On Feb 8, 2008 2:38 PM, Dan Asimov <dasimov@earthlink.net> wrote:
<< Jim asked about the integrability of
2 x sin(1/x) - cos(1/x) if x is not zero, f(x) = 0 if x is zero.
The function is Riemann integrable on [0,1]. Consider the intervals from 0 to \eps, and from \eps to 1. By uniform continuity, there is some \delta such that on the larger interval, any partitions of mesh \delta, on a subinterval of \eps to 1, with any associated Riemann sums, will yield values within \eps of one another. . . . . . .
Huh? Uniform continuuity of what? Because of the cos(1/x) term, f is not continuous on [0,1].
f is uniformly continuous on [\eps, 1], which I think is all that is used in the above argument. -- Andy.Latto@pobox.com