LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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.

Changing a bounded integrand on a content-zero set does not change its Riemann integral

Statement

Let $E \subseteq R^{m}$ be bounded and Jordan measurable, let $f, g : E \to R$ be bounded, and suppose ${x \in E : f (x) \neq g (x)}$ has content zero. Then $f$ is Riemann integrable over $E$ if and only if $g$ is, and when they are integrable their integrals are equal.

Facts & Assumptions

Given: The set and functions of the Statement, a nondegenerate bounding rectangle $Q \supseteq E$ , their zero extensions $\tilde{f}, \tilde{g}$ to $Q$ , and $N : = {x \in E : f (x) \neq g (x)}$ .

[L1]

Riemann integrability over $E$ means integrability of the zero extension on $Q$ (The Riemann integral of a bounded function over a bounded Jordan measurable set).

[L2]

A set has content zero when every positive volume allowance admits a finite closed-cube cover within that allowance (Measure zero and content zero in $R^{m}$ by countable and finite cube covers).

[L3]

If a subset of a rectangle is covered by finitely many rectangles of total volume $V$ , then a grid exists whose cells meeting the set have total volume below $V + η$ (A finite rectangle cover admits grid control with arbitrarily small volume excess).

[L4]

A bounded function on a nondegenerate rectangle is Riemann integrable exactly when grids can make its upper-minus-lower sum arbitrarily small (Riemann's criterion on a nondegenerate rectangle in $R^{m}$ : integrability is equivalent to arbitrarily small Darboux gaps).

[L5]

Proper multidimensional Riemann integrals are linear (Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in $R^{m}$ ).

Proof

technique · cases

1.1L1algebra

Put $h : = \tilde{g} - \tilde{f}$ . It is bounded, vanishes on $Q ∖ N$ , and has some bound $∣ h ∣ \leq M$ .

2.1assume-case zerostep 1.1L4

In the case $M = 0$ , the function $h$ is identically zero, hence integrable with integral zero.

2.2assume-case posstep 1.1L2L3L4choose

In the case $M > 0$ , given $ε > 0$ , [L2] covers $N$ by finitely many cubes of total volume below $ε / (4 M)$ , and [L3] gives a grid whose cells meeting $N$ have total volume below $ε / (2 M)$ . On all other cells $h = 0$ , while on a cell meeting $N$ its oscillation is at most $2 M$ , so the total Darboux gap is below $ε$ . Thus [L4] makes $h$ integrable; the bounds $- M 1_{N} \leq h \leq M 1_{N}$ with the same grids force its integral to be zero.

3.1step 2.1step 2.2L5cases-exhaustive

The two cases exhaust $M \geq 0$ , so $h$ is integrable with integral zero. If $f$ is integrable, then $\tilde{g} = \tilde{f} + h$ is integrable and has the same integral by [L5].

4.1step 1.1step 3.1L5∎

Interchanging $f$ and $g$ applies step 3.1 to $- h$ , proving the reverse integrability implication and the same equality of values.

Changing a bounded integrand on a content-zero set does not change its Riemann integral

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources