TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

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Botsko's theorem: if $F$ is continuous on $[a, b]$ , $F^{'} (x) = f (x)$ off a countable subset of $(a, b)$ , and $f$ is Riemann integrable, then $\int_{a}^{b} f = F (b) - F (a)$

Statement

Let $a < b$ , let $E \subseteq (a, b)$ be at most countable, let $F : [a, b] \to R$ be continuous, and let $f : [a, b] \to R$ be Riemann integrable. If $F$ is differentiable at every $x \in (a, b) ∖ E$ and

$F^{'} (x) = f (x) (x \in (a, b) ∖ E),$

then

$\int_{a}^{b} f = F (b) - F (a) .$

Neither derivatives at the endpoints nor derivatives at points of $E$ are required.

Facts & Assumptions

Given: The data in the statement.

[L1]

An at-most-countable set is empty or, when nonempty, is the range of a surjection $e : N \to E$ ; repetitions are allowed and no choice is required (Finite, countably infinite, countable, uncountable, A nonempty set is at most countable iff it is a surjective image of $N$ ).

[L2]

Continuity at $c$ means that every prescribed positive error bounds $∣ H (x) - H (c) ∣$ throughout some neighbourhood of $c$ (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]

Differentiability at $x$ means that the difference quotients $(H (y) - H (x)) / (y - x)$ tend to $H^{'} (x)$ as $y \to x$ (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]

A nested sequence of nonempty closed bounded intervals whose lengths tend to $0$ has a one-point intersection (A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to $0$ ).

[L5]

The geometric sequence $2^{- n}$ tends to $0$ (For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ ).

[L6]

For a partition $P$ , the lower and upper Darboux sums use the infima and suprema on its subintervals, and an integrable function has its integral between every such pair of sums (For bounded $f$ on $[a, b]$ and a partition $P$ : the infimum $m_{i}$ and supremum $M_{i}$ of $f$ on the $i$ -th subinterval, and the lower and upper Darboux sums $L (f, P) = \sum_{i} m_{i} Δ_{i}$ and $U (f, P) = \sum_{i} M_{i} Δ_{i}$ , The lower and upper Darboux integrals of a bounded $f$ on $[a, b]$ as $\sup_{P} L (f, P)$ and $\inf_{P} U (f, P)$ , Darboux integrability as their equality, and the notation $\int_{a}^{b} f$ , 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).

Proof

technique · squeeze

1.1

It suffices first to prove the countable-exception monotonicity lemma: if $H : [u, v] \to R$ is continuous and $H^{'} (x) \leq 0$ on $(u, v) ∖ E$ , where $E$ is at most countable, then $H (v) \leq H (u)$ .

givensuffices

1.2

Suppose contrariwise that $H (v) > H (u)$ . By continuity choose $u < p < q < v$ with $H (q) > H (p)$ , and put $c = (H (q) - H (p)) / (q - p) > 0$ .

givenL2algebra

1.3

If $E$ is nonempty, fix the surjection $e : N \to E$ from [L1]; if $E$ is empty, put $e_{n} = u$ for every $n$ . Assign stage $n$ the slope-loss budget $δ_{n} = c 2^{- n - 2}$ . The finite geometric-sum identity gives $\sum_{j \leq n} δ_{j} < c / 2$ for every $n$ .

L1algebra

1.4

Fix a partition $P = (t_{0}, \dots, t_{m})$ and let $m_{i}, M_{i}$ be the infimum and supremum of $f$ on $[t_{i}, t_{i + 1}]$ . Off $E$ , the functions $F (x) - M_{i} x$ and $m_{i} x - F (x)$ have derivatives at most $0$ .

givenL6algebra

2.1

Construct nested closed intervals $I_{n} = [p_{n}, q_{n}] \subseteq [p, q]$ . Start with $I_{0} = [p, q]$ . Given $I_{n}$ , one of its two closed halves has secant slope at least the slope of $I_{n}$ , because the latter is the length-weighted average of the two half-slopes. Call that half $J$ . If $e_{n} \notin J$ , take $I_{n + 1} = J$ . If $e_{n} \in J$ , split $J$ at $e_{n}$ ; one nondegenerate side has slope at least the slope of $J$ , and [L2] lets us move its endpoint $e_{n}$ slightly into that side so that the resulting closed interval excludes $e_{n}$ and loses less than $δ_{n}$ in slope. Thus $I_{n + 1} \subseteq I_{n}$ , $∣ I_{n + 1} ∣ \leq ∣ I_{n} ∣ / 2$ , $e_{n} \notin I_{n + 1}$ , and its secant slope is at least $c - \sum_{j \leq n} δ_{j} \geq c / 2$ .

step 1.2step 1.3L2algebra

3.1

By step 2.1 and [L5], the nested intervals have lengths tending to $0$ , so [L4] gives a unique $x$ in their intersection. The initial interval lies in $(u, v)$ , and $x \neq e_{n}$ for every $n$ ; hence $x \in (u, v) ∖ E$ .

step 2.1L1L4L5

4.1

Write $I_{n} = [p_{n}, q_{n}]$ . Both endpoints tend to $x$ . The secant slope on $I_{n}$ is a convex combination of the two difference quotients based at $x$ (omitting a zero-length side), so [L3] makes those slopes tend to $H^{'} (x)$ . Step 2.1 keeps every slope at least $c / 2$ , whence $H^{'} (x) \geq c / 2 > 0$ , contradicting the hypothesis. Therefore $H (v) \leq H (u)$ and the monotonicity lemma is proved.

step 2.1step 3.1L3algebradischarge-contradiction

5.1

Apply the lemma from step 4.1 on $[t_{i}, t_{i + 1}]$ to obtain $m_{i} (t_{i + 1} - t_{i}) \leq F (t_{i + 1}) - F (t_{i}) \leq M_{i} (t_{i + 1} - t_{i})$ .

step 4.1step 1.4

6.1

Summing step 5.1 and telescoping yields $L (f, P) \leq F (b) - F (a) \leq U (f, P)$ .

step 5.1L6

7.1

Since $f$ is integrable, the supremum of all lower sums and the infimum of all upper sums are both $\int_{a}^{b} f$ ; step 6.1 therefore forces $\int_{a}^{b} f = F (b) - F (a)$ .

step 6.1L6∎

Depends on

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Sources

M. W. Botsko, A Fundamental Theorem of Calculus that Applies to All Riemann Integrable Functions (standard reference, not scraped)
C. Swartz, Even More on the Fundamental Theorem of Calculus (standard reference, not scraped)

Botsko's theorem: if FF is continuous on [a,b][a,b], F′(x)=f(x)F′(x)=f(x) off a countable subset of (a,b)(a,b), and ff is Riemann integrable, then ∫abf=F(b)−F(a)∫ab​f=F(b)−F(a)