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FALSE: every bounded function on [a,b][a,b] is Riemann integrable

Statement

False claim: every bounded function f:[a,b]Rf : [a,b] \to \mathbb{R} (Lower bound, bounded below, bounded set) is Riemann integrable on [a,b][a,b] (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).

Boundedness is exactly what is needed for the Darboux sums to exist (For bounded ff on [a,b][a,b] and a partition PP: the infimum mim_i and supremum MiM_i of ff on the ii-th subinterval, and the lower and upper Darboux sums L(f,P)=imiΔiL(f,P) = \sum_i m_i \Delta_i and U(f,P)=iMiΔiU(f,P) = \sum_i M_i \Delta_i), and the claim above confuses that with their two extrema agreeing. The witness below is bounded, takes only the values 00 and 11, and has lower Darboux integral 00 and upper Darboux integral 11: as far apart as the values allow.

Facts & Assumptions

Given: The Dirichlet function 1Q\mathbf{1}_{\mathbb{Q}} restricted to [0,1][0,1], that is g:[0,1]Rg : [0,1] \to \mathbb{R} with g(x)=1g(x) = 1 for rational xx and g(x)=0g(x) = 0 for 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).

[A1]

The false claim: every bounded function on a closed bounded interval with distinct endpoints is Riemann integrable.

[L2]

For a partition P=(n,t)P = (n,t) of [0,1][0,1]: n1n \ge 1, ti<ti+1t_i < t_{i+1}, Δi=ti+1ti>0\Delta_i = t_{i+1}-t_i > 0, i<nΔi=10=1\sum_{i<n}\Delta_i = 1 - 0 = 1, and Ii=[ti,ti+1]I_i = [t_i,t_{i+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).

[L3]

mi=infg[Ii]m_i = \inf g[I_i], Mi=supg[Ii]M_i = \sup g[I_i], L(g,P)=i<nmiΔiL(g,P) = \sum_{i<n}m_i\Delta_i, U(g,P)=i<nMiΔiU(g,P) = \sum_{i<n}M_i\Delta_i; 01g\underline{\int_0^1} g is the supremum of the lower sums and 01g\overline{\int_0^1} g the infimum of the upper sums; gg is integrable exactly when these agree (For bounded ff on [a,b][a,b] and a partition PP: the infimum mim_i and supremum MiM_i of ff on the ii-th subinterval, and the lower and upper Darboux sums L(f,P)=imiΔiL(f,P) = \sum_i m_i \Delta_i and U(f,P)=iMiΔiU(f,P) = \sum_i M_i \Delta_i, 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).

[L4]

A set with a least element has it as its infimum, and one with a greatest element has it as its supremum; the supremum and infimum of {c}\{c\} are both cc (Greatest lower bound (infimum), Maximum and minimum of a set, Complete ordered field (least-upper-bound property)).

[L5]

Finite sums: scaling and i<n0=0\sum_{i<n}0 = 0 (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L6]

Ordered-field arithmetic: the midpoint (u+v)21(u+v)\cdot 2^{-1} of u<vu < v satisfies u<(u+v)21<vu < (u+v)\cdot 2^{-1} < v, and Nρ(c)(u,v)N_\rho(c) \subseteq (u,v) for cc the midpoint and ρ:=(vu)21\rho := (v-u)\cdot 2^{-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), The ε\varepsilon-neighbourhood and the punctured ε\varepsilon-neighbourhood of a point of R\mathbb{R}). 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

gg is bounded: 0g(x)10 \le g(x) \le 1 for every x[0,1]x \in [0,1], since gg takes only the values 00 and 11. So its Darboux sums and integrals are defined by [L3].

givenL3
1.2

Every subinterval of every partition of [0,1][0,1] contains both a rational and an irrational. Let P=(n,t)P = (n,t) be a partition and i<ni < n; then ti<ti+1t_i < t_{i+1} by [L2], so with c:=(ti+ti+1)21c := (t_i + t_{i+1})\cdot 2^{-1} and ρ:=(ti+1ti)21\rho := (t_{i+1}-t_i)\cdot 2^{-1} one has Nρ(c)(ti,ti+1)IiN_\rho(c) \subseteq (t_i,t_{i+1}) \subseteq I_i by [L6], and Nρ(c)N_\rho(c) meets Q\mathbb{Q} and meets RQ\mathbb{R}\setminus\mathbb{Q} by [L1].

L1L2L6
2.1

Hence g[Ii]={0,1}g[I_i] = \{0,1\} for every i<ni < n, so mi=0m_i = 0 and Mi=1M_i = 1 by [L4].

step 1.2givenL4
3.1

Therefore L(g,P)=i<n0Δi=0L(g,P) = \sum_{i<n}0\cdot\Delta_i = 0 and U(g,P)=i<n1Δi=i<nΔi=1U(g,P) = \sum_{i<n}1\cdot\Delta_i = \sum_{i<n}\Delta_i = 1, for every partition PP of [0,1][0,1], by [L3], [L5] and [L2].

step 2.1L2L3L5
4.1

The set of lower sums is {0}\{0\} and the set of upper sums is {1}\{1\}, so 01g=0\underline{\int_0^1} g = 0 and 01g=1\overline{\int_0^1} g = 1 by [L4] and [L3]. Since 010 \ne 1, gg is not Riemann integrable on [0,1][0,1].

step 3.1L3L4
5.1

gg is a bounded function on [0,1][0,1], an interval with 0<10 < 1, and it is not Riemann integrable; so [A1] fails at gg and the claim is false.

step 1.1step 4.1A1

Remarks

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