Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 1 / 12D (Analysis)

For a bounded function, a partition of an interval is $P:0=x_0<x_1<\cdots <x_N=1$. Put $m_j={\inf }_{[x_{j-1},x_j]}f$ and $M_j={\sup }_{[x_{j-1},x_j]}f$. Its lower Darboux sum and upper Darboux sum areThe lower Darboux integral is ${\sup }_{P}L(f,P)$ and the upper Darboux integral is ${\inf }_{P}U(f,P)$. Riemann integrability means that these two real numbers coincide, and their common value is the Riemann integral.

Equivalently, the Riemann integrability criterion says that for every $\epsilon >0$ some partition satisfies $U(f,P)-L(f,P)<\epsilon$. To see equivalence, all lower sums are at most all upper sums: refine any two partitions to their common refinement and use Darboux sum refinement monotonicity. Thus a small gap forces the difference of the two integrals below $\epsilon$. Conversely, if both integrals equal $I$, choose a lower sum above $I-\epsilon /2$ and an upper sum below $I+\epsilon /2$; their common refinement has gap below $\epsilon$.

This also agrees with the tagged Riemann sum definition: there is $I$ such that every tagged sum on a partition of sufficiently small mesh is within any prescribed tolerance of $I$. Each tagged sum lies between the two Darboux sums. From a partition with small gap, an arbitrary small-mesh partition differs from its refinement at only the cells crossing the finitely many original division points; boundedness makes their total contribution tend to zero. Conversely, on a fine partition tags approximating each cell supremum or infimum force both Darboux sums close to the same tagged-sum limit.

Now suppose $f$ is continuous on $[0,1]$. It is bounded by the result of Question 11. It is also uniformly continuous: otherwise there are $x_n,y_n\in [0,1]$ with $|x_n-y_n|\to 0$ but $|f(x_n)-f(y_n)|\ge {\epsilon }_0>0$. The Bolzano-Weierstrass theorem gives a convergent subsequence of $x_n$; the corresponding $y_n$ have the same limit, contradicting continuity. Given $\epsilon >0$, this uniform continuity gives $\delta >0$ such that points less than $\delta$ apart have function values less than $\epsilon /2$ apart. A partition of mesh below $\delta$ then satisfiesHence every continuous function on the interval is Riemann integrable, with the required proof supplied by its oscillation estimates.

For $x>0$, the function is a composition and product of continuous functions. At zero,Thus it is continuous on the whole closed interval, including its assigned value at zero. The continuous-function proof in the introductory solution applies, so this function is Riemann integrable. The vanishing factor $x$ controls the infinitely rapid oscillation at the endpoint.

The PDF defines this function using $\sin (1/x)$; the converted TeX contains a conflicting duplicate $\sin (1/2)$ before its later correct version. Use the PDF function. It is bounded by one and continuous at every positive $x$, although it is discontinuous at zero. Indeed the sequences $x_n=(\pi /2+2\pi n{)}^{-1}$ and $y_n=(3\pi /2+2\pi n{)}^{-1}$ tend to zero with function values $1$ and $-1$.

That single discontinuity does not prevent Riemann integrability. Given $\epsilon >0$, choose $0<\delta <\min \{1,\epsilon /4\}$. The cell $[0,\delta ]$ contributes at most $2\delta <\epsilon /2$ to the difference between the upper Darboux sum and the lower Darboux sum. On $[\delta ,1]$ the function is uniformly continuous, so choose a partition there whose oscillation sum is less than $\epsilon /2$, by the introductory proof. Joining that partition to the cell $[0,\delta ]$ givesThus the PDF's function is Riemann integrable despite its endpoint discontinuity. This is a direct application of bounded functions continuous away from one endpoint are Riemann integrable, proved here with the actual oscillation bound rather than an improper-integral argument.