Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

An open dense set of measure less than 1 is the monotone L1-limit of Riemann integrable indicators, but its indicator is not Riemann integrable

Example

Assume the Axiom of Countable Choice. There exist open sets U1U2(0,1) such that, with U:=n1Un,

  1. each 1Un is Riemann integrable on [0,1];
  2. 1Un1U pointwise and [0,1]1U1Undλ10;
  3. U is open and dense with λ1(U)<1;
  4. 1U is not Riemann integrable on [0,1].

Facts & Assumptions

Given: The Axiom of Countable Choice.

[L1]

The rationals are countably infinite, and both the rationals and the irrationals are dense in R. (Q is countably infinite, Both Q and RQ are dense in R, and every nonempty open subset of R is uncountable)

[L2]

Countable subadditivity bounds the measure of a countable union by the sum of the individual measures. (Finite and countable subadditivity of measures)

[L4]

For an increasing sequence of measurable sets (En), λ1 ⁣(nEn)=supnλ1(En). (Continuity from below for measures)

[L5]

A bounded function on [0,1] that is continuous except at finitely many points is Riemann integrable. (A bounded function on [a,b] that is continuous except at finitely many points is Riemann integrable)

[L6]

A bounded function on [0,1] is Riemann integrable exactly when its discontinuity set has Lebesgue measure 0. (A bounded function on a closed bounded interval, or on a closed nondegenerate rectangle, is Riemann integrable exactly when its discontinuity set has Lebesgue measure zero)

[L7]

If AB are measurable with λ1(A)<+, then λ1(BA)=λ1(B)λ1(A). (Measure of a set difference when the smaller set has finite measure)

[L9]

Every Borel subset of R is Lebesgue measurable. (Assuming countable choice, every Borel subset of Rn is Lebesgue measurable)

Verification

technique · direct
1.1

By [L1], fix an enumeration (qk)k0 of Q(0,1). For each k0, let Ik:=(qk2k4,qk+2k4)(0,1), put Un:=k<nIk for n1, and put U:=k0Ik. Each Un and U is open, and every rational point of (0,1) lies in U, so U is dense in (0,1) and hence in [0,1]. Each interval Ik has length at most 2k3, so [L2], [L3], [L8], and [L9] give λ1(U)k=0λ1(Ik)k=02k3=23k=02k=1/4<1.

L1L2L3L8L9construct
2.1

Each Un is a finite union of open intervals, so 1Un is continuous away from the finitely many endpoints of those intervals. Thus [L5] makes every 1Un Riemann integrable on [0,1]. Also UnU, so 1Un1U pointwise and 1U1Un=1UUn. By [L4] and [L7], [0,1]1U1Undλ1=λ1(UUn)=λ1(U)λ1(Un)0.

step 1.1L4L5L7
3.1

Let C:=[0,1]U. Step 1.1 and [L8] give λ1(C)=λ1([0,1])λ1(U)=1λ1(U)>0. Because U is open, every point of U is a continuity point of 1U. Because U is dense, every point of C is a boundary point of U, hence every neighbourhood of such a point meets both U and C; so 1U is discontinuous at every point of C. Therefore the discontinuity set of 1U contains the positive-measure set C, and [L6] shows that 1U is not Riemann integrable on [0,1].

step 1.1step 2.1L6L7L8

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

94 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources