Question Number 94544 by Ar Brandon last updated on 19/May/20
![1\Show that the function f(x)=[x] is of Riemann for all segments of R 2\Show that the function f(x) defined within x∈[0,1] f(x)= { ((1 if x∈Q∩[0,1])),((0 otherwise)) :} is not of Riemann on x∈[0,1]](Q94544.png)
$$\mathrm{1}\backslash\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)=\left[\mathrm{x}\right]\:\mathrm{is}\:\mathrm{of}\:\mathrm{Riemann} \\ $$$$\mathrm{for}\:\mathrm{all}\:\mathrm{segments}\:\mathrm{of}\:\mathbb{R} \\ $$$$\mathrm{2}\backslash\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{defined}\:\mathrm{within}\:\mathrm{x}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\begin{cases}{\mathrm{1}\:\mathrm{if}\:\mathrm{x}\in\mathbb{Q}\cap\left[\mathrm{0},\mathrm{1}\right]}\\{\mathrm{0}\:\:\mathrm{otherwise}}\end{cases}\:\:\mathrm{is}\:\mathrm{not}\:\mathrm{of}\:\mathrm{Riemann}\:\mathrm{on}\:\mathrm{x}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$