TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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A monotone function on $[a, b]$ is Riemann integrable: for the uniform partition into $N$ parts the upper minus lower sum telescopes to $∣ f (b) - f (a) ∣ (b - a) / ι (N)$

Statement

Let $a < b$ be reals and let $f : [a, b] \to R$ be monotone, that is nondecreasing or nonincreasing on $[a, b]$ (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $R$ , with the dictionary to monotone sequences). Then $f$ is bounded (Lower bound, bounded below, bounded set) and Riemann integrable on $[a, b]$ (The lower and upper Darboux integrals of a bounded $f$ on $[a, b]$ as $\sup_{P} L (f, P)$ and $\inf_{P} U (f, P)$ , Darboux integrability as their equality, and the notation $\int_{a}^{b} f$ ).

Moreover, for the uniform partition $U_{N}$ of $[a, b]$ into $N \geq 1$ parts (Partition of $[a, b]$ as a finite strictly increasing list $a = t_{0} < t_{1} < \dots < t_{n} = b$ , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions),

$U (f, U_{N}) - L (f, U_{N}) = ∣ f (b) - f (a) ∣ \cdot \frac{b - a}{ι (N)},$

where $ι (N)$ is the canonical natural of $N$ in $R$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field). The right-hand side is an equality, not an estimate: the sum $\sum_{i < N} (M_{i} - m_{i})$ telescopes exactly, because on each subinterval a monotone function attains its extremes at the two endpoints.

No continuity is assumed, and none holds in general: a nondecreasing function may be discontinuous at every rational (Converse to Froda: for every at most countable $E \subseteq R$ there is a bounded nondecreasing $f : R \to R$ whose set of discontinuities is exactly $E$ , every one of them a jump), and the companion page of this pair works out the integral of the floor function, the simplest discontinuous monotone integrand.

Facts & Assumptions

Given: Reals $a < b$ and a monotone $f : [a, b] \to R$ . Let $N \geq 1$ be a natural number and let $U_{N} = (N, t)$ be the uniform partition of $[a, b]$ into $N$ parts, with subintervals $I_{i} = [t_{i}, t_{i + 1}]$ and lengths $Δ_{i} = (b - a) / ι (N)$ for $i < N$ .

[L1]

$f$ is nondecreasing, meaning $f (x) \leq f (y)$ whenever $x \leq y$ in $[a, b]$ , or nonincreasing, meaning $f (x) \geq f (y)$ whenever $x \leq y$ ; these two cases are what "monotone" means and they exhaust it (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $R$ , with the dictionary to monotone sequences).

[L2]

For the uniform partition: $t_{0} = a$ , $t_{N} = b$ , $t_{i} < t_{i + 1}$ , every $Δ_{i} = (b - a) / ι (N) > 0$ , and $\sum_{i < N} Δ_{i} = b - a$ (Partition of $[a, b]$ as a finite strictly increasing list $a = t_{0} < t_{1} < \dots < t_{n} = b$ , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions).

[L3]

If a nonempty set $S \subseteq R$ has a greatest element then that element is $\sup S$ , and if it has a least element then that element is $\inf S$ (Maximum and minimum of a set, Greatest lower bound (infimum), Complete ordered field (least-upper-bound property)).

[L4]

$M_{i} = \sup f [I_{i}]$ , $m_{i} = \inf f [I_{i}]$ , $U (f, P) - L (f, P) = \sum_{i < n} (M_{i} - m_{i}) Δ_{i}$ (For bounded $f$ on $[a, b]$ and a partition $P$ : the infimum $m_{i}$ and supremum $M_{i}$ of $f$ on the $i$ -th subinterval, and the lower and upper Darboux sums $L (f, P) = \sum_{i} m_{i} Δ_{i}$ and $U (f, P) = \sum_{i} M_{i} Δ_{i}$ , Laws of finite sums and finite products).

[L5]

Finite sums: scaling and telescoping, $\sum_{i < N} (c_{i + 1} - c_{i}) = c_{N} - c_{0}$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L6]

Riemann's criterion: a bounded $f$ is integrable if and only if for every real $ε > 0$ there is a partition $P$ with $U (f, P) - L (f, P) < ε$ (Riemann's criterion: a bounded $f$ on $[a, b]$ is Darboux integrable if and only if for every real $ε > 0$ there is a partition $P$ with $U (f, P) - L (f, P) < ε$ ).

[L7]

For every real $η > 0$ there is a natural $N \geq 1$ with $1 / ι (N) < η$ , and $ι (N) > 0$ for $N \geq 1$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Every complete ordered field is Archimedean, The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing).

[L8]

Ordered-field arithmetic and the absolute value: adding a constant and multiplying by a positive quantity preserve an inequality; the order is total and transitive; $∣ x ∣ = x$ for $x \geq 0$ and $∣ x ∣ = - x$ for $x \leq 0$ (Basic properties of the absolute value, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property), Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

Proof

technique · cases

1.1givenL1cases

By [L1] there are two cases, $f$ nondecreasing and $f$ nonincreasing, and they exhaust the hypothesis. Every $x \in [a, b]$ satisfies $a \leq x \leq b$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

1.2assume-case upgivenL2L3L8

Case: $f$ is nondecreasing. Then $f (a) \leq f (x) \leq f (b)$ for every $x \in [a, b]$ , so $f$ is bounded by [L8]; and for $i < N$ and $x \in I_{i}$ one has $t_{i} \leq x \leq t_{i + 1}$ , hence $f (t_{i}) \leq f (x) \leq f (t_{i + 1})$ . Since $f (t_{i})$ and $f (t_{i + 1})$ themselves lie in $f [I_{i}]$ , they are its least and greatest elements, so $m_{i} = f (t_{i})$ and $M_{i} = f (t_{i + 1})$ by [L3].

1.3assume-case downgivenL2L3L8

Case: $f$ is nonincreasing. Then $f (b) \leq f (x) \leq f (a)$ for every $x \in [a, b]$ , so $f$ is bounded; and for $i < N$ and $x \in I_{i}$ one has $f (t_{i + 1}) \leq f (x) \leq f (t_{i})$ , so $m_{i} = f (t_{i + 1})$ and $M_{i} = f (t_{i})$ by [L3].

2.1step 1.2L2L4L5L8

In the nondecreasing case, $M_{i} - m_{i} = f (t_{i + 1}) - f (t_{i})$ , so by [L4], [L2] and [L5], $U (f, U_{N}) - L (f, U_{N}) = \sum_{i < N} (f (t_{i + 1}) - f (t_{i})) \frac{b - a}{ι (N)} = \frac{b - a}{ι (N)} (f (t_{N}) - f (t_{0})) = \frac{b - a}{ι (N)} (f (b) - f (a))$ , and $f (b) - f (a) \geq 0$ , so this equals $∣ f (b) - f (a) ∣ (b - a) / ι (N)$ by [L8].

2.2step 1.3L2L4L5L8cases-exhaustive

In the nonincreasing case the same computation gives $U (f, U_{N}) - L (f, U_{N}) = \frac{b - a}{ι (N)} (f (a) - f (b))$ with $f (a) - f (b) \geq 0$ , which is again $∣ f (b) - f (a) ∣ (b - a) / ι (N)$ by [L8]. The two cases of step 1.1 exhaust the hypothesis, so the displayed identity holds for every monotone $f$ , which is also bounded.

3.1step 2.2L7L8choose

Let a real $ε > 0$ be given and put $η : = ε \cdot ((∣ f (b) - f (a) ∣ + 1) (b - a))^{- 1}$ , a positive real by [L8]. By [L7] fix a natural $N \geq 1$ with $1 / ι (N) < η$ .

4.1step 2.1step 2.2step 3.1L8

Then $U (f, U_{N}) - L (f, U_{N}) = ∣ f (b) - f (a) ∣ (b - a) / ι (N) \leq (∣ f (b) - f (a) ∣ + 1) (b - a) / ι (N) < (∣ f (b) - f (a) ∣ + 1) (b - a) η = ε$ : the first inequality because $(b - a) / ι (N) > 0$ and $∣ f (b) - f (a) ∣ \leq ∣ f (b) - f (a) ∣ + 1$ , and the second because $(∣ f (b) - f (a) ∣ + 1) (b - a) > 0$ and $1 / ι (N) < η$ . So $U (f, U_{N}) - L (f, U_{N}) < ε$ .

5.1step 1.2step 1.3step 4.1L6∎

Since the real $ε > 0$ was arbitrary and step 4.1 produced a partition with $U - L < ε$ , criterion [L6] applies: $f$ is bounded by steps 1.2 and 1.3 and Riemann integrable on $[a, b]$ .

Remarks

The telescoping is exact, and that is what makes the proof short. No estimate of $M_{i} - m_{i}$ is needed subinterval by subinterval: the whole sum of the gaps is the total rise of $f$ , however wildly the rise is distributed. This is why a monotone function with infinitely many jumps is no harder than a continuous one here, and it is the same telescoping that Partition of $[a, b]$ as a finite strictly increasing list $a = t_{0} < t_{1} < \dots < t_{n} = b$ , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions uses for the lengths.
Uniformity of the partition is a convenience, not a necessity. For an arbitrary partition $P$ the same identification of $m_{i}$ and $M_{i}$ gives $U (f, P) - L (f, P) \leq ∥ P ∥ \cdot ∣ f (b) - f (a) ∣$ , by bounding each $Δ_{i}$ by the mesh before telescoping. The uniform partition is used above only so that $Δ_{i}$ can be pulled out of the sum as a constant.
No choice principle is used. The partition is given by a formula in $N$ , and $N$ is obtained from one instance of the Archimedean property. See Choice ledger for the Riemann integral: the page results are proved in ZF.
Monotone is strictly weaker than continuous here. The floor function is nondecreasing with jumps at every integer and is integrable (the companion page of this pair); and a monotone function has at most countably many discontinuities (Froda's theorem: the set of discontinuities of a monotone function on an interval is at most countable, the injection into $N$ being built from one fixed enumeration of the rationals by least index, so no choice principle is used), so this theorem is also a special case of A bounded function on $[a, b]$ whose set of discontinuities is at most countable is Riemann integrable once that is available. The direct proof is kept because it is elementary and quantitative, and because it costs no choice at all.

A monotone function on $[a, b]$ is Riemann integrable: for the uniform partition into $N$ parts the upper minus lower sum telescopes to $∣ f (b) - f (a) ∣ (b - a) / ι (N)$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

A monotone function on [a,b][a,b] is Riemann integrable: for the uniform partition into NN parts the upper minus lower sum telescopes to ∣f(b)−f(a)∣ (b−a)/ι(N)∣f(b)−f(a)∣(b−a)/ι(N)

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

A monotone function on $[a, b]$ is Riemann integrable: for the uniform partition into $N$ parts the upper minus lower sum telescopes to $∣ f (b) - f (a) ∣ (b - a) / ι (N)$