TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$

Statement

Let $a < b$ be reals and let $f, g : [a, b] \to R$ be integrable (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 $\int_{a}^{b} f$ ). Then:

Nonnegativity. If $f (x) \geq 0$ for every $x \in [a, b]$ then $\int_{a}^{b} f \geq 0$ .
Monotonicity. If $f (x) \leq g (x)$ for every $x \in [a, b]$ then $\int_{a}^{b} f \leq \int_{a}^{b} g .$
Two-sided bound. If $m \leq f (x) \leq M$ for every $x \in [a, b]$ , with $m, M$ real, then $m (b - a) \leq \int_{a}^{b} f \leq M (b - a) .$

Equality in claim 1 does not force $f$ to vanish. A nonnegative integrable function with integral $0$ may be positive at infinitely many points; that is FALSE: a nonnegative Riemann integrable function on $[a, b]$ with $\int_{a}^{b} f = 0$ is identically zero on the previous page's companion. Under the additional hypothesis of continuity the conclusion does hold, and that is A continuous $f \geq 0$ on $[a, b]$ with $\int_{a}^{b} f = 0$ is identically $0$ below.

Claim 2 is stated for $a < b$ and is not orientation-invariant. With the convention of The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ , $f \leq g$ gives $\int_{u}^{v} f \leq \int_{u}^{v} g$ when $u \leq v$ and the reverse inequality when $u \geq v$ , since both sides change sign together.

Facts & Assumptions

Given: Reals $a < b$ and integrable $f, g : [a, b] \to R$ , with reals $m \leq M$ where claim 3 is concerned.

[A1]

$f (x) \geq 0$ for every $x \in [a, b]$ .

[A2]

$f (x) \leq g (x)$ for every $x \in [a, b]$ .

[A3]

$m \leq f (x) \leq M$ for every $x \in [a, b]$ .

[L1]

If $m^{'} \leq h \leq M^{'}$ on $[a, b]$ then $m^{'} (b - a) \leq L (h, P) \leq \underline{\int_{a}^{b}} h \leq \overline{\int_{a}^{b}} h \leq U (h, P) \leq M^{'} (b - a)$ for every partition $P$ (If $m \leq f \leq M$ on $[a, b]$ then $m (b - a) \leq L (f, P) \leq \underline{\int_{a}^{b}} f \leq \overline{\int_{a}^{b}} f \leq U (f, P) \leq M (b - a)$ for every partition $P$ ; in particular every constant function is integrable, with $\int_{a}^{b} c = c (b - a)$ , 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 $U (f, P) = \sum_{i} M_{i} Δ_{i}$ , Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L2]

If $h$ is integrable then $\int_{a}^{b} h$ is the common value of the lower and upper integrals (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 $\int_{a}^{b} f$ , Greatest lower bound (infimum)).

[L3]

Sums and scalar multiples of integrable functions are integrable, and $\int_{a}^{b} (λ h + ν k) = λ \int_{a}^{b} h + ν \int_{a}^{b} k$ (Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ ).

[L4]

Ordered-field arithmetic: adding a constant to both sides of an inequality preserves it, and the order is total and transitive (Ordered field, Complete ordered field (least-upper-bound property)). The nonstrict forms follow from the strict ones by adjoining the case of equality.

Proof

technique · direct

1.1

Claim 1. Under [A1] the constant $0$ is a lower bound of $f$ on $[a, b]$ , so [L1] applies with $m^{'} : = 0$ and gives $\underline{\int_{a}^{b}} f \geq 0$ .

A1L1

1.2

Claim 2. Under [A2] the function $h : = g - f$ satisfies $h (x) \geq 0$ for every $x \in [a, b]$ , and $h$ is integrable with $\int_{a}^{b} h = \int_{a}^{b} g - \int_{a}^{b} f$ by [L3].

A2L3L4

2.1

Since $f$ is integrable, $\int_{a}^{b} f = \underline{\int_{a}^{b}} f \geq 0$ by [L2].

step 1.1L2

3.1

By claim 1 applied to $h$ , $\int_{a}^{b} g - \int_{a}^{b} f \geq 0$ , that is $\int_{a}^{b} f \leq \int_{a}^{b} g$ .

step 2.1step 1.2L4

4.1

Claim 3. Under [A3], [L1] applied to $f$ with $m^{'} : = m$ and $M^{'} : = M$ gives $m (b - a) \leq \underline{\int_{a}^{b}} f$ and $\overline{\int_{a}^{b}} f \leq M (b - a)$ , and both integrals equal $\int_{a}^{b} f$ by [L2].

A3L1L2∎

Remarks

Claim 3 is cited, not reproved. If $m \leq f \leq M$ on $[a, b]$ then $m (b - a) \leq L (f, P) \leq \underline{\int_{a}^{b}} f \leq \overline{\int_{a}^{b}} f \leq U (f, P) \leq M (b - a)$ for every partition $P$ ; in particular every constant function is integrable, with $\int_{a}^{b} c = c (b - a)$ already proves the five-term chain for every partition, and it is the item that also computes the integral of a constant, $\int_{a}^{b} c = c (b - a)$ . Claim 3 is that chain read at an integrable $f$ ; nothing new is established here.
Claim 2 is proved through claim 1 and linearity, and not by comparing Darboux sums. Comparing sums works too, since $f \leq g$ gives $m_{i} (f) \leq m_{i} (g)$ and $M_{i} (f) \leq M_{i} (g)$ on every subinterval, but the route through $g - f$ is shorter and uses only results already available. Either way the hypothesis is a pointwise inequality on the whole of $[a, b]$ ; an inequality holding off a finite set gives the same conclusion, by Changing an integrable function at finitely many points changes neither its integrability nor its integral, and that is a separate statement.
What claim 2 is for. It is what turns a pointwise estimate on an integrand into an estimate on the integral, and every estimate of that shape on this page and its companion is an application of it. No count of those applications is asserted here; the dependency graph of the page is where that is read off.

Depends on

Used by

Dependency tree · two levels

35 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Riemann integral (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 6 (standard reference, not scraped)
Carnegie Mellon 21-269, Riemann integration notes (standard reference, not scraped)
Encyclopedia of Mathematics, Integral calculus (standard reference, not scraped)

If f≤gf≤g on [a,b][a,b] and both are integrable then ∫abf≤∫abg∫ab​f≤∫ab​g; and m(b−a)≤∫abf≤M(b−a)m(b−a)≤∫ab​f≤M(b−a)