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.

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✓ 13 results · all verified · 7 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 6 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Properties of the Integral and the Working FTC: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29Open item page →

$\int_{0}^{1} x^{m} = 1 / ι (m + 1)$ , computed by the fundamental theorem and checked against the definition

Example

Let $m \in N$ and let $p_{m} : [0, 1] \to R$ be $p_{m} (x) = x^{m}$ (Integer powers $a^{m}$ ). Then $p_{m}$ is integrable on $[0, 1]$ and

$\int_{0}^{1} x^{m} d x = \frac{1}{ι (m + 1)},$

where $ι (m + 1)$ is the canonical natural of $m + 1$ in $R$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field) and is positive because $m + 1 \geq 1$ (Canonical naturals are positive and strictly increasing).

The $ι$ is not decoration. A natural number is a von Neumann natural, that is a set, so $m + 1$ is not an element of $R$ and $1 / (m + 1)$ is not an expression of the field; what the display says is $1 / ι (m + 1)$ , and that is why the reader meets $ι$ here at all.

Two independent checks are carried out below: the value at $m = 0$ against the published formula for the integral of a constant (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 $\int_{a}^{b} c = c (b - a)$ ), and the monotonicity of the answer in $m$ against the pointwise inequality $x^{m + 1} \leq x^{m}$ on $[0, 1]$ .

Facts & Assumptions

Given: A natural number $m$ and the function $p_{m} (x) = x^{m}$ on $[0, 1]$ .

[L1]

For $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable at every real $c$ with derivative $ι (n) c^{n - 1}$ (For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, claim 2); the constant $x \mapsto x^{0} = 1$ has derivative $0$ (claim 1).

[L2]

A scalar multiple of a differentiable function is differentiable with the scaled derivative (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , claim 2).

[L3]

Every polynomial function is continuous on every subset of $R$ (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, claim 5, Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point), and a continuous function on $[0, 1]$ is integrable there (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, 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 $\int_{a}^{b} f$ ).

[L4]

If $H$ is differentiable at every point of $[p, q]$ with $H^{'}$ integrable there, then $\int_{p}^{q} H^{'} = H (q) - H (p)$ (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ ); a continuous function on an interval has a primitive (Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ ).

[L5]

Powers: $1^{n} = 1$ for every $n$ , $0^{n} = 0$ for $n \geq 1$ , $0^{0} = 1$ , and $0 \leq x \leq 1$ gives $0 \leq x^{n + 1} \leq x^{n} \leq 1$ (Integer powers $a^{m}$ , Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , claims 1 and 3).

[L6]

$ι (0) = 0$ , $ι (n + 1) = ι (n) + 1$ , $ι (n) > 0$ for $n \geq 1$ , and $ι$ is increasing on the naturals $\geq 1$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing).

[L7]

$\int_{p}^{q} c = c (q - p)$ for a constant $c$ , and if $u \leq w$ pointwise with both integrable then $\int_{p}^{q} u \leq \int_{p}^{q} w$ (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 $\int_{a}^{b} c = c (b - a)$ , If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ ).

[L8]

Ordered-field arithmetic: a positive real has a positive inverse, and $s \leq t$ with $s, t > 0$ gives $1 / t \leq 1 / s$ (Ordered field, Complete ordered field (least-upper-bound property), Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ , The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

Verification

technique · direct

1.1

$p_{m}$ is continuous on $[0, 1]$ , hence integrable there, by [L3].

1.2

Define $H : [0, 1] \to R$ by $H (x) : = x^{m + 1} / ι (m + 1)$ ; this is legitimate because $ι (m + 1) > 0$ by [L6].

L6L8construct

2.1

By [L1] with $n : = m + 1 \geq 1$ and [L2], $H$ is differentiable at every point of $[0, 1]$ with $H^{'} (x) = ι (m + 1) x^{m} / ι (m + 1) = x^{m} = p_{m} (x)$ .

step 1.2L1L2L6

3.1

By [L4] applied to $H$ on $[0, 1]$ , whose derivative $p_{m}$ is integrable by step 1.1, $\int_{0}^{1} x^{m} = H (1) - H (0)$ .

step 1.1step 2.1L4

4.1

By [L5], $H (1) = 1^{m + 1} / ι (m + 1) = 1 / ι (m + 1)$ and $H (0) = 0^{m + 1} / ι (m + 1) = 0$ , since $m + 1 \geq 1$ ; so $\int_{0}^{1} x^{m} = 1 / ι (m + 1)$ .

step 1.2step 3.1L5L6

5.1

First check, at $m = 0$ . There $p_{0}$ is the constant function $1$ by [L5], so $\int_{0}^{1} p_{0} = 1 \cdot (1 - 0) = 1$ by [L7], while the formula gives $1 / ι (1) = 1 / 1 = 1$ . The two agree.

step 4.1L5L6L7

6.1

Second check, monotonicity in $m$ . By [L5], $x^{m + 1} \leq x^{m}$ for every $x \in [0, 1]$ , so $\int_{0}^{1} x^{m + 1} \leq \int_{0}^{1} x^{m}$ by [L7]; the formula gives $1 / ι (m + 2) \leq 1 / ι (m + 1)$ , which holds by [L6] and [L8]. The two agree.

step 4.1L5L6L7L8∎

Remarks

The primitive is the only input, and the index range inside it is the trap. For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term states its power rule for $n \geq 1$ only, because at $n = 0$ the formula $ι (0) c^{- 1}$ mentions $c^{- 1}$ , which is undefined at $c = 0$ . The exponent used above is $m + 1$ , which is $\geq 1$ for every $m \in N$ including $m = 0$ , so claim 2 of that lemma applies with no case split and the example is correct at $m = 0$ as well.
The same computation from the definition is a different exercise. The previous page's companion evaluates $\int_{0}^{1} x^{2}$ from Darboux sums on uniform partitions and a closed form for $\sum_{k < n} ι (k)^{2}$ ; that is the value $1 / 3$ , which is $1 / ι (3)$ and agrees with the formula above at $m = 2$ . The point of computing by the fundamental theorem is that no closed form for a power sum is needed at any $m$ .
Linearity extends this to every polynomial. For $P (x) = \sum_{j < n} a_{j} x^{j}$ one gets $\int_{0}^{1} P = \sum_{j < n} a_{j} / ι (j + 1)$ from Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ ; that is a routine consequence and is not stated as a separate claim.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29Open item page →

A step function integrated by additivity over subintervals, and the same value from the definition

Example

Let $f : [0, 2] \to R$ be

$f (x) : = {\begin{cases} 1 & 0 \leq x < 1, \\ 3 & 1 \leq x \leq 2. \end{cases}$

Then $f$ is integrable on $[0, 2]$ and

$\int_{0}^{2} f = 1 + 3 = 4.$

Two routes are carried out below. The first splits at the jump with For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ and evaluates each piece with Changing an integrable function at finitely many points changes neither its integrability nor its integral and the constant case of 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 $\int_{a}^{b} c = c (b - a)$ . The second computes lower and upper Darboux sums directly on a three-point partition and squeezes.

Why this is the smallest interesting example. The jump sits at the interior point $1$ , which is a partition point of both routes, and a partition point lies in two subintervals (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). That is exactly the case singled out in the proof of Changing an integrable function at finitely many points changes neither its integrability nor its integral, and it is why the value of $f$ at the single point $1$ is irrelevant to both the integrability and the value while the upper sums still see it.

Facts & Assumptions

Given: The function $f$ above on $[0, 2]$ , and a real $η$ with $0 < η < 1$ .

[L1]

A monotone function on a closed bounded interval with distinct endpoints is bounded and integrable there (A monotone function on $[a, b]$ is Riemann integrable: for the uniform partition into $N$ parts the upper minus lower sum telescopes to $∣ f (b) - f (a) ∣ (b - a) / ι (N)$ , Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $R$ , with the dictionary to monotone sequences, 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 $\int_{a}^{b} f$ ).

[L2]

For $0 < c < 2$ : $f$ is integrable on $[0, 2]$ if and only if it is integrable on $[0, c]$ and on $[c, 2]$ , and then $\int_{0}^{2} f = \int_{0}^{c} f + \int_{c}^{2} f$ (For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ , The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ ).

[L3]

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

[L4]

$\int_{p}^{q} c = c (q - p)$ for a constant $c$ , and the five-term chain $L (u, P) \leq \int_{‾} u \leq \int^{‾} u \leq U (u, P)$ holds for every partition (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 $\int_{a}^{b} c = c (b - a)$ , 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 $\int_{a}^{b} f$ ).

[L5]

For a partition $P = (n, t)$ and bounded $u$ : $L (u, P) = \sum_{i < n} m_{i} Δ_{i}$ and $U (u, P) = \sum_{i < n} M_{i} Δ_{i}$ with $m_{i} = \inf u [I_{i}]$ , $M_{i} = \sup u [I_{i}]$ , and a set with a least element has it as its infimum (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 $U (f, P) = \sum_{i} M_{i} Δ_{i}$ , Finite sums and finite products, by recursion, Laws of finite sums and finite products, Greatest lower bound (infimum), Maximum and minimum of a set, 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).

[L6]

Ordered-field arithmetic: the order is total and transitive, and a real that is $\leq 4 + 2 η$ for every $η$ in $(0, 1)$ is $\leq 4$ (Ordered field, Complete ordered field (least-upper-bound property), Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Verification

technique · direct

1.1

$f$ is nondecreasing on $[0, 2]$ : for $x \leq y$ the only possibility not giving $f (x) = f (y)$ is $x < 1 \leq y$ , where $f (x) = 1 < 3 = f (y)$ . Hence $f$ is bounded and integrable on $[0, 2]$ , and likewise on $[0, 1]$ and on $[1, 2]$ , by [L1].

givenL1

1.2

Second route. Let $P_{η}$ be the partition of $[0, 2]$ with point set ${0, 1 - η, 1, 2}$ , so $n = 3$ with $Δ_{0} = 1 - η$ , $Δ_{1} = η$ and $Δ_{2} = 1$ .

givenL5construct

2.1

First route, the piece $[0, 1]$ . The constant function $1$ on $[0, 1]$ agrees with $f$ at every point of $[0, 1]$ except $x = 1$ , a single point; so by [L3] and [L4], $\int_{0}^{1} f = \int_{0}^{1} 1 = 1 \cdot (1 - 0) = 1$ .

step 1.1L3L4

2.2

The piece $[1, 2]$ . There $f$ is the constant $3$ , so $\int_{1}^{2} f = 3 \cdot (2 - 1) = 3$ by [L4].

step 1.1L4

2.3

On $I_{0} = [0, 1 - η]$ , $f$ is constantly $1$ , so $m_{0} = M_{0} = 1$ . On $I_{1} = [1 - η, 1]$ , $f$ takes the value $1$ at every point below $1$ and $3$ at $1$ , so $m_{1} = 1$ and $M_{1} = 3$ by [L5]. On $I_{2} = [1, 2]$ , $f$ is constantly $3$ , so $m_{2} = M_{2} = 3$ .

step 1.2givenL5

3.1

By [L2] with $c : = 1$ , $\int_{0}^{2} f = \int_{0}^{1} f + \int_{1}^{2} f = 1 + 3 = 4$ .

step 1.1step 2.1step 2.2L2

3.2

Hence $L (f, P_{η}) = 1 (1 - η) + 1 \cdot η + 3 \cdot 1 = 4$ and $U (f, P_{η}) = 1 (1 - η) + 3 η + 3 \cdot 1 = 4 + 2 η$ , by [L5].

step 2.3L5

4.1

By [L4] the number $\int_{0}^{2} f$ lies between $L (f, P_{η}) = 4$ and $U (f, P_{η}) = 4 + 2 η$ for every $η \in (0, 1)$ , so $4 \leq \int_{0}^{2} f \leq 4$ by [L6]; the second route agrees with step 3.1.

step 1.1step 3.1step 3.2L4L6∎

Remarks

The two sums of $P_{η}$ are not symmetric, and that is a statement about these partitions. For each $η \in (0, 1)$ the lower sum $L (f, P_{η})$ is exactly $4$ while the upper sum is $4 + 2 η > 4$ , so this family already locates the integral exactly from below while its upper sums only approach it. Nothing is claimed here about an arbitrary partition: a partition whose points do not include $1$ has a subinterval on which $f$ takes both values, and its lower sum is then strictly below $4$ .
The value at the jump is irrelevant to the integral, but not to the sums. Replacing $f (1) = 3$ by any real leaves $f$ integrable with the same integral, by Changing an integrable function at finitely many points changes neither its integrability nor its integral. What it changes is which partitions realise the extremes: with $f (1) : = 1$ the subinterval $[1 - η, 1]$ carries $m_{1} = M_{1} = 1$ , while $[1, 2]$ now carries $m_{2} = 1$ and $M_{2} = 3$ , so $L (f, P_{η}) = 2$ and $U (f, P_{η}) = 4$ and the asymmetry moves to the other side. The integral is $4$ either way.
Additivity is what makes the first route short. Without For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ the only route is the second one, which needs the Darboux sums of a function that is not constant on any partition containing the jump in the interior of a subinterval.

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

The integral test applied to $\sum 1 / ι (k + 1)^{p}$ for rational $p > 0$ , cross-checked against the published $p$ -series theorem

Example

Let $p \in Q$ with $p > 0$ (Order on the rationals) and define

$f_{p} : [0, \infty) \to R, f_{p} (t) : = (t + 1)^{- p},$

the rational power of the positive base $t + 1 \geq 1$ (Rational powers $a^{r}$ of a positive base). Then $f_{p}$ is nonnegative and nonincreasing, so The integral test: for $f \geq 0$ nonincreasing on $[0, \infty)$ , $\sum_{k} f (k)$ converges if and only if the sequence $(\int_{0}^{N} f)_{N}$ is bounded, with $\int_{0}^{N} f \leq \sum_{k < N} f (k) \leq f (0) + \int_{0}^{N} f$ applies, and its terms are

$f_{p} (k) = \frac{1}{ι (k + 1)^{p}} (k \in N) .$

The series $\sum_{k} f_{p} (k)$ is exactly the $p$ -series $\sum_{k \geq 1} 1 / ι (k)^{p}$ in the sense of Series, partial sums, convergence and the sum, divergence, and the tail series, which converges if and only if $p > 1$ (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ). The integral test therefore delivers, with no primitive computed anywhere:

$(\int_{0}^{N} f_{p})_{N \in N} is bounded above ⟺ p > 1.$

The cross-check. At $p = 2$ the integral can also be computed directly: the primitive $G (t) = - (t + 1)^{- 1}$ gives

$\int_{0}^{N} (t + 1)^{- 2} d t = 1 - \frac{1}{ι (N + 1)} < 1,$

so the sequence is bounded by $1$ , in agreement with the verdict above at $p = 2 > 1$ . At $p = 1$ the verdict is that $(\int_{0}^{N} (t + 1)^{- 1})_{N}$ is unbounded, since the harmonic series diverges. No named logarithmic primitive is available from the current dependency vocabulary, and none is needed for this conclusion.

The exponent must be rational. Real exponents do not exist in this library at this point in the reading order (Why real exponents are deferred on the rational-powers page), so "for $p \in [1, \infty)$ " is not a statement that can be made here.

Facts & Assumptions

Given: A rational $p > 0$ , the function $f_{p} (t) = (t + 1)^{- p}$ on $[0, \infty)$ , and a natural number $N$ .

[L1]

For $a > 0$ and rationals $r, s$ : $a^{r} > 0$ , $a^{r + s} = a^{r} a^{s}$ , $a^{- r} = 1 / a^{r}$ , and $a^{0} = 1$ (Laws of rational exponents, Rational powers $a^{r}$ of a positive base).

[L2]

For rational $r > 0$ and $0 < a < b$ : $a^{r} < b^{r}$ (Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ , claim 2); the nonstrict form follows by adjoining equality.

[L3]

$ι (k + 1) \geq 1 > 0$ for $k \in N$ , $ι (k + 1) = ι (k) + 1$ , and $ι$ is nondecreasing (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing).

[L4]

Integral test: for $f \geq 0$ nonincreasing on $[0, \infty)$ , $\sum_{k} f (k)$ converges if and only if $(\int_{0}^{N} f)_{N}$ is bounded above (The integral test: for $f \geq 0$ nonincreasing on $[0, \infty)$ , $\sum_{k} f (k)$ converges if and only if the sequence $(\int_{0}^{N} f)_{N}$ is bounded, with $\int_{0}^{N} f \leq \sum_{k < N} f (k) \leq f (0) + \int_{0}^{N} f$ , Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $R$ , with the dictionary to monotone sequences, Lower bound, bounded below, bounded set).

[L5]

$\sum_{k \geq 1} 1 / ι (k)^{p}$ converges if and only if $p > 1$ , and by Series, partial sums, convergence and the sum, divergence, and the tail series that series is by definition the series of the sequence $j \mapsto 1 / ι (j + 1)^{p}$ on $N$ (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ).

[L6]

$a^{1 / 1} = a$ , so for a negative integer exponent the rational power of Rational powers $a^{r}$ of a positive base is the integer power of Integer powers $a^{m}$ (Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ ).

[L7]

For $n \geq 1$ the map $y \mapsto y^{- n}$ has derivative $- ι (n) y^{- n - 1}$ at every $y \neq 0$ ; sums, scalar multiples and composites of differentiable functions differentiate by the usual rules (For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, claim 3, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

[L8]

If $H$ is differentiable at every point of $[p^{'}, q^{'}]$ with $H^{'}$ integrable there, then $\int_{p^{'}}^{q^{'}} H^{'} = H (q^{'}) - H (p^{'})$ ; a continuous function on a closed bounded interval is integrable; a continuous function on an interval has a primitive (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ , Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point, The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ ).

[L10]

A quotient of continuous functions is continuous where the denominator does not vanish, and every polynomial function is continuous (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, claims 4 and 5).

[L9]

Ordered-field arithmetic: a positive real has a positive inverse, $0 < s \leq t$ gives $1 / t \leq 1 / s$ , and the order is total and transitive (Ordered field, Complete ordered field (least-upper-bound property), Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Verification

technique · direct

1.1

For $t \geq 0$ the base $t + 1$ is $\geq 1 > 0$ , so $f_{p} (t) = (t + 1)^{- p}$ is defined and positive by [L1].

givenL1

1.2

The cross-check at $p = 2$ . By [L6], $f_{2} (t) = (t + 1)^{- 2}$ is the integer power, and by [L7] the function $G (t) : = - (t + 1)^{- 1}$ is differentiable at every $t \geq 0$ with $G^{'} (t) = - (- (t + 1)^{- 2}) \cdot 1 = (t + 1)^{- 2} = f_{2} (t)$ .

givenL6L7

2.1

$f_{p}$ is nonincreasing: for $0 \leq t \leq u$ one has $0 < t + 1 \leq u + 1$ , so $(t + 1)^{p} \leq (u + 1)^{p}$ by [L2], and taking reciprocals reverses the inequality by [L9], giving $f_{p} (u) \leq f_{p} (t)$ by [L1].

step 1.1L1L2L9

2.2

$f_{p} (k) = (ι (k) + 1)^{- p} = 1 / ι (k + 1)^{p}$ by [L1] and [L3], so the sequence $k \mapsto f_{p} (k)$ is the one named in [L5].

step 1.1L1L3

2.3

$f_{2} (t) = 1 / (t + 1)^{2}$ is a quotient of polynomial functions whose denominator does not vanish on $[0, N]$ , hence continuous there by [L10], hence integrable there by [L8]; so [L8] applied to $G$ gives $\int_{0}^{N} f_{2} = G (N) - G (0) = - 1 / ι (N + 1) + 1$ .

step 1.2L3L8

3.1

By [L4], $\sum_{k} f_{p} (k)$ converges if and only if $(\int_{0}^{N} f_{p})_{N}$ is bounded above.

step 1.1step 2.1L4

4.1

Hence, by [L5] and step 3.1, $(\int_{0}^{N} f_{p})_{N}$ is bounded above if and only if $p > 1$ .

step 3.1step 2.2L5

5.1

Since $ι (N + 1) \geq 1 > 0$ , $0 < 1 / ι (N + 1) \leq 1$ , so $0 \leq \int_{0}^{N} f_{2} < 1$ for every $N$ : the sequence is bounded above by $1$ , which agrees with step 4.1 at $p = 2 > 1$ .

step 2.3L3L9

6.1

The verdict at $p = 1$ . By [L5] the series $\sum_{k \geq 1} 1 / ι (k)$ diverges, so by step 4.1 the sequence $(\int_{0}^{N} (t + 1)^{- 1} d t)_{N}$ is not bounded above. No primitive of $(t + 1)^{- 1}$ is exhibited, and none is needed for this conclusion.

step 4.1L5∎

Remarks

The test is run backwards here, and that is the point. The usual textbook order computes $\int_{1}^{N} t^{- p} d t$ with a primitive and reads off the convergence of the series. That route is unavailable at $p = 1$ in this library, because the primitive is the logarithm and the logarithm is built on a later page. Running the equivalence in the other direction costs nothing: the published For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ settles the series for every rational $p > 0$ , and The integral test: for $f \geq 0$ nonincreasing on $[0, \infty)$ , $\sum_{k} f (k)$ converges if and only if the sequence $(\int_{0}^{N} f)_{N}$ is bounded, with $\int_{0}^{N} f \leq \sum_{k < N} f (k) \leq f (0) + \int_{0}^{N} f$ transfers the verdict to the integrals.
The index shift is real and is checked in step 2.2. The $p$ -series starts at $k = 1$ because $0^{p} = 0$ has no reciprocal, while a sequence in this library is a function on $N$ , which contains $0$ ; the integrand $(t + 1)^{- p}$ is shifted by exactly one for that reason, and $f_{p} (k) = 1 / ι (k + 1)^{p}$ . Substituting $t^{- p}$ for $(t + 1)^{- p}$ would put an undefined value at $t = 0$ and make $\int_{0}^{N}$ improper.
An independent elementary route to the series verdict. For a nonincreasing nonnegative sequence, $\sum a_{k}$ converges iff $\sum 2^{k} a_{2^{k}}$ converges applies to the nonnegative nonincreasing family $(1 / ι (k)^{p})_{k \geq 1}$ and turns it into $\sum_{j \geq 0} ι (2)^{j} / ι (2^{j})^{p}$ , a geometric series; that is the standard elementary route to the same verdict, and it is noted here for orientation only. No claim is made about how For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ is itself proved, and nothing above depends on this remark.

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

$H (x) = 2 \sqrt{x}$ on $[0, 1]$ : $H$ is continuous, $H^{'}$ is unbounded on $(0, 1]$ , and $H^{'}$ is therefore not Riemann integrable

Example

Write $\sqrt{x} : = x^{1 / 2}$ for the unique nonnegative square root of $x \geq 0$ (Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ , Rational powers $a^{r}$ of a positive base) and put

$H : [0, 1] \to R, H (x) : = 2 \sqrt{x} .$

Then:

$H$ is continuous on $[0, 1]$ ;
$H$ is differentiable at every $x \in (0, 1]$ , with $H^{'} (x) = 1 / \sqrt{x}$ , and it is not differentiable at $0$ ;
$H^{'}$ is unbounded on $(0, 1]$ : $H^{'} (1 / ι (n + 1)^{2}) = ι (n + 1)$ for every $n \in N$ ;
consequently no function on $[0, 1]$ agreeing with $H^{'}$ on $(0, 1]$ is Riemann integrable on $[0, 1]$ , because 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 $U (f, P) = \sum_{i} M_{i} Δ_{i}$ ).

So the second fundamental theorem does not apply on $[0, 1]$ , even though $H$ is continuous there and differentiable on $(0, 1]$ : both hypotheses of The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ fail, differentiability at $0$ and integrability of the derivative.

What is available, and what is not. On $[η, 1]$ with $0 < η < 1$ everything works: $H^{'}$ is continuous there, hence integrable, and

$\int_{η}^{1} H^{'} = H (1) - H (η) = 2 - 2 \sqrt{η} .$

The value that the right-hand side approaches as $η$ shrinks is not computed here and is not called an integral: $\int_{0}^{1} H^{'}$ is undefined, and the object that repairs it is the improper integral, which belongs to a later page.

Facts & Assumptions

Given: The function $H (x) = 2 x^{1 / 2}$ on $[0, 1]$ , a real $η$ with $0 < η < 1$ , and a natural number $n$ .

[L1]

For $a \geq 0$ there is a unique $s \geq 0$ with $s^{2} = a$ , written $a^{1 / 2}$ ; $a^{1 / 2} > 0$ when $a > 0$ , and $0^{1 / 2} = 0$ (Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ , Rational powers $a^{r}$ of a positive base, Laws of rational exponents).

[L2]

$q : [0, \infty) \to R$ , $q (x) = x^{2}$ , is continuous and injective on $[0, \infty)$ , and $q (x) \geq 0$ there (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, claim 5, Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , claim 2, Injection, surjection, bijection, Integer powers $a^{m}$ ).

[L3]

A continuous injective function on an order-convex set has a continuous inverse on its image (Continuous inverse theorem: a continuous injective $f$ on an interval $I$ is a bijection onto the order-convex set $f [I]$ , and the inverse $g : f [I] \to I$ is continuous and strictly monotone in the same sense as $f$ , claims 3 and 5, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $R$ , with the dictionary to monotone sequences).

[L4]

Derivative of an inverse: with $f$ continuous and injective on an order-convex $I$ having at least two elements, $g$ its inverse, $c \in I$ and $b = f (c)$ : if $f$ is differentiable at $c$ with $f^{'} (c) \neq 0$ then $g$ is differentiable at $b$ with $g^{'} (b) = 1 / f^{'} (c)$ (Derivative of an inverse: if $f$ is continuous and injective on a nondegenerate interval $I$ and differentiable at $c \in I$ with $f^{'} (c) \neq 0$ , then the inverse $g$ is differentiable at $f (c)$ with $g^{'} (f (c)) = 1 / f^{'} (c)$ ; and if $f^{'} (c) = 0$ then $g$ is not differentiable at $f (c)$ , The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

[L5]

$q^{'} (c) = ι (2) c$ for every real $c$ , and a scalar multiple of a differentiable function is differentiable with the scaled derivative (For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, claim 2, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , claim 2, The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[L6]

A quotient of continuous functions is continuous where the denominator does not vanish; a continuous function on a closed bounded interval with distinct endpoints is integrable there (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, claim 4, A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point, 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 $\int_{a}^{b} f$ ).

[L7]

If $G$ is differentiable at every point of $[p, q]$ with $G^{'}$ integrable there, then $\int_{p}^{q} G^{'} = G (q) - G (p)$ (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ ).

[L8]

Darboux sums, and therefore Riemann integrability, 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, 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 $\int_{a}^{b} f$ ).

[L9]

$ι (n + 1) \geq 1 > 0$ , and for every real $w$ there is a natural $n$ with $w < ι (n + 1)$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing, Every complete ordered field is Archimedean, For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L10]

Ordered-field arithmetic: a positive real has a positive inverse, the order is total and transitive, and multiplying an inequality by a positive real preserves it (Ordered field, Complete ordered field (least-upper-bound property), The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ ).

Verification

technique · direct

1.1

By [L2] the map $q (x) = x^{2}$ is continuous and injective on the order-convex $[0, \infty)$ , and its image is exactly $[0, \infty)$ : every value is $\geq 0$ , and every $a \geq 0$ is $q (a^{1 / 2})$ by [L1].

L1L2

2.1

By [L3] the inverse $g : [0, \infty) \to [0, \infty)$ of $q$ is continuous, and $g (a) = a^{1 / 2}$ by the uniqueness in [L1]. Hence $H = 2 g$ restricted to $[0, 1]$ is continuous, which is claim 1.

step 1.1L1L3

2.2

Let $x \in (0, 1]$ and put $c : = x^{1 / 2} > 0$ by [L1]. By [L5], $q$ is differentiable at $c$ with $q^{'} (c) = ι (2) c = 2 c \neq 0$ ; so [L4] gives $g$ differentiable at $x = q (c)$ with $g^{'} (x) = 1 / (2 c) = 1 / (2 \sqrt{x})$ .

step 1.1L1L4L5L10

3.1

Hence $H = 2 g$ is differentiable at every $x \in (0, 1]$ with $H^{'} (x) = 2 / (2 \sqrt{x}) = 1 / \sqrt{x}$ , by [L5].

step 2.2L5

4.1

Claim 3. For $n \in N$ put $x_{n} : = 1 / ι (n + 1)^{2}$ , a real in $(0, 1]$ by [L9]. Then $\sqrt{x_{n}} = 1 / ι (n + 1)$ , since that number is positive with square $x_{n}$ and [L1] gives uniqueness; so $H^{'} (x_{n}) = ι (n + 1)$ by step 3.1.

step 3.1L1L9L10

5.1

$H^{'}$ is unbounded on $(0, 1]$ : given a real $M \geq 0$ , [L9] supplies $n$ with $M < ι (n + 1) = H^{'} (x_{n})$ .

step 4.1L9

5.2

$H$ is not differentiable at $0$ . The difference quotient of $H$ at $0$ is $x \mapsto 2 \sqrt{x} / x = 2 / \sqrt{x}$ for $x \in (0, 1]$ , by [L1] and [L10]; at $x_{n}$ it takes the value $2 ι (n + 1)$ , which exceeds every real by [L9]. So no real $L$ can satisfy the $ε$ - $δ$ condition with $ε = 1$ : any $δ > 0$ admits some $x_{n} < δ$ , again by [L9], at which the quotient exceeds $L + 1$ .

step 4.1L1L9L10

6.1

Claim 4. Let $u : [0, 1] \to R$ agree with $H^{'}$ on $(0, 1]$ . By step 5.1, $u$ is unbounded on $[0, 1]$ , so it has no Darboux sums and is not Riemann integrable there, by [L8].

step 5.1L8

7.1

The hypotheses of the second fundamental theorem both fail on $[0, 1]$ , by step 5.2 and step 6.1; so [L7] gives nothing there, and $\int_{0}^{1} H^{'}$ is an undefined symbol.

step 5.2step 6.1L7L8

8.1

On $[η, 1]$ everything works. There $\sqrt{\cdot}$ is continuous and does not vanish, so $H^{'} = 1 / \sqrt{\cdot}$ is continuous on $[η, 1]$ by [L6] and integrable there; $H$ is differentiable at every point of $[η, 1]$ by step 3.1; so [L7] gives $\int_{η}^{1} H^{'} = H (1) - H (η) = 2 - 2 \sqrt{η}$ .

step 2.1step 3.1L1L6L7∎

Remarks

Boundedness, not continuity, is what fails. $H^{'}$ is continuous at every point of $(0, 1]$ ; what defeats Riemann integrability on $[0, 1]$ is that 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 $U (f, P) = \sum_{i} M_{i} Δ_{i}$ takes a supremum of $f$ over each subinterval and the subinterval containing $0$ has none. No choice of value at $0$ repairs this, which is why claim 4 quantifies over every extension of $H^{'}$ .
This is the standard motivation for the improper integral, and it is left standing here. The numbers $2 - 2 \sqrt{η}$ of step 8.1 approach $2$ as $η$ shrinks, and a later page defines an object whose value is $2$ . Calling that object $\int_{0}^{1} H^{'}$ before it has been defined would be the error this item exists to avoid; nothing above computes it.
The square root is obtained from published material and not assumed. Existence and uniqueness of the nonnegative $n$ -th root is Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ ; continuity of $x \mapsto \sqrt{x}$ comes from Continuous inverse theorem: a continuous injective $f$ on an interval $I$ is a bijection onto the order-convex set $f [I]$ , and the inverse $g : f [I] \to I$ is continuous and strictly monotone in the same sense as $f$ applied to $x \mapsto x^{2}$ on $[0, \infty)$ ; and the derivative comes from Derivative of an inverse: if $f$ is continuous and injective on a nondegenerate interval $I$ and differentiable at $c \in I$ with $f^{'} (c) \neq 0$ , then the inverse $g$ is differentiable at $f (c)$ with $g^{'} (f (c)) = 1 / f^{'} (c)$ ; and if $f^{'} (c) = 0$ then $g$ is not differentiable at $f (c)$ , whose second clause also explains the failure at $0$ : there $q^{'} (0) = 0$ , so the inverse is not differentiable at $q (0) = 0$ .

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-08-01Open item page →

The sign function is Riemann integrable on $[- 1, 1]$ and has no primitive there

Statement refuted

False claim: every Riemann integrable function on a closed bounded interval has a primitive there, that is, is the derivative of some function on that interval.

Let $s : [- 1, 1] \to R$ be the sign function,

$s (x) : = {\begin{cases} - 1 & - 1 \leq x < 0, \\ 0 & x = 0, \\ 1 & 0 < x \leq 1. \end{cases}$

Then $s$ is Riemann integrable on $[- 1, 1]$ with $\int_{- 1}^{1} s = 0$ , and there is no $G : [- 1, 1] \to R$ differentiable at every point of $[- 1, 1]$ with $G^{'} = s$ .

The integral function of $s$ is $F (x) = ∣ x ∣ - 1$ , which is differentiable at every point of $[- 1, 1]$ except $x = 0$ — exactly the one point where $s$ is discontinuous. That is consistent with The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive, which claims $F^{'} (c) = s (c)$ only at points of continuity of $s$ , and it is what the refutation below turns into a contradiction.

Facts & Assumptions

Given: The sign function $s$ on $[- 1, 1]$ and its integral function $F (x) = \int_{- 1}^{x} s$ .

[A1]

There is $G : [- 1, 1] \to R$ , differentiable at every point of $[- 1, 1]$ as a function on $[- 1, 1]$ , with $G^{'} (x) = s (x)$ for every $x \in [- 1, 1]$ .

[L1]

A bounded function on $[p, q]$ with at most finitely many discontinuities is Riemann integrable there (A bounded function on $[a, b]$ that is continuous except at finitely many points is Riemann integrable, Lower bound, bounded below, bounded set, 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 $\int_{a}^{b} f$ , Discontinuity of $f$ at a point of its domain, and its classification: removable discontinuity, jump discontinuity and essential discontinuity, equivalently Rudin's discontinuities of the first and of the second kind).

[L2]

Changing an integrable function at finitely many points changes neither its integrability nor its integral, and $\int_{p}^{q} c = c (q - p)$ for a constant (Changing an integrable function at finitely many points changes neither its integrability nor its integral, 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 $\int_{a}^{b} c = c (b - a)$ ).

[L3]

Additivity, in the oriented form valid for arbitrary points (For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ , claim 3, The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ , The integral function $F (x) : = \int_{a}^{x} f$ of an integrable $f$ ).

[L4]

If $u$ is continuous on an order-convex $I$ and differentiable with $u^{'} = 0$ at every interior point of $I$ , then $u$ is constant on $I$ (A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ ; consequently two such functions with the same derivative differ by a constant, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L5]

A function differentiable at a point is continuous there, and a restriction to a subinterval still having the point as a limit point is differentiable there with the same derivative; every point of a nondegenerate interval is a limit point of it (A function differentiable at $c$ is continuous at $c$ , The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Limit point, isolated point, adherent point, derived set, and dense subset of $R$ , Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point).

[L6]

If both one-sided limits of a function at $c$ exist and differ, the two-sided limit does not exist (If $c$ is a limit point of the domain from both sides, the limit exists iff both one-sided limits exist and agree, The left and right limits of $f$ at $c$ , as limits of the restrictions of $f$ to $A \cap (- \infty, c)$ and $A \cap (c, \infty)$ , The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ ).

[L7]

Absolute value: $∣ x ∣ = x$ for $x \geq 0$ and $∣ x ∣ = - x$ for $x \leq 0$ ; $∣ x ∣ / x = 1$ for $x > 0$ and $= - 1$ for $x < 0$ (Absolute value in an ordered field, Basic properties of the absolute value, Ordered field, Complete ordered field (least-upper-bound property)).

Counterexample

technique · contradiction

1.1

$s$ is bounded, with $∣ s ∣ \leq 1$ , and continuous at every point of $[- 1, 1]$ other than $0$ : near a point $x \neq 0$ it is locally constant. So $s$ is integrable on $[- 1, 1]$ , and likewise on $[- 1, 0]$ and on $[0, 1]$ , by [L1].

givenL1

1.2

Assume [A1]. Then $G^{'} = s$ everywhere on $[- 1, 1]$ .

assume-contraA1

2.1

On $[- 1, 0]$ the function $s$ agrees with the constant $- 1$ except at the single point $0$ , so $\int_{- 1}^{0} s = - 1 \cdot (0 - (- 1)) = - 1$ by [L2]; on $[0, 1]$ it agrees with the constant $1$ except at $0$ , so $\int_{0}^{1} s = 1$ .

step 1.1L2

2.2

On $I_{+} : = [0, 1]$ the function $u_{+} (x) : = G (x) - x$ is continuous, by [L5], and differentiable at every interior point $x \in (0, 1)$ with $u_{+}^{'} (x) = s (x) - 1 = 0$ ; so $u_{+}$ is constant on $[0, 1]$ by [L4], say $G (x) = x + c_{1}$ there.

step 1.2L4L5choose

2.3

On $I_{-} : = [- 1, 0]$ the function $u_{-} (x) : = G (x) + x$ is continuous and differentiable at every $x \in (- 1, 0)$ with $u_{-}^{'} (x) = s (x) + 1 = 0$ ; so $G (x) = - x + c_{2}$ on $[- 1, 0]$ .

step 1.2L4L5choose

3.1

By [L3], $\int_{- 1}^{1} s = \int_{- 1}^{0} s + \int_{0}^{1} s = - 1 + 1 = 0$ .

step 2.1L3

3.2

By [L3] again, $F (x) = \int_{- 1}^{x} s$ equals $- 1 + x$ for $x \in [0, 1]$ and equals $- (x - (- 1)) = - x - 1$ for $x \in [- 1, 0]$ , by [L2] applied on the relevant piece; in both cases $F (x) = ∣ x ∣ - 1$ by [L7].

step 2.1L2L3L7

3.3

Evaluating both formulas at $x = 0$ gives $c_{1} = G (0) = c_{2}$ ; write $c$ for the common value, so $G (x) = ∣ x ∣ + c$ for every $x \in [- 1, 1]$ by [L7].

step 2.2step 2.3L7

4.1

The difference quotient of $G$ at $0$ is $x \mapsto (G (x) - G (0)) / x = ∣ x ∣ / x$ , which equals $1$ for $x > 0$ and $- 1$ for $x < 0$ by [L7]; so its right-hand limit at $0$ is $1$ and its left-hand limit is $- 1$ .

step 3.3L7

5.1

By [L6] the two-sided limit of that quotient at $0$ does not exist, so $G$ is not differentiable at $0$ ; this contradicts [A1].

step 4.1A1L6

6.1

Hence no such $G$ exists: $s$ is integrable on $[- 1, 1]$ by step 1.1 and has no primitive there, so the claim is false.

step 1.1step 3.1step 5.1discharge-contradiction∎

Remarks

Darboux's theorem is not used. The slick argument — a derivative has the intermediate value property, and $s$ does not — rests on Darboux's theorem on the intermediate value property of derivatives. The refutation above uses only A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ ; consequently two such functions with the same derivative differ by a constant and the one-sided limits of a difference quotient, both already published.
The tension with the first fundamental theorem is only apparent. $F$ is differentiable with $F^{'} = s$ at every point of $[- 1, 1]$ except $0$ , which is exactly what The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive promises, since $s$ is continuous exactly off $0$ . What fails is the existence of any function differentiable at $0$ too with derivative $s (0)$ there, and step 4.1 shows the obstruction is the jump of $s$ at $0$ , not a defect of $F$ .
The value $s (0)$ is irrelevant to both halves. Changing it changes neither the integral, by Changing an integrable function at finitely many points changes neither its integrability nor its integral, nor the conclusion of step 5.1, which shows $G$ is not differentiable at $0$ at all and therefore cannot have any prescribed derivative there.

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

A function differentiable on $[0, 1]$ whose derivative is unbounded, hence not Riemann integrable

Statement refuted

False claim: if $G : [0, 1] \to R$ is differentiable at every point of $[0, 1]$ (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set), then $G^{'}$ is Riemann integrable on $[0, 1]$ (so that $\int_{0}^{1} G^{'} = G (1) - G (0)$ makes sense).

The claim is false. Put

$ϕ (t) : = t^{2} (1 - t)^{2} (t \in R),$

a polynomial with $ϕ (0) = ϕ (1) = 0$ and $ϕ^{'} (0) = ϕ^{'} (1) = 0$ , and for $n \in N$ set

$α_{n} : = \frac{1}{ι (n + 2)}, h_{n} : = \frac{1}{ι (n + 2)^{4}}, c_{n} : = \frac{1}{ι (n + 2)^{2}}, β_{n} : = α_{n} + h_{n}, I_{n} : = [α_{n}, β_{n}]$

(The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Integer powers $a^{m}$ ). The intervals $I_{n}$ are pairwise disjoint and lie in $(0, 1]$ , and

$G : [0, 1] \to R, G (x) : = {\begin{cases} c_{n} ϕ ⁣ (\frac{x - α_{n}}{h_{n}}) & x \in I_{n} for some n, \\ 0 & otherwise \end{cases}$

is differentiable at every point of $[0, 1]$ , while

$G^{'} ⁣ (α_{n} + \frac{1}{4} h_{n}) = \frac{3}{16} ι (n + 2)^{2},$

so $G^{'}$ is unbounded on $[0, 1]$ and therefore has no Darboux sums at all (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 $U (f, P) = \sum_{i} M_{i} Δ_{i}$ ) and is not Riemann integrable.

The construction is entirely polynomial, and deliberately so. The classical witness is $x^{2} \sin (1 / x^{2})$ ; the trigonometric functions are built on a later page of this library, so a bump glued from a single quartic is used instead. Only one bump is nonzero near any point of $(0, 1]$ , so no series converges anywhere in the argument and no limit function is formed.

Facts & Assumptions

Given: The polynomial $ϕ (t) = t^{2} (1 - t)^{2}$ , the numbers $α_{n}, h_{n}, c_{n}, β_{n}$ and intervals $I_{n}$ above, the function $G$ above, and a real $ε > 0$ .

[L1]

Polynomial calculus: every polynomial function is differentiable at every real and continuous there, with $(t^{n})^{'} = ι (n) t^{n - 1}$ for $n \geq 1$ , and sums, scalar multiples and products differentiate by the usual rules (For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Integer powers $a^{m}$ ).

[L2]

Chain rule, and the derivative of an affine reparametrisation: $y \mapsto ψ ((y - α) / h)$ has derivative $ψ^{'} ((y - α) / h) / h$ for $h \neq 0$ (The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[L3]

$ι (n + 2) \geq 2 > 1$ , $ι$ is increasing on $N$ , $ι (m) < ι (n)$ for $m < n$ , and for every real $w$ there is a natural $n$ with $w < ι (n + 2)$ ; also $ι (n)^{k} \geq ι (n)$ for $k \geq 1$ when $ι (n) \geq 1$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing, Every complete ordered field is Archimedean, For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , claims 3 and 4).

[L4]

If both one-sided limits of a function at $c$ exist and are equal to $L$ , the two-sided limit exists and equals $L$ (If $c$ is a limit point of the domain from both sides, the limit exists iff both one-sided limits exist and agree, The left and right limits of $f$ at $c$ , as limits of the restrictions of $f$ to $A \cap (- \infty, c)$ and $A \cap (c, \infty)$ , The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ ).

[L5]

Locality of limits: two functions on $A$ agreeing on a punctured neighbourhood of $c$ have the same limit behaviour at $c$ ; and a derivative survives restricting the domain provided the smaller domain still accumulates at the point (The limit at $c$ depends only on the restriction of $f$ to a punctured neighbourhood of $c$ , and passes to any subset of the domain having $c$ as a limit point, The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Limit point, isolated point, adherent point, derived set, and dense subset of $R$ ).

[L6]

Every point of a nondegenerate interval is a limit point of it (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Limit point, isolated point, adherent point, derived set, and dense subset of $R$ ).

[L7]

Darboux sums, and therefore Riemann integrability, require a bounded function (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, 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 $\int_{a}^{b} f$ ).

[L8]

A nonempty finite set of reals has a least element (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L9]

Ordered-field arithmetic: $t (1 - t) \leq 1 / 4$ for every real $t$ , since $(t - 1 / 2)^{2} \geq 0$ ; a positive real has a positive inverse; multiplying an inequality by a positive real preserves it; the order is total and transitive (Ordered field, Complete ordered field (least-upper-bound property), Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , claim 1).

[L10]

Counterexample

technique · direct

1.1

By [L1], $ϕ$ is differentiable everywhere with $ϕ^{'} (t) = 2 t (1 - t)^{2} - 2 t^{2} (1 - t) = 2 t (1 - t) (1 - 2 t)$ ; in particular $ϕ (0) = ϕ (1) = 0$ and $ϕ^{'} (0) = ϕ^{'} (1) = 0$ , and $ϕ^{'} (1 / 4) = 2 \cdot \frac{1}{4} \cdot \frac{3}{4} \cdot \frac{1}{2} = \frac{3}{16}$ .

L1L9

1.2

For $t \in [0, 1]$ , $0 \leq ϕ (t) = (t (1 - t))^{2} \leq (1 / 4)^{2} = 1 / 16$ , by [L9].

1.3

Each $α_{n}, h_{n}, c_{n}$ is a positive real, $h_{n} \leq α_{n}$ and $c_{n} \leq α_{n}$ because $ι (n + 2) > 1$ , and $α_{n} \leq 1 / 2$ ; so $I_{n} \subseteq (0, 1]$ , since $β_{n} = α_{n} + h_{n} \leq 2 α_{n} \leq 1$ .

givenL3L9

2.1

The $I_{n}$ are pairwise disjoint. For $n \geq 1$ , $α_{n - 1} - α_{n} = 1 / (ι (n + 1) ι (n + 2))$ while $h_{n} = 1 / ι (n + 2)^{4}$ , and $ι (n + 2)^{4} > ι (n + 1) ι (n + 2)$ because $ι (n + 2)^{3} \geq 2 ι (n + 2) > ι (n + 1)$ by [L3]; so $β_{n} < α_{n - 1}$ . Since $(α_{n})$ is strictly decreasing, $m < n$ gives $β_{n} < α_{n - 1} \leq α_{m}$ , and $I_{n}$ lies strictly below $I_{m}$ .

step 1.3L3L9

2.2

For each $n$ define the polynomial $Φ_{n} (y) : = c_{n} ϕ ((y - α_{n}) / h_{n})$ on $R$ . By [L1] and [L2] it is differentiable everywhere with $Φ_{n}^{'} (y) = (c_{n} / h_{n}) ϕ^{'} ((y - α_{n}) / h_{n})$ , and $Φ_{n} (α_{n}) = Φ_{n} (β_{n}) = 0$ , $Φ_{n}^{'} (α_{n}) = Φ_{n}^{'} (β_{n}) = 0$ by step 1.1.

step 1.1L1L2construct

2.3

Differentiability at a point of $(0, 1]$ outside every $I_{n}$ . Let $x \in (0, 1]$ with $x \notin I_{m}$ for every $m$ . By [L3] fix $N$ with $2 / x < ι (N + 2)$ , so $2 / ι (N + 2) < x$ ; for $m \geq N$ , $β_{m} \leq 2 α_{m} = 2 / ι (m + 2) \leq 2 / ι (N + 2) < x$ by step 1.3 and [L3].

step 1.3L3L9choose

3.1

So $G$ is a well-defined function on $[0, 1]$ , no $x$ lying in two of the $I_{n}$ .

step 2.1given

3.2

Gaps around the endpoints. For each $n$ , $(β_{n + 1}, α_{n})$ meets no $I_{m}$ : for $m \leq n$ one has $α_{m} \geq α_{n}$ by step 2.1, and for $m \geq n + 1$ one has $β_{m} \leq β_{n + 1}$ , again by step 2.1. Likewise $(β_{n}, α_{n - 1})$ meets no $I_{m}$ for $n \geq 1$ , and $(β_{0}, 1]$ meets none.

step 2.1L3

3.3

The finitely many closed intervals $I_{0}, \dots, I_{N - 1}$ do not contain $x$ , so each of the positive reals $∣ x - α_{m} ∣$ and $∣ x - β_{m} ∣$ for $m < N$ , together with $x - 2 / ι (N + 2)$ , forms a nonempty finite set of positive reals; let $ρ$ be its least element, which is positive by [L8].

step 2.3L8L9choose

4.1

$G$ agrees with $Φ_{n}$ on $I_{n}$ and with the zero function off $⋃_{m} I_{m}$ ; in particular $G (α_{n}) = G (β_{n}) = 0$ and $G \geq 0$ everywhere.

step 3.1step 2.2L9

5.1

Differentiability at an interior point of a bump. Let $x \in (α_{n}, β_{n})$ . The difference quotients of $G$ and of $Φ_{n}$ at $x$ agree on the punctured neighbourhood $0 < ∣ y - x ∣ < \min {x - α_{n}, β_{n} - x}$ inside $[0, 1]$ , so by [L5] and step 2.2, $G$ is differentiable at $x$ with $G^{'} (x) = Φ_{n}^{'} (x)$ .

step 2.2step 4.1L5L6L8

5.2

Differentiability at a left endpoint $α_{n}$ . On the right, $G$ agrees with $Φ_{n}$ on $[α_{n}, β_{n}]$ and $G (α_{n}) = Φ_{n} (α_{n})$ , so the right-hand limit of the difference quotient is $Φ_{n}^{'} (α_{n}) = 0$ by [L5] and step 2.2. On the left, $G$ vanishes on $(β_{n + 1}, α_{n}]$ by step 3.2 and step 4.1, so the quotient is identically $0$ there and its left-hand limit is $0$ . By [L4], $G^{'} (α_{n}) = 0$ .

step 2.2step 4.1step 3.2L4L5L6

5.3

Differentiability at a right endpoint $β_{n}$ . Symmetrically: on the left $G$ agrees with $Φ_{n}$ , giving limit $Φ_{n}^{'} (β_{n}) = 0$ ; on the right $G$ vanishes on $[β_{n}, α_{n - 1})$ when $n \geq 1$ and on $[β_{0}, 1]$ when $n = 0$ , by step 3.2, giving limit $0$ . By [L4], $G^{'} (β_{n}) = 0$ .

step 2.2step 4.1step 3.2L4L5L6

5.4

Then $G$ vanishes on $(x - ρ, x + ρ) \cap [0, 1]$ , so its difference quotient at $x$ is identically $0$ there and $G^{'} (x) = 0$ by [L5] and [L6].

step 2.3step 3.3step 4.1L5L6

5.5

Differentiability at $0$ . For $y \in (0, 1]$ : if $y \notin ⋃_{m} I_{m}$ then $G (y) / y = 0$ ; and if $y \in I_{m}$ then $0 \leq G (y) \leq c_{m} / 16$ by step 1.2 and step 4.1 and $y \geq α_{m}$ , so $0 \leq G (y) / y \leq c_{m} / (16 α_{m}) = 1 / (16 ι (m + 2))$ .

step 1.2step 4.1L9

6.1

Given $ε > 0$ , fix by [L3] a natural $N$ with $1 / ε < ι (N + 2)$ and put $δ : = α_{N} > 0$ . If $0 < y < δ$ and $y \in I_{m}$ then $α_{m} \leq y < α_{N}$ , so $ι (N + 2) < ι (m + 2)$ by [L3], and step 5.5 gives $∣ G (y) / y ∣ \leq 1 / (16 ι (m + 2)) < ε$ ; otherwise $G (y) / y = 0$ .

step 5.5L3L9choose

6.2

$G^{'}$ is unbounded. Put $u_{n} : = α_{n} + h_{n} / 4$ , an interior point of $I_{n}$ ; by step 5.1 and step 2.2, $G^{'} (u_{n}) = (c_{n} / h_{n}) ϕ^{'} (1 / 4) = ι (n + 2)^{2} \cdot \frac{3}{16}$ , using $c_{n} / h_{n} = ι (n + 2)^{4} / ι (n + 2)^{2} = ι (n + 2)^{2}$ . Given a real $M \geq 0$ , [L3] supplies $n$ with $\frac{16}{3} M < ι (n + 2) \leq ι (n + 2)^{2}$ , so $G^{'} (u_{n}) > M$ .

step 1.1step 2.2step 5.1L3L9

7.1

So the difference quotient of $G$ at $0$ , which is $y \mapsto G (y) / y$ on $(0, 1]$ , tends to $0$ ; hence $G$ is differentiable at $0$ with $G^{'} (0) = 0$ by [L6].

step 6.1L6

8.1

By steps 5.1, 5.2, 5.3, 5.4 and 7.1, $G$ is differentiable at every point of $[0, 1]$ : every $x$ is either $0$ , or an interior point of some $I_{n}$ , or an endpoint of some $I_{n}$ , or a point of $(0, 1]$ outside every $I_{n}$ .

step 5.1step 5.2step 5.3step 5.4step 7.1L9

9.1

Hence $G^{'}$ is a function on $[0, 1]$ that is not bounded, so it has no Darboux sums and is not Riemann integrable on $[0, 1]$ by [L7]; the claim is false, and $\int_{0}^{1} G^{'}$ is an undefined symbol, so [L10] gives nothing here.

step 8.1step 6.2L7L10∎

Remarks

Only one bump is active near any point of $(0, 1]$ , and that is what makes every step finite. The intervals $I_{n}$ accumulate only at $0$ , so a point of $(0, 1]$ has a neighbourhood meeting at most one of them (steps 3.2 and 3.3), and the only place where infinitely many bumps are seen at once is the origin, where step 5.5 controls all of them by a single estimate. No series is summed anywhere.
The two exponents are what the construction turns on. Differentiability at $0$ needs $c_{n} / α_{n} \to 0$ , and unboundedness of $G^{'}$ needs $c_{n} / h_{n} \to \infty$ ; with $α_{n} = 1 / ι (n + 2)$ the choices $c_{n} = α_{n}^{2}$ and $h_{n} = α_{n}^{4}$ give $c_{n} / α_{n} = α_{n} \to 0$ and $c_{n} / h_{n} = α_{n}^{- 2} \to \infty$ . Any pair of exponents with the same two properties would do; these are verified explicitly in steps 6.1 and 6.2 because the construction is only as good as those two inequalities.
What this refutes and what it does not. It refutes the claim that every derivative is Riemann integrable, hence the naive reading of The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ with its integrability hypothesis deleted. It says nothing about whether $G^{'}$ has a primitive — it does, namely $G$ — and nothing about the sharp class of functions for which $\int_{a}^{b} G^{'} = G (b) - G (a)$ holds, which this library records but does not prove (Conventions of this page, and which sharpenings of the integral are taken up later in the reading order).

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Integrable $φ$ and integrable $f$ with $φ \circ f$ not integrable: the order of the hypotheses in the composition theorem cannot be reversed

Statement refuted

False claim: if $f : [a, b] \to [m, M]$ is Riemann integrable and $φ : [m, M] \to R$ is Riemann integrable, then $φ \circ f$ is Riemann integrable on $[a, b]$ .

That is If $f$ is integrable on $[a, b]$ with values in $[m, M]$ and $φ$ is continuous on $[m, M]$ , then $φ \circ f$ is integrable with "continuous" weakened to "integrable" on the outer function, and it is false. On $[0, 1]$ take $f : = t$ , Thomae's function (The Dirichlet function $1_{Q}$ , and Thomae's function $t$ with $t (x) = 1 / q$ at a rational $x = p / q$ in lowest terms with $q \geq 1$ and $t (x) = 0$ at every irrational $x$ ), whose values lie in $[0, 1]$ , and

$φ : [0, 1] \to R, φ (u) : = {\begin{cases} 0 & u = 0, \\ 1 & 0 < u \leq 1. \end{cases}$

Both are Riemann integrable, $t$ because its discontinuity set is at most countable and $φ$ because it is nondecreasing. But

$φ \circ t = 1_{Q} on [0, 1],$

the Dirichlet function, which is not Riemann integrable.

Exactly one hypothesis of the composition theorem fails, and it is named: $φ$ is not continuous, being discontinuous at $0$ . The theorem's hypothesis is continuous after integrable; here the outer function is merely integrable, and that is not enough.

Facts & Assumptions

Given: Thomae's function $t$ on $[0, 1]$ , with $t (x) = 1 / ι (q (x))$ at a rational $x$ of least denominator $q (x) \geq 1$ and $t (x) = 0$ at an irrational $x$ , and the function $φ$ above.

[L1]

$0 \leq t (x) \leq 1$ for every $x$ , and $t (x) > 0$ exactly at the rational $x$ , since $ι (q (x)) \geq 1 > 0$ (The Dirichlet function $1_{Q}$ , and Thomae's function $t$ with $t (x) = 1 / q$ at a rational $x = p / q$ in lowest terms with $q \geq 1$ and $t (x) = 0$ at every irrational $x$ , The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing, Ordered field).

[L2]

$t$ is continuous at every irrational and discontinuous at every rational, and $1_{Q}$ is continuous at no real (The Dirichlet function is continuous at no point of $R$ , 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 $c$ equals $t (c)$ , claims 1 and 3, Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point).

[L3]

$Q$ is countably infinite and every subset of an at most countable set is at most countable ( $Q$ 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 $[p, q]$ whose set of discontinuities is at most countable is Riemann integrable there (A bounded function on $[a, b]$ whose set of discontinuities is at most countable is Riemann integrable, Lower bound, bounded below, bounded set, 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 $\int_{a}^{b} f$ ).

[L5]

[L6]

A bounded function on $[p, q]$ is integrable if and only if its discontinuity set has measure zero; a nondegenerate interval does not have measure zero; an at most countable set does (Lebesgue's criterion for Riemann integrability: a bounded $f$ on $[a, b]$ is Riemann integrable if and only if its set of discontinuities has measure zero, A sequence of intervals covering $[a, b]$ has total length at least $b - a$ , so no interval of positive length has measure zero, Every at most countable subset of $R$ has measure zero, Measure zero (a countable cover by intervals of total length below every $ε$ ) and content zero (a finite such cover)).

[L7]

Ordered-field arithmetic and intervals: the order is total and transitive (Ordered field, Complete ordered field (least-upper-bound property), Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Counterexample

technique · direct

1.1

$t$ is bounded on $[0, 1]$ with values in $[0, 1]$ , by [L1].

givenL1

1.2

The set of discontinuities of $t$ in $[0, 1]$ is $Q \cap [0, 1]$ by [L2], which is at most countable by [L3].

L2L3

1.3

$φ$ is nondecreasing on $[0, 1]$ : for $u \leq v$ the only case not giving $φ (u) = φ (v)$ is $u = 0 < v$ , where $φ (u) = 0 < 1 = φ (v)$ . So $φ$ is bounded and integrable on $[0, 1]$ by [L5].

givenL5L7

2.1

Hence $t$ is Riemann integrable on $[0, 1]$ by [L4].

step 1.1step 1.2L4

2.2

For $x \in [0, 1]$ : if $x$ is rational then $t (x) > 0$ by [L1], so $φ (t (x)) = 1$ ; if $x$ is irrational then $t (x) = 0$ , so $φ (t (x)) = 0$ . Hence $φ \circ t$ agrees with $1_{Q}$ at every point of $[0, 1]$ .

step 1.1givenL1

3.1

$1_{Q}$ is bounded on $[0, 1]$ and continuous at no point of it by [L2], so its discontinuity set is $[0, 1]$ , which does not have measure zero by [L6]; therefore $φ \circ t$ is not Riemann integrable on $[0, 1]$ by [L6].

step 2.2L2L6

4.1

So $t$ and $φ$ are integrable while $φ \circ t$ is not, and the claim is false. The hypothesis of If $f$ is integrable on $[a, b]$ with values in $[m, M]$ and $φ$ is continuous on $[m, M]$ , then $φ \circ f$ is integrable that fails here is the continuity of the outer function: $φ$ is discontinuous at $0$ , since $φ (0) = 0$ while $φ (u) = 1$ for every $u \in (0, 1]$ and every neighbourhood of $0$ in $[0, 1]$ contains such a $u$ .

step 2.1step 1.3step 3.1L7∎

Remarks

The order of composition is the whole point. The published theorem is continuous after integrable, and it is true; the reversal, integrable after integrable, is refuted here. A reader who remembers only "composition preserves integrability" will get this wrong in one direction out of two, and the direction that fails is the one this item names.
Nothing is claimed about the inner function being continuous. If $f$ is continuous and $φ$ merely integrable, the composite may still fail to be integrable; that case is not covered above and no claim is made about it here.
Thomae is the right inner function because its positivity set is exactly $Q$ . Any integrable $f$ that is positive exactly on a dense set of measure zero and zero elsewhere would do; Thomae is the published one, and The Dirichlet function is continuous at no point of $R$ , 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 $c$ equals $t (c)$ is what supplies its continuity set without recomputing it.

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A function that is not Riemann integrable although $∣ f ∣$ is

Statement refuted

False claim: if $∣ f ∣$ is Riemann integrable on $[a, b]$ then so is $f$ ; that is, the first clause of If $f, g$ are integrable on $[a, b]$ then so are $∣ f ∣$ , $f^{2}$ , $f g$ , $\max (f, g)$ and $\min (f, g)$ , and $∣ \int_{a}^{b} f ∣ \leq \int_{a}^{b} ∣ f ∣$ has a converse.

Let $1_{Q}$ be the Dirichlet function (The Dirichlet function $1_{Q}$ , and Thomae's function $t$ with $t (x) = 1 / q$ at a rational $x = p / q$ in lowest terms with $q \geq 1$ and $t (x) = 0$ at every irrational $x$ ) and put

$f : [0, 1] \to R, f (x) : = 2 1_{Q} (x) - 1 = {\begin{cases} 1 & x rational, \\ - 1 & x irrational . \end{cases}$

Then $∣ f ∣$ is the constant function $1$ , integrable with $\int_{0}^{1} ∣ f ∣ = 1$ , while $f$ is not Riemann integrable on $[0, 1]$ : every lower Darboux sum of $f$ is $- 1$ and every upper Darboux sum is $1$ , so the lower and upper integrals are $- 1$ and $1$ .

Facts & Assumptions

Given: The function $f = 21_{Q} - 1$ on $[0, 1]$ , and a partition $P = (n, t)$ of $[0, 1]$ .

[L1]

$1_{Q} (x) = 1$ at a rational $x$ and $0$ at an irrational $x$ (The Dirichlet function $1_{Q}$ , and Thomae's function $t$ with $t (x) = 1 / q$ at a rational $x = p / q$ in lowest terms with $q \geq 1$ and $t (x) = 0$ at every irrational $x$ ).

[L2]

Both $Q$ and the irrationals are dense in $R$ , so every nonempty open interval contains a rational and an irrational (Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable).

[L3]

For a partition $P = (n, t)$ of $[0, 1]$ : $n \geq 1$ , $Δ_{i} = t_{i + 1} - t_{i} > 0$ , $\sum_{i < n} Δ_{i} = 1$ , and the open interval $(t_{i}, t_{i + 1})$ is nonempty (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, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L4]

$L (u, P) = \sum_{i < n} m_{i} Δ_{i}$ and $U (u, P) = \sum_{i < n} M_{i} Δ_{i}$ with $m_{i} = \inf u [I_{i}]$ and $M_{i} = \sup u [I_{i}]$ ; a set with a least element has it as its infimum and with a greatest element has it as its supremum (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 $U (f, P) = \sum_{i} M_{i} Δ_{i}$ , Greatest lower bound (infimum), Maximum and minimum of a set).

[L5]

$\underline{\int_{0}^{1}} u = \sup_{P} L (u, P)$ and $\overline{\int_{0}^{1}} u = \inf_{P} U (u, P)$ ; $u$ is integrable exactly when the two agree (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 $\int_{a}^{b} f$ , Lower bound, bounded below, bounded set).

[L6]

Finite sums: scaling and $\sum_{i < n} λ Δ_{i} = λ \sum_{i < n} Δ_{i}$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products, clause 2).

[L7]

$\int_{p}^{q} c = c (q - p)$ for a constant $c$ (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 $\int_{a}^{b} c = c (b - a)$ ).

[L8]

Absolute value and ordered-field arithmetic: $∣ 1 ∣ = ∣ - 1 ∣ = 1$ , and the order is total (Absolute value in an ordered field, Ordered field, Complete ordered field (least-upper-bound property)).

Counterexample

technique · direct

1.1

$∣ f ∣$ is the constant function $1$ on $[0, 1]$ : at a rational $x$ , $f (x) = 2 \cdot 1 - 1 = 1$ , and at an irrational $x$ , $f (x) = 2 \cdot 0 - 1 = - 1$ , and $∣ 1 ∣ = ∣ - 1 ∣ = 1$ by [L1] and [L8]. Hence $∣ f ∣$ is integrable with $\int_{0}^{1} ∣ f ∣ = 1$ by [L7].

givenL1L7L8

1.2

Let $P = (n, t)$ be any partition of $[0, 1]$ and let $i < n$ . The open interval $(t_{i}, t_{i + 1})$ is nonempty by [L3], so it contains a rational and an irrational by [L2]; both lie in $I_{i}$ , so $1 \in f [I_{i}]$ and $- 1 \in f [I_{i}]$ .

givenL2L3

2.1

$f$ is bounded, with values in ${- 1, 1}$ , so its Darboux sums are defined by [L4] and [L5].

step 1.1givenL4L5

3.1

Since $f [I_{i}] \subseteq {- 1, 1}$ and both values occur, $m_{i} = - 1$ and $M_{i} = 1$ by [L4].

step 2.1step 1.2L4

4.1

Hence $L (f, P) = \sum_{i < n} (- 1) Δ_{i} = - 1$ and $U (f, P) = \sum_{i < n} 1 \cdot Δ_{i} = 1$ , by [L4], [L6] and [L3].

step 3.1L3L4L6

5.1

That holds for every partition $P$ , so the set of lower sums is ${- 1}$ and the set of upper sums is ${1}$ ; by [L5], $\underline{\int_{0}^{1}} f = - 1 \neq 1 = \overline{\int_{0}^{1}} f$ and $f$ is not integrable.

step 4.1L5

6.1

So $∣ f ∣$ is integrable on $[0, 1]$ while $f$ is not, and the claim is false.

step 1.1step 5.1∎

Remarks

The Lebesgue criterion says the same thing. $f$ agrees with $21_{Q} - 1$ , and $1_{Q}$ is continuous at no point of $R$ (The Dirichlet function is continuous at no point of $R$ , 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 $c$ equals $t (c)$ , claim 1), so the discontinuity set of $f$ in $[0, 1]$ is the whole of $[0, 1]$ , which is not null (A sequence of intervals covering $[a, b]$ has total length at least $b - a$ , so no interval of positive length has measure zero); by Lebesgue's criterion for Riemann integrability: a bounded $f$ on $[a, b]$ is Riemann integrable if and only if its set of discontinuities has measure zero $f$ is not integrable. The direct computation above is given because it also locates both Darboux integrals, which the criterion does not.
Why no converse can be expected. Passing to $∣ f ∣$ destroys all sign information, and integrability is a statement about the oscillation of $f$ ; here $∣ f ∣$ has oscillation $0$ on every subinterval while $f$ has oscillation $2$ on every subinterval. The corollary's implication runs only in the direction the composition theorem provides, because $t \mapsto ∣ t ∣$ is continuous and no continuous $ψ$ satisfies $ψ (∣ t ∣) = t$ .
The same example separates $\int ∣ f ∣$ from $∣ \int f ∣$ in the strongest possible way: the right-hand side is not even defined.

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Continuous $f_{n} \to 0$ pointwise on $[0, 1]$ with $\int_{0}^{1} f_{n} = 1$ for every $n$

Statement refuted

False claim: if $(f_{n})$ is a sequence of Riemann integrable functions on $[0, 1]$ converging pointwise to $f$ , and $f$ is integrable, then $\int_{0}^{1} f_{n} \to \int_{0}^{1} f$ .

For $n \in N$ write $c_{n} : = ι (n + 1) \geq 1$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing) and define the tent $f_{n} : [0, 1] \to R$ by

$f_{n} (x) : = {\begin{cases} 4 c_{n}^{2} x & 0 \leq x \leq \frac{1}{2 c_{n}}, \\ 4 c_{n}^{2} (\frac{1}{c_{n}} - x) & \frac{1}{2 c_{n}} \leq x \leq \frac{1}{c_{n}}, \\ 0 & \frac{1}{c_{n}} \leq x \leq 1. \end{cases}$

Each $f_{n}$ is continuous on $[0, 1]$ , hence integrable, with

$\int_{0}^{1} f_{n} = 1 for every n,$

while $f_{n} (x) \to 0$ for every $x \in [0, 1]$ . So the pointwise limit is the zero function, whose integral is $0$ , and the integrals do not converge to it.

The heights are unbounded: $f_{n}$ attains the value $2 c_{n}$ at $x = 1 / (2 c_{n})$ , and $2 c_{n} \to \infty$ . That is what the example refutes and what it does not: it refutes the interchange for pointwise convergence, and it says nothing whatever about sequences that are uniformly bounded, for which no theorem is stated on this page in any direction.

Facts & Assumptions

Given: For $n \in N$ , $c_{n} = ι (n + 1)$ and the function $f_{n}$ above; a point $x \in [0, 1]$ and a real $ε > 0$ .

[L1]

$ι (n + 1) \geq 1 > 0$ , $ι$ is nondecreasing on $N$ , and for every real $η > 0$ there is a natural $N$ with $1 / ι (N + 1) < η$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing, Every complete ordered field is Archimedean, For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L2]

Every polynomial function is continuous, and a function agreeing with continuous functions on the pieces of a finite subdivision of $[0, 1]$ , with matching values at the shared endpoints, is continuous on $[0, 1]$ (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, claim 5, Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point).

[L3]

A continuous function on a closed bounded interval with distinct endpoints is integrable there (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, 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 $\int_{a}^{b} f$ ).

[L4]

Additivity over adjacent subintervals (For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ , The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ ).

[L5]

If $H$ is differentiable at every point of $[p, q]$ with $H^{'}$ integrable there, then $\int_{p}^{q} H^{'} = H (q) - H (p)$ ; a continuous function on an interval has a primitive (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ ).

[L6]

For $n \geq 1$ the map $x \mapsto x^{n}$ has derivative $ι (n) x^{n - 1}$ ; sums and scalar multiples of differentiable functions are differentiable with the corresponding derivatives; and $\int_{p}^{q} c = c (q - p)$ for a constant (For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , 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 $\int_{a}^{b} c = c (b - a)$ , The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Integer powers $a^{m}$ ).

[L7]

A sequence of reals converges to $L$ when for every real $ε > 0$ there is $N$ with $∣ a_{n} - L ∣ < ε$ for all $n \geq N$ (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L8]

Ordered-field arithmetic and linearity of the integral: multiplying inequalities by positive reals, transitivity, and $\int_{p}^{q} (λ u) = λ \int_{p}^{q} u$ (Ordered field, Complete ordered field (least-upper-bound property), Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Counterexample

technique · direct

1.1

$c_{n} \geq 1$ by [L1], so $0 < 1 / (2 c_{n}) < 1 / c_{n} \leq 1$ and the three pieces of the definition subdivide $[0, 1]$ .

givenL1

2.1

The three formulas agree at the shared endpoints: at $x = 1 / (2 c_{n})$ both give $4 c_{n}^{2} / (2 c_{n}) = 2 c_{n}$ , and at $x = 1 / c_{n}$ the second gives $0$ , which is the third. So $f_{n}$ is a well-defined function and is continuous on $[0, 1]$ by [L2], hence integrable by [L3].

step 1.1L2L3

2.2

On $[1 / c_{n}, 1]$ the function $f_{n}$ is constantly $0$ , so $\int_{1 / c_{n}}^{1} f_{n} = 0$ by [L6]; when $1 / c_{n} = 1$ this piece is degenerate and the integral is $0$ by [L4].

step 1.1L4L6

2.3

Pointwise convergence to $0$ . At $x = 0$ every $f_{n} (0) = 0$ . For $x > 0$ , [L1] gives $N$ with $1 / ι (N + 1) < x$ , and for $n \geq N$ one has $c_{n} = ι (n + 1) \geq ι (N + 1)$ , hence $1 / c_{n} \leq 1 / ι (N + 1) < x$ , so $x$ lies in the third piece and $f_{n} (x) = 0$ .

step 1.1givenL1L8

3.1

On $[0, 1 / (2 c_{n})]$ the function $H (x) : = 2 c_{n}^{2} x^{2}$ has $H^{'} (x) = 4 c_{n}^{2} x = f_{n} (x)$ by [L6], so $\int_{0}^{1 / (2 c_{n})} f_{n} = H (1 / (2 c_{n})) - H (0) = 2 c_{n}^{2} / (4 c_{n}^{2}) = 1 / 2$ by [L5].

step 2.1L5L6

3.2

On $[1 / (2 c_{n}), 1 / c_{n}]$ the function $H_{2} (x) : = 4 c_{n}^{2} (x / c_{n} - x^{2} / 2)$ has $H_{2}^{'} (x) = 4 c_{n}^{2} (1 / c_{n} - x) = f_{n} (x)$ by [L6], and $H_{2} (1 / c_{n}) = 4 c_{n}^{2} (1 / c_{n}^{2} - 1 / (2 c_{n}^{2})) = 2$ while $H_{2} (1 / (2 c_{n})) = 4 c_{n}^{2} (1 / (2 c_{n}^{2}) - 1 / (8 c_{n}^{2})) = 3 / 2$ ; so $\int_{1 / (2 c_{n})}^{1 / c_{n}} f_{n} = 2 - 3 / 2 = 1 / 2$ by [L5].

step 2.1L5L6

3.3

Hence $∣ f_{n} (x) - 0 ∣ = 0 < ε$ for all $n \geq N$ , so $f_{n} (x) \to 0$ for every $x \in [0, 1]$ by [L7].

step 2.3L7

4.1

By [L4] applied twice, $\int_{0}^{1} f_{n} = 1 / 2 + 1 / 2 + 0 = 1$ for every $n$ .

step 3.1step 3.2step 2.2L4

5.1

The pointwise limit is the zero function, which is integrable with integral $0$ by [L6], while $\int_{0}^{1} f_{n} = 1$ for every $n$ by step 4.1; so the integrals do not converge to the integral of the limit and the claim is false.

step 4.1step 3.3L6L7∎

Remarks

What this refutes, stated exactly. It refutes the interchange of a limit with an integral under pointwise convergence alone, even when every $f_{n}$ is continuous and the limit function is as regular as possible. It does not refute, and does not address, any statement about uniformly convergent sequences or about uniformly bounded ones; no such statement is proved on this page, and none is contradicted here.
Unboundedness of the heights is essential to the construction and is stated as a feature, not hidden. $\sup_{[0, 1]} f_{n} = 2 c_{n}$ grows without bound, and the mass $1$ escapes into a spike of shrinking width. A reader who wants a theorem in this direction should note that the hypothesis to look for is a bound on the whole sequence, and that whichever theorem supplies it is not on this page.
The integral of the limit exists here. The failure is not that the limit function is non-integrable — it is the zero function — but that the numbers $\int_{0}^{1} f_{n}$ simply do not converge to $\int_{0}^{1} 0$ . A separate failure, in which the pointwise limit of integrable functions is not integrable at all, is recorded as a false statement on the companion page of The Riemann Integral.

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29Open item page →

Continuous $f$ and integrable sign-changing $g$ with $\int_{a}^{b} f g \neq f (ξ) \int_{a}^{b} g$ for every $ξ$

Statement refuted

False claim: if $f$ is continuous on $[a, b]$ and $g$ is integrable on $[a, b]$ , then there is $ξ \in [a, b]$ with $\int_{a}^{b} f g = f (ξ) \int_{a}^{b} g$ .

That is If $f$ is continuous on $[a, b]$ and $g$ is integrable with $g \geq 0$ , there is $ξ \in [a, b]$ with $\int_{a}^{b} f g = f (ξ) \int_{a}^{b} g$ with the hypothesis $g \geq 0$ deleted, and it is false. On $[- 1, 1]$ take

$f (t) = t, g (t) = t .$

Both are continuous, hence integrable, and $g$ changes sign. Then

$\int_{- 1}^{1} f g = \int_{- 1}^{1} t^{2} d t = \frac{2}{ι (3)} > 0, \int_{- 1}^{1} g = \int_{- 1}^{1} t d t = 0,$

so $f (ξ) \int_{- 1}^{1} g = 0$ for every $ξ \in [- 1, 1]$ , while the left-hand side is positive. No $ξ$ works.

Facts & Assumptions

Given: The functions $f (t) = g (t) = t$ on $[- 1, 1]$ , and $ξ \in [- 1, 1]$ arbitrary.

[L1]

Every polynomial function is continuous, and a continuous function on $[- 1, 1]$ is integrable there (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, claim 5, Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point, A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, 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 $\int_{a}^{b} f$ ).

[L2]

A product of two integrable functions on $[- 1, 1]$ is integrable (If $f, g$ are integrable on $[a, b]$ then so are $∣ f ∣$ , $f^{2}$ , $f g$ , $\max (f, g)$ and $\min (f, g)$ , and $∣ \int_{a}^{b} f ∣ \leq \int_{a}^{b} ∣ f ∣$ , claim 1).

[L3]

For $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable at every real $c$ with derivative $ι (n) c^{n - 1}$ , and a scalar multiple of a differentiable function is differentiable with the scaled derivative (For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, claim 2, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , claim 2, The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

[L4]

[L5]

Powers and canonical naturals: $1^{2} = 1$ , $(- 1)^{2} = 1$ , $1^{3} = 1$ , $(- 1)^{3} = - 1$ , $ι (2) = 2 > 0$ and $ι (3) = 3 > 0$ (Integer powers $a^{m}$ , The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing, Ordered field).

[L6]

If $u \geq 0$ is integrable then $\int_{p}^{q} u \geq 0$ , and $u \cdot 0 = 0$ for every real $u$ (If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ , Ordered field, Complete ordered field (least-upper-bound property), Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ ).

Counterexample

technique · direct

1.1

$f$ and $g$ are continuous on $[- 1, 1]$ , hence integrable there by [L1], and $f g$ , the function $t \mapsto t^{2}$ , is integrable by [L1] or [L2].

givenL1L2

1.2

The function $H_{1} (t) : = t^{3} / ι (3)$ is differentiable at every real with $H_{1}^{'} (t) = ι (3) t^{2} / ι (3) = t^{2}$ , by [L3] and [L5].

L3L5construct

1.3

The function $H_{2} (t) : = t^{2} / ι (2)$ is differentiable at every real with $H_{2}^{'} (t) = ι (2) t / ι (2) = t$ , by [L3] and [L5].

L3L5construct

2.1

By [L4] applied to $H_{1}$ on $[- 1, 1]$ , $\int_{- 1}^{1} t^{2} d t = H_{1} (1) - H_{1} (- 1) = 1 / ι (3) - (- 1) / ι (3) = 2 / ι (3)$ , a positive real by [L5].

step 1.1step 1.2L4L5

2.2

By [L4] applied to $H_{2}$ on $[- 1, 1]$ , $\int_{- 1}^{1} t d t = H_{2} (1) - H_{2} (- 1) = 1 / ι (2) - 1 / ι (2) = 0$ .

step 1.1step 1.3L4L5

3.1

For every $ξ \in [- 1, 1]$ , $f (ξ) \int_{- 1}^{1} g = ξ \cdot 0 = 0$ by step 2.2 and [L6], while $\int_{- 1}^{1} f g = 2 / ι (3) > 0$ by step 2.1.

step 2.1step 2.2L5L6

4.1

Hence $\int_{- 1}^{1} f g \neq f (ξ) \int_{- 1}^{1} g$ for every $ξ \in [- 1, 1]$ , and the claim fails at this pair.

step 3.1L5∎

Remarks

What survives when the weight changes sign. The bound $∣ \int_{a}^{b} f g ∣ \leq (\sup_{[a, b]} ∣ f ∣) \int_{a}^{b} ∣ g ∣$ still holds, by If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ applied to $- \sup ∣ f ∣ ∣ g ∣ \leq f g \leq \sup ∣ f ∣ ∣ g ∣$ together with If $f, g$ are integrable on $[a, b]$ then so are $∣ f ∣$ , $f^{2}$ , $f g$ , $\max (f, g)$ and $\min (f, g)$ , and $∣ \int_{a}^{b} f ∣ \leq \int_{a}^{b} ∣ f ∣$ . What is lost is the identity: the weighted average $\int f g / \int g$ need not be a value of $f$ , and here it is not even defined.
Where the proof of the theorem breaks. With $g \geq 0$ the pointwise inequalities $m g \leq f g \leq M g$ survive integration and bracket $\int_{a}^{b} f g$ between $m \int_{a}^{b} g$ and $M \int_{a}^{b} g$ (the pointwise inequalities are step 2.1 of If $f$ is continuous on $[a, b]$ and $g$ is integrable with $g \geq 0$ , there is $ξ \in [a, b]$ with $\int_{a}^{b} f g = f (ξ) \int_{a}^{b} g$ and the bracket is its step 3.1). Multiplying $m \leq f \leq M$ by a $g$ that changes sign reverses the inequality where $g < 0$ , so no such bracket is available, and the pair above shows that no weaker bracket can rescue the conclusion: the left-hand side is positive and the right-hand side is $0$ whatever $ξ$ is.
The theorem with $g \leq 0$ holds too, by applying If $f$ is continuous on $[a, b]$ and $g$ is integrable with $g \geq 0$ , there is $ξ \in [a, b]$ with $\int_{a}^{b} f g = f (ξ) \int_{a}^{b} g$ to $- g$ and using linearity; what cannot be dropped is that $g$ has one sign.

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-29verified 2026-09-09 (gpt-6-astra)Open item page →

FALSE: if $u$ and $v$ are differentiable on $[a, b]$ then $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$

Statement

False claim: if $u, v : [a, b] \to R$ are differentiable at every point of $[a, b]$ (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set), then

$\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v .$

That is If $u, v$ are differentiable on $[a, b]$ with $u^{'}, v^{'}$ integrable, then $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ with the hypothesis " $u^{'}$ and $v^{'}$ are integrable" deleted, and it is false.

The falsity is undefinedness, not a wrong number. Take $[a, b] = [0, 1]$ , let $u : = G$ be the everywhere-differentiable function of A function differentiable on $[0, 1]$ whose derivative is unbounded, hence not Riemann integrable, and let $v (x) : = x$ . Then $u$ and $v$ are differentiable at every point of $[0, 1]$ , and $u v^{'} = G$ is continuous hence integrable, so the left-hand side exists. But $u^{'} v$ is the function $x \mapsto x G^{'} (x)$ , which is unbounded on $[0, 1]$ , hence has no Darboux sums at all (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 $U (f, P) = \sum_{i} M_{i} Δ_{i}$ ) and is not Riemann integrable: the symbol $\int_{0}^{1} u^{'} v$ on the right-hand side does not denote. An equation one of whose sides is undefined is not a true equation.

The correct hypothesis, and when it is automatic. If $u, v$ are differentiable on $[a, b]$ with $u^{'}, v^{'}$ integrable, then $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ asks that $u^{'}$ and $v^{'}$ be integrable, which is what makes $(u v)^{'} = u^{'} v + u v^{'}$ integrable and lets the second fundamental theorem be applied to $u v$ . It holds automatically when $u$ and $v$ are continuously differentiable, since a continuous function on $[a, b]$ is integrable (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion).

Facts & Assumptions

Given: The function $G : [0, 1] \to R$ of A function differentiable on $[0, 1]$ whose derivative is unbounded, hence not Riemann integrable, differentiable at every point of $[0, 1]$ , together with the points $u_{n} : = α_{n} + \frac{1}{4} h_{n}$ of that item, where $α_{n} = 1 / ι (n + 2)$ , and $v (x) : = x$ on $[0, 1]$ .

[A1]

The false claim above.

[L1]

$G$ is differentiable at every point of $[0, 1]$ , $G^{'}$ is unbounded there, and $G^{'} (u_{n}) = \frac{3}{16} ι (n + 2)^{2}$ with $u_{n} > α_{n} = 1 / ι (n + 2)$ (A function differentiable on $[0, 1]$ whose derivative is unbounded, hence not Riemann integrable).

[L2]

$v (x) = x$ is differentiable at every real with $v^{'} (x) = 1$ , and every polynomial function is differentiable (For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, claim 2, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Integer powers $a^{m}$ ).

[L3]

A function differentiable at every point of $[0, 1]$ is continuous there, and a continuous function on $[0, 1]$ is integrable there (A function differentiable at $c$ is continuous at $c$ , A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point, 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 $\int_{a}^{b} f$ ).

[L4]

Darboux sums, and hence Riemann integrability, 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, 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 $\int_{a}^{b} f$ ).

[L5]

$ι (n + 2) \geq 2 > 0$ , and for every real $w$ there is a natural $n$ with $w < ι (n + 2)$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing, Every complete ordered field is Archimedean).

[L6]

Ordered-field arithmetic: multiplying an inequality by a positive real preserves it, the order is total and transitive, and a positive real has a positive inverse (Ordered field, Complete ordered field (least-upper-bound property), Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Refutation

technique · direct

1.1

$u : = G$ and $v$ are differentiable at every point of $[0, 1]$ , by [L1] and [L2]; so the hypothesis of [A1] is satisfied by this pair.

givenA1L1L2

2.1

$u v^{'} = G \cdot 1 = G$ , which is continuous on $[0, 1]$ by [L3] and therefore integrable there; so the left-hand side of [A1] exists.

step 1.1L2L3

2.2

$u^{'} v$ is the function $x \mapsto x G^{'} (x)$ on $[0, 1]$ . At the point $u_{n}$ its value is $u_{n} G^{'} (u_{n}) > α_{n} \cdot \frac{3}{16} ι (n + 2)^{2} = \frac{3}{16} ι (n + 2)$ , using $α_{n} = 1 / ι (n + 2) > 0$ and [L1].

step 1.1L1L5L6

3.1

Given a real $M \geq 0$ , [L5] supplies $n$ with $\frac{16}{3} M < ι (n + 2)$ , so $u_{n} G^{'} (u_{n}) > M$ by step 2.2; hence $u^{'} v$ is unbounded on $[0, 1]$ .

step 2.2L5L6

4.1

By [L4] the function $u^{'} v$ has no Darboux sums and is not Riemann integrable on $[0, 1]$ , so the symbol $\int_{0}^{1} u^{'} v$ appearing in [A1] does not denote a real number.

step 3.1L4

5.1

So [A1] fails at this pair: its left-hand side is defined by step 2.1 and its right-hand side is not, by step 4.1, and the asserted identity is therefore not a true statement about them.

step 2.1step 4.1A1∎

Remarks

A second, even simpler witness. Taking $v \equiv 1$ instead, so $v^{'} = 0$ , makes the left-hand side $\int_{0}^{1} 0 = 0$ and the right-hand side $G (1) \cdot 1 - G (0) \cdot 1 - \int_{0}^{1} G^{'}$ , whose last term is undefined for the same reason. That version is the observation that the second fundamental theorem itself needs its integrability hypothesis; the witness in the refutation is given instead because it keeps both integrands genuinely non-constant.
This is the same defect as in the fundamental theorem, seen through a product. If $u, v$ are differentiable on $[a, b]$ with $u^{'}, v^{'}$ integrable, then $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ is proved by applying The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ to $u v$ , and that theorem needs $(u v)^{'}$ integrable. Deleting the hypothesis here deletes it there.
Nothing is claimed about the identity holding whenever both sides happen to exist. If $u^{'}$ and $v^{'}$ are integrable the identity is If $u, v$ are differentiable on $[a, b]$ with $u^{'}, v^{'}$ integrable, then $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ ; what happens when $u^{'} v$ and $u v^{'}$ are integrable without $u^{'}$ and $v^{'}$ being so is not addressed anywhere on this page.

False statementConstruction: Literature-sourcedVerification: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

FALSE: in the substitution theorem the continuity of $f$ may be weakened to integrability, $f \circ φ$ still being integrable

Statement

False claim: let $c < d$ be reals, let $φ : [c, d] \to R$ be differentiable at every point of $[c, d]$ with $φ^{'}$ integrable, and let $f$ be Riemann integrable on an interval $J$ containing $φ [[c, d]]$ . Then $f \circ φ$ is Riemann integrable on $[c, d]$ — so the hypothesis " $f$ is continuous" in Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ may be weakened to " $f$ is integrable" without the right-hand side $\int_{c}^{d} (f \circ φ) φ^{'}$ losing its meaning.

The claim is false. Let $S \subseteq [0, 1]$ be the Smith-Volterra-Cantor set (The Smith-Volterra-Cantor set: the same construction removing, at stage $n \geq 1$ , an open middle interval of length $4^{- n}$ from each of the $2^{n - 1}$ remaining intervals), which is compact, nowhere dense and not null (The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero), and let

$d_{S} (u) : = \inf {∣ u - s ∣ : s \in S}, φ (x) : = \int_{0}^{x} d_{S} (x \in [0, 1]) .$

Then $φ$ is differentiable at every point of $[0, 1]$ with $φ^{'} = d_{S}$ continuous, hence integrable; $φ$ is strictly increasing; and $φ [S]$ has measure zero. Taking

$f : = 1_{φ [S]} on J : = [0, φ (1)]$

gives an integrable $f$ , because its discontinuity set is contained in the null closed set $φ [S]$ , while

$f \circ φ = 1_{S} on [0, 1],$

whose discontinuity set is exactly $S$ and is not null. So $f \circ φ$ is not Riemann integrable.

What this does and does not show. It shows that continuity of $f$ in Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ cannot simply be weakened to integrability: the composite in the right-hand side need not be integrable. It does not exhibit a pair for which both sides of the substitution identity exist and differ, and no such pair is claimed here.

Facts & Assumptions

Given: The Smith-Volterra-Cantor set $S \subseteq [0, 1]$ , the function $d_{S}$ and the function $φ$ above, a real $ε > 0$ and a natural number $N \geq 1$ .

[L1]

$S$ is a compact, nowhere dense subset of $[0, 1]$ ; it is nonempty; and no cover of $S$ by intervals has total length below $2^{- 1}$ , so $S$ does not have measure zero (The Smith-Volterra-Cantor set: the same construction removing, at stage $n \geq 1$ , an open middle interval of length $4^{- n}$ from each of the $2^{n - 1}$ remaining intervals, The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero, Nowhere dense, meager (first category), residual, and second category subsets of $R$ , Measure zero (a countable cover by intervals of total length below every $ε$ ) and content zero (a finite such cover), A subset of $R$ is compact if and only if it is closed and bounded, Open cover, subcover, compact subset of $R$ (every open cover has a finite subcover), and sequentially compact subset).

[L2]

Nowhere dense plus closed means empty interior: no nonempty open interval is contained in $S$ (Nowhere dense, meager (first category), residual, and second category subsets of $R$ , Interior, closure, boundary and exterior of a subset of $R$ , Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen).

[L3]

$u \mapsto d_{S} (u)$ is defined for every real (the set is nonempty and bounded below by $0$ ) and is $1$ -Lipschitz, hence continuous ( $∣ d (x, A) - d (y, A) ∣ \leq d (x, y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded, Dictionary: for $A \subseteq R$ with the metric $d (x, y) = ∣ x - y ∣$ , continuity and uniform continuity of $f : A \to R$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $R$ is compact in the open-cover sense of $R$ exactly when it is a compact metric subspace, clauses 3 and 6, Greatest lower bound (infimum), Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point).

[L4]

[L5]

First fundamental theorem: for $u$ integrable on $[p, q]$ and continuous at every point, $x \mapsto \int_{p}^{x} u$ is differentiable with derivative $u$ (The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive, The integral function $F (x) : = \int_{a}^{x} f$ of an integrable $f$ , For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ , The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ ).

[L6]

A continuous $w \geq 0$ on $[p, q]$ with $p < q$ and $\int_{p}^{q} w = 0$ vanishes identically (A continuous $f \geq 0$ on $[a, b]$ with $\int_{a}^{b} f = 0$ is identically $0$ ).

[L7]

If $κ \leq w \leq μ$ on $[p, q]$ with $w$ integrable then $κ (q - p) \leq \int_{p}^{q} w \leq μ (q - p)$ (If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ , 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 $\int_{a}^{b} c = c (b - a)$ ).

[L8]

The continuous image of a compact set is compact, and a compact subset of $R$ is closed and bounded (The image of a compact subset of $R$ under a continuous real function is compact, A subset of $R$ is compact if and only if it is closed and bounded).

[L9]

Lebesgue's criterion: a bounded function on $[p, q]$ is integrable if and only if its discontinuity set has measure zero; a set of content zero has measure zero (Lebesgue's criterion for Riemann integrability: a bounded $f$ on $[a, b]$ is Riemann integrable if and only if its set of discontinuities has measure zero, A set of content zero has measure zero, Measure zero (a countable cover by intervals of total length below every $ε$ ) and content zero (a finite such cover), Lower bound, bounded below, bounded set).

[L10]

Finite sums: monotonicity in the terms and $\sum_{i < N} λ = ι (N) λ$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products, clauses 2 and 4); $ι (N) \geq 1 > 0$ for $N \geq 1$ , and for every real $η > 0$ there is $N \geq 1$ with $1 / ι (N) < η$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing, 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).

[L11]

Ordered-field arithmetic and suprema: a nonempty bounded set has a supremum and an infimum; $\min {s, t}$ is at most the average of $s$ and $t$ when $s + t$ is fixed; multiplying inequalities by positive reals preserves them; the order is total and transitive (Complete ordered field (least-upper-bound property), Ordered field, Greatest lower bound (infimum), Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Injection, surjection, bijection).

Refutation

technique · direct

1.1

$d_{S} \geq 0$ everywhere, $d_{S} (u) = 0$ for $u \in S$ , and $d_{S} (u) > 0$ for $u \notin S$ : $S$ is closed by [L1], so some $ρ > 0$ has $(u - ρ, u + ρ) \cap S = \emptyset$ , whence $∣ u - s ∣ \geq ρ$ for every $s \in S$ and $d_{S} (u) \geq ρ$ .

givenL1L3L11

2.1

$d_{S}$ is continuous by [L3], hence integrable on every $[0, x]$ with $x > 0$ by [L4]; so $φ$ is defined on $[0, 1]$ , and by [L5] it is differentiable at every point of $[0, 1]$ with $φ^{'} = d_{S}$ , which is integrable by [L4].

step 1.1L3L4L5

3.1

$φ$ is strictly increasing, hence injective. For $0 \leq x < y \leq 1$ , $φ (y) - φ (x) = \int_{x}^{y} d_{S} \geq 0$ by [L5] and [L7]; if it were $0$ then [L6] would force $d_{S} \equiv 0$ on $[x, y]$ , so $[x, y] \subseteq S$ by step 1.1, contradicting [L2]. Hence $φ (x) < φ (y)$ . In particular $φ (0) = 0 < φ (1)$ .

step 1.1step 2.1L2L5L6L7

3.2

A quadratic contraction on $S$ . Let $x < y$ both lie in $S$ . For $u \in [x, y]$ one has $d_{S} (u) \leq \min {u - x, y - u} \leq (y - x) \cdot 2^{- 1}$ , since $x, y \in S$ ; so by [L5] and [L7], $0 \leq φ (y) - φ (x) = \int_{x}^{y} d_{S} \leq (y - x)^{2} \cdot 2^{- 1}$ .

step 2.1L5L7L11

4.1

$φ [S]$ has content zero. Fix $N \geq 1$ and for $i < N$ put $J_{i} : = [ι (i) / ι (N), ι (i + 1) / ι (N)]$ , so the $J_{i}$ cover $[0, 1]$ and each has length $1 / ι (N)$ . If $S \cap J_{i} \neq \emptyset$ let $E_{i} : = φ [S \cap J_{i}]$ , a nonempty bounded set, and put $a_{i} : = \inf E_{i}$ , $b_{i} : = \sup E_{i}$ ; otherwise put $a_{i} : = b_{i} : = 0$ .

step 3.2L10L11construct

5.1

For $z, w \in E_{i}$ one has $∣ z - w ∣ \leq (1 / ι (N))^{2} \cdot 2^{- 1}$ by step 3.2, the two preimages lying in $S \cap J_{i}$ ; hence $b_{i} \leq a_{i} + (1 / ι (N))^{2} \cdot 2^{- 1}$ , since every $z \in E_{i}$ is at most $w + (1 / ι (N))^{2} 2^{- 1}$ for each fixed $w$ , and then $w \geq b_{i} - (1 / ι (N))^{2} 2^{- 1}$ for every $w$ .

step 3.2step 4.1L11

6.1

Every point of $φ [S]$ lies in some $[a_{i}, b_{i}]$ , because every point of $S$ lies in some $J_{i}$ ; and $\sum_{i < N} (b_{i} - a_{i}) \leq ι (N) \cdot (1 / ι (N))^{2} \cdot 2^{- 1} = 1 / (2 ι (N))$ by [L10].

step 4.1step 5.1L10

7.1

Given $ε > 0$ , [L10] supplies $N \geq 1$ with $1 / (2 ι (N)) \leq ε$ ; so $φ [S]$ has content zero and therefore measure zero by [L9].

step 6.1L9L10

8.1

$f : = 1_{φ [S]}$ is integrable on $J = [0, φ (1)]$ . It is bounded, with values in ${0, 1}$ . $S$ is compact by [L1] and $φ$ is continuous by [L5] and [L3], so $φ [S]$ is compact, hence closed, by [L8]; therefore at every $z \in J ∖ φ [S]$ some neighbourhood misses $φ [S]$ and $f$ vanishes on it, so $f$ is continuous there. The discontinuity set of $f$ is thus contained in $φ [S]$ , which is null by step 7.1, so $f$ is integrable by [L9].

step 2.1step 7.1L1L3L8L9

9.1

$f \circ φ = 1_{S}$ on $[0, 1]$ . For $x \in [0, 1]$ : if $x \in S$ then $φ (x) \in φ [S]$ and $f (φ (x)) = 1$ ; if $x \notin S$ then $φ (x) \notin φ [S]$ , since $φ$ is injective by step 3.1, and $f (φ (x)) = 0$ . Also $φ [[0, 1]] \subseteq J$ by step 3.1.

step 3.1step 8.1

10.1

$1_{S}$ is discontinuous at every point of $S$ . Let $x \in S$ and $ρ > 0$ ; the set $(x - ρ, x + ρ) \cap (0, 1)$ contains a nonempty open interval, which by [L2] is not contained in $S$ , so some $y$ in it has $1_{S} (y) = 0$ while $1_{S} (x) = 1$ ; no $δ$ works for $ε = 2^{- 1}$ . At $x \notin S$ the function vanishes on a neighbourhood, $S$ being closed, so it is continuous there.

step 9.1L1L2L11

11.1

The discontinuity set of $f \circ φ$ on $[0, 1]$ is therefore exactly $S$ , which is not null by [L1]; so $f \circ φ$ is bounded and not Riemann integrable, by [L9].

step 9.1step 10.1L1L9

12.1

So $φ$ is differentiable on $[0, 1]$ with $φ^{'}$ integrable, $f$ is integrable on an interval containing $φ [[0, 1]]$ , and $f \circ φ$ is not integrable: the claim is false, and the continuity hypothesis on $f$ in Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ cannot be weakened to integrability.

step 2.1step 8.1step 9.1step 11.1∎

Remarks

Why $φ$ has to be built and cannot be a familiar function. The set on which $f \circ φ$ misbehaves is $φ^{- 1}$ of a null set, and for a Lipschitz $φ$ with $φ^{'}$ bounded away from $0$ that preimage is again null. What makes the witness work is that $φ^{'} = d_{S}$ vanishes on the whole of $S$ , so $φ$ crushes a set of positive measure onto a null set while remaining injective; step 3.2 is the quantitative form of that crushing.
The substitution identity itself is not refuted here, and that is stated rather than glossed over. For this very pair one has $(f \circ φ) φ^{'} = 1_{S} \cdot d_{S}$ , which is the zero function because $d_{S}$ vanishes on $S$ ; so the right-hand side of the substitution formula does exist for this pair and equals $0$ , as does the left-hand side. What fails is the intermediate claim that the composite $f \circ φ$ is integrable, which is what a proof of the identity with $f$ merely integrable would have to establish, and which is what Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ 's continuity hypothesis delivers through If $f$ is integrable on $[a, b]$ with values in $[m, M]$ and $φ$ is continuous on $[m, M]$ , then $φ \circ f$ is integrable. No pair with both sides defined and unequal is exhibited anywhere on this page.
The fat Cantor set is doing exactly one job. It supplies a closed set with empty interior that is not null. Any such set would serve; $S$ is the published one, and The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero is what supplies all three properties without reproving them.

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29Open item page →

FALSE: for every integrable $f$ on $[a, b]$ , the integral function $F (x) = \int_{a}^{x} f$ satisfies $F^{'} = f$ on $[a, b]$

Statement

False claim: let $a < b$ be reals and let $f : [a, b] \to R$ be Riemann 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 $\int_{a}^{b} f$ ). Then its integral function $F (x) = \int_{a}^{x} f$ (The integral function $F (x) : = \int_{a}^{x} f$ of an integrable $f$ ) is differentiable at every point of $[a, b]$ with $F^{'} (x) = f (x)$ there.

The claim fails in two independent ways, and both are exhibited below.

$F^{'}$ may fail to exist. For the sign function $s$ on $[- 1, 1]$ (The sign function is Riemann integrable on $[- 1, 1]$ and has no primitive there) one has $F (x) = ∣ x ∣ - 1$ , which is not differentiable at $0$ .
$F^{'}$ may exist and differ from $f$ . On $[0, 1]$ let $f (x) : = 0$ for $x \neq 1 / 2$ and $f (1 / 2) : = 1$ . Then $F$ is the zero function, so $F^{'} (1 / 2) = 0$ while $f (1 / 2) = 1$ .

The second witness shows the failure is not exotic: any integrable $f$ that differs from a continuous $g$ at a single point has the same integral function as $g$ , by Changing an integrable function at finitely many points changes neither its integrability nor its integral, and therefore has $F^{'} = g \neq f$ at that point. Continuity of $f$ at the point is what the true theorem The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive asks for, and it asks for nothing more. It is not claimed here to be necessary: what the conclusion needs is the equality $F^{'} (c) = f (c)$ , and the last Remark below exhibits an $f$ discontinuous at a point where that equality nevertheless holds.

Facts & Assumptions

Given: The sign function $s$ on $[- 1, 1]$ of The sign function is Riemann integrable on $[- 1, 1]$ and has no primitive there, with $s (x) = - 1$ for $x < 0$ , $s (0) = 0$ and $s (x) = 1$ for $x > 0$ ; and the function $f$ on $[0, 1]$ with $f (1 / 2) = 1$ and $f (x) = 0$ otherwise.

[A1]

The false claim: for every integrable $u$ on $[p, q]$ , the integral function of $u$ is differentiable everywhere on $[p, q]$ with derivative $u$ .

[L1]

$s$ is Riemann integrable on $[- 1, 1]$ (The sign function is Riemann integrable on $[- 1, 1]$ and has no primitive there, A bounded function on $[a, b]$ that is continuous except at finitely many points is Riemann integrable, Discontinuity of $f$ at a point of its domain, and its classification: removable discontinuity, jump discontinuity and essential discontinuity, equivalently Rudin's discontinuities of the first and of the second kind, Lower bound, bounded below, bounded set).

[L2]

[L3]

Additivity in the oriented form for arbitrary points, and $\int_{p}^{p} u = 0$ (For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ , claim 3, The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ , The integral function $F (x) : = \int_{a}^{x} f$ of an integrable $f$ ).

[L4]

If both one-sided limits of a function at $c$ exist and differ, the two-sided limit does not exist, so the derivative there does not exist (If $c$ is a limit point of the domain from both sides, the limit exists iff both one-sided limits exist and agree, The left and right limits of $f$ at $c$ , as limits of the restrictions of $f$ to $A \cap (- \infty, c)$ and $A \cap (c, \infty)$ , The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ , The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

[L5]

Absolute value: $∣ x ∣ = x$ for $x \geq 0$ , $∣ x ∣ = - x$ for $x \leq 0$ , and $∣ x ∣ / x$ is $1$ for $x > 0$ and $- 1$ for $x < 0$ (Absolute value in an ordered field, Basic properties of the absolute value).

[L6]

First fundamental theorem: if $u$ is integrable on $[p, q]$ and continuous at $c$ , then the integral function of $u$ has derivative $u (c)$ at $c$ (The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive, Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point).

[L7]

Ordered-field arithmetic: the order is total and transitive (Ordered field, Complete ordered field (least-upper-bound property), Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Refutation

technique · direct

1.1

First witness. $s$ is integrable on $[- 1, 1]$ by [L1], and its integral function is $F (x) = \int_{- 1}^{x} s$ .

givenL1

1.2

Second witness. The function $f$ on $[0, 1]$ is bounded and agrees with the constant $0$ off the single point $1 / 2$ , so it is integrable with $\int_{0}^{x} f = \int_{0}^{x} 0 = 0$ for every $x \in [0, 1]$ by [L2] and [L3]; hence its integral function is the zero function.

givenL2L3

2.1

For $x \in [0, 1]$ : by [L3], $F (x) = \int_{- 1}^{0} s + \int_{0}^{x} s$ , and $s$ agrees with the constant $- 1$ on $[- 1, 0]$ off the single point $0$ and with the constant $1$ on $[0, x]$ off the single point $0$ , so [L2] gives $F (x) = - 1 + x$ . For $x \in [- 1, 0]$ : $s$ agrees with the constant $- 1$ on $[x, 0]$ off $0$ , so $\int_{- 1}^{x} s = - (x - (- 1)) = - x - 1$ by [L2] and [L3]. In both cases $F (x) = ∣ x ∣ - 1$ by [L5].

step 1.1L2L3L5

2.2

The zero function is differentiable everywhere with derivative $0$ , so its derivative at $1 / 2$ is $0$ , while $f (1 / 2) = 1 \neq 0$ . Here $F^{'}$ exists at the point and differs from $f$ there, so [A1] fails again, in a different way.

step 1.2givenL2L7

3.1

The difference quotient of $F$ at $0$ is $x \mapsto (F (x) - F (0)) / x = ∣ x ∣ / x$ , which is $1$ for $x > 0$ and $- 1$ for $x < 0$ by [L5]; so its one-sided limits at $0$ are $1$ and $- 1$ .

step 2.1L5

4.1

By [L4] the limit of that quotient at $0$ does not exist, so $F$ is not differentiable at $0$ and [A1] fails at $s$ : the claim is false.

step 3.1A1L4

5.1

Both failures occur exactly at a discontinuity of the integrand: $s$ is discontinuous at $0$ and $f$ at $1 / 2$ . Off those points [L6] applies and gives $F^{'} = f$ , so the correct statement is The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive, whose hypothesis is continuity of the integrand at the point in question.

step 4.1step 2.2L6∎

Remarks

The two witnesses are genuinely different failures. In the first, $F$ has no derivative at the bad point at all; in the second, $F$ is as smooth as could be wished and simply computes a different number. A repair attempting to weaken the conclusion to " $F$ is differentiable wherever it can be" would still be refuted by the second witness.
What is always true of $F$ is one dimension weaker. For every integrable $f$ the integral function is Lipschitz, hence uniformly continuous (The integral function of a bounded integrable $f$ is Lipschitz, hence uniformly continuous); differentiability is exactly what continuity of the integrand buys, and nothing more is available.
The failure set can be much larger than a point. For Thomae's function $t$ on $[0, 1]$ the integral function is identically $0$ , so $F^{'} = 0$ while $t$ is positive at every rational: the claim above then fails at every point of an infinite set, not merely at finitely many. No general statement is made here about an arbitrary integrable $f$ — at a discontinuity where $f$ happens to take the value $F^{'}$ does, the two agree, and $f$ vanishing off ${1 / ι (n + 1)}$ is such a case at the point $0$ .

Properties of the Integral and the Working FTC: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

∫01xm=1/ι(m+1)∫01​xm=1/ι(m+1), computed by the fundamental theorem and checked against the definition

Example

Facts & Assumptions

Verification

Remarks

A step function integrated by additivity over subintervals, and the same value from the definition

Example

Facts & Assumptions

Verification

Remarks

The integral test applied to ∑1/ι(k+1)p∑1/ι(k+1)p for rational p>0p>0, cross-checked against the published pp-series theorem

Example

Facts & Assumptions

Verification

Remarks

H(x)=2xH(x)=2x​ on [0,1][0,1]: HH is continuous, H′H′ is unbounded on (0,1](0,1], and H′H′ is therefore not Riemann integrable

Example

Facts & Assumptions

Verification

Remarks

The sign function is Riemann integrable on [−1,1][−1,1] and has no primitive there

Statement refuted

Facts & Assumptions

Counterexample

Remarks

A function differentiable on [0,1][0,1] whose derivative is unbounded, hence not Riemann integrable

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Integrable φφ and integrable ff with φ∘fφ∘f not integrable: the order of the hypotheses in the composition theorem cannot be reversed

Statement refuted

Facts & Assumptions

Counterexample

Remarks

A function that is not Riemann integrable although ∣f∣∣f∣ is

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Continuous fn→0fn​→0 pointwise on [0,1][0,1] with ∫01fn=1∫01​fn​=1 for every nn

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Continuous ff and integrable sign-changing gg with ∫abfg≠f(ξ)∫abg∫ab​fg=f(ξ)∫ab​g for every ξξ

Statement refuted

Facts & Assumptions

Counterexample

Remarks

FALSE: if uu and vv are differentiable on [a,b][a,b] then ∫abuv′=u(b)v(b)−u(a)v(a)−∫abu′v∫ab​uv′=u(b)v(b)−u(a)v(a)−∫ab​u′v

Statement

Facts & Assumptions

Refutation

Remarks

FALSE: in the substitution theorem the continuity of ff may be weakened to integrability, f∘φf∘φ still being integrable

Statement

Facts & Assumptions

Refutation

Remarks

FALSE: for every integrable ff on [a,b][a,b], the integral function F(x)=∫axfF(x)=∫ax​f satisfies F′=fF′=f on [a,b][a,b]

Statement

Facts & Assumptions

Refutation

Remarks

Sources

$\int_{0}^{1} x^{m} = 1 / ι (m + 1)$ , computed by the fundamental theorem and checked against the definition

The integral test applied to $\sum 1 / ι (k + 1)^{p}$ for rational $p > 0$ , cross-checked against the published $p$ -series theorem

$H (x) = 2 \sqrt{x}$ on $[0, 1]$ : $H$ is continuous, $H^{'}$ is unbounded on $(0, 1]$ , and $H^{'}$ is therefore not Riemann integrable

The sign function is Riemann integrable on $[- 1, 1]$ and has no primitive there

A function differentiable on $[0, 1]$ whose derivative is unbounded, hence not Riemann integrable

Integrable $φ$ and integrable $f$ with $φ \circ f$ not integrable: the order of the hypotheses in the composition theorem cannot be reversed

A function that is not Riemann integrable although $∣ f ∣$ is

Continuous $f_{n} \to 0$ pointwise on $[0, 1]$ with $\int_{0}^{1} f_{n} = 1$ for every $n$

Continuous $f$ and integrable sign-changing $g$ with $\int_{a}^{b} f g \neq f (ξ) \int_{a}^{b} g$ for every $ξ$

FALSE: if $u$ and $v$ are differentiable on $[a, b]$ then $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$

FALSE: in the substitution theorem the continuity of $f$ may be weakened to integrability, $f \circ φ$ still being integrable

FALSE: for every integrable $f$ on $[a, b]$ , the integral function $F (x) = \int_{a}^{x} f$ satisfies $F^{'} = f$ on $[a, b]$