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FALSE: a nonnegative Riemann integrable function on [a,b][a,b] with abf=0\int_a^b f = 0 is identically zero

Statement

False claim: if f:[a,b]Rf : [a,b] \to \mathbb{R} is Riemann integrable (The lower and upper Darboux integrals of a bounded ff on [a,b][a,b] as supPL(f,P)\sup_P L(f,P) and infPU(f,P)\inf_P U(f,P), Darboux integrability as their equality, and the notation abf\int_a^b f) with f(x)0f(x) \ge 0 for every x[a,b]x \in [a,b] and abf=0\int_a^b f = 0, then f(x)=0f(x) = 0 for every x[a,b]x \in [a,b].

The claim is true under the additional hypothesis that ff 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][0,1] (The Dirichlet function 1Q1_{\mathbb{Q}}, and Thomae's function tt with t(x)=1/qt(x) = 1/q at a rational x=p/qx = p/q in lowest terms with q1q \ge 1 and t(x)=0t(x) = 0 at every irrational xx) is nonnegative, integrable, has integral 00, and is positive at every rational point of [0,1][0,1], of which there are infinitely many.

Facts & Assumptions

Given: Thomae's function restricted to [0,1][0,1], that is t:[0,1]Rt : [0,1] \to \mathbb{R} with t(x)=1/ι(q(x))t(x) = 1/\iota(q(x)) at a rational xx with least denominator q(x)1q(x) \ge 1, and t(x)=0t(x) = 0 at an irrational xx (The Dirichlet function 1Q1_{\mathbb{Q}}, and Thomae's function tt with t(x)=1/qt(x) = 1/q at a rational x=p/qx = p/q in lowest terms with q1q \ge 1 and t(x)=0t(x) = 0 at every irrational xx, The canonical natural ι(n)=n1F\iota(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 00 vanishes identically.

[L1]

t(x)=1/ι(q(x))t(x) = 1/\iota(q(x)) with ι(q(x))1>0\iota(q(x)) \ge 1 > 0 at a rational xx, and t(x)=0t(x) = 0 at an irrational xx; hence 0t(x)10 \le t(x) \le 1 everywhere and t(x)>0t(x) > 0 at every rational xx (The Dirichlet function 1Q1_{\mathbb{Q}}, and Thomae's function tt with t(x)=1/qt(x) = 1/q at a rational x=p/qx = p/q in lowest terms with q1q \ge 1 and t(x)=0t(x) = 0 at every irrational xx, The canonical natural ι(n)=n1F\iota(n) = n \cdot 1_F of a field, Canonical naturals are positive and strictly increasing).

[L3]

Q\mathbb{Q} is countably infinite, and every subset of an at most countable set is at most countable (Q\mathbb{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][a,b] with an at most countable set of discontinuities is Riemann integrable (A bounded function on [a,b][a,b] whose set of discontinuities is at most countable is Riemann integrable, Lower bound, bounded below, bounded set).

[L6]

For a partition P=(n,tP)P = (n,t^{P}) of [0,1][0,1]: n1n \ge 1, tiP<ti+1Pt^{P}_i < t^{P}_{i+1}, Δi>0\Delta_i > 0, and Ii=[tiP,ti+1P][0,1]I_i = [t^{P}_i, t^{P}_{i+1}] \subseteq [0,1] (Partition of [a,b][a,b] as a finite strictly increasing list a=t0<t1<<tn=ba = 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\mathbb{R}: the nine order-convex forms, nondegeneracy, and length).

[L8]

A set with a least element has it as its infimum; the supremum of {0}\{0\} is 00 (Greatest lower bound (infimum), Maximum and minimum of a set, Complete ordered field (least-upper-bound property)).

[L9]
[L10]

Ordered-field arithmetic: the order is total and transitive; a reciprocal of a positive quantity is positive; 0<10 < 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
1.1

tt is nonnegative and bounded on [0,1][0,1], with 0t(x)10 \le t(x) \le 1 for every xx, by [L1].

givenL1L10
1.2

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

givenL2L3
1.3

Separately, and independently of everything below, tt does not vanish identically: 212^{-1} is a rational point of [0,1][0,1], so t(21)>0t(2^{-1}) > 0 by [L1] and [L10].

L1L10
2.1

By [L4] applied to [0,1][0,1], with 0<10 < 1, the function tt is Riemann integrable on [0,1][0,1].

step 1.1step 1.2L4
2.2

Every lower Darboux sum of tt is 00: let PP be a partition of [0,1][0,1] and i<ni < n; the open interval (tiP,ti+1P)(t^{P}_i, t^{P}_{i+1}) is nonempty by [L6] and contains an irrational yy by [L5], and yIi[0,1]y \in I_i \subseteq [0,1], so t(y)=0t(y) = 0 by [L1]. Since t0t \ge 0 by [L1], the value 00 is the least element of t[Ii]t[I_i] and mi=0m_i = 0 by [L8]. Hence L(t,P)=i<n0Δi=0L(t,P) = \sum_{i<n}0\cdot\Delta_i = 0 by [L7] and [L9].

step 1.1L1L5L6L7L8L9
3.1

The set of lower sums is therefore {0}\{0\}, so 01t=0\underline{\int_0^1} t = 0 by [L8], and since tt is integrable by step 2.1 its integral is 01t=0\int_0^1 t = 0 by [L7].

step 2.1step 2.2L7L8
4.1

So tt is a nonnegative Riemann integrable function on [0,1][0,1] with 01t=0\int_0^1 t = 0 that is not identically zero; [A1] fails at tt and the claim is false.

step 1.1step 3.1step 1.3A1

Remarks

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