Monotone Functions, Discontinuities, and Continuity Sets: Examples and Counterexamples

Let $t$ be Thomae's function (The Dirichlet function $1_{\mathbb{Q}}$, and Thomae's function $t$ with $t(x) = 1/q$ at a rational $x = p/q$ in lowest terms with $q \ge 1$ and $t(x) = 0$ at every irrational $x$), so that $t(x) = 1/\iota(q(x))$ at a rational $x$ with least denominator $q(x)$ and $t(x) = 0$ at an irrational $x$. Then:

1. $t(0) = 1$ and $t(m) = 1$ for every integer $m$;
2. $t(1/2) = 1/2$, and more generally $t(1/\iota(q)) = 1/\iota(q)$ for every natural $q \ge 1$;
3. $t(2/3) = 1/3$;
4. $t(x) = 0$ at every irrational $x$;
5. $\omega_{t}(c) = t(c)$ at every real $c$ (The oscillation $\omega_f(S) = \sup\{\,|f(x) - f(y)| : x, y \in S\,\}$ of $f$ on a set and the oscillation $\omega_f(c) = \inf_{\delta > 0} \omega_f(A \cap N_\delta(c))$ at a point, both taken in the extended reals), so $\omega_{t}$ is $1$ at every integer, $1/2$ at every half-integer that is not an integer, and $0$ at every irrational.

Claim 5 is claim 2 of The Dirichlet function is continuous at no point of $\mathbb{R}$, and Thomae's function is continuous at every irrational and at no rational, so its set of continuity points is exactly the set of irrationals and its oscillation at $c$ equals $t(c)$ evaluated at the points computed here; nothing new is proved about the oscillation, and the point of the example is to see the numbers.

• The least denominator is what the values record. $t$ is large exactly at the rationals with small denominators, and those are sparse: every point with least denominator $q$ is a multiple of $1/\iota(q)$, and consecutive multiples of $1/\iota(q)$ are $1/\iota(q)$ apart. The graph is the familiar picture of tall spikes at the integers, half as tall at the half-integers, and so on down.

• Every value $1/\iota(q)$ is attained, by step 1.2, so the range of $t$ is exactly $\{0\} \cup \{\, 1/\iota(q) : q \in \mathbb{N},\ q \ge 1 \,\}$; the value $0$ is attained at every irrational.

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

$f(x) \;=\; \sum_{k=0}^{\infty} a_{k}(x), \qquad a_{k}(x) = \begin{cases} 1/2^{\,k+1} & \text{if } e(k) < x,\\ 0 & \text{otherwise,}\end{cases}$

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 $|r| < 1$, $\sum_{k \ge 0} r^k = 1/(1-r)$, and for $|r| \ge 1$ 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 $1/2^{\,k+1}$, 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 $\sum_{k \ge 0} 1/2^{\,k+1} = 1$, 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.

Put

$E \;:=\; \Bigl\{\, 1 - \frac{1}{\iota(k+1)} \;:\; k \in \mathbb{N} \,\Bigr\} \;=\; \Bigl\{\, 0,\ \tfrac12,\ \tfrac23,\ \tfrac34,\ \dots \Bigr\} \;\subseteq\; [0,1)$

(The canonical natural $\iota(n) = n \cdot 1_F$ of a field, Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length). Then:

1. $E$ is countably infinite (Finite, countably infinite, countable, uncountable);
2. there is a nondecreasing $f : \mathbb{R} \to \mathbb{R}$ with $0 \le f \le 1$ whose set of discontinuities is exactly $E$, every one of them a jump (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, Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $\mathbb{R}$, with the dictionary to monotone sequences, 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);
3. $E$ is contained in the bounded interval $[0,1)$, so a monotone function may have infinitely many discontinuities inside a bounded interval.

Indexing. $\mathbb{N}$ contains $0$, so the points are $1 - 1/\iota(k+1)$ for $k \in \mathbb{N}$ and never $1 - 1/\iota(k)$, which is undefined at $k = 0$; the first point of $E$ is $1 - 1/\iota(1) = 0$.

The point $1$ is not in $E$ and $f$ is continuous there. $E$ has $1$ as a limit point but does not contain it, and claim 2 asserts continuity at every point outside $E$, so in particular at $1$: a monotone function may be continuous at a limit point of its own discontinuity set.

• What the example is for. Froda's theorem bounds the discontinuity set of a monotone function by countability and by nothing else; in particular it does not bound it by finiteness, even inside a bounded interval. The set $E$ above is the simplest witness: infinitely many jumps accumulating at a single point, all within $[0,1)$.

• The accumulation point is a point of continuity. The real $1$ is not a member of $E$, so claim 2 gives continuity of $f$ at $1$, even though every neighbourhood of $1$ contains infinitely many discontinuities of $f$. Being a limit of discontinuities is not itself an obstruction to continuity.

• A denser example is available. Taking $E = \mathbb{Q}$ instead gives a monotone function discontinuous on a dense set (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); the present example is the smaller and more concrete one, and it is the one where the points can be listed.

Let $n \in \mathbb{N}$ with $n \ge 1$ and put $I := [0,\infty)$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) and $p : I \to \mathbb{R}$, $p(x) := x^{n}$ (Integer powers $a^m$). Then:

1. $p$ is continuous on $I$ (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$: the $\varepsilon$-$\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point);
2. $p$ is increasing on $I$ (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $\mathbb{R}$, with the dictionary to monotone sequences), hence injective (Injection, surjection, bijection);
3. $p[I] = I$;
4. consequently the inverse map $g : I \to I$ of $p$ is continuous on $I$ and increasing, and $g(a) = a^{1/n}$ is the unique nonnegative $n$-th root of $a$ (Existence and uniqueness of $n$-th roots: a unique $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$).

So the $n$-th root function is continuous, and it is obtained from Continuous inverse theorem: a continuous injective $f$ on an interval $I$ is a bijection onto the order-convex set $f[I]$, and the inverse $g : f[I] \to I$ is continuous and strictly monotone in the same sense as $f$ rather than from a direct $\varepsilon$-$\delta$ estimate.

• Why $n \ge 1$. At $n = 0$ the map $x \mapsto x^{0}$ is constantly $1$ (Integer powers $a^m$), so it is neither injective nor surjective onto $[0,\infty)$, and no inverse exists. Every claim above is stated for $n \ge 1$ and the hypothesis is used in step 1.2.

• Why the domain is $[0,\infty)$ and not $\mathbb{R}$. For even $n$ the map $x \mapsto x^{n}$ is not injective on $\mathbb{R}$, since $(-x)^{n} = x^{n}$, so the continuous inverse theorem does not apply there; restricting to the nonnegative reals is what makes it injective, and it is also where Existence and uniqueness of $n$-th roots: a unique $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$ provides the roots.

• What is gained over the root theorem alone. Existence and uniqueness of $n$-th roots: a unique $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$ produces the number $a^{1/n}$ for each $a$ separately and says nothing about how it varies with $a$. Claim 4 is the statement that $a \mapsto a^{1/n}$ is a continuous increasing function, and it comes from the structure of the situation rather than from any estimate on roots.

Let $C$ be the Cantor set (The Cantor middle-thirds set as the intersection of the sets $C_n$ obtained by removing open middle thirds) and let $c : [0,1] \to \mathbb{R}$ be the Cantor function (The Cantor function on $[0,1]$, defined on the Cantor set through ternary digits and extended constantly across each removed interval). Then:

1. $C$ has measure zero (Measure zero (a countable cover by intervals of total length below every $\varepsilon$) and content zero (a finite such cover));
2. $c[C] = [0,1]$: the Cantor function maps the Cantor set onto the whole of $[0,1]$ (Injection, surjection, bijection);
3. $[0,1]$ does not have measure zero (A sequence of intervals covering $[a,b]$ has total length at least $b - a$, so no interval of positive length has measure zero).

So a continuous function can carry a set of measure zero onto a set that is not of measure zero, and indeed onto an interval of length $1$: being null is not preserved by continuous images.

• Nothing here contradicts any theorem about null sets. Measure zero is preserved by countable unions (A countable union of measure-zero sets has measure zero, by countable choice) and by passing to subsets, both of which are statements about covers. What this example shows is that it is not preserved by continuous images, and that is not a gap in any theorem above: no result in this library asserts that a continuous image of a null set is null.

• The image is as large as it could possibly be. $c$ takes values in $[0,1]$ (The Cantor function is well defined, satisfies $c(x) \le c(y)$ whenever $x \le y$, is surjective onto $[0,1]$, and is constant on every interval removed from the Cantor set), so $c[C] \subseteq [0,1]$ always; claim 2 says the inclusion is an equality. So $C$, which is null and nowhere dense (The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points), surjects onto an interval of length $1$; that $C$ is uncountable is proved independently as claim 4 of The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points.

• Where the increase happens. The remark The Cantor function is continuous and nondecreasing, climbs from $0$ to $1$, and is constant on every interval removed in the construction of the Cantor set, so all of its increase happens on a set of measure zero records the complementary fact: $c$ is locally constant off $C$, so all of its climb from $0$ to $1$ takes place on the null set $C$, and this example says that the climb is complete.

Let $q(x)$ be the least denominator of a rational $x$ (The Dirichlet function $1_{\mathbb{Q}}$, and Thomae's function $t$ with $t(x) = 1/q$ at a rational $x = p/q$ in lowest terms with $q \ge 1$ and $t(x) = 0$ at every irrational $x$) and define $h : \mathbb{R} \to \mathbb{R}$ by

$h(x) := \iota(q(x)) \ \text{ for } x \in \mathbb{Q}, \qquad h(x) := 0 \ \text{ for } x \notin \mathbb{Q},$

where $\iota(q)$ is the canonical natural (The canonical natural $\iota(n) = n \cdot 1_F$ of a field) and $\mathbb{Q}$ is the canonical copy of the rationals inside $\mathbb{R}$ (The rationals embed densely in the reals). Equivalently $h(x) = 1/t(x)$ at a rational $x$, where $t$ is Thomae's function. Then:

1. $h(x)$ is a real number for every real $x$: $h$ is finite at every point;
2. $h$ is unbounded on every nondegenerate interval (Lower bound, bounded below, bounded set, Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length): for all reals $a < b$ and every real $M$ there is $x \in (a,b)$ with $h(x) > M$.

So a function may be finite at every single point and yet fail to be bounded on every interval, however short. In particular $h$ is bounded on no neighbourhood of any point.

• Finiteness at a point says nothing about local boundedness. The two notions are often conflated; this example separates them as sharply as possible, since $h$ is unbounded on every nondegenerate interval and yet takes only real values.

• $h$ is nowhere continuous, and the oscillation is infinite everywhere. Continuity at $c$ would give a neighbourhood on which $|h - h(c)| < 1$, hence a neighbourhood on which $h$ is bounded, which claim 2 forbids; equivalently $\omega_{h}(c) = +\infty$ at every real $c$ (The oscillation $\omega_f(S) = \sup\{\,|f(x) - f(y)| : x, y \in S\,\}$ of $f$ on a set and the oscillation $\omega_f(c) = \inf_{\delta > 0} \omega_f(A \cap N_\delta(c))$ at a point, both taken in the extended reals, Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$: the $\varepsilon$-$\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point). This is a different pathology from Thomae's function, whose values are bounded by $1$ and which is continuous at every irrational (The Dirichlet function is continuous at no point of $\mathbb{R}$, and Thomae's function is continuous at every irrational and at no rational, so its set of continuity points is exactly the set of irrationals and its oscillation at $c$ equals $t(c)$).

• The separation estimate is the only arithmetic used. Step 1.2 is the standard fact that two distinct rationals with denominators $q_{1}$ and $q_{2}$ are at least $1/(q_{1}q_{2})$ apart, and it is proved from nothing more than "a nonzero integer has absolute value at least $1$".

Let $D : [0,1] \to \mathbb{R}$ be the restriction to $[0,1]$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) of the Dirichlet function (The Dirichlet function $1_{\mathbb{Q}}$, and Thomae's function $t$ with $t(x) = 1/q$ at a rational $x = p/q$ in lowest terms with $q \ge 1$ and $t(x) = 0$ at every irrational $x$), so $D(x) = 1$ at a rational $x$ and $D(x) = 0$ at an irrational $x$. Then: