Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 1 / 12E (Analysis)

For a bounded function, the lower and upper integrals are the supremum of lower Darboux sums and infimum of upper Darboux sums. It is Riemann integrable when these agree.

Here $u(x)=1/x-\lfloor 1/x\rfloor$ for $x>0$, so $0\le u<1$. On $[\delta ,1]$ it has only finitely many discontinuities and is piecewise continuous, hence integrable. On $[0,\delta ]$ its upper-minus-lower contribution is at most $\delta$. Taking $\delta \downarrow 0$ proves integrability on $[0,1]$.

Changing variables givesThe average of a continuous function over $[y,y+h]$ tends to $f(y)$ as $h\to 0$. The two terms therefore tend to $f(x)$ and $f(a)$, proving the stated limit.

The continuous approximation of a Riemann-integrable function follows here by choosing partitions $P_n$ with upper-minus-lower sum below $1/n$. On each subinterval choose the midpoint of the infimum and supremum to form a step function $s_n$. ThenReplace each of the finitely many jumps of $s_n$ by a linear transition on intervals of sufficiently small total length, obtaining a continuous ${\varphi }_n$ with $\int |s_n-{\varphi }_n|<1/n$. Hence $\int |g-{\varphi }_n|<2/n$, and uniformly for every subinterval $[\alpha ,\beta ]$,