LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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 an integrable function at finitely many points changes neither its integrability nor its integral

Statement

Let $a < b$ be reals, let $f : [a, b] \to R$ be integrable (The lower and upper Darboux integrals of a bounded $f$ on $[a, b]$ as $\sup_{P} L (f, P)$ and $\inf_{P} U (f, P)$ , Darboux integrability as their equality, and the notation $\int_{a}^{b} f$ ), let $E \subseteq [a, b]$ be finite (Finite, countably infinite, countable, uncountable, Equinumerous sets, $A \approx B$ and $A ⪯ B$ ), and let $g : [a, b] \to R$ satisfy

$g (x) = f (x) for every x \in [a, b] ∖ E .$

Then $g$ is integrable on $[a, b]$ and

$\int_{a}^{b} g = \int_{a}^{b} f .$

In particular the values of an integrand at the endpoints of the interval, and at any finite set of points, are irrelevant to both questions.

Facts & Assumptions

Given: Reals $a < b$ , an integrable $f : [a, b] \to R$ , a finite $E \subseteq [a, b]$ , and $g : [a, b] \to R$ agreeing with $f$ off $E$ . Finite means: there are $r \in N$ and a bijection $e$ from ${j \in N : j < r}$ onto $E$ (Finite, countably infinite, countable, uncountable, Equinumerous sets, $A \approx B$ and $A ⪯ B$ ).

[L1]

Riemann's criterion (Riemann's criterion: a bounded $f$ on $[a, b]$ is Darboux integrable if and only if for every real $ε > 0$ there is a partition $P$ with $U (f, P) - L (f, P) < ε$ ), and: an integrable function is bounded, since Darboux sums are defined only for bounded functions (For bounded $f$ on $[a, b]$ and a partition $P$ : the infimum $m_{i}$ and supremum $M_{i}$ of $f$ on the $i$ -th subinterval, and the lower and upper Darboux sums $L (f, P) = \sum_{i} m_{i} Δ_{i}$ and $U (f, P) = \sum_{i} M_{i} Δ_{i}$ , Lower bound, bounded below, bounded set).

[L2]

For a partition $R = (n, t)$ and a bounded function $h$ on the interval: $L (h, R) = \sum_{i < n} m_{i} Δ_{i}$ , $U (h, R) = \sum_{i < n} M_{i} Δ_{i}$ with $m_{i} = \inf h [I_{i}] \leq M_{i} = \sup h [I_{i}]$ , and $L (h, R) \leq \underline{\int_{a}^{b}} h \leq \overline{\int_{a}^{b}} h \leq U (h, R)$ (For bounded $f$ on $[a, b]$ and a partition $P$ : the infimum $m_{i}$ and supremum $M_{i}$ of $f$ on the $i$ -th subinterval, and the lower and upper Darboux sums $L (f, P) = \sum_{i} m_{i} Δ_{i}$ and $U (f, P) = \sum_{i} M_{i} Δ_{i}$ , The lower and upper Darboux integrals of a bounded $f$ on $[a, b]$ as $\sup_{P} L (f, P)$ and $\inf_{P} U (f, P)$ , Darboux integrability as their equality, and the notation $\int_{a}^{b} f$ ).

[L3]

The uniform partition $U_{N}$ of $[a, b]$ into $N \geq 1$ parts has $n = N$ and every $Δ_{i}$ equal to $(b - a) / ι (N)$ , and its subintervals cover $[a, b]$ ; the index list $t$ is strictly increasing on indices $\leq N$ , hence injective there (Partition of $[a, b]$ as a finite strictly increasing list $a = t_{0} < t_{1} < \dots < t_{n} = b$ , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions, The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L4]

Finite sums: monotonicity in the terms, scaling, additivity, and splitting; consequently, if $x_{i} = 0$ for every $i < n$ except $i = j$ , then $\sum_{i < n} x_{i} = x_{j}$ , by splitting at $j$ and at $j + 1$ and $\sum 0 = 0$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products, clauses 1 to 4).

[L5]

Sums and scalar multiples of integrable functions are integrable, with the corresponding identity for the integrals (Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ ); and the constant function $0$ is integrable with integral $0$ (If $m \leq f \leq M$ on $[a, b]$ then $m (b - a) \leq L (f, P) \leq \underline{\int_{a}^{b}} f \leq \overline{\int_{a}^{b}} f \leq U (f, P) \leq M (b - a)$ for every partition $P$ ; in particular every constant function is integrable, with $\int_{a}^{b} c = c (b - a)$ ).

[L6]

For every real $η > 0$ there is a natural $N \geq 1$ with $1 / ι (N) < η$ , and $ι (N) > 0$ for $N \geq 1$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Every complete ordered field is Archimedean, Canonical naturals are positive and strictly increasing).

[L7]

Induction on $N$ (The principle of mathematical induction).

[L8]

Ordered-field arithmetic: multiplying an inequality by a nonnegative quantity and adding constants preserve it, the order is total and transitive, and a real of absolute value below every positive real is $0$ (Ordered field, Complete ordered field (least-upper-bound property)).

Proof

technique · induction

1.1givenL1

The one-point case is proved first, for a function called $w$ so that no symbol is reused. Let $p \in [a, b]$ and let $w : [a, b] \to R$ satisfy $w (x) = 0$ for every $x \neq p$ ; put $K : = ∣ w (p) ∣$ , so $∣ w (x) ∣ \leq K$ for every $x$ and $w$ is bounded.

1.2L3construct

Fix $N \geq 1$ and write $U_{N} = (N, t)$ with subintervals $I_{i}$ and lengths $Δ_{i} = (b - a) / ι (N)$ . Define $θ_{i} : = 1$ if $p \in I_{i}$ and $θ_{i} : = 0$ otherwise, for $i < N$ .

1.3givenconstruct

Setting up the induction. Put $h : = g - f$ , so that $h (x) = 0$ for every $x \notin E$ , and for $j < r$ define $h_{j} : [a, b] \to R$ by $h_{j} (x) : = h (x)$ if $x = e (j)$ and $h_{j} (x) : = 0$ otherwise. Each $h_{j}$ vanishes off the single point $e (j)$ .

2.1step 1.2L3choose

At most two indices $i < N$ have $θ_{i} = 1$ , and they are consecutive: if $p \in I_{i} \cap I_{j}$ with $i < j$ then $t_{j} \leq p \leq t_{i + 1}$ and $t_{i + 1} \leq t_{j}$ , so $t_{i + 1} = t_{j}$ and $j = i + 1$ by injectivity of $t$ . Also some index has $θ_{i} = 1$ , since the subintervals cover $[a, b]$ ; let $i_{0}$ be the least such.

2.2step 1.3L4

For every $x \in [a, b]$ and every $k \leq r$ , $\sum_{j < k} h_{j} (x) = h (x)$ when $x = e (j_{0})$ for some $j_{0} < k$ , and $\sum_{j < k} h_{j} (x) = 0$ otherwise: in the first case all terms with $j \neq j_{0}$ vanish, because $e$ is injective, and [L4] evaluates the sum; in the second every term is $0$ .

3.1step 1.2step 2.1L2L8

For every $i < N$ : if $θ_{i} = 0$ then $w$ vanishes on $I_{i}$ , so $m_{i} = M_{i} = 0$ , where $m_{i}$ and $M_{i}$ are the extreme values of $w$ on $I_{i}$ ; and always $- K θ_{i} \leq m_{i} \leq M_{i} \leq K θ_{i}$ .

3.2step 1.2step 2.1L3L4construct

Define $y_{i} : = Δ_{i}$ for $i = i_{0}$ and $y_{i} : = 0$ otherwise, and $z_{i} : = Δ_{i}$ for $i = i_{0} + 1$ with $i_{0} + 1 < N$ , and $z_{i} : = 0$ otherwise. Then $θ_{i} Δ_{i} \leq y_{i} + z_{i}$ for every $i < N$ by step 2.1, and $\sum_{i < N} (y_{i} + z_{i}) \leq 2 (b - a) / ι (N)$ by [L4] and [L3].

3.3step 2.2construct

Let $Q (k)$ , for $k \leq r$ , be the statement that the function $H_{k} : x \mapsto \sum_{j < k} h_{j} (x)$ is integrable on $[a, b]$ with $\int_{a}^{b} H_{k} = 0$ .

4.1step 3.1step 3.2L2L4L8

By step 3.1 and monotonicity of finite sums, $U (w, U_{N}) - L (w, U_{N}) = \sum_{i < N} (M_{i} - m_{i}) Δ_{i} \leq \sum_{i < N} 2 K θ_{i} Δ_{i} \leq 4 K (b - a) / ι (N)$ , and likewise $- 2 K (b - a) / ι (N) \leq L (w, U_{N})$ and $U (w, U_{N}) \leq 2 K (b - a) / ι (N)$ .

4.2basestep 3.3L5

Base. $H_{0}$ is the constant function $0$ , integrable with integral $0$ by [L5], so $Q (0)$ holds.

4.3ihstep 3.3

Induction hypothesis. Fix $k < r$ and assume $Q (k)$ .

5.1step 4.1L1L6L8

Given a real $η > 0$ , [L6] supplies $N$ with $4 K (b - a) / ι (N) < η$ , so $w$ satisfies Riemann's criterion and is integrable by [L1].

5.2step 4.1L2L6L8

Moreover $∣ \int_{a}^{b} w ∣ \leq 2 K (b - a) / ι (N)$ for every $N \geq 1$ by step 4.1 and [L2], and the right-hand side is below every positive real by [L6]; hence $\int_{a}^{b} w = 0$ . Steps 1.1 to 5.2 therefore prove: every function on $[a, b]$ vanishing off a single point is integrable with integral $0$ .

6.1step 5.1step 5.2step 1.3step 4.3L4L5

$H_{k + 1} = H_{k} + h_{k}$ pointwise by [L4], and $h_{k}$ is integrable with integral $0$ by steps 5.1 and 5.2 applied to $w : = h_{k}$ and $p : = e (k)$ ; so $H_{k + 1}$ is integrable with $\int_{a}^{b} H_{k + 1} = 0 + 0 = 0$ by [L5], which is $Q (k + 1)$ .

7.1step 2.2step 4.2step 4.3step 6.1L7

By [L7] with steps 4.2 and 6.1, $Q (k)$ holds for every $k \leq r$ ; at $k = r$ , and by step 2.2, $H_{r} = h$ , so $h$ is integrable with $\int_{a}^{b} h = 0$ .

8.1step 7.1L5discharge-induction∎

Hence $g = f + h$ is integrable with $\int_{a}^{b} g = \int_{a}^{b} f + 0 = \int_{a}^{b} f$ by [L5].

Remarks

The subtle point is that a partition point lies in two subintervals. The subintervals of a partition are closed and overlap at their shared endpoints (Partition of $[a, b]$ as a finite strictly increasing list $a = t_{0} < t_{1} < \dots < t_{n} = b$ , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions), so the exceptional point $p$ may belong to two of them; step 2.1 proves that it belongs to at most two and that they are consecutive, and step 3.2 is what pays for that possibility, with the factor $2$ that survives into step 4.1.
Nothing here is an index-subset sum. The indicator sequence $θ$ of step 1.2 is a genuine sequence on ${i : i < N}$ , and every estimate is applied term by term and then summed by the monotonicity clause of Laws of finite sums and finite products, whose clauses are stated for $\sum_{i < n}$ ; no sum over a subset of the index range is used.
The contrast with the derivative is the point of the lemma. Changing a function at one point can destroy differentiability at that point, and changes nothing at any other point, since a limit at $x \neq p$ may be taken over a neighbourhood excluding $p$ ; it changes no integral at all. This is also why the integral function of an integrable $f$ cannot detect a change of $f$ at a point, which is what FALSE: for every integrable $f$ on $[a, b]$ , the integral function $F (x) = \int_{a}^{x} f$ satisfies $F^{'} = f$ on $[a, b]$ ↗ on the companion page turns into a refutation.

Changing an integrable function at finitely many points changes neither its integrability nor its integral

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources