) an object, at ¿rst, we appear to be “out of luck.” However, when the quantities given are well enough behaved, we can obtain bounds on what we 275 276 CHAPTER 7. RIEMANN-STIELTJES INTEGRATION wish to study, by making certain assumptions and applying the known formulas incrementally. Note 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) varies. Let’s get this more down to earth! Suppose that you build a rocket launcher as part of a physics project. Your launcher ¿res rockets with an initial velocity of 25 ft/min, and, due to various forces, travels at a rate ) t given by )t 25 t2 ft/min where t is the time given in minutes. We want to know how far the rocket travels in the ¿rst three minutes after launch. The only formula that we have is d r t,but to use it, we need a constant rate of speed. We can make use of the formula to obtain bounds or estimates on the distance travelled. To do this, we can take increments in the time from 0 minutes to 3 minutes and “pick a relevant rate” to compute a bound on the distance travelled in each section of time. For example, over the entire three minutes, the velocity of the rocket is never more that 25 ftmin. What does this tell us about the product 25 ft/min 3min compared 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 0 minutes to 3 minutes) and choosing a different “estimating velocity” on each subin- terval. For example, using increments of 15 minutes and the maximum velocity that is achieved in each subinterval as the estimate for a constant rate through each 7.1. RIEMANN SUMS AND INTEGRABILITY 277 subinterval, yields an estimate of tt u u 9 573 25 ft/min 15min 25 ft/min 15 min ft. 4 8 Excursion 7.0.1 Find the estimate for the distance travelled taking increments of one minute (which is not small for the purposes of calculus) and using the minimum velocity achieved in each subinterval as the “estimating velocity.” ***Hopefully, you obtained 61 feet.*** Notice that none of the work done actually gave us the answer to the original problem. Using Calculus, we can develop the appropriate tools to solve the problem as an appropriate limit. This motivates the development of the very important and useful theory of integration. We start with some formal de¿nitions that enable us to carry the “closing in on process” to its logical conclusion. 7.1 Riemann Sums and Integrability De¿nition 7.1.1 Given a closed interval I [a b],apartition of I is any ¿nite strictly increasing sequence of points S x0 x1xn1 xn such that a x0 and b xn.Themesh of the partition x0 x1xn1 xn is de¿ned by b c mesh S max x j x j1 1n jnn dEach partitione of I, x0 x1xn1 xn , decomposes I into n subintervals I j x j1 x j ,j 1 2 n, such that I j D Ik x j if and only if k j 1 and is empty for k / jork/ j 1. Each such decomposition of I into subintervals is called a subdivision of I. Notation 7.1.2 Given a partition Sb c x0 x1xnb1 xn of anc interval I [a b], the two notations x j and ( I j will be used for x j x j1 , the length of the j th subinterval in the partition. The symbol or I will be used to denote an arbitrary subdivision of an interval I. 278 CHAPTER 7. RIEMANN-STIELTJES INTEGRATION If f is a function whose domain contains the closed interval I and f is bounded on the interval I , we know that f has both a least upper bound and a greatest lower bound on I as well as on each interval of any subdivision of I. De¿nition 7.1.3 Given a function f that is bounded and de¿nedontheintervald e I and a partition S x0 x1xn1 xn of I, let I j x j1 x j ,Mj sup f x and m j inf f x for j 1 2n. Then the upper Riemann sum of x+I x+I j j f with respect to the partition S, denoted by U S f ,isde¿ned by ;n U S f M j x j j1 and the lower Riemann sum of f with respect to the partition S, denoted by L S f ,isde¿ned by ;n L S f m j x j j1 b c where x j x j x j1 . Notation 7.1.4 With the subdivision notation the upper and lower Riemann sums for f are denoted by U f and L f , respectively. | } 1 1 3 Example 7.1.5 For f x 2x 1 in I [0 1] and S 0 1 , t u t 4 2u4 1 3 5 9 1 3 5 7 U S f 2 3 and L S f 1 2 . 4 2 2 4 4 2 2 4 0 , for x + TD [0 2] Example 7.1.6 For g x 1 , for x + TD [0 2] U I g 2 and L I g 0 for any subdivision of [0 2]. To build on the motivation that constructed some Riemann sums to estimate a distance travelled, we want to introduce the idea of re¿ning or adding points to partitions in an attempt to obtain better estimates. 7.1. RIEMANN SUMS AND INTEGRABILITY 279 De¿nition 7.1.7 For a partition Sk x0 x1 xk1 xk of an interval I [a b],letk denote to corresponding subdivision of [a b].IfSn and Sm are partitions of [a b] having n 1 and m 1 points, respectively, and Sn t Sm, then Sm is a re¿nement of Sn or m is a re¿nement of n. If the partitions Sn and Sm are independently chosen, then the partition Sn C Sm is a common re¿nement of Sn and Sm and the resulting Sn C Sm is called a common re¿nement of n and m. | } | } 1 3 1 1 1 5 3 Excursion 7.1.8 Let S 0 1 and S` 0 1 . 2 4 4 3 2 8 4 ` ` (a) If and are the subdivisions|v w v of I w v[0 1]w}that correspond S and S , 1 1 3 3 respectively, then 0 1 .Find`. 2 2 4 4 v w v w v w 1 1 3 3 (b) Set I 0 ,I ,andI 1 .Fork 1 2 3,letk be 1 2 2 2 4 3 4 ` the subdivision of Ik that consists of all the elements of that are contained in Ik.Find k for k 1 2 and 3. (c) For f x x2 and the notation established in parts (a) and (b), ¿nd each of the following. (i) m inf f x x+I 280 CHAPTER 7. RIEMANN-STIELTJES INTEGRATION (ii) m j inf f x for j 1 2 3 x+I j | } (iii) m` inf inf f x : J + j j x+J (iv) M sup f x x+I (v) M j sup f x for j 1 2 3 x+I j | } ` + (vi) M j sup sup f x : J j x+J ` ` (d) Note how the values m, m j ,mj ,M,Mj , and M j compare. What you ob- served is a special case of the general situation. Let S x0 a x1 xn1 xn b be a partition of an interval I [a b], be the corresponding subdivision of [a b] and S` denote a re¿nement of S with corresponding subdivision de- ` noted by .Fork 1 2 n, let k be the subdivision of Ik consisting ` of the elements of that are contained in Ik. Justify each of the following claims for any function that is de¿ned and bounded on I.