False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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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 $f : [a,b] \to \mathbb{R}$ 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 $x \ne 1/2$ 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 $F' = g \ne f$ 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]

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 $m \le f \le M$ on $[a,b]$ then $m(b-a) \le L(f,P) \le \underline{\int_a^b} f \le \overline{\int_a^b} f \le U(f,P) \le M(b-a)$ 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 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 $\varepsilon$ - $\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$ , The derivative $f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}$ of $f : A \to \mathbb{R}$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

[L5]

Absolute value: $|x| = x$ for $x \ge 0$ , $|x| = -x$ for $x \le 0$ , 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 $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ 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 $\mathbb{R}$ : 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 $f(1/2) = 1 \ne 0$ . 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 $\{1/\iota(n+1)\}$ is such a case at the point $0$ .

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Sources

Fundamental theorem of calculus (Wikipedia) (standard reference, not scraped)
Sign function (Wikipedia) (standard reference, not scraped)
J. Lebl, Basic Analysis I, Fundamental theorem of calculus (standard reference, not scraped)

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