Past exam of the mathematics course of the University of Cambridge / 2009 / ia / Paper 1 / 12E (Analysis I)

A Riemann integrable function on a closed interval is bounded and has equal upper and lower Darboux integrals. More explicitly, for a partition of an interval $P:a=x_0<x_1<\cdots <x_m=b$, define the Darboux sumsThen Riemann integrability meansand this common value is the Riemann integral. Equivalently, the Riemann integrability criterion requires that for every $\epsilon >0$ some partition of an interval satisfies $U(f,P)-L(f,P)<\epsilon$. Indeed, the common partition refinement of two partitions increases each lower Darboux sum and decreases each upper Darboux sum, so equality of the two extremal values is equivalent to arbitrarily small gaps.

Let $|f|\le M$. If $M=0$ the result is immediate, so take $M>0$. Given $\epsilon >0$, choose $0<\delta <\epsilon /(4M)$ and the finitely many intervals supplied by the hypothesis. On each $[a_i,b_i]$, choose a partition of an interval whose Darboux sum gap is less than $\epsilon /(2n)$, using Riemann integrability there.

Combine all these partitions with $0,1$ and all interval endpoints to obtain a global partition of an interval $P$. On the chosen intervals the gaps add to less than $\epsilon /2$. The complementary intervals have total lengthand on each the difference between the supremum and infimum is at most $2M$. ConsequentlyThe Riemann integrability criterion proves $f$ is Riemann integrable on $[0,1]$. Shared endpoints cause no problem: the Darboux sums use closed intervals, and the complementary interval estimate includes every endpoint value. This proves Riemann integrability from almost full interval coverage.

Let $K$ be the set where the indicator function $f$ equals one. Fix a positive integer $N$. Each permitted length-$N$ decimal prefix $w=(w_1,\dots ,w_N)\in \{2,7{\}}^N$ gives the intervalThere are $2^N$ such intervals, each of length $10^{-N}$, and they cover $K$. Every point of $K$ is strictly inside the interval for its prefix, since its remaining tail lies between $2/(9\cdot 10^N)$ and $7/(9\cdot 10^N)$. Thus none of the interval endpoints belongs to $K$. The intervals are disjoint and separated: their distinct prefix integers cannot be consecutive because their final digits are only $2$ and $7$.

The complement consists of finitely many closed intervals on which $f=0$, of total length $1-2^{N}10^{-N}=1-5^{-N}$. Part (b) therefore already proves Riemann integrability, since this length tends to one.

To compute the Riemann integral directly, make a partition of an interval using $0,1$ and all the endpoints of the $I_w$. On complementary cells the upper Darboux sum contribution is zero, and on the $I_w$ it is at most their total length. Moreover every nonempty cell contains a terminating decimal, which is not in $K$: neither its terminating expansion with a zero tail nor its alternative expansion with a nine tail consists entirely of twos and sevens. Such decimals are dense, so every cell has infimum zero. ThereforeLetting $N\to \infty$ gives the exact answerThis is an example of an indicator of a restricted-digit decimal set: infinitely many, indeed uncountably many, points can support an indicator function whose Riemann integral is zero.