Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 1 / 12F (Analysis I)

For a partition of an interval $P:0=x_0<\cdots <x_m=1$, putThe upper Darboux sum and lower Darboux sum areThe upper and lower integrals areThe bounded function is Riemann integrable when these values agree, and their common value is its Riemann integral.

For the indicator of the rational numbers, every nondegenerate interval contains both a rational number and an irrational number. Hence every upper sum is one and every lower sum is zero. The function is not Riemann integrable.

Let $C_n$ be the union of the $2^n$ closed intervals remaining after $n$ steps of the Cantor set construction. Their total length is $(2/3{)}^n$. Any point outside $C_n$ has a ternary expansion with a $1$ among its first $n$ digits, so $A^c\subseteq C_n$. Endpoints of deleted intervals also belong to $A$ because they have an alternative ternary expansion containing a $1$, as in $2/3=0.1222\dots {}_3$.

A partition using all endpoints of $C_n$ has lower Darboux sum at leastSince $A$ is dense, every upper Darboux sum is one. Letting $n\to \infty$ gives equal upper and lower integrals, and therefore

Both $B$ and its complement are dense. Indeed, inside any sufficiently small ternary cylinder, extend its finite digit prefix by a tail containing infinitely many $1$s to obtain a point of $B$, or by a tail of all zeros to obtain a point outside $B$. Therefore every nondegenerate interval has supremum one and infimum zero for the indicator function. Its upper integral is one and its lower integral is zero, so