Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 1 / 12D / b

No. DefineIt is continuous at every point of $(0,1)$ and is unbounded. Its truncation is Riemann integrable, butThus it is not improperly integrable with a finite integral.

No. Enumerate the rationals in $[0,1]$ asand defineThe function is nonnegative and unbounded, while it is identically zero on $[0,1]\setminus A$. For any $r>0$, however, $f_r$ is positive on the dense set of rationals and zero on the dense set of irrationals. Every interval therefore has positive oscillation, and the upper and lower Darboux integrals differ. Thus even the truncations need not be Riemann integrable.

For $k\ge 1$, let $E_k$ be the set of numbers whose first digit $5$ occurs in position $k$. The first $k-1$ digits each have nine allowed values, while the $k$th is fixed, so $E_k$ is a union of decimal intervals of total lengthThe set $E_{\infty }$ of numbers having no digit $5$ can, after $N$ places, be covered by intervals of total length $(9/10{)}^N$; hence it has length zero.

For each fixed truncation level, the function is constant on finitely many decimal cylinders apart from the remaining set, whose covering length can be made arbitrarily small. The Riemann integrability criterion therefore shows that every truncation is Riemann integrable.

For integer $N$,The final term tends to zero, and monotonicity handles noninteger truncation levels. Therefore the first occurrence of a decimal digit calculation gives