HOMEWORK 1 MATH 4121 (1) Textbook, I.3.B, page 34. Show that if f :[a; b] ! R is a Rie- mann integrable function defined on the interval [a; b], then f is bounded. Solution: Assume that f is not bounded. We will show that f is not Riemann integrable. To do so, we will show that for any partition P , there exists two choices of sample points ξj and ηj so that the Riemann sum S(P ; f; ξ) and S(P ; f; η) are not close to each other. If f is not bounded, then its values are either not bounded above or not bounded below (or both). Assume without loss of generality that the values of f are not bounded above. Now let P be a partition, and let δ be the length of the smallest interval of P . Let ξj be an arbitrary choice of sample points, and let M = maxff(ξ1); : : : ; f(ξn)g. Because f is not bounded above, there exists some x∗ such that f(x∗) ≥ M + δ−1: We now choose our second set of sample points ηj as follows. Let ∗ ∗ ∗ j be the index of an interval of P that contains x . Set ηj∗ = x . For all other intervals, set ηj = ξj. Because ηj agrees with ξj on all but one of the intervals, it is easy to compute S(P ; f; η) − S(P ; f; ξ). Indeed, ∗ S(P ; f; η) − S(P ; f; ξ) = (f(x ) − f(ξj))(xj∗ − xj∗−1) ≥ (M + δ−1) − M δ = 1: On the other hand, if f were Riemann integrable, there would exist an L such that for some partition P , we would have jS(P ; f; ξ) − Lj < 1 1 2 and jS(P ; f; η) − Lj < 2 . This would imply that jS(P ; f; ξ) − S(P ; f; η)j < 1; which contradicts our construction where S(P ; f; η) − S(P ; f; ξ) ≥ 1. (2) Textbook, I.3.C, page 34. Show that the Dirichlet function is not Riemann integrable. Recall that the Dirichlet function D : [0; 1] ! R is defined as ( 1 if x 2 D(x) = Q 0 if x2 = Q: 1 2 MATH 4121 Solution: Because every nonempty open interval in [0; 1] must con- tain both rational and irrational numbers, for an arbitrary parti- tion P the upper and lower Darboux sums satisfy S(P; f) = 1 and S(P; f) = 0. Therefore, Z b Z b D(x) dx − D(x) dx = 1 a a so D(x) cannot be Riemann integrable. (3) Textbook, I.3.D, page 34. Prove that every continuous function on [a; b] is Riemann integrable on [a; b]. Solution: If f :[a; b] ! R is continutous, then f is uniformly contin- uous. (This is Heine's theorem 0.6.6, on page 17 of the textbook.) Hence for each > 0 there is δ > 0 so that jf(x) − f(x0)j < when- ever jx − x0j < δ. If P is a partition of [a; b] with jP j < δ then it follows that fi − fi ≤ for each i, so X X S(P; f) − S(P; f) = fi − fi (xi − xi−1) ≤ (xi − xi−1) = (b − a). i i Therefore, infP S(P; f) − supP S(P; f) ≤ infP S(P; f) − S(P; f) = 0, which means that f is Riemann-integrable. (4) Textbook, I.3.H, page 35. Let f : [0; 1] ! R be defined by ( 1=q if x = p=q where p; q are coprime positive integers and q 6= 0 f(x) = 0 if x2 = Q or x = 0; 1: (a) Prove that f is continuous at every irrational point of [0; 1] and discontinuous where f(x) 6= 0. R 1 (b) Prove that f is Riemann integrable on [0; 1] and 0 f(x) dx = 0. Solution: (a) It is clear that f is discontinuous at every point where it is non-zero. In fact, let x0 = p=q 2 (0; 1) be such a point. Then f(x0) > 0 while arbitrarily near x0 there are irrational points, where f is zero. In order to show that f is continuous at any irrational x 2 [0; 1], 1 let > 0 be given and choose q 2 N big enough so that q < . Let S be the finite set consisting of all rational numbers between m 0 and 1 that can be written as n for m; n 2 N and n ≤ q. Since x is irrational, it can't be an element of S, hence there are consecutive numbers s; s0 2 S, s < s0, such that x 2 (s; s0). By consecutive I mean that no element of S lies in (s; s0). Now let δ > 0 be sufficiently small that (x − δ; x + δ) ⊂ (s; s0). It follows that for all y such that jx − yj < δ, either y is irrational, in which case f(y) = f(x) = 0, or y is a rational number that cannot be written as a fraction with denominator less than q, HOMEWORK 1 3 1 in which case jf(x) − f(y)j = f(y) < q < . This shows that f is continuous at x. (b) The claim that f is Riemann integrable is an immediate conse- quence of Lebesgue's theorem (I.5.1, page 41), which says that a bounded function on [a; b] is Riemann integrable if and only if it is continuous almost everywhere. We can also prove this directly as follows. For any given > 0 1 let us fix q 2 N such that q < . Let S = fs1; s2; : : : ; sN g be the same finite set of rational numbers with denominators less than or equal to q introduced in the first part of this exercise, where s1 < s2 < ··· < sN . We now define a partition P of [0; 1] + − ± consisting of 0; 1 and the points xi and xi given by xi = si± M for i = 1; 2;:::;N where M ≥ 2N is big enough in order to + − guarantee that xi−1 < xi for each i. + − Notice that f ≤ 1=q on each interval of the form xi−1; xi and f is bounded by 1 on each interval (around si) of the form − + xi ; xi . The sum of the lengths of the intervals of the second kind is less than and the sum of the lengths of the intervals of first kind is less than 1. Thus we obtain 1 S(P; f) ≤ + < 2. q Therefore, the upper Dirichlet integral is 0, and since the lower Dirichlet integral is not negative for a non-negative function, we must have Z 1 Z 1 f(x) dx = f(x) dx: 0 0 R 1 Thus we conclude that f is Riemann-integrable with 0 f(x) dx = 0: (5) Riemann's example. To demonstrate the value of his formulation of the integral, Riemann (1854) proposed the following example of a function which is discontinuous on a dense set of points but still integrable. (The previous problem gives a simpler but less interesting example of such a function.) First define hxi = the integer closest to x (and for definiteness let us require it to be right-continuous, so that h1=2i = 1 for example. It is not important how it is defined on the points of discontinuity.) Now set ( x − hxi if x 6= k=2 for any k 2 B(x) = Z 0 if x = k=2 for some k 2 Z 4 MATH 4121 and finally define Riemann's function as 1 X B(nx) f(x) = : n2 n=1 The graph of this function over [0; 1] is shown in figure 1. (a) Show that f is discontinuous at every rational point with an even denominator (and odd numerator). (b) Show that f is integrable over [0; 1]. (c) Show that 1 Z 1=2 1 X 1 f(x) dx = ≈ 0:131475: 8 (2k − 1)3 0 k=1 Solution: (a) B(x) is clearly discontinuous at every point of the form (2m + 1)=2, for m 2 Z. It follows that B(nx) is discontinuous at each point x = (2m + 1)=2n, m 2 Z. At such point, the jump (the difference between the right-limit and the left-limit) is negative: B(nx+) − B(nx−) < 0, so there cannot be a cancellation of Pn B(kx) jumps in the partial sums fn(x) = k=1 k2 . Therefore, each point of discontinuity of B(kx) for each k 2 N is also a point of discontinuity for fn(x) for n ≥ k, and the jump in fn at x is at B(kx) 1 least as big as the jump in k2 at x, namely at least k2 in the negative direction. The limit of the partial sums fn is the function f, and we would like to conclude that because the fn are discontinuous at the point x, so is f. To do so, we must use the fact that the fn converge to f uniformly, proved in part (b) below. As discussed, since x is a point of discontinuity for B(kx), it is also a point 1 of discontinuity for fn for n ≥ k, with a jump at least k2 . By uniform convergence, we can choose an n large enough so that 1 we also have jf(t) − fn(t)j < 5k2 for all t. We know that there exists a δ such that for all y with x−δ < y < 1 x, we have that fn(y) is within 5k2 of the left-hand limit fn(x−).