Thread: Question 4
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Old October 24th, 2008, 08:38 AM
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Originally Posted by chiph588@ View Post
for the second one, couldn't you just divide [0,1] into a partition P and say that
\lim_{n \to \infty} \sum_{i=0}^n D(x_{i max})(x_{i+1}-x_{i}) = \lim_{n \to \infty} \sum_{i=0}^n 1(x_{i+1}-x_{i}) = 1

but \lim_{n \to \infty} \sum_{i=0}^n D(x_{i min})(x_{i+1}-x_{i}) = \lim_{n \to \infty} \sum_{i=0}^n 0(x_{i+1}-x_{i}) = 0

Therefore D(x) is not Riemann integrable.
Yes chiph588@, that's exactly what I've done.
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