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Thomae's function is Riemann integrable on [0,1][0,1] with integral 00: it is continuous at every irrational, so its discontinuity set is countable, and every lower Darboux sum is 00

Example

Let t:[0,1]Rt : [0,1] \to \mathbb{R} be Thomae's function restricted to [0,1][0,1]: at a rational xx with least denominator q(x)1q(x) \ge 1 it takes the value 1/ι(q(x))1/\iota(q(x)), and at an irrational xx the value 00 (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). Then tt is Riemann integrable on [0,1][0,1] and

01t  =  0.\int_0^1 t \;=\; 0 .

Two ingredients, and they pull in opposite directions. tt is continuous at every irrational and discontinuous at every rational (The Dirichlet function is continuous at no point of R\mathbb{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 cc equals t(c)t(c)), so its discontinuity set is Q[0,1]\mathbb{Q}\cap[0,1], which is infinite and dense — and countable, which is what A bounded function on [a,b][a,b] whose set of discontinuities is at most countable is Riemann integrable needs. The value is then read off the lower sums, every one of which is 00 because every subinterval contains an irrational (Both Q\mathbb{Q} and RQ\mathbb{R} \setminus \mathbb{Q} are dense in R\mathbb{R}, and every nonempty open subset of R\mathbb{R} is uncountable).

Every upper sum, by contrast, is strictly positive, since every subinterval contains a rational; the upper integral is nevertheless 00, an infimum of positive numbers.

Facts & Assumptions

Given: Thomae's function t:[0,1]Rt : [0,1] \to \mathbb{R} as above.

[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 a<ba < b whose set of discontinuities is at most countable 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,s)P = (n,s) of [0,1][0,1]: n1n \ge 1, si<si+1s_i < s_{i+1}, Δi>0\Delta_i > 0, Ii=[si,si+1][0,1]I_i = [s_i,s_{i+1}] \subseteq [0,1], and (si,si+1)(s_i,s_{i+1}) is a nonempty open interval (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, and a reciprocal of a positive quantity is positive (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.

Verification

technique · direct
1.1

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

givenL1L10
1.2

By [L2], tt is continuous at every irrational point of [0,1][0,1], 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]; a subset of it is then at most countable as well.

givenL2L3
2.1

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

step 1.1step 1.2L4
2.2

Every lower sum is 00. Let P=(n,s)P = (n,s) be a partition of [0,1][0,1] and i<ni < n. By [L6] the interval (si,si+1)(s_i,s_{i+1}) is nonempty and open, so by [L5] it contains an irrational yy, 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)=0L(t,P) = 0 by [L7] and [L9].

step 1.1L1L5L6L7L8L9
3.1

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

step 2.1step 2.2L7L8

Remarks

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