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Suppose that $f:[0,1] \to \mathbb{R}$ is defined by $f(x) = 1$ when $x = \frac{1}{n}$ for some positive integer $n$ and $f(x) = 0$ otherwise. How can I prove that $f$ is Riemann integrable on $[0,1]$?

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    Cover each $x = \frac1n$ by interval of size $\frac{\epsilon}{2^n}$ i.e. consider intervals around $\frac1n$ as $\left(\frac1n - \frac{\epsilon}{2^{n+1}}, \frac1n + \frac{\epsilon}{2^{n+1}} \right)$2011-12-09
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    @Sivaram Perhaps I am missing something, but I am unsure how to proceed from your hint. Doesn't Riemann integration consider only finite partitions of $[0,1]$?2011-12-09
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    @Srivatsan: This should go with the previous hint. Choose $N$ such that $\frac1N < \frac{\epsilon}{2}$. For all $n < N$, cover each $x = \frac1n$ by interval of size $\frac{\epsilon}{2^{n+1}}$ i.e. consider intervals around $\frac1n$ as $\left(\frac1n - \frac{\epsilon}{2^{n+2}}, \frac1n + \frac{\epsilon}{2^{n+2}} \right)$ and for the rest use an interval of the form $[0,\epsilon/2)$2011-12-09
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    You need to use these intervals to show that the set of discontinuities has measure zero. http://www.math.ncku.edu.tw/~rchen/Advanced%20Calculus/Lebesgue%20Criterion%20for%20Riemann%20Integrability.pdf2011-12-09

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