Introduction to Integration Learning goals: Set up the basic ideas of the Riemann Integral In the realm of analysis there are a large number of different kinds of integrals. The kind we studied in BC calculus is probably the second-simplest kind. We will actually define the simplest kind here and then show it is equivalent to what we discussed in BC. The integral from BC is called the Riemann integral. But first we will define the Darboux integral. (Pronouned Dar-boo.) You may recall that what we did in BC was to take the interval, split it into pieces, evaluate the function somewhere in the interval, and add up the areas|make a so-called Riemann sum. Through some kind of mathematical voodoo we crossed our fingers and said these kinds of sums converge in some kind of way to give us a number which is the area under the curve. Then we forgot that integrals really meant area under a curve and conflated the ideas of integrals and antiderivatives once we proved the Fundamental Theorem of Calculus. We're going to be a lot more careful now! To this end, we need some notation and restrictions on the functions we deal with. At first, we must have bounded functions, though they need not have any other special properties, and we'll even relax boundedness later. We will be dealing with integrals on some finite closed interval [a; b]. Our function must be defined at each point of [a; b] or our definition won't make any sense. A partition P of the interval [a; b] is a finite set of points, fx0; x1; : : : xng with a = x0 < x1 < ··· < xn = b. For partition, ∆xk = xk − xk−1 is the width of the kth interval. Notice Pn that k=1 ∆xk = x1 − x0 + x2 − x2 + ··· + xn − xn−1 which telescopes to b − a. The norm or mesh of a partition is the width of the largest interval. We can write this as jjP jj = max(∆xk). We will declare partition Pe to be a refinement of P , or say that Pe is finer than P , if P ⊆ Pe. Notice that refining a partitions can only decrease its norm. Given any two partitions P and Q, the partition P [ Q is simultaneously a refinement of both. The Definition Consider a partition of [a; b] and a bounded function f defined on [a; b]. Since f is bounded on [a; b] it is bounded on any subinterval [xk−1; xk], and therefore has a well-defined infimum and supremum on that interval. Let mk = infff(x): xk−1 ≤ x ≤ xkg and Mk = supff(x): xk−1 ≤ x ≤ xkg. So mk is not larger than any value of f on the interval [xk−1; x − k] and Mk is not smaller than any such value. So if there is a meaningful area under f(x) on this interval, it is clearly between the values of mk∆xk and Mk∆xk. Pn So define the upper and lower Darboux sums to be L(P; f) = k=1 mk∆xk and U(P; f) = Pn k=1 Mk∆xk. Again, if there is a meaningful area underneath f then certainly L(P; f) and U(P; f) must be lower and upper bounds for it. Now let since f is bounded, choose any m < f(x) for any x 2 [a; b] and M > f(x). Then given any partition P we have m(b − a) ≤ L(P; f) ≤ U(P; f) ≤ M(b − a). For clearly 1 P P P m ≤ mk ≤ Mk ≤ M for all k. Then m(b − a) = m ∆xk = m∆xk ≤ mk∆xk ≤ P P P Mk∆xk ≤ M∆xk = M ∆xk = M(b − a). This means that the set of all Darboux sums|upper and lower|is bounded both above and below. If we have a partition P and a refinement widetildeP it should be pretty clear (we'll prove it soon) that L(P; f) ≤ L(Pe ; f) and U(P; f) ≥ U(Pe ; f) because when we subdivide an interval the infimum over each part of the subinterval can't be smaller but could be larger than that over the entire interval, and vice versa for suprema. So if we take finer and finer partitions, the lower Darboux sums can only go up and the upper sums can only come down. We've already shown that the lower sums can't get larger then the upper for a particular partition (although it is worth considering whether there are partitions P and Q with L(P; f) < U(Q; f). ), but we do have to hope they meet in the middle! So let's make the definitions: Z b f(x) dx = supfL(P; f): P is a partition of [a; b]g a Z b f(x) dx = inffU(P; f): P is a partition of [a; b]g a which we will call the lower Darboux integral and upper Darboux integral respectively. (No- tationally, we sometimes leave off the dx when the variable is obvious or irrelevant.) Ideally, these two integrals are equal to each other, but that is not always the case. R 1 R 1 Example. The Dirichlet function f(x) = 1Q has 0 f = 0 and 0 f = 1. This is because on each interval of any partition, mk = 0 and Mk = 1 since there are both rational and irrational numbers in every interval. Let's prove the claim we made a few paragraphs ago, that refining a partition can only increase the lower sums and decrease the upper. Theorem. Given bounded function f :[a; b] ! R, partition P , and refinement Pe, then L(P; f) ≤ L (Pe ; f) while U(P; f) ≥ U(Pe ; f). Proof. Let P = fx0; x1; : : : xng be a partition and Pe = fx~0; x~1;::: x~mg a refinement. Thus P ⊆ Pe so there are integers k1 < k2 < ··· < kn such thatx ~ki = xi. Consider ∆xi = xi − xi−1: k Xi ∆xi = xi − xi−1 =x ~ki − x~ki−1 +x ~ki−1 − x~ki−2 + ··· +x ~ki−1+1 − x~ki−1 = ∆~xj: j=ki−1+1 ~ Now let mi and Mi be defined as before. Note that mi ≤ m~ j and Mi ≥ Mj for any j between ki−1 + 1 and ki since each of thosex ~ subintervals are subsets of [xi−1; xi]. Thus for this subinterval, k k k Xi Xi Xi mi∆xi = mi ∆~xj = mi∆~xj ≤ m~ j∆~xj j=ki−1+1 j=ki−1+1 j=ki−1+1 with a reversed inequality for Mi. Summing this over all i gives the desired result. 2 Next, we answer the question we asked a bit ago|could the lower sum for some partition exceed the upper sum for another? The answer is, fortunately, no! Theorem. Let P and Q be partitions of [a; b] and f a bounded function of [a; b]. Then L(P; f) ≤ U(Q; f). Proof. Consider the partition R = P [ Q which is a refinement of both P and Q. From the previous theorem, we have L(P; f) ≤ L(R; f) ≤ U(R; f) ≤ U(Q; f). Theorem. If f is bounded on [a; b], m < f(x) for any x 2 [a; b] and M > f(x), then Z b Z b m(b − a) ≤ f ≤ f ≤ M(b − a): a a Proof. We've already shown m(b − a) ≤ L(P; f) for any P , so it is certainly less than their supremum, which is the lower Darboux integral, and the first inequality is established. Similarly, M(b − a) ≥ U(P; f) establishes the last inequaltiy. But since L(P; f) ≤ U(Q; f) for any P and Q by the previous theorem, the supremum of the left must be at most the infimum of the right hand side and the proof is complete. So we've done a lot of work to show that all these Darboux sums and integrals exist and have certain relationships. By taking refinements, lower sums keep getting bigger and upper sums keep getting smaller, but a lower sum is never larger than an upper sum. So maybe they meet in the middle? We already know that doesn't always happen, but when it does things are happy! R b R b Definition. If f :[a; b] ! R is a bounded function and a f = a f then we say that f is R b Riemann integrable, denoted f 2 R[a; b] and we define a f(x) dx to be the common value of the lower and upper Darboux integral. Comparing to BC At this point, we take a breath and show that what we did in BC is equivalent to all of this, even though it might look more general. We need to make sure the calculus we know is on firm ground! So what was the definition of a Riemann integral in BC? We again started with any P partition and chose a ti 2 [xi−1; xi] and formed the Riemann sum f(ti)∆xi. That is, we allowed the point to vary any where on the subinterval|left-hand endpoint, right-hand endpoint, minimum or maximum value of f on the subinterval (if there is one!), midpoint, random point, whatever. We kind of waved out hands in BC about what happened next, but it is not too complicated. Definition. We say that f is integrable on [a; b] if there is some number I such that given > 0 there is a partition P0 such that for any refinement P of P0 and any choice of ti on P , the corresponding Riemann sum is within of I.