Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics II / 7

By the Heine-Cantor theorem, a continuous function on the compact interval $[a,b]$ is uniformly continuous. Given $\epsilon >0$, choose $\delta >0$ such that $|f(s)-f(t)|<\epsilon$ whenever $s,t\in [a,b]$ and $|s-t|<\delta$.

Choose a finite partition of an interval $a=t_0<t_1<\cdots <t_m=b$ with every width below $\delta$. Define the step functionFor each $t<b$, its distance from the chosen left endpoint is below $\delta$, so $|f(t)-\varphi (t)|<\epsilon$; at $b$ the error is zero. HenceA step function may have separately assigned values at its finitely many partition points; these values do not affect its Riemann integral. If $a=b$, simply take $\varphi =f$ on that singleton.

Put $E_n={\sup }_{t\in [a,b]}|f(t)-{\varphi }_n(t)|\to 0$, so the hypothesis is uniform convergence. Every step function is bounded and Riemann integrable: away from its finitely many partition points it is constant, and those exceptional points can be covered by intervals of arbitrarily small total length. A partition of an interval containing their boundaries makes the difference of its upper Darboux sum and lower Darboux sum arbitrarily small.

Choose one $n$ with $E_n<\infty$. Boundedness of ${\varphi }_n$ and $|f|\le |{\varphi }_n|+E_n$ show that $f$ is a bounded function. Assume $a<b$, since for $a=b$ all the integrals are zero. Given $\eta >0$, choose $n$ so that $2E_n(b-a)<\eta /2$, and then a partition of an interval $P$ such that $U({\varphi }_n,P)-L({\varphi }_n,P)<\eta /2$. On every partition subinterval, $|f-{\varphi }_n|\le E_n$ givesThusThe Riemann integrability criterion proves that $f$ is Riemann integrable. Finally, linearity of the integral and monotonicity of the Riemann integral giveThereforeThis proves the relevant case of uniform convergence preserves integrals on a compact interval.

The printed assertion needs an endpoint qualification: the enumeration includes zero, but $1_{[0,0)}$ is identically zero. In fact $f$ is continuous at zero; the claimed discontinuities occur at every $r_j>0$.

Define the finite step function sumsThe assumed convergence of ${\sum }_{k}1/k^2$ and the bound on each indicator function giveThus the defining series converges everywhere and has uniform convergence. Part (b) proves that $f$ is Riemann integrable; it also gives, if desired,For $r_j>0$ and $0\le t<r_j$, every indicator function is nonincreasing, and the $j$th drops by one between $t$ and $r_j$. HenceTaking inputs from the left tending to $r_j$ proves a discontinuity there.

For completeness, continuity at zero follows directly from a finite-head and tail argument. Given $\epsilon >0$, choose $N$ with ${\sum }_{k>N}1/k^2<\epsilon$. Choose $\delta >0$ below every positive $r_k$ with $k\le N$. For $0\le t<\delta$, the first $N$ indicator functions have exactly their values at zero; the one with $r_k=0$, if present, is always zero. Thus $|f(t)-f(0)|\le {\sum }_{k>N}1/k^2<\epsilon$. This confirms that the endpoint exception is real. Replacing the printed enumeration of $[0,1)$ by one of $(0,1)$ would remove it.

Finally, the density of the rational numbers supplies a positive rational $r_j$ in every nonempty open subinterval of $[0,1]$. Thus a Riemann integrable function can have a discontinuity in every open subinterval: dense jumps do not prevent Riemann integrability.