Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28
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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 cc is a discontinuity exactly when limxcf(x)<limxc+f(x)\lim_{x \to c^{-}} f(x) < \lim_{x \to c^{+}} f(x)

Statement

Let IRI \subseteq \mathbb{R} be order-convex (Intervals of R\mathbb{R}: the nine order-convex forms, nondegeneracy, and length) and let f:IRf : I \to \mathbb{R} be nondecreasing (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of R\mathbb{R}, with the dictionary to monotone sequences). Write I=I(,c)I^{-} = I \cap (-\infty,c) and I+=I(c,)I^{+} = I \cap (c,\infty) for cIc \in I.

  1. At every cIc \in I, each of the two one-sided limits that is well posed exists (The left and right limits of ff at cc, as limits of the restrictions of ff to A(,c)A \cap (-\infty, c) and A(c,)A \cap (c, \infty)). Consequently ff has no discontinuity of the second kind (Discontinuity of ff 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 ff is of the first kind.
  2. Call cIc \in I an interior point of II when both II^{-} and I+I^{+} are nonempty. At such a point limxcf(x)    f(c)    limxc+f(x),\lim_{x \to c^{-}} f(x) \;\le\; f(c) \;\le\; \lim_{x \to c^{+}} f(x), and ff is continuous at cc (Continuity of f:ARf : A \to \mathbb{R} at a point of AA and on AA: the ε\varepsilon-δ\delta condition, its agreement with limxcf(x)=f(c)\lim_{x \to c} f(x) = f(c) at a limit point, and continuity at an isolated point) if and only if limxcf(x)=limxc+f(x)\lim_{x \to c^{-}} f(x) = \lim_{x \to c^{+}} f(x).
  3. Hence an interior point cc is a discontinuity of ff exactly when limxcf(x)  <  limxc+f(x),\lim_{x \to c^{-}} f(x) \;<\; \lim_{x \to c^{+}} f(x), and every such discontinuity is a jump, of jump limxc+f(x)limxcf(x)>0\lim_{x \to c^{+}} f(x) - \lim_{x \to c^{-}} f(x) > 0.

The same three claims hold for a nonincreasing ff, with the two one-sided limits exchanged and all inequalities reversed, by applying the above to f-f, which is nondecreasing (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of R\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)|(-f)(x) - (-f)(c)| = |f(x) - f(c)|.

A point of II that is not interior is an endpoint, and there are at most two. I=I^{-} = \varnothing says that cc is a least element of II and I+=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.

Facts & Assumptions

Given: An order-convex IRI \subseteq \mathbb{R}, a nondecreasing f:IRf : I \to \mathbb{R}, and cIc \in I.

[L1]

If II^{-} \ne \varnothing then cc is a limit point of II^{-} and limxcf(x)=sup{f(x):xI,x<c}f(c)\lim_{x \to c^{-}} f(x) = \sup\{f(x) : x \in I, x < c\} \le f(c); if I+I^{+} \ne \varnothing then cc is a limit point of I+I^{+} and limxc+f(x)=inf{f(x):xI,x>c}f(c)\lim_{x \to c^{+}} f(x) = \inf\{f(x) : x \in I, x > c\} \ge f(c) (One-sided limits of a monotone function always exist: for ff nondecreasing on an interval II and cIc \in I, limxcf(x)=sup{f(x):xI, x<c}\lim_{x \to c^{-}} f(x) = \sup\{f(x) : x \in I,\ x < c\} whenever II has points below cc, limxc+f(x)=inf{f(x):xI, x>c}\lim_{x \to c^{+}} f(x) = \inf\{f(x) : x \in I,\ x > c\} whenever it has points above cc, and these satisfy limxcf(x)f(c)limxc+f(x)\lim_{x \to c^{-}} f(x) \le f(c) \le \lim_{x \to c^{+}} f(x)).

[L2]

If cc is a limit point of both II^{-} and I+I^{+}, then limxcf(x)=L\lim_{x \to c} f(x) = L holds if and only if both one-sided limits at cc exist and equal LL; in particular the two-sided limit exists exactly when the two one-sided limits exist and agree (If cc is a limit point of the domain from both sides, the limit exists iff both one-sided limits exist and agree).

[L4]

A discontinuity at a two-sided point is of the second kind when at least one one-sided limit fails to exist, of the first kind otherwise, and is a jump when the two one-sided limits exist and differ; at a one-sided point it is of the first kind when the one available one-sided limit exists (Discontinuity of ff 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).

Proof

technique · direct
1.1

Let cIc \in I. If II^{-} \ne \varnothing then limxcf(x)\lim_{x \to c^{-}} f(x) exists, and if I+I^{+} \ne \varnothing then limxc+f(x)\lim_{x \to c^{+}} f(x) exists; if one of the two sets is empty the corresponding symbol is not defined and there is nothing to prove for it.

L1
2.1

Claim 1 follows: at every point of II every well-posed one-sided limit of ff exists, so no discontinuity of ff can be of the second kind, and every discontinuity is therefore of the first kind.

step 1.1L4
2.2

Now let cc be an interior point of II, and write L:=limxcf(x)L^{-} := \lim_{x \to c^{-}} f(x) and L+:=limxc+f(x)L^{+} := \lim_{x \to c^{+}} f(x), both of which exist by step 1.1. Then Lf(c)L+L^{-} \le f(c) \le L^{+}, which is the displayed inequality of claim 2.

step 1.1L1
3.1

Suppose L=L+L^{-} = L^{+}. Then Lf(c)L+=LL^{-} \le f(c) \le L^{+} = L^{-} forces L=f(c)=L+L^{-} = f(c) = L^{+}, so both one-sided limits equal f(c)f(c); hence limxcf(x)\lim_{x \to c} f(x) exists and equals f(c)f(c), and ff is continuous at cc.

step 2.2L2L3
3.2

Suppose conversely that ff is continuous at cc. Since cc is a limit point of II^{-} and hence of II, continuity gives limxcf(x)=f(c)\lim_{x \to c} f(x) = f(c), and then both one-sided limits exist and equal f(c)f(c); in particular L=L+L^{-} = L^{+}.

step 2.2L1L2L3
4.1

Claim 2 is proved by steps 3.1 and 3.2 together with step 2.2.

step 2.2step 3.1step 3.2
5.1

Claim 3: at an interior point cc, ff is discontinuous exactly when LL+L^{-} \ne L^{+}, and since Lf(c)L+L^{-} \le f(c) \le L^{+} the only way for them to differ is L<L+L^{-} < L^{+}. Both one-sided limits exist and differ, so the discontinuity is a jump, of jump L+L>0L^{+} - L^{-} > 0.

step 2.2step 4.1L4

Remarks

  • 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 N\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 (limxcf(x),limxc+f(x))(\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+L^{-} = L^{+} already forces continuity, because the inequality Lf(c)L+L^{-} \le f(c) \le L^{+} pins f(c)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 II 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 II 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.

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