Monotone Functions, Discontinuities, and Continuity Sets

Throughout, $\mathbb{R}$ is the complete ordered field (Complete ordered field (least-upper-bound property), Ordered field) with its order (Order on the reals). Let $A \subseteq \mathbb{R}$ and let $f : A \to \mathbb{R}$. Then $f$ is:

• nondecreasing when $f(x) \le f(y)$ for all $x, y \in A$ with $x \le y$;
• increasing, or strictly increasing, when $f(x) < f(y)$ for all $x, y \in A$ with $x < y$;
• nonincreasing when $f(x) \ge f(y)$ for all $x, y \in A$ with $x \le y$;
• decreasing, or strictly decreasing, when $f(x) > f(y)$ for all $x, y \in A$ with $x < y$;
• monotone when it is nondecreasing or nonincreasing;
• strictly monotone when it is increasing or decreasing.

The naming follows the convention of Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences, which is the convention of this library throughout: increasing is the strict notion and nondecreasing the weak one.

An increasing function is nondecreasing, and a decreasing function is nonincreasing. For $x \le y$ either $x < y$, and then $f(x) < f(y)$, hence $f(x) \le f(y)$; or $x = y$, and then $f(x) = f(y)$. The same argument with the inequalities reversed gives the second claim. So strictly monotone implies monotone.

A strictly monotone function is injective (Injection, surjection, bijection). Let $f$ be increasing and let $x, y \in A$ with $x \ne y$. By trichotomy either $x < y$, and then $f(x) < f(y)$, or $y < x$, and then $f(y) < f(x)$; in both cases $f(x) \ne f(y)$. The decreasing case is the same argument. The converse fails, and the failure is not exotic: a continuous injection on an interval is strictly monotone (A continuous injective function on an interval is strictly monotone), but on a domain that is not an interval it need not be.

Negation exchanges the two directions. For $g := -f$, that is $g(x) := -f(x)$, the four conditions above are exchanged in pairs: $f$ is nondecreasing exactly when $g$ is nonincreasing, and $f$ is increasing exactly when $g$ is decreasing, because $u \le v$ holds exactly when $-v \le -u$ (Ordered field). Several proofs below use this to reduce a nonincreasing case to a nondecreasing one.

Monotone on a set, not at a point. All six conditions are conditions on the whole of $A$; unlike continuity (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) there is no pointwise version, and none is used in this library. The domain $A$ is an arbitrary subset of $\mathbb{R}$; where a result needs $A$ to be an interval (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) it says so, and the hypothesis is never decoration.

The dictionary to monotone sequences

A sequence of reals is a function $x : \mathbb{N} \to \mathbb{R}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), and Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences calls it nondecreasing when $x_j \le x_k$ for all $j \le k$, increasing when $x_j < x_k$ for all $j < k$, and so on. Those are the same four conditions as above, read with the ordered set $\mathbb{N}$ in place of the ordered subset $A \subseteq \mathbb{R}$ and with the comparison of indices in place of the comparison of arguments. So nothing new is introduced here for sequences, and the two vocabularies may be used interchangeably: the words nondecreasing, increasing, nonincreasing, decreasing, monotone and strictly monotone mean the corresponding condition on the domain at hand.

One consequence is used repeatedly, and it has to be stated carefully because composition does not simply preserve the four words. Let $(x_k)$ be a nondecreasing sequence with $x_k \in A$ for every $k$, so that $j \le k$ gives $x_j \le x_k$. Then:

• if $f$ is nondecreasing, $(f(x_k))$ is nondecreasing, since $f(x_j) \le f(x_k)$;
• if $f$ is nonincreasing, $(f(x_k))$ is nonincreasing, since $f(x_j) \ge f(x_k)$.

So along a nondecreasing sequence the composite inherits the direction of $f$; and with $(x_k)$ increasing and $f$ increasing, $(f(x_k))$ is increasing, while with $(x_k)$ increasing and $f$ decreasing, $(f(x_k))$ is decreasing.

Along a nonincreasing sequence the direction is reversed, not inherited. If $(x_k)$ is nonincreasing and $f$ is nonincreasing, then $j \le k$ gives $x_j \ge x_k$ and hence $f(x_j) \le f(x_k)$: the composite is nondecreasing. The witness is $f(x) = -x$ on $A = \mathbb{R}$ with $x_k = -k$, where both $f$ and $(x_k)$ are decreasing and $f(x_k) = k$ is increasing. Two order-reversing maps compose to an order-preserving one, exactly as for the four words applied to functions.

Let $I \subseteq \mathbb{R}$ be order-convex (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length), let $f : I \to \mathbb{R}$ be 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 let $c \in I$. Write

$I^{-} := I \cap (-\infty, c), \qquad I^{+} := I \cap (c, \infty)$

(The left and right limits of $f$ at $c$, as limits of the restrictions of $f$ to $A \cap (-\infty, c)$ and $A \cap (c, \infty)$).

1. Left. If $I^{-} \ne \varnothing$ then $c$ is a limit point of $I^{-}$ (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$), the set $\{\, f(x) : x \in I^{-} \,\}$ is nonempty and bounded above by $f(c)$, and $\lim_{x \to c^{-}} f(x) \;=\; \sup\{\, f(x) : x \in I,\ x < c \,\} \;\le\; f(c) .$
2. Right. If $I^{+} \ne \varnothing$ then $c$ is a limit point of $I^{+}$, the set $\{\, f(x) : x \in I^{+} \,\}$ is nonempty and bounded below by $f(c)$, and $\lim_{x \to c^{+}} f(x) \;=\; \inf\{\, f(x) : x \in I,\ x > c \,\} \;\ge\; f(c) .$
3. Together. If both $I^{-}$ and $I^{+}$ are nonempty then $\lim_{x \to c^{-}} f(x) \;\le\; f(c) \;\le\; \lim_{x \to c^{+}} f(x) .$

In particular a nondecreasing function on an interval has, at every point of that interval, every one-sided limit that is well posed at all: no hypothesis of continuity, of boundedness, or of any other kind is needed.

The nonincreasing case is not a separate theorem. If $g : I \to \mathbb{R}$ is nonincreasing then $-g$ 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 a real $L$ is the left limit of $-g$ at $c$ exactly when $-L$ is the left limit of $g$ at $c$, since $|(-g)(x) - L| = |g(x) - (-L)|$; so claims 1 to 3 hold for $g$ with the suprema and infima exchanged and the inequalities reversed.

Order-convexity of $I$ is what makes the limits well posed. Without it the symbol $\lim_{x \to c^{-}} f(x)$ need not be defined even though $I^{-}$ is nonempty: for $I = \{0\} \cup [1,2]$ and $c = 1$ the set $I^{-} = \{0\}$ is nonempty but $1$ is not a limit point of it, and The left and right limits of $f$ at $c$, as limits of the restrictions of $f$ to $A \cap (-\infty, c)$ and $A \cap (c, \infty)$ leaves the symbol undefined there for exactly that reason.

• Where completeness is spent. Exactly once on each side, in the existence of $\sup S^{-}$ and of $\inf S^{+}$; the rest of the proof is the definition of a one-sided limit and the monotonicity hypothesis. Over an ordered field that is not complete the statement fails, because the supremum need not exist.

• The two one-sided limits need not agree, and that is the point. When both are defined they satisfy $\lim_{x \to c^{-}} f(x) \le f(c) \le \lim_{x \to c^{+}} f(x)$, and a strict inequality between the outer two is exactly a jump discontinuity; A monotone function on an interval has no discontinuity of the second kind: at every point both relevant one-sided limits exist, and an interior point $c$ is a discontinuity exactly when $\lim_{x \to c^{-}} f(x) < \lim_{x \to c^{+}} f(x)$ turns that observation into the classification of the discontinuities of a monotone function, 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 counts them.

Let $A \subseteq \mathbb{R}$, let $f : A \to \mathbb{R}$ and let $c \in A$. Then $f$ is discontinuous at $c$, and $c$ is a discontinuity of $f$, when $f$ is not continuous at $c$ (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). As in The left and right limits of $f$ at $c$, as limits of the restrictions of $f$ to $A \cap (-\infty, c)$ and $A \cap (c, \infty)$ write

$A^{-} := A \cap (-\infty, c), \qquad A^{+} := A \cap (c, \infty)$

(Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length), and recall that $\lim_{x \to c^{-}} f(x)$ is defined only when $c$ is a limit point of $A^{-}$, and $\lim_{x \to c^{+}} f(x)$ only when $c$ is a limit point of $A^{+}$ (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$).

At an isolated point there is nothing to classify. If $c$ is an isolated point of $A$ (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$), so that $A \cap N_{\rho}(c) = \{c\}$ for some real $\rho > 0$, then $f$ is continuous at $c$: the $\varepsilon$-$\delta$ condition of 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 is satisfied by $\delta := \rho$, since the only $x \in A$ with $|x - c| < \rho$ is $c$ itself and $|f(c) - f(c)| = 0$. So every discontinuity is a limit point of $A$, and the classification below covers every case that occurs.

Two-sided points

Suppose $c$ is a limit point of both $A^{-}$ and $A^{+}$, so that both one-sided limits are well posed. Say that $c$ is a discontinuity

• of the first kind when both one-sided limits exist;
• of the second kind, also called essential, when at least one of the two one-sided limits fails to exist.

A discontinuity of the first kind is further

• removable when $\lim_{x \to c^{-}} f(x) = \lim_{x \to c^{+}} f(x)$; the common value is then different from $f(c)$, for otherwise If $c$ is a limit point of the domain from both sides, the limit exists iff both one-sided limits exist and agree would give $\lim_{x \to c} f(x) = f(c)$ and $f$ would be continuous at $c$ (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);
• a jump when $\lim_{x \to c^{-}} f(x) \ne \lim_{x \to c^{+}} f(x)$; the difference $\lim_{x \to c^{+}} f(x) - \lim_{x \to c^{-}} f(x)$ is then called the jump of $f$ at $c$.

The three cases removable, jump, essential are mutually exclusive and exhaust the two-sided discontinuities of $f$: either both one-sided limits exist, and then they are equal or not, or one of them does not exist.

Removable is a name for what can be repaired. If $c$ is a removable discontinuity with common one-sided value $L$, then the function agreeing with $f$ off $c$ and taking the value $L$ at $c$ is continuous at $c$, again by If $c$ is a limit point of the domain from both sides, the limit exists iff both one-sided limits exist and agree and 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: changing the single value $f(c)$ removes the discontinuity. No such repair is available at a jump or at an essential discontinuity, since there the two-sided limit does not exist at all and no choice of value at $c$ can create it.

One-sided points

If $c$ is a limit point of exactly one of $A^{-}$ and $A^{+}$, only that side is defined and only that side is used: $c$ is a discontinuity of the first kind when the one-sided limit on the side in question exists, and of the second kind otherwise. When it exists it is different from $f(c)$, since on such a point the one-sided condition and the continuity condition are the same condition; and there is no jump case, there being nothing to compare the value with. The endpoints of an interval are the typical instance.

On the two vocabularies. First kind and second kind are Rudin's terms and are recorded because the literature uses them; removable, jump and essential are the names used in the rest of this library. They name the same three cases and no third classification is introduced.

Let $I \subseteq \mathbb{R}$ be order-convex (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) and let $f : I \to \mathbb{R}$ be 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). Write $I^{-} = I \cap (-\infty,c)$ and $I^{+} = I \cap (c,\infty)$ for $c \in I$.

1. At every $c \in I$, each of the two one-sided limits that is well posed exists (The left and right limits of $f$ at $c$, as limits of the restrictions of $f$ to $A \cap (-\infty, c)$ and $A \cap (c, \infty)$). Consequently $f$ has no discontinuity of the second kind (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): every discontinuity of $f$ is of the first kind.
2. Call $c \in I$ an interior point of $I$ when both $I^{-}$ and $I^{+}$ are nonempty. At such a point $\lim_{x \to c^{-}} f(x) \;\le\; f(c) \;\le\; \lim_{x \to c^{+}} f(x),$ and $f$ is continuous at $c$ (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) if and only if $\lim_{x \to c^{-}} f(x) = \lim_{x \to c^{+}} f(x)$.
3. Hence an interior point $c$ is a discontinuity of $f$ exactly when $\lim_{x \to c^{-}} f(x) \;<\; \lim_{x \to c^{+}} f(x),$ and every such discontinuity is a jump, of jump $\lim_{x \to c^{+}} f(x) - \lim_{x \to c^{-}} f(x) > 0$.

The same three claims hold for a nonincreasing $f$, with the two one-sided limits exchanged and all inequalities reversed, by applying the above to $-f$, which 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 has exactly the same points of continuity, since $|(-f)(x) - (-f)(c)| = |f(x) - f(c)|$.

A point of $I$ that is not interior is an endpoint, and there are at most two. $I^{-} = \varnothing$ says that $c$ is a least element of $I$ and $I^{+} = \varnothing$ that it is a greatest one, and a set has at most one of each. Those two points are excluded from claims 2 and 3 only because a comparison of two one-sided limits is not available there; claim 1 covers them.

• Nothing here counts the discontinuities. Claim 3 says only what a discontinuity of a monotone function looks like at an interior point. That the set of them is at most countable is a further 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, and its proof is exactly the observation that the open intervals $(\lim_{x \to c^{-}} f(x), \lim_{x \to c^{+}} f(x))$ attached to distinct discontinuities are disjoint.

• Why no interior discontinuity of a monotone function is removable. Claim 2 rules them out at interior points: there $L^{-} = L^{+}$ already forces continuity, because the inequality $L^{-} \le f(c) \le L^{+}$ pins $f(c)$ between the two one-sided values. That inequality is special to monotone functions, and it is what makes jump the only kind of interior discontinuity available. At a point of $I$ that is not interior the inequality is one-sided too and the argument does not apply, so a monotone function may fail to be continuous at an endpoint of $I$ while having its one one-sided limit; that failure is a discontinuity of the first kind and it is not a jump, there being only one side to compare.