ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-09-05

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.

The function x plus summable rational jumps decomposes as its continuous part x and its jump part

Example

Let $(q_{n})_{n \geq 1}$ enumerate $Q \cap (0, 1]$ without repetitions, and define, for $x \in [0, 1]$ ,

F (x) : = x + \sum_{q_{n} \leq x} 2^{- n} .

Then the continuous part of $F : [0, 1] \to R$ is $x$ , and the jump part is $x \mapsto \sum_{q_{n} \leq x} 2^{- n}$ .

Facts & Assumptions

Given: An enumeration $(q_{n})$ without repetitions of $Q \cap (0, 1]$ and the function $F : [0, 1] \to R$ above.

[A1]

The symbols are those of the statement.

Verification

technique · direct

1.1given

Each summand $x \mapsto 1_{{q_{n} \leq x}} 2^{- n}$ is nondecreasing, and the geometric tail $\sum_{n > N} 2^{- n}$ tends to $0$ . Consequently $J (x) : = \sum_{q_{n} \leq x} 2^{- n}$ is well defined and nondecreasing. Since $x \mapsto x$ is continuous and increasing, $F = x + J$ is nondecreasing.

2.1step 1.1

Fix $N$ . Away from the finite set ${q_{1}, \dots, q_{N}}$ , the first $N$ summands defining $J$ are locally constant, while the remaining summands have total size at most $\sum_{n > N} 2^{- n}$ . Letting $N \to \infty$ shows that $J$ is continuous at every point outside the enumeration. At $q_{m}$ , take $N \geq m$ : the same tail estimate shows that the left limit differs from $J (q_{m})$ by exactly $2^{- m}$ and that the right limit equals $J (q_{m})$ . It also shows that $J (x) \to 0 = J (0)$ as $x ↓ 0$ . Thus $J$ has no endpoint defect at $0$ , has an interior left jump of size $2^{- n}$ at each $q_{n} \in (0, 1)$ , has the left jump $2^{- n}$ at the unique $q_{n} = 1$ , and has no right jumps.

3.1step 1.1step 2.1

The continuous summand $x \mapsto x$ does not change these jump data. Claim 2 of A nondecreasing function splits uniquely into a jump part and a continuous part therefore computes the jump function of $F$ on $[0, 1]$ as precisely $J$ . The continuous remainder is $F - J = x$ .

4.1step 3.1∎

This is the claimed decomposition.

Depends on

Used by

Nothing in the library uses this result yet.

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