2. Let J R R be Thomaes function and let gR R be the sign function.pdf

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2. Let J R R be Thomae\'s function and let g:R R be the sign function ie. if a 0 g(z) if a 0 -1 if a Solution Z 1 0 f dx = 0. To show this, let P = {I1, I2, . . . , In} be a partition of [0, 1]. Then, since f(x) = 0 for x > 0, Mk = sup Ik f = 0, mk = inf Ik f = 0 for k = 2, . . . , n. The first interval in the partition is I1 = [0, x1], where 0 < x1 1, and M1 = 1, m1 = 0, since f(0) = 1 and f(x) = 0 for 0 < x x1. It follows that U(f; P) = x1, L(f; P) = 0. Thus, L(f) = 0 and U(f) = inf{x1 : 0 < x1 1} = 0, so U(f) = L(f) = 0 are equal, and the integral of f is 0. In this example, the infimum of the upper Riemann sums is not attained and U(f; P) > U(f) for every partition P. A similar argument shows that a function f : [a, b] R that is zero except at finitely many points in [a, b] is Riemann integrable with integral 0..

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