False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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

Statement

Throughout, $N \subseteq Z \subseteq Q \subseteq R$ denotes the canonical copies. Natural numbers, integers and rationals are written directly when used in real arithmetic.

False claim: if $f : [a, b] \to R$ is Riemann 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 $f (x) \geq 0$ for every $x \in [a, b]$ and $\int_{a}^{b} f = 0$ , then $f (x) = 0$ for every $x \in [a, b]$ .

The claim is true under the additional hypothesis that $f$ is continuous, and that is the version worth remembering; without it the integral simply cannot see a function that is positive on a null set. Thomae's function on $[0, 1]$ (The Dirichlet function $1_{Q}$ , and Thomae's function $t$ with $t (x) = 1 / q$ at a rational $x = p / q$ in lowest terms with $q \geq 1$ and $t (x) = 0$ at every irrational $x$ ) is nonnegative, integrable, has integral $0$ , and is positive at every rational point of $[0, 1]$ , of which there are infinitely many.

Facts & Assumptions

Given: Thomae's function restricted to $[0, 1]$ , that is $t : [0, 1] \to R$ with $t (x) = 1 / q (x)$ at a rational $x$ with least denominator $q (x) \geq 1$ , and $t (x) = 0$ at an irrational $x$ (The Dirichlet function $1_{Q}$ , and Thomae's function $t$ with $t (x) = 1 / q$ at a rational $x = p / q$ in lowest terms with $q \geq 1$ and $t (x) = 0$ at every irrational $x$ , The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[A1]

The false claim: a nonnegative Riemann integrable function on a closed bounded interval with distinct endpoints whose integral is $0$ vanishes identically.

[L1]

$t (x) = 1 / q (x)$ with $q (x) \geq 1 > 0$ at a rational $x$ , and $t (x) = 0$ at an irrational $x$ ; hence $0 \leq t (x) \leq 1$ everywhere and $t (x) > 0$ at every rational $x$ (The Dirichlet function $1_{Q}$ , and Thomae's function $t$ with $t (x) = 1 / q$ at a rational $x = p / q$ in lowest terms with $q \geq 1$ and $t (x) = 0$ at every irrational $x$ , The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing).

[L2]

Thomae's function on $R$ is continuous at every irrational and discontinuous at every rational (The Dirichlet function is continuous at no point of $R$ , and Thomae's function is continuous at every irrational and at no rational, so its set of continuity points is exactly the set of irrationals and its oscillation at $c$ equals $t (c)$ ); a restriction of a function continuous at a point of the smaller domain is continuous there, the same $δ$ serving a condition quantified over fewer points (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).

[L3]

$Q$ is countably infinite, and every subset of an at most countable set is at most countable ( $Q$ is countably infinite, Every subset of an at most countable set is at most countable, Finite, countably infinite, countable, uncountable).

[L4]

A bounded function on $[a, b]$ with an at most countable set of discontinuities is Riemann integrable (A bounded function on $[a, b]$ whose set of discontinuities is at most countable is Riemann integrable, Lower bound, bounded below, bounded set).

[L5]

The irrationals are dense in $R$ , so every nonempty open interval contains an irrational (Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable, The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points, Interior, closure, boundary and exterior of a subset of $R$ , The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ).

[L6]

For a partition $P = (n, t^{P})$ of $[0, 1]$ : $n \geq 1$ , $t_{i}^{P} < t_{i + 1}^{P}$ , $Δ_{i} > 0$ , and $I_{i} = [t_{i}^{P}, t_{i + 1}^{P}] \subseteq [0, 1]$ (Partition of $[a, b]$ as a finite strictly increasing list $a = t_{0} < t_{1} < \dots < t_{n} = b$ , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L7]

$L (t, P) = \sum_{i < n} m_{i} Δ_{i}$ with $m_{i} = \inf t [I_{i}]$ ; $\underline{\int_{0}^{1}} t$ is the supremum of the lower sums, $\overline{\int_{0}^{1}} t$ the infimum of the upper sums, and the integral is their common value when they agree (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 $U (f, P) = \sum_{i} M_{i} Δ_{i}$ , 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 $\int_{a}^{b} f$ ).

[L8]

A set with a least element has it as its infimum; the supremum of ${0}$ is $0$ (Greatest lower bound (infimum), Maximum and minimum of a set, Complete ordered field (least-upper-bound property)).

[L9]

Finite sums: $\sum_{i < n} 0 = 0$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L10]

Ordered-field arithmetic: the order is total and transitive; a reciprocal of a positive quantity is positive; $0 < 1$ (Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property)). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

Refutation

technique · direct

1.1givenL1L10

$t$ is nonnegative and bounded on $[0, 1]$ , with $0 \leq t (x) \leq 1$ for every $x$ , by [L1].

1.2givenL2L3

$t$ is continuous at every irrational point of $[0, 1]$ by [L2], so its set of discontinuities in $[0, 1]$ is contained in $Q \cap [0, 1]$ , which is at most countable by [L3]; every subset of it is then at most countable by [L3] as well.

1.3L1L10

Separately, and independently of everything below, $t$ does not vanish identically: $2^{- 1}$ is a rational point of $[0, 1]$ , so $t (2^{- 1}) > 0$ by [L1] and [L10].

2.1step 1.1step 1.2L4

By [L4] applied to $[0, 1]$ , with $0 < 1$ , the function $t$ is Riemann integrable on $[0, 1]$ .

2.2step 1.1L1L5L6L7L8L9

Every lower Darboux sum of $t$ is $0$ : let $P$ be a partition of $[0, 1]$ and $i < n$ ; the open interval $(t_{i}^{P}, t_{i + 1}^{P})$ is nonempty by [L6] and contains an irrational $y$ by [L5], and $y \in I_{i} \subseteq [0, 1]$ , so $t (y) = 0$ by [L1]. Since $t \geq 0$ by [L1], the value $0$ is the least element of $t [I_{i}]$ and $m_{i} = 0$ by [L8]. Hence $L (t, P) = \sum_{i < n} 0 \cdot Δ_{i} = 0$ by [L7] and [L9].

3.1step 2.1step 2.2L7L8

The set of lower sums is therefore ${0}$ , so $\underline{\int_{0}^{1}} t = 0$ by [L8], and since $t$ is integrable by step 2.1 its integral is $\int_{0}^{1} t = 0$ by [L7].

4.1step 1.1step 3.1step 1.3A1∎

So $t$ is a nonnegative Riemann integrable function on $[0, 1]$ with $\int_{0}^{1} t = 0$ that is not identically zero; [A1] fails at $t$ and the claim is false.

Remarks

The correct statement, and the hypothesis that repairs it. If $f$ is continuous on $[a, b]$ , nonnegative and $\int_{a}^{b} f = 0$ , then $f \equiv 0$ : a point with $f (c) > 0$ would, by continuity, force $f > f (c) \cdot 2^{- 1}$ on a whole subinterval, and $\int_{a}^{b} f$ would then be at least $f (c) \cdot 2^{- 1}$ times the length of that subinterval, by 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 $\int_{a}^{b} c = c (b - a)$ applied there. That argument is not carried out here, because the additivity of the integral over subintervals is not available at this point in the reading order; what is asserted by this item is only that the hypothesis of continuity cannot simply be dropped.
The exceptional set is exactly the rationals. $t$ is positive precisely on $Q \cap [0, 1]$ , a countable dense set, which is null (Every at most countable subset of $R$ has measure zero). The integral is blind to a null set of positive values, and the Lebesgue criterion explains why: what governs integrability, and here the value, is the size of the set where the function misbehaves, measured by interval covers (Lebesgue's criterion for Riemann integrability: a bounded $f$ on $[a, b]$ is Riemann integrable if and only if its set of discontinuities has measure zero).
Every upper sum, by contrast, is positive. Each subinterval contains a rational, so $M_{i} > 0$ for every $i$ and $U (t, P) > 0$ for every partition $P$ ; the upper integral is nevertheless $0$ , since the infimum of a set of positive numbers may be $0$ . That is the one place where this example is worth pausing over, and it is the reason the lower sums, not the upper ones, are what step 2.2 computes.

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

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources

FALSE: a nonnegative Riemann integrable function on [a,b][a,b] with ∫abf=0∫ab​f=0 is identically zero

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources

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