Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 1 / 12D (Analysis I)

Assume $a_n\to 0$. The function is bounded, and every interval has lower Darboux sum zero because it contains an irrational. Given $\epsilon >0$, choose $N$ so that $a_n<\epsilon /2$ for $n>N$. Put small intervals around the finitely many points $q_1,\dots ,q_N$ whose total length makes their contribution to the upper sum less than $\epsilon /2$, and refine their endpoints to a partition. On every remaining subinterval the function is at most $\epsilon /2$, so the upper sum is less than $\epsilon$. The Riemann integrability criterion proves that $f$ is Riemann integrable with integral zero.

The condition $a_n\to 0$ is sufficient but not necessary. Assign $a_n=1$ when $q_n=1/k$ for some positive integer $k$, and $a_n=1/n$ otherwise. Then $(a_n)$ does not tend to zero. Outside a short interval near zero only finitely many unit spikes occur and can be enclosed in intervals of arbitrarily small total length; all remaining values can be made uniformly small except for finitely many further points. The same Darboux-sum argument proves that this $f$ is Riemann integrable with integral zero.

The fundamental theorem of calculus says first that if $h$ is continuous on $[a,b]$ and $H(x)={\int }_a^{x}h(t)dt$, then $H^{\prime }(x)=h(x)$. Indeed,whose absolute value tends to zero by continuity. Consequently, if $F$ is differentiable with continuous derivative, applying the first part to $h=F^{\prime }$ shows that $F(x)-{\int }_a^x{F}^{\prime }(t)dt$ has zero derivative and is constant. Hence

A derivative need not be Riemann integrable. Define $f(0)=0$ and $f(x)=x^2\sin (x^{-2})$ for $x\ne 0$. Then $f$ is differentiable at zero with $f^{\prime }(0)=0$, while for $x\ne 0$,which is unbounded near zero. Since every Riemann-integrable function on a compact interval is bounded, $g=f^{\prime }{|}_{[0,1]}$ need not be Riemann integrable.