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

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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.

Statement references

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 $\int_{a}^{b} f$

FALSE: a nonnegative Riemann integrable function on $[a, b]$ with $\int_{a}^{b} f = 0$ is identically zero

A continuous $f \geq 0$ on $[a, b]$ with $\int_{a}^{b} f = 0$ is identically $0$

The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$

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

1.1A1L1

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

1.2A2L3L4

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].

2.1step 1.1L2

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

3.1step 2.1step 1.2L4

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

4.1A3L1L2∎

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].

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.

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

Statement references

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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)

Statement

Statement references

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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