Let$\displaystyle f(x)=1$ if $\displaystyle \frac{1}{2^{2k+1}}<x\leq \frac{1}{2^{2k}}$ and $\displaystyle f(x)=0$ $\displaystyle otherwise$.

Show that$\displaystyle f$ is Reimann integrable by finding, $\displaystyle \forall$ $\displaystyle \epsilon$, a partition $\displaystyle P_\epsilon$ of $\displaystyle [0,1]$ and $\displaystyle n=n(\epsilon)$ such that for any partition $\displaystyle P$ finer than $\displaystyle P_\epsilon$, the difference between the upper and the lower sums is less than $\displaystyle \epsilon$, that is $\displaystyle U(P,f)-L(P,f)<\epsilon$.