A bounded nondecreasing $f:\mathbb{R}\to \mathbb{R}$ whose set of discontinuities is exactly $\mathbb{Q}$, obtained from the prescribed-jump construction applied to one fixed enumeration of the rationals

Write $\mathbb{Q}$ for the canonical copy of the rationals inside $\mathbb{R}$ (The rationals embed densely in the reals). There is a function $f:\mathbb{R}\to \mathbb{R}$ with all of the following properties:

1. $f$ is nondecreasing (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $\mathbb{R}$, with the dictionary to monotone sequences) and $0\le f(x)\le 1$ for every real $x$;
2. $f$ is discontinuous at every rational and continuous at every irrational, so its discontinuity set is exactly $\mathbb{Q}$;
3. every discontinuity of $f$ is a jump (Discontinuity of $f$ at a point of its domain, and its classification: removable discontinuity, jump discontinuity and essential discontinuity, equivalently Rudin's discontinuities of the first and of the second kind).

Explicitly, fixing a bijection $e:\mathbb{N}\to \mathbb{Q}$ ($\mathbb{Q}$ is countably infinite), one may take

$f(x)={\sum }_{k=0}^{\mathrm{\infty }}{a}_k(x),a_k(x)=\{\begin{array}{ll}1\mathrm{/}{2}^{k+1}& \text{if }e(k)<x,\\ 0& \text{otherwise,}\end{array}$

which is the construction of Converse to Froda: for every at most countable $E\subseteq \mathbb{R}$ there is a bounded nondecreasing $f:\mathbb{R}\to \mathbb{R}$ whose set of discontinuities is exactly $E$, every one of them a jump applied to $E:=\mathbb{Q}$ (Series, partial sums, convergence and the sum, divergence, and the tail series, For $\mathrm{\mid }r\mathrm{\mid }<1$, ${\sum }_{k\ge 0}{r}^k=1\mathrm{/}(1-r)$, and for $\mathrm{\mid }r\mathrm{\mid }\ge 1$ the series diverges).

This is the extreme case allowed by Froda's theorem. Froda's theorem: the set of discontinuities of a monotone function on an interval is at most countable, the injection into $\mathbb{N}$ being built from one fixed enumeration of the rationals by least index, so no choice principle is used says that a monotone function on an interval has at most countably many discontinuities; $\mathbb{Q}$ is countable and dense, so the bound is attained by a set that meets every interval. A monotone function can therefore be discontinuous on a dense set, and it is nevertheless continuous on a set whose complement is countable.

• The jump at a rational $r$ is at least $1\mathrm{/}{2}^{k+1}$, where $k$ is the index with $e(k)=r$. That lower bound is what the construction of Converse to Froda: for every at most countable $E\subseteq \mathbb{R}$ there is a bounded nondecreasing $f:\mathbb{R}\to \mathbb{R}$ whose set of discontinuities is exactly $E$, every one of them a jump establishes, and it is what makes $r$ a discontinuity; the total mass available is ${\sum }_{k\ge 0}1\mathrm{/}{2}^{k+1}=1$, which is why $f$ stays inside $[0,1]$. A different enumeration gives a different function with the same discontinuity set.

• Continuity at every irrational is not an accident of this construction. The complement of a countable set is where a monotone function built this way must be continuous, and Froda's theorem: the set of discontinuities of a monotone function on an interval is at most countable, the injection into $\mathbb{N}$ being built from one fixed enumeration of the rationals by least index, so no choice principle is used says the same thing in general: the discontinuities of a monotone function can never fill an uncountable set. The companion statement in the other direction, that no function whatever is continuous exactly at the rationals (No function $\mathbb{R}\to \mathbb{R}$ is continuous at every rational and discontinuous at every irrational, because $\mathbb{Q}$ is not $G_{\delta }$), shows that the roles of $\mathbb{Q}$ and its complement cannot be exchanged here.