TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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A continuous $f \ge 0$ on $[a,b]$ with $\int_a^b f = 0$ is identically $0$

Statement

Let $a < b$ be reals and 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) with $f(x) \ge 0$ for every $x \in [a,b]$ and

$\int_a^b f \;=\; 0 .$

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

This is the exact repair of a published false statement. Without continuity the conclusion fails: FALSE: a nonnegative Riemann integrable function on $[a,b]$ with $\int_a^b f = 0$ is identically zero, on the companion page of The Riemann Integral, exhibits a nonnegative integrable function with integral $0$ that is positive at every rational point. The remark there says that the continuous case is true and that its proof was not available at that point in the reading order, because additivity over subintervals had not been proved. It is proved now (For $a<c<b$ : $f$ is integrable on $[a,b]$ if and only if it is integrable on $[a,c]$ and on $[c,b]$ , and then $\int_a^b f = \int_a^c f + \int_c^b f$ ; with the oriented form for arbitrary $a,b,c$ ), and this item is that proof.

Facts & Assumptions

Given: Reals $a<b$ and a continuous $f : [a,b] \to \mathbb{R}$ with $f \ge 0$ on $[a,b]$ and $\int_a^b f = 0$ .

[A1]

There is $c \in [a,b]$ with $f(c) > 0$ .

[L1]

$f$ is integrable on $[a,b]$ and on every closed subinterval with distinct endpoints (A continuous function on $[a,b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, A function integrable on $[a,b]$ is integrable on every closed subinterval, 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 $\int_a^b f$ ).

[L2]

Continuity at $c$ : for every real $\eta>0$ there is a real $\delta>0$ such that every $x \in [a,b]$ with $|x-c|<\delta$ satisfies $|f(x)-f(c)| < \eta$ (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).

[L3]

Additivity: for $a \le p \le q \le b$ , $\int_a^b f = \int_a^p f + \int_p^q f + \int_q^b f$ , the degenerate pieces being $0$ (For $a<c<b$ : $f$ is integrable on $[a,b]$ if and only if it is integrable on $[a,c]$ and on $[c,b]$ , and then $\int_a^b f = \int_a^c f + \int_c^b f$ ; with the oriented form for arbitrary $a,b,c$ , claim 3, The integral with oriented limits: $\int_a^a f := 0$ and $\int_b^a f := -\int_a^b f$ ).

[L4]

If $u \ge 0$ is integrable on $[p,q]$ with $p<q$ then $\int_p^q u \ge 0$ ; and if $u \ge \kappa$ there then $\int_p^q u \ge \kappa(q-p)$ (If $f \le g$ on $[a,b]$ and both are integrable then $\int_a^b f \le \int_a^b g$ ; and $m(b-a) \le \int_a^b f \le M(b-a)$ , 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 $\int_a^b c = c(b-a)$ ).

[L5]

Ordered-field arithmetic and minima: the order is total and transitive, $\min\{s,t\}$ and $\max\{s,t\}$ are reals lying appropriately, a product of two positive reals is positive, and adding constants preserves inequalities (Maximum and minimum of a set, Ordered field, Complete ordered field (least-upper-bound property), Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

Proof

technique · contradiction

1.1

Suppose, for contradiction, that $f$ does not vanish identically; since $f \ge 0$ , this gives $c \in [a,b]$ with $f(c) > 0$ , which is [A1].

assume-contragiven

1.2

By [L2] with $\eta := f(c)\cdot 2^{-1} > 0$ , fix a real $\delta > 0$ such that every $x \in [a,b]$ with $|x-c| < \delta$ satisfies $|f(x)-f(c)| < f(c)\cdot 2^{-1}$ , hence $f(x) > f(c)\cdot 2^{-1}$ .

A1L2L5choose

2.1

Put $p := \max\{a,\ c - \delta\cdot 2^{-1}\}$ and $q := \min\{b,\ c + \delta\cdot2^{-1}\}$ . Then $a \le p \le c \le q \le b$ , and $[p,q] \subseteq [a,b]$ .

step 1.2A1L5construct

3.1

$p < q$ : indeed $p \le c \le q$ , and $p = q$ would force $p = c = q$ , hence $c = \max\{a, c-\delta\cdot2^{-1}\}$ and $c = \min\{b, c+\delta\cdot 2^{-1}\}$ , so $c = a$ and $c = b$ , contradicting $a<b$ .

step 2.1L5

3.2

Every $x \in [p,q]$ satisfies $|x - c| \le \delta\cdot 2^{-1} < \delta$ , so $f(x) > f(c)\cdot2^{-1}$ there by step 1.2.

step 1.2step 2.1L5

4.1

Hence $\int_p^q f \ge f(c)\cdot 2^{-1}\,(q-p) > 0$ by [L4] and step 3.1.

step 3.1step 3.2L1L4L5

5.1

By [L3] and [L4], $\int_a^b f = \int_a^p f + \int_p^q f + \int_q^b f \ge \int_p^q f > 0$ , the first and third pieces being $\ge 0$ because $f \ge 0$ there, or $0$ when degenerate.

step 4.1L1L3L4L5

6.1

This contradicts the hypothesis $\int_a^b f = 0$ , so no such $c$ exists and $f(x) = 0$ for every $x \in [a,b]$ .

step 5.1givendischarge-contradiction∎

Remarks

The nonnegativity on the two outer pieces is cited, not assumed away. The usual one-line version writes "so $\int_a^b f \ge \int_p^q f$ " without saying why; what makes that step legitimate is that $f \ge 0$ on $[a,p]$ and on $[q,b]$ too, so both of those integrals are $\ge 0$ (If $f \le g$ on $[a,b]$ and both are integrable then $\int_a^b f \le \int_a^b g$ ; and $m(b-a) \le \int_a^b f \le M(b-a)$ ). Without a sign hypothesis outside $[p,q]$ the argument would fail.
The case where $c$ is an endpoint is covered by the construction, not by a case split. Taking $p$ and $q$ as a maximum and a minimum with $a$ and $b$ makes $[p,q]$ a one-sided neighbourhood of $c$ when $c = a$ or $c = b$ , and step 3.1 is what checks that it is still nondegenerate.
Continuity is used only at the single point $c$ . The proof needs no uniform continuity and no continuity anywhere else, so the statement could be sharpened to: a nonnegative integrable $f$ with $\int_a^b f = 0$ vanishes at every point of continuity. That sharpening is not asserted as a separate clause because nothing on this page uses it.

Depends on

Used by

FALSE: in the substitution theorem the continuity of f may be weakened to integrability, f∘φ still being integrable False statement

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 77 results over 16 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

Riemann integral (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 6 (standard reference, not scraped)
MIT 18.100, problem-set solutions on the Riemann integral (standard reference, not scraped)
J. Lebl, Basic Analysis I, Properties of the Riemann integral (standard reference, not scraped)

A continuous f≥0f \ge 0f≥0 on [a,b][a,b][a,b] with ∫abf=0\int_a^b f = 0∫ab​f=0 is identically 000