Past exam of the mathematics course of the University of Cambridge / 2001 / ia / Paper 1 / 12D (Analysis I)

For a bounded function on a nondegenerate compact interval $[a,b]$, a partition $P$ has Darboux sumsThe Riemann integrability criterion isEquivalently the supremum of lower sums and infimum of upper sums agree; their common value is the Riemann integral. A bounded nonintegrable example is the Dirichlet function $\boxed{f=1_{\mathbb{Q}}}$. Every positive-length interval contains rational and irrational points, so every lower sum is zero and every upper sum is $b-a>0$.

Suppose the continuous nonnegative function has $f(t_0)>0$ somewhere. Continuity supplies a positive-length interval within $[a,b]$ on which $f\ge f(t_0)/2$; at an endpoint use a one-sided interval. By monotonicity of the Riemann integral, its contribution to the integral is strictly positive, a contradiction. Therefore $\boxed{f\equiv 0}$ under the stated continuous-function assumptions.

For the general integrable case, choose $c\in (a,b)$ and let $f(c)=1$ while $f(t)=0$ for $t\ne c$. This nonnegative indicator function is Riemann integrable with integral zero: isolate $c$ in partition intervals of arbitrarily small total length, making the upper sum arbitrarily small while the lower sum stays zero. Yet $f(c)=1$. This is the single-point changes preserve Riemann integrals phenomenon and supplies the counterexample.

Put $G(t)={\int }_a^{t}f(x)dx$. At an interior point and for $h\ne 0$ small enough,By continuity at $t$, the modulus is bounded by the supremum of $|f(x)-f(t)|$ between $t$ and $t+h$, which tends to zero. Thus the fundamental theorem of calculus givesAt the interval endpoints the same argument supplies the corresponding one-sided derivatives.

For an integrable discontinuous function, the single-point spike from part (i) has $G\equiv 0$, so $G^{\prime }(c)=0\ne f(c)=1$; this disproves the derivative identity even when the primitive is differentiable. To see that differentiability itself may fail, take the step function $f(x)=0$ for $x<c$ and $f(x)=1$ for $x\ge c$. It is Riemann integrable, but its primitive is $G(t)=\max (0,t-c)$ and has left derivative zero and right derivative one at $c$. Hence the full assertion is false for general Riemann-integrable functions.