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Question Number 94544 by Ar Brandon last updated on 19/May/20

$$1\mathrm{\setminus }\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{function}\mathrm{f}\left(\mathrm{x}\right)=\left[\mathrm{x}\right]\mathrm{is}\mathrm{of}\mathrm{Riemann}$$$$\mathrm{for}\mathrm{all}\mathrm{segments}\mathrm{of}\mathbb{R}$$$$2\mathrm{\setminus }\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{function}\mathrm{f}\left(\mathrm{x}\right)\mathrm{defined}\mathrm{within}\mathrm{x}\in [0,1]$$$$\mathrm{f}\left(\mathrm{x}\right)=\{\begin{array}{l}1\mathrm{if}\mathrm{x}\in \mathbb{Q}\cap [0,1]\\ 0\mathrm{otherwise}\end{array}\mathrm{is}\mathrm{not}\mathrm{of}\mathrm{Riemann}\mathrm{on}\mathrm{x}\in [0,1]$$

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