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

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If $f$ is continuous on $[a, b]$ and $g$ is integrable with $g \geq 0$ , there is $ξ \in [a, b]$ with $\int_{a}^{b} f g = f (ξ) \int_{a}^{b} g$

Statement

Let $a < b$ be reals, let $f : [a, b] \to R$ be continuous on $[a, b]$ (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point) and let $g : [a, b] \to R$ 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 $\int_{a}^{b} f$ ) with $g (x) \geq 0$ for every $x \in [a, b]$ . Then $f g$ is integrable and there is $ξ \in [a, b]$ with

$\int_{a}^{b} f g = f (ξ) \int_{a}^{b} g .$

The special case $g \equiv 1$ is the familiar statement that a continuous function attains its average value: there is $ξ \in [a, b]$ with

$\int_{a}^{b} f = f (ξ) (b - a),$

and it is this clause that the fundamental theorem below is usually derived from in other treatments.

The hypothesis $g \geq 0$ is essential. For a sign-changing integrable $g$ the conclusion fails, and the witness is the counterexample with a sign-changing weight on the companion page.

Facts & Assumptions

Given: Reals $a < b$ , a continuous $f : [a, b] \to R$ , and an integrable $g : [a, b] \to R$ with $g \geq 0$ on $[a, b]$ .

[L1]

$[a, b]$ is compact, and a continuous real function on a nonempty compact set attains a minimum and a maximum there (Heine-Borel by bisection: every closed bounded interval $[a, b]$ is compact, Open cover, subcover, compact subset of $R$ (every open cover has a finite subcover), and sequentially compact subset, Extreme value theorem: a continuous real function on a nonempty compact subset of $R$ attains a greatest and a least value, Maximum and minimum of a set, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L2]

For continuous $f$ on $[a, b]$ with $a \leq b$ , $f [[a, b]] = [\min f [[a, b]], \max f [[a, b]]]$ (The image of an interval under a continuous real function is order-convex, hence an interval, and the image of a closed bounded interval is a closed bounded interval, claim 2).

[L3]

A continuous function on $[a, b]$ is integrable there (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion).

[L4]

A product of two integrable functions on $[a, b]$ is integrable (If $f, g$ are integrable on $[a, b]$ then so are $∣ f ∣$ , $f^{2}$ , $f g$ , $\max (f, g)$ and $\min (f, g)$ , and $∣ \int_{a}^{b} f ∣ \leq \int_{a}^{b} ∣ f ∣$ , claim 1).

[L5]

If $u \leq w$ pointwise and both are integrable then $\int_{a}^{b} u \leq \int_{a}^{b} w$ ; and if $u \geq 0$ is integrable then $\int_{a}^{b} u \geq 0$ (If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ ).

[L6]

Scalar multiples: $\int_{a}^{b} λ u = λ \int_{a}^{b} u$ (Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ ), and $\int_{a}^{b} 1 = b - a$ (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 $\int_{a}^{b} c = c (b - a)$ ).

[L7]

Ordered-field arithmetic: multiplying an inequality by a nonnegative quantity preserves it, a positive real has a positive inverse, and the order is total and transitive (Ordered field, Complete ordered field (least-upper-bound property)).

Proof

technique · direct

1.1givenL3L4

$f$ is integrable by [L3], so $f g$ is integrable by [L4].

1.2givenL1choose

By [L1] fix $p, q \in [a, b]$ with $f (q) = \min f [[a, b]] = : m$ and $f (p) = \max f [[a, b]] = : M$ , so $m \leq f (x) \leq M$ for every $x \in [a, b]$ .

1.3givenL5

By [L5], $\int_{a}^{b} g \geq 0$ .

2.1step 1.1step 1.2givenL6L7

Since $g \geq 0$ , multiplying the inequalities of step 1.2 by $g (x)$ gives $m g (x) \leq f (x) g (x) \leq M g (x)$ for every $x \in [a, b]$ , and all three functions are integrable by step 1.1 and [L6].

3.1step 2.1L5L6

By [L5] and [L6] applied to step 2.1, $m \int_{a}^{b} g \leq \int_{a}^{b} f g \leq M \int_{a}^{b} g$ .

4.1step 1.3step 3.1L7

The case $\int_{a}^{b} g = 0$ . Then step 3.1 reads $0 \leq \int_{a}^{b} f g \leq 0$ , so $\int_{a}^{b} f g = 0 = f (a) \int_{a}^{b} g$ , and $ξ : = a$ works.

4.2step 1.3step 3.1L7construct

The case $\int_{a}^{b} g > 0$ . Then $λ : = (\int_{a}^{b} f g) / (\int_{a}^{b} g)$ is a real satisfying $m \leq λ \leq M$ , by step 3.1 divided by the positive $\int_{a}^{b} g$ .

5.1step 1.2step 4.2L2L7choose

By step 1.2 and [L2], $f [[a, b]] = [m, M]$ , so $λ = f (ξ)$ for some $ξ \in [a, b]$ ; then $\int_{a}^{b} f g = λ \int_{a}^{b} g = f (ξ) \int_{a}^{b} g$ .

6.1step 1.3step 4.1step 5.1L7

The two cases $\int_{a}^{b} g = 0$ and $\int_{a}^{b} g > 0$ are exhaustive by step 1.3, so the theorem holds.

7.1step 6.1L6∎

The clause $g \equiv 1$ . The constant $1$ is integrable, nonnegative, with $\int_{a}^{b} 1 = b - a$ by [L6], so step 6.1 gives $ξ \in [a, b]$ with $\int_{a}^{b} f = f (ξ) (b - a)$ .

Remarks

The case $\int_{a}^{b} g = 0$ is handled first because the usual proof divides by it. There the conclusion is trivially true for every $ξ$ , and nothing is claimed about the location of a distinguished point; the theorem asserts only that some $ξ$ works.
No intermediate value theorem is invoked directly. What is needed is that the continuous image of $[a, b]$ is exactly $[\min, \max]$ , which is claim 2 of The image of an interval under a continuous real function is order-convex, hence an interval, and the image of a closed bounded interval is a closed bounded interval; that item is itself proved from the intermediate and extreme value theorems, and citing it here saves repeating the argument.
$ξ$ can be forced to lie in the open interval only under extra hypotheses, and none is claimed. The standard refinement puts $ξ$ in the open interval when $g \equiv 1$ ; it is not proved here, it is not needed anywhere on this page, and the theorem does not assert it.
Forward reference, orientation only. The witness showing that $g \geq 0$ cannot be dropped is Continuous $f$ and integrable sign-changing $g$ with $\int_{a}^{b} f g \neq f (ξ) \int_{a}^{b} g$ for every $ξ$ ↗ on the companion page; nothing above depends on it.

If $f$ is continuous on $[a, b]$ and $g$ is integrable with $g \geq 0$ , there is $ξ \in [a, b]$ with $\int_{a}^{b} f g = f (ξ) \int_{a}^{b} g$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

If ff is continuous on [a,b][a,b] and gg is integrable with g≥0g≥0, there is ξ∈[a,b]ξ∈[a,b] with ∫abfg=f(ξ)∫abg∫ab​fg=f(ξ)∫ab​g

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

If $f$ is continuous on $[a, b]$ and $g$ is integrable with $g \geq 0$ , there is $ξ \in [a, b]$ with $\int_{a}^{b} f g = f (ξ) \int_{a}^{b} g$