Show-that-the-function-defined-within-0-1-by-f-x-1-if-x-Q-0-1-0-otherwise-is-not-Riemann-integrable-within-0-1-

Question Number 102763 by Ar Brandon last updated on 10/Jul/20

$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{function}\:\mathrm{defined}\:\mathrm{within}\:\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\mathrm{by}\:\mathrm{f}\left(\mathrm{x}\right)=\begin{cases}{\mathrm{1}\:\mathrm{if}\:\mathrm{x}\in\mathbb{Q}\cap\left[\mathrm{0},\mathrm{1}\right]}\\{\mathrm{0}\:\mathrm{otherwise}}\end{cases}\:\:\mathrm{is}\:\mathrm{not}\:\mathrm{Riemann}\:\mathrm{integrable} \\ $$$$\mathrm{within}\:\left[\mathrm{0},\mathrm{1}\right] \\ $$

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