If a function $f$ is Riemann-integrable on a compact interval $[a,b]$, must it be of bounded variation ?
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No, $[0,1] \ni t \mapsto \begin{cases} 0, & t=0, \\ t \sin(1/t), & t >0, \end{cases}$ is continuous, so it is Riemann-integrable, but it is not of bounded variation.
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For the sake of completeness, $t\sin(1/t)$ is Riemann Integrable. Further, it has unbounded variation – Kumar Dec 03 '20 at 08:02