Lecture 10 Riemann-Stieltjes Integration In this section we will develop the basic definitions and some of the properties of the Riemann-Stieltjes Integral. The development will follow that of the Darboux integral that we have just finished. In fact, what we do today will serve as a review of the major points there, because the proofs are very nearly the same. The Stieltjes integral is based on the notion of replacing the length of the subintervals [t k 1, t k ] by the difference of function values for an increasing function F (t); (t k t k 1 ) is replaced by F (t k ) F (t k 1 ). We motivate the introduction of the Stieltjes integral by something I mentioned previously about how integration by parts can be used to define a derivative in a weak sense. If φ is an infinitely differentiable function which is zero outside some bounded interval [a, b] (depending on the function), then φ will be called a test function. For an integrable function F, if we had the integration by parts formula, then b a b φf = φ F a (because the boundary terms are zero by the nature of φ). In this way, the derivative F can be associated with the linear mapping F : φ φ F Let s examine this for the Heaviside function introduced in the last lecture: { 1, if t > 0; H(t) = 0, if t 0. Let φ be a test function which vanishes outside [a, b] and suppose 0 [a, b). Let s look at some partition P of [a, b]. The following argument is for heuristic purposes only (you will see some holes in it): Suppose that for each subinterval [t k 1, t k ], we select x k so that φ (x k )(t k t k 1 ) = φ(t k ) φ(t k 1 ) and we approximate the integral b a φ F by the sum φ (x k )F (t k 1 )(t k t k 1 ) = = = = (φ(t k ) φ(t k 1 ))F (t k 1 ) φ(t k )F (t k 1 ) + φ(t k 1 )F (t k 1 ) n 1 φ(t k )F (t k 1 ) + φ(t k )F (t k ) φ(t k )[F (t k ) F (t k 1 )]. 1
If F = H, then the only nonzero term in the last sum is for the interval [t k0 1, t k0 ] that contains 0, and the sum reduces to φ(t k0 ). As the mesh of the partition goes to zero, φ(t k0 ) φ(0). In this way, the derivative H of the Heaviside function is associated with the linear functional δ 0 : φ φ(0), (sometimes called a impulse function by engineers). Since F is increasing, the term [F (t k ) F (t k 1 )] will be nonnegative, which was an important property of the length that we used implicitly all the time in the previous discussion. But increasing functions may have jumps at some of the points t k. To take jumps into account in a natural way, we use the one-sided limits from inside the interval [t k 1, t k ] to define the values for F to be used at t k and t k 1. Notation: We assume F is an increasing function on [a, b]. We adopt the notation: F (t ) = lim x t F (x), and F (t+ ) = lim x t + F (x). We also take F (a) = F (a ) < F (b) = F (b + ) to allow use of the notation at the endpoints. Note that always F (t ) F (t + ) and if the difference F (t + ) F (t ) > 0, then its value is called the jump of F at t. The treatment of jumps is a distinguishing feature of the development given in the textbook by Ross. Definition 10.1: For a bounded function f on [a, b] and a partition P = {a = t 0 < t 1 <... < t n = b} of the interval [a, b] the jump sum of f relative to F and P is J F (f, P ) := f(t k )[F (t + k ) F (t k )] The upper and lower Darboux-Stieltjes sums for f relative to F are U F (f, P ) = J F (f, P ) + L F (f, P ) = J F (f, P ) + M(f, (t k 1, t k ))[F (t k ) F (t+ k 1 )] m(f, (t k 1, t k ))[F (t k ) F (t+ k 1 )], and respectively. Note the use of the open interval for M, m in this case. The information carried by f at these two points is carried in the jump term. Example 10.: Let f and F be given by x(x 1), if 0 x <.5; f(x) = 4x 1 if.5 x <.75; (4x 3) 3, if.75 x 1. { x, if x <.5; F (x) = x + 1, if.5 x 1.
Let P = {0 <.5 <.5 <.75 < 1}. Then J F (f, P ) = 0(0 0) + 9 64 (.5.5) + 1(1.5.5) + 0(1.75 1.75) + 1( ) U(f, P ) = J F (f, P ) + 9 64 (.5 0) + 4 7 (.5) + (1.75 1.5) + 1( 1.75) L F (f, P ) = J F (f, P ) + 0(.5 0) + 1 8 (.5) + 1(1.75 1.5) + 0( 1.75). Let us now review the major steps of the development of the Darboux and Riemann integrals. Some of these can be carried over in a straightforward manner, and others cannot. Also there will be some new results later without analogs stated here. 1. Inequalities between upper sums and lower sums for partitions P Q. Idea: See the effect of adding one more point to an upper (or lower) sum. Have to check this here.. Lower sums are always less than upper sums. Idea: Uses the fact that P Q contains both partitions P and Q and the last step. Would not require anything new after last step. 3. Upper integral always greater than or equal lower integral. Follows from the last step and definitions as infs and sups. Would not change. 4. Definition of Darboux integrable if upper and lower integrals are equal. Would not change. 5. Cauchy Criterion: Existence of a partition P for which the upper and lower sums are within a given tolerance of each other. Just used the properties of infs and sups and the definitions of the upper and lower integral as an inf and a sup respectively. Would not change. 6. Cauchy Criterion on Mesh size: Upper and lower sums are always within a given tolerance of each other if the mesh size is small enough. This used 3
the previous Cauchy criterion to get a partition P 0 with m points for which upper and lower sums were close. Then it sprinkled these points into an arbitrary partition P and noted the points fell into at most m intervals. The argument then proceeded to examine the difference that injection of points made in the lower sums and in the upper sums. The boundedness of the function f was used as well as the telescoping of the lengths of the subintervals. It is this last argument that must change and it is tied to just what we mean by mesh in this case. 7. Definition of Riemann sums and Riemann integrable. We will have to make the appropriate definition. 8. Equivalence of Darboux and Riemann integrals. We will show the equivalence of various definitions. 9. Monotone Functions are integrable. The idea here was to take a uniform mesh size so that the length of the intervals were constant and could be pulled out of the upper and lower sums. An new argument will have to be done here. 10. Continuous Functions are integrable. This used the uniform continuity of f and the proof will not be much different here. 11. Linearity of the integral. This used the Cauchy Criterion, the relation sup(s) = inf( S) and inequalities inf(a + B) inf(a) + inf(b), sup(a + B) sup(a) + sup(b). The proof in this case will be essentially the same. 1. Order properties of the integral. The proofs here are completely analogous as well. 13. Adjoining intervals theorem. The proof here will almost be the same except now we have to take into account the special ways endpoints are now treated. More care is needed here. 14. Intermediate Value Theorem for integrals. We will not give an analog to this theorem. 15. Fundamental Theorem of Calculus I. The best we will do toward this is to show that sometimes the Riemann-Stieltjes integral can be written as an ordinary Riemann integral. 16. Integration by parts. There is an analog here in spite of the fact that we don t have a fundamental theorem. 17. Fundamental Theorem of Calculus II. No analog given. 18. Change of variables in integration. No analog given. For several of the steps listed above, most of the changes come about as a result of how the new definitions effect the differences U F (f, P ) L F (f, P ) and in the telescoping of F (t k ) F (t+ k 1 ) as opposed to t k t k 1. Let us first examine these: U F (f, P ) L F (f, P ) = ( ) M(f, (t k 1, t k )) m(f, (t k 1, t k )) [F (t k ) F (t+ k 1 )]. (10.1) If the difference M(f, (t k 1, t k )) m(f, (t k 1, t k )) were replaced by a constant in an estimation, then we would have the sum [F (t k ) F (t+ k 1 )] 4
which does not (necessarily) telescope because of possible jumps of F at the t k. However, we do have telescoping if we add in the similar part of the jump sum n [F (t k ) F (t+ k 1 )] + [F (t + k ) F (t k )] = (10.) F (t + n ) F (t 0 ) = F (b) F (a). In particular, then when M(f, (t k 1, t k )) and f(t k ) are replaced by M(f, [a, b]) and m(f, (t k 1, t k )) and f(t k ) are replaced by m(f, [a, b]), we derive m(f, [a, b])[f (b) F (a)] L F (f, P ) U F (f, P ) M(f, [a, b])[f (b) F (a)]. (10.3) These three inequalities are enough to make good progress in achieving the steps above for the Stieltjes integral. Indeed, the inequality (10.3) means that the upper and lower sums are bounded so that we can define the upper and lower integrals as before: is Definition 10.3. The upper Darboux-Stieltjes integral of f with respect to F U F (f) := inf{u F (f, P ) : P is a partition of [a, b]}, and the lower Darboux-Stieltjes integral of f with respect to F is L F (f) := sup{l F (f, P ) : P is a partition of [a, b]}. Step 1. above is achieved with Lemma 10.4. Let f be a bounded function on [a, b] and let P and Q be partitions with P Q. Then L F (f, P ) L F (f, Q) U F (f, Q) U F (f, P ). Proof. Suppose Q is formed from P by adding the point u with t k 1 < u < t k for a fixed k. We look at the differences U F (f, P ) U F (f, Q) and L F (f, Q) L F (f, P ). Foe the first of these, everything cancels except for terms where the new point is introduced. We get a contribution from the jump sum as well in this case: M(f, (t k 1, t k ))[F (t k ) F (t+ k 1 )] M(f, (t k 1, u))[f (u ) F (t + k 1 )] M(f, (u, t k ))[F (t k ) F (u+ )] f(u)[f (u + ) F (u )] ( M(f, (t k 1, t k )) [F (t k ) F (t+ k 1 )] [F (u ) F (t + k 1 )] ) [F (t k ) F (u+ )] [F (u + ) F (u )] = 0. 5
Similarly, for the difference L F (f, Q) L F (f, P ), we find f(u)[f (u + ) F (u )] + m(f, (t k 1, u))[f (u ) F (t + k 1 )] + m(f, (u, t k ))[F (t k ) F (u+ )] m(f, (t k 1, t k ))[F (t k ) F (t+ k 1 ( )] m(f, (t k 1, t k )) [F (u + ) F (u )] + [F (u ) F (t + k 1 )] ) + [F (t k ) F (u+ )] [F (t k ) F (t+ k 1 )] = 0. Therefore, the lemma is true if Q = P {u}. The general case follows inductively. As was mentioned in the outline above, the proofs of the next statements would not change from the earlier case. Lemma 10.5. If f is bounded on [a, b] and if P and Q are partitions of [a, b], then L F (f, P ) U F (f, Q). Theorem 10.6. For a bounded function f on [a, b], Therefore, we can define Step 4: Definition 10.7. L F (f) = U F (f). L F (f) U F (f). A bounded function f is F integrable on [a, b] if and only if As mentioned above, the Cauchy criterion now follows as before Theorem 10.8. (Cauchy Criterion) A bounded function f on [a, b] is F -integrable, if and only if, for every ε > 0 there is a partition P = P ε so that U F (f, P ) L F (f, P ) < ε. We finish today with the analog of Step 8: Theorem 10.9. If f is continuous on [a, b], then f is F -integrable. Proof. This uses the uniform continuity of f on [a, b] and an estimate derived from (10.). Given ε > 0, choose δ = δ ε > 0 so that x y < δ = f(x) f(y) < ε F (b) F (a). Then, as in the proof of the ordinary case, by the properties of continuous functions on a closed interval and (10.1) and (10.), we obtain U F (f, P ) L F (f, P ) ε F (b) F (a) [F (t k ) F (t+ k 1 )] < ε provided mesh(p ) < δ. Thus, the Cauchy Criterion is fulfilled. 6