= Solution
For a <bounded function>, the lower and upper <integrals> are the supremum of lower Darboux sums and infimum of upper Darboux sums. It is Riemann integrable when these agree.
Here $u(x)=1/x-\lfloor1/x\rfloor$ for $x>0$, so $0\leq u<1$. On $[\delta,1]$ it has only finitely many discontinuities and is piecewise continuous, hence integrable. On $[0,\delta]$ its upper-minus-lower contribution is at most $\delta$. Taking $\delta\downarrow0$ proves integrability on $[0,1]$.
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