Let f(x)=1 if \frac{1}{2^{2k+1}}<x\leq \frac{1}{2^{2k}} and f(x)=0 otherwise.
Show that f is Reimann integrable by finding, \forall \epsilon, a partition P_\epsilon of [0,1] and n=n(\epsilon) such that for any partition P finer than P_\epsilon, the difference between the upper and the lower sums is less than \epsilon, that is U(P,f)-L(P,f)<\epsilon.