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If $f$ is continuous on $[a,b]$ and $g$ is integrable with $g \ge 0$ , there is $\xi \in [a,b]$ with $\int_a^b fg = f(\xi)\int_a^b g$

Statement

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

$\int_a^b f\,g \;=\; f(\xi)\,\int_a^b g .$

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

$\int_a^b f \;=\; f(\xi)\,(b-a) ,$

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

The hypothesis $g \ge 0$ 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 $f : [a,b] \to \mathbb{R}$ , and an integrable $g : [a,b] \to \mathbb{R}$ with $g \ge 0$ 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 $\mathbb{R}$ (every open cover has a finite subcover), and sequentially compact subset, Extreme value theorem: a continuous real function on a nonempty compact subset of $\mathbb{R}$ attains a greatest and a least value, Maximum and minimum of a set, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[L2]

For continuous $f$ on $[a,b]$ with $a \le b$ , $f[\,[a,b]\,] = [\min f[\,[a,b]\,],\ \max f[\,[a,b]\,]]$ (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 $\lvert f\rvert$ , $f^{2}$ , $fg$ , $\max(f,g)$ and $\min(f,g)$ , and $\bigl\lvert\int_a^b f\bigr\rvert \le \int_a^b\lvert f\rvert$ , claim 1).

[L5]

If $u \le w$ pointwise and both are integrable then $\int_a^b u \le \int_a^b w$ ; and if $u \ge 0$ is integrable then $\int_a^b u \ge 0$ (If $f \le g$ on $[a,b]$ and both are integrable then $\int_a^b f \le \int_a^b g$ ; and $m(b-a) \le \int_a^b f \le M(b-a)$ ).

[L6]

Scalar multiples: $\int_a^b \lambda u = \lambda\int_a^b u$ (Integrable functions on $[a,b]$ form a set closed under sums and scalar multiples, and $\int_a^b(\lambda f+\mu g) = \lambda\int_a^b f + \mu\int_a^b g$ ), and $\int_a^b 1 = b-a$ (If $m \le f \le M$ on $[a,b]$ then $m(b-a) \le L(f,P) \le \underline{\int_a^b} f \le \overline{\int_a^b} f \le U(f,P) \le M(b-a)$ 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

1.1

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

givenL3L4

1.2

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

givenL1choose

1.3

By [L5], $\int_a^b g \ge 0$ .

givenL5

2.1

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

step 1.1step 1.2givenL6L7

3.1

By [L5] and [L6] applied to step 2.1, $m\int_a^b g \;\le\; \int_a^b fg \;\le\; M\int_a^b g$ .

step 2.1L5L6

4.1

The case $\int_a^b g = 0$ . Then step 3.1 reads $0 \le \int_a^b fg \le 0$ , so $\int_a^b fg = 0 = f(a)\int_a^b g$ , and $\xi := a$ works.

step 1.3step 3.1L7

4.2

The case $\int_a^b g > 0$ . Then $\lambda := \bigl(\int_a^b fg\bigr)\big/\bigl(\int_a^b g\bigr)$ is a real satisfying $m \le \lambda \le M$ , by step 3.1 divided by the positive $\int_a^b g$ .

step 1.3step 3.1L7construct

5.1

By step 1.2 and [L2], $f[\,[a,b]\,] = [m,M]$ , so $\lambda = f(\xi)$ for some $\xi \in [a,b]$ ; then $\int_a^b fg = \lambda\int_a^b g = f(\xi)\int_a^b g$ .

step 1.2step 4.2L2L7choose

6.1

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

step 1.3step 4.1step 5.1L7

7.1

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 $\xi \in [a,b]$ with $\int_a^b f = f(\xi)(b-a)$ .

step 6.1L6∎

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 $\xi$ , and nothing is claimed about the location of a distinguished point; the theorem asserts only that some $\xi$ 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.
$\xi$ can be forced to lie in the open interval only under extra hypotheses, and none is claimed. The standard refinement puts $\xi$ 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 $g \ge 0$ cannot be dropped is Continuous $f$ and integrable sign-changing $g$ with $\int_a^b fg \ne f(\xi)\int_a^b g$ for every $\xi$ ↗ on the companion page; nothing above depends on it.

Depends on

Used by

Continuous f and integrable sign-changing g with ∫ₐᵇ fg ≠ f(ξ)∫ₐᵇ g for every ξ Counterexample

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 105 results over 18 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Mean value theorem (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 6 (standard reference, not scraped)
Encyclopedia of Mathematics, Integral calculus (standard reference, not scraped)

If fff is continuous on [a,b][a,b][a,b] and ggg is integrable with g≥0g \ge 0g≥0, there is ξ∈[a,b]\xi \in [a,b]ξ∈[a,b] with ∫abfg=f(ξ)∫abg\int_a^b fg = f(\xi)\int_a^b g∫ab​fg=f(ξ)∫ab​g