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Suppose $f: [-2,3] \longrightarrow \mathbb{R}$ is defined by $$ f(x) = \left\{ \begin{array}{l l} 2|x| + 1, & \text{if $x$ is rational}, \\ 0, & \text{if $x$ is irrational}. \end{array} \right.$$

Prove that $f$ is not Riemann integrable.

We know that the lower integral is $0$ and the upper integral is $18$, then because they are not equal $f$ is not Riemann integrable.

Is this correct? Thanks!

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Yes, that's correct. – David Mitra Apr 19 '12 at 2:32
You might want to show some work to justify your calculations of the lower and upper integrals, but that's the right approach for sure. – Brett Frankel Apr 19 '12 at 2:35
Hello Math Friends. I went the the professors office hours and she explained the correct way about solving this problem. You have to rigorously prove that the upper and lower bounds are always x away from each other. So something like, mK= 0 and MK=1 so the upper bound is greater or equal to 5 and the lower bound is always 0. Just thought I'd update. – Mark Apr 23 '12 at 21:46

Set of points of discontinuity is not measure zero

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