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The Riemann Integral: Definition and Integrability: 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-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28Open item page →

$\int_{0}^{1} x^{2} = 1 / 3$ , computed from the Darboux definition with uniform partitions and the closed form $\sum_{k < n} k^{2} = n (n - 1) (2 n - 1) / 6$

Example

Let $f : [0, 1] \to R$ be $f (x) = x^{2}$ (Integer powers $a^{m}$ ). Then $f$ is Riemann integrable on $[0, 1]$ and

$\int_{0}^{1} x^{2} d x = \frac{1}{3} .$

Everything is computed from the definition. For the uniform partition $U_{N}$ of $[0, 1]$ into $N \geq 1$ parts (Partition of $[a, b]$ as a finite strictly increasing list $a = t_{0} < t_{1} < \dots < t_{n} = b$ , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions) the two Darboux 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 $U (f, P) = \sum_{i} M_{i} Δ_{i}$ ) are

$L (f, U_{N}) = \frac{(N - 1) (2 N - 1)}{6 N^{2}}, U (f, U_{N}) = \frac{(N + 1) (2 N + 1)}{6 N^{2}}, U (f, U_{N}) - L (f, U_{N}) = \frac{1}{N},$

where a natural $N$ multiplying or dividing a real stands for its canonical natural $ι (N)$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field), as in clause 2 of Laws of finite sums and finite products. Both expressions converge to $1 / 3$ as $N$ grows, the lower sums from below and the upper sums from above, and the gap $1 / N$ is what Archimedes' property drives to $0$ .

The arithmetic rests on one closed form, valid for every $n \in N$ :

$6 \sum_{k < n} ι (k)^{2} = ι (n) (ι (n) - 1) (2 ι (n) - 1),$

which at $n = 0$ reads $0 = 0$ and is the familiar $n (n - 1) (2 n - 1) / 6$ written so that the empty sum is the case $n = 0$ rather than an exception. Note the indexing: the sum runs over $k < n$ , so its last term is $ι (n - 1)^{2}$ and not $ι (n)^{2}$ .

Facts & Assumptions

Given: $f : [0, 1] \to R$ with $f (x) = x^{2}$ , and for $N \geq 1$ the uniform partition $U_{N} = (N, t)$ of $[0, 1]$ with $t_{i} = ι (i) / ι (N)$ for $i \leq N$ and lengths $Δ_{i} = 1 / ι (N)$ .

[L1]

For $U_{N}$ : $t_{0} = 0$ , $t_{N} = 1$ , $t_{i} < t_{i + 1}$ , $Δ_{i} = 1 / ι (N)$ for $i < N$ , $\sum_{i < N} Δ_{i} = 1$ , and $I_{i} = [t_{i}, t_{i + 1}]$ (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).

[L2]

$m_{i} = \inf f [I_{i}]$ , $M_{i} = \sup f [I_{i}]$ , $L (f, P) = \sum_{i < N} m_{i} Δ_{i}$ , $U (f, P) = \sum_{i < N} M_{i} Δ_{i}$ , and $L (f, P) \leq \underline{\int_{0}^{1}} f \leq \overline{\int_{0}^{1}} f \leq U (f, P)$ for every partition $P$ ; when $f$ is integrable, $\int_{0}^{1} f$ is the common value of the two integrals (For bounded $f$ on $[a, b]$ and a partition $P$ : the infimum $m_{i}$ and supremum $M_{i}$ of $f$ on the $i$ -th subinterval, and the lower and upper Darboux sums $L (f, P) = \sum_{i} m_{i} Δ_{i}$ and $U (f, P) = \sum_{i} M_{i} Δ_{i}$ , The lower and upper Darboux integrals of a bounded $f$ on $[a, b]$ as $\sup_{P} L (f, P)$ and $\inf_{P} U (f, P)$ , Darboux integrability as their equality, and the notation $\int_{a}^{b} f$ ).

[L3]

For $0 \leq a < b$ one has $a^{2} < b^{2}$ ; a set with a least element has it as its infimum and one with a greatest element has it as its supremum (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , Integer powers $a^{m}$ , Greatest lower bound (infimum), Maximum and minimum of a set, Complete ordered field (least-upper-bound property)).

[L4]

The closed form. For every $n \in N$ , $6 \sum_{k < n} ι (k)^{2} = ι (n) (ι (n) - 1) (2 ι (n) - 1)$ . This is an induction on $n$ (The principle of mathematical induction) from the recursion clause $\sum_{k < n + 1} = \sum_{k < n} + ι (n)^{2}$ of Finite sums and finite products, by recursion and the identity $x (x - 1) (2 x - 1) + 6 x^{2} = x (x + 1) (2 x + 1)$ , valid in any commutative ring and in particular for $x = ι (n)$ (Ordered field, Canonical naturals are positive and strictly increasing).

[L5]

Finite sums: scaling, additivity, telescoping $\sum_{i < N} (c_{i + 1} - c_{i}) = c_{N} - c_{0}$ , and $ι (i + 1) = ι (i) + 1$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products, The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing).

[L6]

$x \mapsto x^{2}$ is continuous on $[0, 1]$ , being a polynomial function, and a continuous function on a closed bounded interval with distinct endpoints is Riemann integrable (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, 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).

[L7]

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

[L8]

Ordered-field arithmetic and the absolute value: adding a constant and multiplying by a positive quantity preserve an inequality; the order is total and transitive; $∣ u - v ∣ \leq w$ whenever $c \leq u \leq d$ , $c \leq v \leq d$ and $d - c \leq w$ (Basic properties of the absolute value, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property)). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

Verification

technique · direct

1.1

$f$ is continuous on $[0, 1]$ and $0 < 1$ , so $f$ is Riemann integrable on $[0, 1]$ by [L6]; write $J : = \int_{0}^{1} f$ .

givenL6

1.2

On $I_{i} = [t_{i}, t_{i + 1}]$ with $0 \leq t_{i} < t_{i + 1}$ the function $f$ takes its least value at $t_{i}$ and its greatest at $t_{i + 1}$ , by [L3]; both values are attained, so $m_{i} = t_{i}^{2} = ι (i)^{2} / ι (N)^{2}$ and $M_{i} = t_{i + 1}^{2} = ι (i + 1)^{2} / ι (N)^{2}$ by [L3] and [L1].

givenL1L3

2.1

By [L2], [L5] and [L1], $L (f, U_{N}) = \sum_{i < N} \frac{ι (i)^{2}}{ι (N)^{2}} \cdot \frac{1}{ι (N)} = \frac{1}{ι (N)^{3}} \sum_{i < N} ι (i)^{2}$ , and by [L4] this equals $\frac{ι (N) (ι (N) - 1) (2 ι (N) - 1)}{6 ι (N)^{3}} = \frac{(ι (N) - 1) (2 ι (N) - 1)}{6 ι (N)^{2}}$ .

step 1.2L1L2L4L5

2.2

Likewise $U (f, U_{N}) = \frac{1}{ι (N)^{3}} \sum_{i < N} ι (i + 1)^{2}$ . Since $\sum_{i < N} ι (i + 1)^{2} = \sum_{k < N + 1} ι (k)^{2} - ι (0)^{2} = \sum_{k < N + 1} ι (k)^{2}$ by [L5] and $ι (0) = 0$ , [L4] applied at $N + 1$ gives $U (f, U_{N}) = \frac{(ι (N) + 1) (2 ι (N) + 1)}{6 ι (N)^{2}}$ .

step 1.2L1L2L4L5

2.3

Directly, $U (f, U_{N}) - L (f, U_{N}) = \frac{1}{ι (N)^{3}} \sum_{i < N} (ι (i + 1)^{2} - ι (i)^{2}) = \frac{ι (N)^{2} - ι (0)^{2}}{ι (N)^{3}} = \frac{1}{ι (N)}$ , by telescoping in [L5].

step 1.2L1L2L5

3.1

Expanding by [L8], $\frac{1}{3} - L (f, U_{N}) = \frac{3 ι (N) - 1}{6 ι (N)^{2}} > 0$ and $U (f, U_{N}) - \frac{1}{3} = \frac{3 ι (N) + 1}{6 ι (N)^{2}} > 0$ for $N \geq 1$ , so $L (f, U_{N}) < \frac{1}{3} < U (f, U_{N})$ .

step 2.1step 2.2L7L8

3.2

By [L2], $L (f, U_{N}) \leq J \leq U (f, U_{N})$ as well.

step 1.1step 2.1step 2.2L2

4.1

Both $J$ and $1 / 3$ lie in the interval with endpoints $L (f, U_{N})$ and $U (f, U_{N})$ , whose length is $1 / ι (N)$ by step 2.3, so $∣ J - 1 / 3 ∣ \leq 1 / ι (N)$ for every natural $N \geq 1$ , by [L8].

step 2.3step 3.1step 3.2L8

5.1

If $J \neq 1 / 3$ then $∣ J - 1 / 3 ∣ > 0$ , and [L7] supplies $N \geq 1$ with $1 / ι (N) < ∣ J - 1 / 3 ∣$ , contradicting step 4.1. Hence $J = 1 / 3$ , that is $\int_{0}^{1} x^{2} d x = 1 / 3$ .

step 4.1L7L8∎

Remarks

The first index is $0$ and the last term of the lower sum is not $1$ . The lower sum uses $m_{i} = t_{i}^{2}$ with $i$ running from $0$ , so its first term is $0$ and its last is $((N - 1) / N)^{2} \cdot (1 / N)$ ; the upper sum uses $M_{i} = t_{i + 1}^{2}$ and so ends at $1 \cdot (1 / N)$ . Reading the closed form with the other convention, $\sum_{k = 1}^{n} k^{2} = n (n + 1) (2 n + 1) / 6$ , and attaching it to the wrong sum is the standard way to lose the factor and land on $1 / 2$ or $1 / 6$ instead of $1 / 3$ .
The gap is exactly $1 / N$ , not merely $O (1 / N)$ . Step 2.3 computes it by telescoping without evaluating either sum, which is also the cheapest route to integrability through Riemann's criterion: a bounded $f$ on $[a, b]$ is Darboux integrable if and only if for every real $ε > 0$ there is a partition $P$ with $U (f, P) - L (f, P) < ε$ : the continuity of $x^{2}$ is not needed for that, only for the shortcut taken in step 1.1.
What this example does not do. No antiderivative and no fundamental theorem of calculus is used or available at this point in the reading order; the value $1 / 3$ is extracted from the two sums and the Archimedean property alone. The same computation with $x^{3}$ needs $\sum_{k < n} k^{3}$ and is no harder, but it is not carried out here.

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One refinement worked out for $f (x) = x^{2}$ on $[0, 1]$ : adding the point $1 / 2$ to the trivial partition raises the lower sum from $0$ to $1 / 8$ and lowers the upper sum from $1$ to $5 / 8$

Example

Let $f : [0, 1] \to R$ be $f (x) = x^{2}$ (Integer powers $a^{m}$ ). Let $P_{0}$ be the trivial partition of $[0, 1]$ , with point set ${0, 1}$ , and let $P_{1} : = P_{0} + \frac{1}{2}$ be the partition obtained by inserting the point $1 / 2$ , with point set ${0, \frac{1}{2}, 1}$ (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). Then

$L (f, P_{0}) = 0, L (f, P_{1}) = \frac{1}{8}, U (f, P_{1}) = \frac{5}{8}, U (f, P_{0}) = 1,$

so that

$0 = L (f, P_{0}) \leq L (f, P_{1}) \leq U (f, P_{1}) \leq U (f, P_{0}) = 1,$

which is claim 1 of Refining a partition raises the lower Darboux sum and lowers the upper one, and every lower sum is at most every upper sum: $L (f, P) \leq L (f, P^{'}) \leq U (f, P^{'}) \leq U (f, P)$ when $P^{'}$ refines $P$ , and $L (f, P) \leq U (f, Q)$ for arbitrary partitions $P$ and $Q$ ; moreover the two changes are at most $2 M (n^{'} - n) ∥ P ∥$ with both inequalities strict. The gap drops from $U (f, P_{0}) - L (f, P_{0}) = 1$ to $U (f, P_{1}) - L (f, P_{1}) = \frac{1}{2}$ : exactly one refinement halves it, and $\int_{0}^{1} x^{2} = 1 / 3$ , computed from the Darboux definition with uniform partitions and the closed form $\sum_{k < n} k^{2} = n (n - 1) (2 n - 1) / 6$ shows the uniform partitions drive it to $0$ .

Facts & Assumptions

Given: $f : [0, 1] \to R$ with $f (x) = x^{2}$ ; the partition $P_{0} = (1, t)$ with $t_{0} = 0$ and $t_{k} = 1$ for $k \geq 1$ ; and $P_{1} = (2, s)$ with $s_{0} = 0$ , $s_{1} = 1 / 2$ and $s_{k} = 1$ for $k \geq 2$ .

[L1]

Both $P_{0}$ and $P_{1}$ are partitions of $[0, 1]$ , and $P_{1}$ refines $P_{0}$ , since $pts (P_{0}) = {0, 1} \subseteq {0, \frac{1}{2}, 1} = pts (P_{1})$ ; the subintervals of $P_{0}$ are $[0, 1]$ with length $1$ , and those of $P_{1}$ are $[0, \frac{1}{2}]$ and $[\frac{1}{2}, 1]$ , each of length $\frac{1}{2}$ (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).

[L2]

$m_{i} = \inf f [I_{i}]$ , $M_{i} = \sup f [I_{i}]$ , $L (f, P) = \sum_{i < n} m_{i} Δ_{i}$ and $U (f, P) = \sum_{i < n} M_{i} Δ_{i}$ (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 $U (f, P) = \sum_{i} M_{i} Δ_{i}$ ).

[L3]

For $0 \leq a < b$ one has $a^{2} < b^{2}$ , so on an interval $[c, d] \subseteq [0, 1]$ with $c < d$ the function $f$ has least value $c^{2}$ and greatest value $d^{2}$ , both attained; a set with a least element has it as its infimum and one with a greatest element has it as its supremum (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , Integer powers $a^{m}$ , Greatest lower bound (infimum), Maximum and minimum of a set, Complete ordered field (least-upper-bound property)).

[L4]

If $P^{'}$ refines $P$ then $L (f, P) \leq L (f, P^{'}) \leq U (f, P^{'}) \leq U (f, P)$ (Refining a partition raises the lower Darboux sum and lowers the upper one, and every lower sum is at most every upper sum: $L (f, P) \leq L (f, P^{'}) \leq U (f, P^{'}) \leq U (f, P)$ when $P^{'}$ refines $P$ , and $L (f, P) \leq U (f, Q)$ for arbitrary partitions $P$ and $Q$ ; moreover the two changes are at most $2 M (n^{'} - n) ∥ P ∥$ ).

[L5]

Finite sums of one and of two terms: $\sum_{i < 1} a_{i} = a_{0}$ and $\sum_{i < 2} a_{i} = a_{0} + a_{1}$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L6]

Ordered-field arithmetic: $(\frac{1}{2})^{2} = \frac{1}{4}$ , $\frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}$ , $\frac{1}{8} + \frac{1}{2} = \frac{5}{8}$ , and $0 < \frac{1}{8} < \frac{5}{8} < 1$ (Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property)). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

Verification

technique · direct

1.1

For $P_{0}$ the single subinterval is $[0, 1]$ , so by [L3] $m_{0} = 0^{2} = 0$ and $M_{0} = 1^{2} = 1$ , and by [L2] and [L5], $L (f, P_{0}) = 0 \cdot 1 = 0$ and $U (f, P_{0}) = 1 \cdot 1 = 1$ .

givenL1L2L3L5

1.2

For $P_{1}$ the two subintervals are $[0, \frac{1}{2}]$ and $[\frac{1}{2}, 1]$ , each of length $\frac{1}{2}$ , so by [L3] the extreme values are $m_{0} = 0$ , $M_{0} = (\frac{1}{2})^{2} = \frac{1}{4}$ on the first and $m_{1} = \frac{1}{4}$ , $M_{1} = 1$ on the second.

givenL1L3L6

2.1

Hence by [L2], [L5] and [L6], $L (f, P_{1}) = 0 \cdot \frac{1}{2} + \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}$ and $U (f, P_{1}) = \frac{1}{4} \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} = \frac{1}{8} + \frac{1}{2} = \frac{5}{8}$ .

step 1.2L2L5L6

3.1

Comparing with step 1.1 and using [L6]: $L (f, P_{0}) = 0 < \frac{1}{8} = L (f, P_{1})$ and $U (f, P_{1}) = \frac{5}{8} < 1 = U (f, P_{0})$ , while $L (f, P_{1}) = \frac{1}{8} < \frac{5}{8} = U (f, P_{1})$ . This is the chain of [L4] for the refinement $P_{1}$ of $P_{0}$ , here with every inequality strict.

step 1.1step 2.1L1L4L6∎

Remarks

Refinement is an improvement, never a deterioration. That is the content of Refining a partition raises the lower Darboux sum and lowers the upper one, and every lower sum is at most every upper sum: $L (f, P) \leq L (f, P^{'}) \leq U (f, P^{'}) \leq U (f, P)$ when $P^{'}$ refines $P$ , and $L (f, P) \leq U (f, Q)$ for arbitrary partitions $P$ and $Q$ ; moreover the two changes are at most $2 M (n^{'} - n) ∥ P ∥$ and it is what makes 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 $\int_{a}^{b} f$ well posed: the lower sums increase and the upper sums decrease, so the supremum of the one and the infimum of the other are the right things to take.
A single insertion cannot close the gap. Here it halves it, from $1$ to $\frac{1}{2}$ , and no finite number of insertions makes $U - L$ zero for a non-constant $f$ : each $M_{i} - m_{i}$ is positive whenever $f$ is non-constant on $I_{i}$ . Integrability is the statement that the gap can be made arbitrarily small, not zero, which is exactly the form of Riemann's criterion: a bounded $f$ on $[a, b]$ is Darboux integrable if and only if for every real $ε > 0$ there is a partition $P$ with $U (f, P) - L (f, P) < ε$ .
The values are exact rationals, and worth checking by hand. The lower sum of the refined partition is $\frac{1}{8} = 0.125$ and the upper is $\frac{5}{8} = 0.625$ ; the true integral is $\frac{1}{3}$ ( $\int_{0}^{1} x^{2} = 1 / 3$ , computed from the Darboux definition with uniform partitions and the closed form $\sum_{k < n} k^{2} = n (n - 1) (2 n - 1) / 6$ ), which indeed lies strictly between them.

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$\int_{0}^{3} ⌊ x ⌋ = 3$ : the floor function is nondecreasing, hence integrable, and the integral is computed from the uniform partitions

Example

Let $g : [0, 3] \to R$ be $g (x) = ⌊ x ⌋$ , the integer part (Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ ). Then $g$ is nondecreasing (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $R$ , with the dictionary to monotone sequences), hence Riemann integrable on $[0, 3]$ (A monotone function on $[a, b]$ is Riemann integrable: for the uniform partition into $N$ parts the upper minus lower sum telescopes to $∣ f (b) - f (a) ∣ (b - a) / ι (N)$ ), and

$\int_{0}^{3} ⌊ x ⌋ d x = 3.$

$g$ is discontinuous at $x = 1$ , $x = 2$ and $x = 3$ and continuous elsewhere on $[0, 3]$ (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), so this is an integrable function with genuine jumps, not a continuous one in disguise; the value $3$ is $0 \cdot 1 + 1 \cdot 1 + 2 \cdot 1$ , the three constant pieces weighted by their lengths.

The computation below uses the uniform partition into $N = 3 M$ parts, $M \geq 1$ , for which the lower sum is exactly $3$ at every $M$ and the upper sum is $3 + 3 / M$ . So the lower sums do not merely approach the integral, they attain it.

Facts & Assumptions

Given: $g : [0, 3] \to R$ with $g (x) = ⌊ x ⌋$ ; a natural $M \geq 1$ ; $N : = 3 M$ ; and the uniform partition $U_{N} = (N, t)$ of $[0, 3]$ with $t_{i} = 3 ι (i) / ι (N) = ι (i) / ι (M)$ for $i \leq N$ and lengths $Δ_{i} = 3 / ι (N) = 1 / ι (M)$ .

[L1]

For every real $x$ there is exactly one integer $⌊ x ⌋$ with $⌊ x ⌋ \leq x < ⌊ x ⌋ + 1$ (Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ ).

[L2]

$g$ is nondecreasing: for $x \leq y$ , $⌊ x ⌋ \leq x \leq y < ⌊ y ⌋ + 1$ , and $⌊ x ⌋$ , $⌊ y ⌋$ are integers, so $⌊ x ⌋ \leq ⌊ y ⌋$ , no integer lying strictly between $⌊ y ⌋$ and $⌊ y ⌋ + 1$ (Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ , Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $R$ , with the dictionary to monotone sequences).

[L3]

A monotone function on a closed bounded interval with distinct endpoints is bounded and Riemann integrable (A monotone function on $[a, b]$ is Riemann integrable: for the uniform partition into $N$ parts the upper minus lower sum telescopes to $∣ f (b) - f (a) ∣ (b - a) / ι (N)$ , 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]

For $U_{N}$ : $t_{0} = 0$ , $t_{N} = 3$ , $t_{i} < t_{i + 1}$ , $Δ_{i} = 1 / ι (M)$ , $\sum_{i < N} Δ_{i} = 3$ (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).

[L5]

For a nondecreasing $g$ and a subinterval $I_{i} = [t_{i}, t_{i + 1}]$ : $m_{i} = g (t_{i})$ and $M_{i} = g (t_{i + 1})$ , both attained; $L (g, P) = \sum_{i < N} m_{i} Δ_{i}$ , $U (g, P) = \sum_{i < N} M_{i} Δ_{i}$ , and $L (g, P) \leq \underline{\int_{0}^{3}} g \leq \overline{\int_{0}^{3}} g \leq U (g, P)$ (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 $\int_{a}^{b} f$ , Greatest lower bound (infimum), Maximum and minimum of a set, Complete ordered field (least-upper-bound property)).

[L6]

Finite sums: splitting a sum over $i < 3 M$ into the three blocks $i < M$ , $M \leq i < 2 M$ and $2 M \leq i < 3 M$ ; scaling; and $\sum_{i < M} c = ι (M) c$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L7]

For $i \in N$ : $⌊ ι (i) / ι (M) ⌋ = 0$ when $i < M$ , $= 1$ when $M \leq i < 2 M$ , and $= 2$ when $2 M \leq i < 3 M$ ; and $⌊ ι (j) / ι (M) ⌋ = 3$ for $j = 3 M$ . Each case is [L1] applied to the displayed inequalities $0 \leq ι (i) / ι (M) < 1$ , $1 \leq ι (i) / ι (M) < 2$ , $2 \leq ι (i) / ι (M) < 3$ and $ι (3 M) / ι (M) = 3$ , which follow from $ι$ being strictly increasing and additive (Canonical naturals are positive and strictly increasing, The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Order is preserved by adding a constant and by adding inequalities). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used here follow by adjoining the equality case, in which the two sides coincide.

[L8]

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

[L9]

Ordered-field arithmetic and the absolute value: adding a constant and multiplying by a positive quantity preserve an inequality; the order is total and transitive (Basic properties of the absolute value, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property)). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

Verification

technique · direct

1.1

$g$ is nondecreasing by [L2], and $0 < 3$ , so $g$ is Riemann integrable on $[0, 3]$ by [L3]; write $J : = \int_{0}^{3} g$ .

givenL2L3

1.2

By [L5] and [L7], for $i < N = 3 M$ the lower value is $m_{i} = g (t_{i}) = ⌊ ι (i) / ι (M) ⌋$ , which is $0$ for $i < M$ , $1$ for $M \leq i < 2 M$ and $2$ for $2 M \leq i < 3 M$ ; and the upper value is $M_{i} = g (t_{i + 1}) = ⌊ ι (i + 1) / ι (M) ⌋$ , which is $0$ for $i + 1 < M$ , $1$ for $M \leq i + 1 < 2 M$ , $2$ for $2 M \leq i + 1 < 3 M$ and $3$ for $i + 1 = 3 M$ .

givenL1L4L5L7

2.1

By [L6] and step 1.2, $L (g, U_{N}) = \frac{1}{ι (M)} (\sum_{i < M} 0 + \sum_{M \leq i < 2 M} 1 + \sum_{2 M \leq i < 3 M} 2) = \frac{1}{ι (M)} (0 + ι (M) + 2 ι (M)) = 3$ .

step 1.2L4L5L6L9

2.2

By [L6] and step 1.2, the upper values run over $j = i + 1$ from $1$ to $3 M$ , giving $ι (M) - 1$ indices with value $0$ , then $ι (M)$ with value $1$ , then $ι (M)$ with value $2$ , and the single index $j = 3 M$ with value $3$ ; hence $U (g, U_{N}) = \frac{1}{ι (M)} (0 + ι (M) + 2 ι (M) + 3) = 3 + \frac{3}{ι (M)}$ .

step 1.2L4L5L6L9

3.1

By [L5], $3 = L (g, U_{N}) \leq \underline{\int_{0}^{3}} g \leq J \leq \overline{\int_{0}^{3}} g \leq U (g, U_{N}) = 3 + 3 / ι (M)$ for every natural $M \geq 1$ .

step 1.1step 2.1step 2.2L5

4.1

Hence $0 \leq J - 3 \leq 3 / ι (M)$ for every $M \geq 1$ . If $J > 3$ then $(J - 3) \cdot 3^{- 1} > 0$ and [L8] supplies $M \geq 1$ with $1 / ι (M) < (J - 3) \cdot 3^{- 1}$ , that is $3 / ι (M) < J - 3$ , contradicting step 3.1. So $J = 3$ , that is $\int_{0}^{3} ⌊ x ⌋ d x = 3$ .

step 3.1L8L9∎

Remarks

Why $N$ is taken to be a multiple of $3$ . For a general $N$ the partition points $ι (i) \cdot 3 / ι (N)$ do not land on the integers $1$ and $2$ , where $⌊ \cdot ⌋$ jumps, and both sums acquire a boundary term. Restricting to $N = 3 M$ costs nothing, since integrability is already known from A monotone function on $[a, b]$ is Riemann integrable: for the uniform partition into $N$ parts the upper minus lower sum telescopes to $∣ f (b) - f (a) ∣ (b - a) / ι (N)$ and only one sequence of partitions is needed to pin the value.
The lower sums are exactly $3$ , not merely close to it. This is a feature of the step function, not of the method: $g$ takes its value $⌊ x ⌋$ at the left endpoint of each subinterval of $U_{3 M}$ , so the lower sum is the exact area of the three rectangles. The upper sums overshoot by $3 / M$ , the three jumps of size $1$ each spread over one subinterval of length $1 / M$ .
The general identity for a monotone integrand. By A monotone function on $[a, b]$ is Riemann integrable: for the uniform partition into $N$ parts the upper minus lower sum telescopes to $∣ f (b) - f (a) ∣ (b - a) / ι (N)$ , $U (g, U_{N}) - L (g, U_{N}) = ∣ g (3) - g (0) ∣ \cdot 3 / N$ ; here $g (3) = 3$ , $g (0) = 0$ and $N = 3 M$ , so the gap is $3 / M$ , which is what steps 2.1 and 2.2 compute directly.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28Open item page →

The indicator of the Cantor set is discontinuous exactly on the Cantor set, which is null, so it is Riemann integrable with integral $0$ even though it is discontinuous at uncountably many points

Example

Let $C \subseteq [0, 1]$ be the Cantor middle-thirds set (The Cantor middle-thirds set as the intersection of the sets $C_{n}$ obtained by removing open middle thirds) and let $1_{C} : [0, 1] \to R$ be its indicator, $1_{C} (x) = 1$ for $x \in C$ and $1_{C} (x) = 0$ otherwise. Then:

$1_{C}$ is discontinuous at every point of $C$ and continuous at every point of $[0, 1] ∖ C$ , so its set of discontinuities is exactly $C$ (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, 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);
$1_{C}$ is Riemann integrable on $[0, 1]$ , because $C$ has measure zero (The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points, 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);
$\int_{0}^{1} 1_{C} = 0$ .

The point of the example is claim 2 against claim 1. The discontinuity set is uncountable (The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points, Finite, countably infinite, countable, uncountable), so no cardinality argument such as A bounded function on $[a, b]$ whose set of discontinuities is at most countable is Riemann integrable applies; what makes the function integrable is that $C$ can be covered by intervals of arbitrarily small total length, and nothing else.

Only the implication "measure zero $\Rightarrow$ integrable" of 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 is used, so no choice principle is involved.

Facts & Assumptions

Given: The Cantor set $C \subseteq [0, 1]$ and its indicator $1_{C} : [0, 1] \to R$ .

[L1]

$C$ is closed and bounded, has measure zero, is uncountable, and contains no interval with two distinct endpoints; in particular $C$ is nowhere dense, so the interior of $C$ is empty and every nonempty open subset of $R$ contains a point outside $C$ (The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points, The Cantor middle-thirds set as the intersection of the sets $C_{n}$ obtained by removing open middle thirds, Measure zero (a countable cover by intervals of total length below every $ε$ ) and content zero (a finite such cover), Nowhere dense, meager (first category), residual, and second category subsets of $R$ , Interior, closure, boundary and exterior of a subset of $R$ , Finite, countably infinite, countable, uncountable).

[L2]

A set is closed exactly when every point outside it has a neighbourhood missing it (Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen, The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ).

[L3]

A bounded $f$ on $[a, b]$ with $a < b$ is Riemann integrable if and only if its set of discontinuities has measure zero; the implication from "measure zero" to "integrable" uses no choice principle (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, Lower bound, bounded below, bounded set).

[L4]

For a partition $P = (n, t)$ of $[0, 1]$ : $n \geq 1$ , $t_{i} < t_{i + 1}$ , $Δ_{i} > 0$ , $I_{i} = [t_{i}, t_{i + 1}] \subseteq [0, 1]$ , and $(t_{i}, t_{i + 1})$ is a nonempty open subset of $[0, 1]$ (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, Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen).

[L5]

$m_{i} = \inf f [I_{i}]$ , $L (f, P) = \sum_{i < n} m_{i} Δ_{i}$ , $\underline{\int_{0}^{1}} f$ is the supremum of the lower sums, $\overline{\int_{0}^{1}} f$ the infimum of the upper sums, and the integral is their common value when they agree (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 $\int_{a}^{b} f$ ).

[L6]

A set with a least element has it as its infimum; the supremum of ${0}$ is $0$ (Greatest lower bound (infimum), Maximum and minimum of a set, Complete ordered field (least-upper-bound property)).

[L7]

Finite sums: $\sum_{i < n} 0 = 0$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L8]

Ordered-field arithmetic: for $0 \leq x \leq 1$ and a real $ρ > 0$ the reals $u : = \max {0, x - ρ}$ and $v : = \min {1, x + ρ}$ satisfy $u < v$ and $(u, v) \subseteq N_{ρ} (x) \cap [0, 1]$ (Maximum and minimum of a set, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property), The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

Verification

technique · direct

1.1

$1_{C}$ takes only the values $0$ and $1$ , so it is bounded on $[0, 1]$ and its Darboux sums and integrals are defined by [L5].

givenL5

1.2

Discontinuity on $C$ . Let $x \in C$ , so $1_{C} (x) = 1$ , and let a real $ρ > 0$ be given. By [L8] the set $(u, v) \subseteq N_{ρ} (x) \cap [0, 1]$ is a nonempty open interval, so by [L1] it contains a point $y \notin C$ ; then $y \in [0, 1]$ , $∣ y - x ∣ < ρ$ and $∣ 1_{C} (x) - 1_{C} (y) ∣ = 1$ . So the continuity condition fails at $x$ for $ε : = 1$ .

givenL1L8

1.3

Continuity off $C$ . Let $x \in [0, 1]$ with $x \notin C$ . Since $C$ is closed, [L2] gives a real $ρ > 0$ with $N_{ρ} (x) \cap C = \emptyset$ , so $1_{C}$ vanishes on $N_{ρ} (x) \cap [0, 1]$ and $∣ 1_{C} (y) - 1_{C} (x) ∣ = 0 < ε$ there for every $ε > 0$ .

givenL1L2

2.1

So the set of discontinuities of $1_{C}$ in $[0, 1]$ is exactly $C$ , which has measure zero by [L1]; by [L3] and $0 < 1$ , $1_{C}$ is Riemann integrable on $[0, 1]$ .

step 1.1step 1.2step 1.3L1L3

2.2

Every lower sum is $0$ . Let $P = (n, t)$ be a partition of $[0, 1]$ and $i < n$ . By [L4] the interval $(t_{i}, t_{i + 1})$ is a nonempty open subset of $[0, 1]$ , so by [L1] it contains a point outside $C$ , at which $1_{C}$ takes the value $0$ ; since $1_{C} \geq 0$ , the value $0$ is the least element of $1_{C} [I_{i}]$ and $m_{i} = 0$ by [L6]. Hence $L (1_{C}, P) = 0$ by [L5] and [L7].

step 1.1L1L4L5L6L7

3.1

The set of lower sums is ${0}$ , so $\underline{\int_{0}^{1}} 1_{C} = 0$ by [L6]; and $1_{C}$ is integrable by step 2.1, so $\int_{0}^{1} 1_{C} = 0$ by [L5].

step 2.1step 2.2L5L6∎

Remarks

Uncountably many discontinuities, and the integral does not notice. $C$ is uncountable (The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points), so this function is outside the reach of A bounded function on $[a, b]$ whose set of discontinuities is at most countable is Riemann integrable and is the standard demonstration that the Lebesgue criterion is strictly stronger than the countable one.
Every upper sum is at least the total length of the subintervals meeting $C$ , and that total goes to $0$ . The proof above does not need this, since integrability comes from the criterion and the value from the lower sums alone; but it is the reason the upper sums also converge to $0$ , and it is precisely where the Smith-Volterra-Cantor set behaves differently (The indicator of the Smith-Volterra-Cantor set is discontinuous exactly on a nowhere dense set, and is not Riemann integrable, because that set does not have measure zero).
The two hypotheses of claim 1 are the two properties of $C$ that matter. Closedness gives continuity off $C$ ; empty interior gives discontinuity on $C$ . A set with both is exactly a closed nowhere dense set, and any such set is the discontinuity set of its own indicator. Whether that indicator is integrable then depends only on whether the set is null, which is what FALSE: a bounded function on $[a, b]$ is Riemann integrable exactly when its set of discontinuities is nowhere dense settles in the negative for category.

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Thomae's function is Riemann integrable on $[0, 1]$ with integral $0$ : it is continuous at every irrational, so its discontinuity set is countable, and every lower Darboux sum is $0$

Example

Let $t : [0, 1] \to R$ be Thomae's function restricted to $[0, 1]$ : at a rational $x$ with least denominator $q (x) \geq 1$ it takes the value $1 / ι (q (x))$ , and at an irrational $x$ the value $0$ (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). Then $t$ is Riemann integrable on $[0, 1]$ and

$\int_{0}^{1} t = 0.$

Two ingredients, and they pull in opposite directions. $t$ is continuous at every irrational and discontinuous at every rational (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 $t (c)$ ), so its discontinuity set is $Q \cap [0, 1]$ , which is infinite and dense — and countable, which is what A bounded function on $[a, b]$ whose set of discontinuities is at most countable is Riemann integrable needs. The value is then read off the lower sums, every one of which is $0$ because every subinterval contains an irrational (Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable).

Every upper sum, by contrast, is strictly positive, since every subinterval contains a rational; the upper integral is nevertheless $0$ , an infimum of positive numbers.

Facts & Assumptions

Given: Thomae's function $t : [0, 1] \to R$ as above.

[L1]

$t (x) = 1 / ι (q (x))$ with $ι (q (x)) \geq 1 > 0$ at a rational $x$ , and $t (x) = 0$ at an irrational $x$ ; hence $0 \leq t (x) \leq 1$ everywhere and $t (x) > 0$ at every rational $x$ (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 $x$ , The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing).

[L2]

Thomae's function on $R$ is continuous at every irrational and discontinuous at every rational (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 $t (c)$ ); a restriction is continuous at every point of the smaller domain at which the original is continuous, the same $δ$ serving a condition quantified over fewer points (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 $[a, b]$ with $a < b$ whose set of discontinuities is at most countable is Riemann integrable (A bounded function on $[a, b]$ whose set of discontinuities is at most countable is Riemann integrable, Lower bound, bounded below, bounded set).

[L5]

The irrationals are dense in $R$ , so every nonempty open interval contains an irrational (Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable, The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points, Interior, closure, boundary and exterior of a subset of $R$ , The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ).

[L6]

For a partition $P = (n, s)$ of $[0, 1]$ : $n \geq 1$ , $s_{i} < s_{i + 1}$ , $Δ_{i} > 0$ , $I_{i} = [s_{i}, s_{i + 1}] \subseteq [0, 1]$ , and $(s_{i}, s_{i + 1})$ is a nonempty open interval (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).

[L7]

$m_{i} = \inf t [I_{i}]$ , $L (t, P) = \sum_{i < n} m_{i} Δ_{i}$ , $\underline{\int_{0}^{1}} t$ is the supremum of the lower sums and $\overline{\int_{0}^{1}} t$ the infimum of the upper sums, and the integral is their common value when they agree (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 $\int_{a}^{b} f$ ).

[L8]

A set with a least element has it as its infimum; the supremum of ${0}$ is $0$ (Greatest lower bound (infimum), Maximum and minimum of a set, Complete ordered field (least-upper-bound property)).

[L9]

Finite sums: $\sum_{i < n} 0 = 0$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L10]

Ordered-field arithmetic: the order is total and transitive, and a reciprocal of a positive quantity is positive (Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property)). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

Verification

technique · direct

1.1

$t$ is bounded on $[0, 1]$ , with $0 \leq t (x) \leq 1$ for every $x$ , by [L1].

givenL1L10

1.2

By [L2], $t$ is continuous at every irrational point of $[0, 1]$ , so its set of discontinuities in $[0, 1]$ is contained in $Q \cap [0, 1]$ , which is at most countable by [L3]; a subset of it is then at most countable as well.

givenL2L3

2.1

By [L4] applied on $[0, 1]$ , with $0 < 1$ , $t$ is Riemann integrable on $[0, 1]$ .

step 1.1step 1.2L4

2.2

Every lower sum is $0$ . Let $P = (n, s)$ be a partition of $[0, 1]$ and $i < n$ . By [L6] the interval $(s_{i}, s_{i + 1})$ is nonempty and open, so by [L5] it contains an irrational $y$ , and $y \in I_{i} \subseteq [0, 1]$ , so $t (y) = 0$ by [L1]. Since $t \geq 0$ by [L1], the value $0$ is the least element of $t [I_{i}]$ and $m_{i} = 0$ by [L8]. Hence $L (t, P) = 0$ by [L7] and [L9].

step 1.1L1L5L6L7L8L9

3.1

The set of lower sums is therefore ${0}$ and $\underline{\int_{0}^{1}} t = 0$ by [L8]; since $t$ is integrable by step 2.1, $\int_{0}^{1} t = 0$ by [L7].

step 2.1step 2.2L7L8∎

Remarks

The discontinuity set is dense and the function is still integrable. $Q \cap [0, 1]$ meets every subinterval of $[0, 1]$ , so no partition isolates the bad points; what saves the function is that the set is countable, hence null (Every at most countable subset of $R$ has measure zero). This is the cleanest witness that "small" for integrability means small in measure and not small in category or in closure.
The same function refutes a plausible converse. $t$ is nonnegative, integrable with integral $0$ , and positive at every rational, so a vanishing integral does not force a nonnegative integrand to vanish (Thomae's function is nonnegative, Riemann integrable on $[0, 1]$ with integral $0$ , and nonzero at every rational, so a vanishing integral does not force a nonnegative integrand to vanish, FALSE: a nonnegative Riemann integrable function on $[a, b]$ with $\int_{a}^{b} f = 0$ is identically zero).
Contrast with the Dirichlet function. $1_{Q}$ is discontinuous everywhere and not integrable (The Dirichlet function on $[0, 1]$ has lower Darboux integral $0$ and upper Darboux integral $1$ , so it is bounded and not Riemann integrable), yet it is nonzero at exactly the same points as $t$ . Only the values differ, and they differ in a way that makes $t$ continuous at every irrational; that is the whole of the difference.

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28Open item page →

The Dirichlet function on $[0, 1]$ has lower Darboux integral $0$ and upper Darboux integral $1$ , so it is bounded and not Riemann integrable

Statement refuted

Refuted: that every bounded function on a closed bounded interval with distinct endpoints is Riemann integrable (FALSE: every bounded function on $[a, b]$ is 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 $\int_{a}^{b} f$ ).

The witness is the Dirichlet function $1_{Q}$ restricted to $[0, 1]$ (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 $x$ ). It takes only the values $0$ and $1$ , so it is bounded (Lower bound, bounded below, bounded set); every lower Darboux sum is $0$ and every upper Darboux sum is $1$ ; hence

$\underline{\int_{0}^{1}} 1_{Q} = 0 \neq 1 = \overline{\int_{0}^{1}} 1_{Q},$

and the function is not Riemann integrable. The two Darboux integrals are as far apart as the range of the function allows.

Facts & Assumptions

Given: $g : [0, 1] \to R$ with $g (x) = 1$ for rational $x$ and $g (x) = 0$ for 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 $x$ ).

[A1]

The refuted claim: every bounded function on such an interval is Riemann integrable (FALSE: every bounded function on $[a, b]$ is Riemann integrable).

[L1]

Both $Q$ and $R ∖ Q$ 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, The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points, Interior, closure, boundary and exterior of a subset of $R$ , The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ).

[L2]

For a partition $P = (n, t)$ of $[0, 1]$ : $n \geq 1$ , $t_{i} < t_{i + 1}$ , $Δ_{i} > 0$ , $\sum_{i < n} Δ_{i} = 1$ , $I_{i} = [t_{i}, t_{i + 1}]$ , and $(t_{i}, t_{i + 1})$ is a nonempty open interval contained in $I_{i}$ (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).

[L3]

$m_{i} = \inf g [I_{i}]$ , $M_{i} = \sup g [I_{i}]$ , $L (g, P) = \sum_{i < n} m_{i} Δ_{i}$ , $U (g, P) = \sum_{i < n} M_{i} Δ_{i}$ ; $\underline{\int_{0}^{1}} g$ is the supremum of the lower sums and $\overline{\int_{0}^{1}} g$ the infimum of the upper sums; $g$ is integrable exactly when they agree (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 $\int_{a}^{b} f$ ).

[L4]

A set with a least element has it as its infimum and one with a greatest element has it as its supremum; the supremum and infimum of ${c}$ are both $c$ (Greatest lower bound (infimum), Maximum and minimum of a set, Complete ordered field (least-upper-bound property)).

[L5]

Finite sums: scaling and $\sum_{i < n} 0 = 0$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L6]

Ordered-field arithmetic: $0 \neq 1$ , and the order is total and transitive (Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property)). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

Counterexample

technique · direct

1.1

$g$ is bounded, with $0 \leq g (x) \leq 1$ for every $x \in [0, 1]$ , so its Darboux sums and integrals are defined by [L3].

givenL3

1.2

Let $P = (n, t)$ be any partition of $[0, 1]$ and $i < n$ . By [L2] the interval $(t_{i}, t_{i + 1})$ is nonempty and open, so by [L1] it contains a rational and an irrational, both lying in $I_{i}$ . Hence $g [I_{i}] = {0, 1}$ and, by [L4], $m_{i} = 0$ and $M_{i} = 1$ .

givenL1L2L4

2.1

Therefore $L (g, P) = \sum_{i < n} 0 \cdot Δ_{i} = 0$ and $U (g, P) = \sum_{i < n} 1 \cdot Δ_{i} = \sum_{i < n} Δ_{i} = 1$ , for every partition $P$ of $[0, 1]$ , by [L3], [L5] and [L2].

step 1.2L2L3L5

3.1

The set of lower sums is ${0}$ and the set of upper sums is ${1}$ , so $\underline{\int_{0}^{1}} g = 0$ and $\overline{\int_{0}^{1}} g = 1$ by [L4] and [L3]. Since $0 \neq 1$ by [L6], $g$ is not Riemann integrable on $[0, 1]$ .

step 2.1L3L4L6

4.1

So $g$ is bounded on $[0, 1]$ , an interval with $0 < 1$ , and is not Riemann integrable: [A1] is refuted.

step 1.1step 3.1A1∎

Remarks

The failure is uniform over partitions. No partition does better than any other here: $U (g, P) - L (g, P) = 1$ for every $P$ , so Riemann's criterion (Riemann's criterion: a bounded $f$ on $[a, b]$ is Darboux integrable if and only if for every real $ε > 0$ there is a partition $P$ with $U (f, P) - L (f, P) < ε$ ) fails as badly as it can. Contrast the Cantor-set indicator, where the gap does go to $0$ (The indicator of the Cantor set is discontinuous exactly on the Cantor set, which is null, so it is Riemann integrable with integral $0$ even though it is discontinuous at uncountably many points).
The Lebesgue criterion gives the same verdict for the same reason. $g$ is discontinuous at every point of $[0, 1]$ (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 $t (c)$ ), and $[0, 1]$ 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), so 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 refuses it too. The direct computation above is kept because it is elementary and costs no choice principle.
The Riemann sums are not even a warning. Every uniform partition with rational tags gives Riemann sum $1$ , so along that one sequence of tagged partitions the sums converge; this is precisely why the Riemann definition quantifies over all tagged partitions of small mesh (For the Dirichlet function every uniform partition with rational tags gives Riemann sum $1$ , so the sums converge along that sequence of tagged partitions although the function is not integrable: the mesh condition of the Riemann definition quantifies over all tagged partitions and cannot be weakened to one sequence).

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28Open item page →

The indicator of the Smith-Volterra-Cantor set is discontinuous exactly on a nowhere dense set, and is not Riemann integrable, because that set does not have measure zero

Statement refuted

Refuted: that a bounded function on $[a, b]$ is Riemann integrable whenever its set of discontinuities is nowhere dense (FALSE: a bounded function on $[a, b]$ is Riemann integrable exactly when its set of discontinuities is nowhere dense, Nowhere dense, meager (first category), residual, and second category subsets of $R$ ).

The witness is the indicator $g$ of the Smith-Volterra-Cantor set $S \subseteq [0, 1]$ (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). Its set of discontinuities is exactly $S$ , which is closed, nowhere dense and not of measure zero (The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero); and $g$ is not Riemann integrable, with

$\underline{\int_{0}^{1}} g = 0, \overline{\int_{0}^{1}} g \geq \frac{1}{2} .$

The contrast with the Cantor set is the whole point. The Cantor set is also closed and nowhere dense, and its indicator is integrable, with integral $0$ (The indicator of the Cantor set is discontinuous exactly on the Cantor set, which is null, so it is Riemann integrable with integral $0$ even though it is discontinuous at uncountably many points). The two sets differ only in measure, and that is what decides (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).

The proof below is direct, from claim 4 of The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero; it does not go through the forward half of 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 and so uses no choice principle.

Facts & Assumptions

Given: The Smith-Volterra-Cantor set $S \subseteq [0, 1]$ and its indicator $g : [0, 1] \to R$ , with $g (x) = 1$ for $x \in S$ and $g (x) = 0$ otherwise.

[A1]

The refuted claim: a bounded function on a closed bounded interval with distinct endpoints whose set of discontinuities is nowhere dense is Riemann integrable.

[L1]

$S$ is closed, bounded and nowhere dense, so $S$ contains no nonempty open set; and if $(a_{k})$ , $(b_{k})$ are sequences of reals with $a_{k} \leq b_{k}$ , $S \subseteq ⋃_{k} [a_{k}, b_{k}]$ and $\sum_{k < i} (b_{k} - a_{k}) \leq M$ for every $i \in N$ , then $M \geq 2^{- 1}$ (The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and 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, Nowhere dense, meager (first category), residual, and second category subsets of $R$ , Interior, closure, boundary and exterior of a subset of $R$ , Measure zero (a countable cover by intervals of total length below every $ε$ ) and content zero (a finite such cover)).

[L2]

[L3]

For a partition $P = (n, t)$ of $[0, 1]$ : $n \geq 1$ , $t_{i} < t_{i + 1}$ , $Δ_{i} > 0$ , $I_{i} = [t_{i}, t_{i + 1}]$ , $[0, 1] = ⋃_{i < n} I_{i}$ , and $(t_{i}, t_{i + 1})$ is a nonempty open subset of $[0, 1]$ (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, Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen).

[L4]

[L5]

A set with a least element has it as its infimum and one with a greatest element has it as its supremum; an infimum of a set all of whose members are $\geq c$ is $\geq c$ (Greatest lower bound (infimum), Maximum and minimum of a set, Complete ordered field (least-upper-bound property)).

[L6]

Finite sums: scaling, splitting, monotonicity in the terms, $\sum_{i < n} 0 = 0$ ; and a finite list of closed intervals extends to a sequence by degenerate intervals of length $0$ without changing any partial total (Finite sums and finite products, by recursion, Laws of finite sums and finite products, Measure zero (a countable cover by intervals of total length below every $ε$ ) and content zero (a finite such cover)).

[L7]

Ordered-field arithmetic: for $0 \leq x \leq 1$ and a real $ρ > 0$ the reals $u : = \max {0, x - ρ}$ and $v : = \min {1, x + ρ}$ satisfy $u < v$ and $(u, v) \subseteq N_{ρ} (x) \cap [0, 1]$ ; and $0 < 2^{- 1} < 1$ (Maximum and minimum of a set, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property), The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

Counterexample

technique · direct

1.1

$g$ takes only the values $0$ and $1$ , so it is bounded on $[0, 1]$ and its Darboux sums and integrals are defined by [L4].

givenL4

1.2

Discontinuity on $S$ . Let $x \in S$ , so $g (x) = 1$ , and let a real $ρ > 0$ be given. By [L7] the interval $(u, v) \subseteq N_{ρ} (x) \cap [0, 1]$ is nonempty and open, so by [L1] it contains a point $y \notin S$ ; then $∣ y - x ∣ < ρ$ , $y \in [0, 1]$ and $∣ g (x) - g (y) ∣ = 1$ , so continuity fails at $x$ for $ε : = 1$ (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, 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).

givenL1L7

1.3

Continuity off $S$ . Let $x \in [0, 1]$ with $x \notin S$ . Since $S$ is closed, [L2] gives a real $ρ > 0$ with $N_{ρ} (x) \cap S = \emptyset$ , so $g$ vanishes on $N_{ρ} (x) \cap [0, 1]$ and continuity holds at $x$ .

givenL1L2

2.1

So the set of discontinuities of $g$ in $[0, 1]$ is exactly $S$ , which is nowhere dense by [L1].

step 1.2step 1.3L1

2.2

Every lower sum is $0$ . Let $P = (n, t)$ be a partition of $[0, 1]$ and $i < n$ . By [L3] the interval $(t_{i}, t_{i + 1})$ is a nonempty open subset of $[0, 1]$ , so by [L1] it contains a point outside $S$ , at which $g$ takes the value $0$ ; as $g \geq 0$ , that value is the least element of $g [I_{i}]$ and $m_{i} = 0$ by [L5]. Hence $L (g, P) = 0$ by [L4] and [L6].

step 1.1L1L3L4L5L6

2.3

Every upper sum is at least $2^{- 1}$ . With $P$ as above put $B : = {i < n : I_{i} \cap S \neq \emptyset}$ . For $i \in B$ the set $g [I_{i}]$ contains $1$ , so $M_{i} = 1$ by [L5]; for $i \notin B$ one has $g [I_{i}] = {0}$ and $M_{i} = 0$ . Hence $U (g, P)$ is the sum of the $Δ_{i}$ with $i \in B$ , by [L4] and [L6].

step 1.1L4L5L6

3.1

The intervals $I_{i}$ with $i \in B$ cover $S$ , since $S \subseteq [0, 1] = ⋃_{i < n} I_{i}$ by [L3]. Extending that finite list to a sequence by degenerate intervals $[0, 0]$ ([L6]) gives a cover of $S$ all of whose partial total lengths are at most $U (g, P)$ , so [L1] gives $U (g, P) \geq 2^{- 1}$ .

step 2.3L1L3L6

4.1

By [L5] and step 2.2, $\underline{\int_{0}^{1}} g = 0$ ; by [L5] and step 3.1, $\overline{\int_{0}^{1}} g \geq 2^{- 1} > 0$ . The two differ, so $g$ is not Riemann integrable by [L4].

step 2.2step 3.1L4L5L7

5.1

So $g$ is bounded on $[0, 1]$ , an interval with $0 < 1$ , its set of discontinuities is nowhere dense by step 2.1, and it is not Riemann integrable: [A1] is refuted.

step 2.1step 4.1A1∎

Remarks

Nowhere dense and null are independent, and only the second matters here. $S$ is nowhere dense and not null; $Q \cap [0, 1]$ is null and dense. Thomae's function has the second as its discontinuity set and is integrable (Thomae's function is Riemann integrable on $[0, 1]$ with integral $0$ : it is continuous at every irrational, so its discontinuity set is countable, and every lower Darboux sum is $0$ ); $g$ has the first and is not. Neither notion of smallness implies the other, and 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 names the one that decides.
The constant $2^{- 1}$ is what this library can state, and it is enough. No outer measure is defined here, so "the measure of $S$ is $1 / 2$ " is not a statement available; claim 4 of The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero gives the quantitative form actually used, that no interval cover of $S$ has total length below $2^{- 1}$ . The upper integral is therefore at least $2^{- 1}$ ; whether it equals $2^{- 1}$ is not asserted.
What the argument uses about $S$ , and nothing more. Only three properties enter: $S$ is closed, $S$ has empty interior, and no interval cover of $S$ has total length below $2^{- 1}$ . Any set with those three properties would serve as a witness in exactly the same way, and the argument is written so that the particular construction of 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 is used only through The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero.

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28Open item page →

For every $F_{σ}$ subset $E$ of $[0, 1]$ of measure zero there is a bounded Riemann integrable function on $[0, 1]$ whose set of discontinuities is exactly $E$

Example

Let $E \subseteq [0, 1]$ be an $F_{σ}$ subset of $R$ ( $F_{σ}$ and $G_{δ}$ subsets of $R$ ) of measure zero (Measure zero (a countable cover by intervals of total length below every $ε$ ) and content zero (a finite such cover)). Then there is a bounded function $h : [0, 1] \to R$ , with values in $[0, 1]$ , that is Riemann integrable on $[0, 1]$ and whose set of discontinuities is exactly $E$ (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, 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).

The construction. Fix closed sets $F_{0}, F_{1}, F_{2}, \dots$ with $E = ⋃_{n \in N} F_{n}$ and put

$h (x) : = 2^{- n (x)} for x \in E, h (x) : = 0 for x \in [0, 1] ∖ E,$

where $n (x) : = \min {n \in N : x \in F_{n}}$ is the least index of a closed set containing $x$ (The well-ordering principle). Nothing is selected: $n (x)$ is the least element of a set determined by $x$ and the fixed sequence $(F_{n})$ .

Why this is worth stating. Together with For $f : A \to R$ the set of points of $A$ at which $f$ is discontinuous is the intersection with $A$ of an $F_{σ}$ subset of $R$ , and the set of points at which $f$ is continuous is the intersection with $A$ of a $G_{δ}$ subset; for $A = R$ the two sets are $F_{σ}$ and $G_{δ}$ outright, which shows that a discontinuity set is always the trace of an $F_{σ}$ set, and with 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, which shows that an integrable function has a null discontinuity set, the example says that the two necessary conditions are also jointly sufficient: null and $F_{σ}$ is exactly what a discontinuity set of a Riemann integrable function on $[0, 1]$ can be. The Cantor set and any at most countable subset of $[0, 1]$ are instances.

Choice. The construction uses none; the only choice principle in the statement comes from the direction of 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 used at the end, and that direction, "null $\Rightarrow$ integrable", is a theorem of ZF.

Facts & Assumptions

Given: An $F_{σ}$ set $E \subseteq [0, 1]$ of measure zero, and a sequence $(F_{n})_{n \in N}$ of closed subsets of $R$ with $E = ⋃_{n} F_{n}$ .

[L1]

$E$ is a union of a sequence of closed sets ( $F_{σ}$ and $G_{δ}$ subsets of $R$ , Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L2]

Every nonempty subset of $N$ has a least element (The well-ordering principle).

[L3]

A set $F$ is closed exactly when every point outside $F$ has a neighbourhood missing $F$ (Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen, The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ).

[L4]

$E$ has measure zero and a subset of a null set is null, so $E$ contains no interval with two distinct endpoints: such an interval would be null, contradicting A sequence of intervals covering $[a, b]$ has total length at least $b - a$ , so no interval of positive length has measure zero (Measure zero (a countable cover by intervals of total length below every $ε$ ) and content zero (a finite such cover)).

[L5]

Powers: $2^{- n} > 0$ for every $n \in N$ , $2^{0} = 1$ , $2^{- n} \leq 1$ , and $m < n$ implies $2^{- n} < 2^{- m}$ (Integer powers $a^{m}$ , Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

[L6]

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

[L7]

For every real $η > 0$ there is a natural $N \geq 1$ with $1 / ι (N) < η$ , and $2^{- N} \leq 1 / ι (N)$ for $N \geq 1$ , since $ι (N) \leq 2^{N}$ by induction (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, The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing, Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

[L8]

A bounded function on $[a, b]$ with $a < b$ whose set of discontinuities has measure zero is Riemann integrable, and that implication uses no choice principle (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, 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]

Ordered-field arithmetic and the absolute value: the order is total and transitive; for $0 \leq x \leq 1$ and a real $ρ > 0$ the reals $u : = \max {0, x - ρ}$ and $v : = \min {1, x + ρ}$ satisfy $u < v$ and $(u, v) \subseteq N_{ρ} (x) \cap [0, 1]$ ; a nonempty open interval is a nondegenerate interval (Basic properties of the absolute value, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Maximum and minimum of a set, Ordered field, Complete ordered field (least-upper-bound property), Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ , Interior, closure, boundary and exterior of a subset of $R$ , The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

Verification

technique · constructive

1.1

Fix the sequence $(F_{n})$ of [L1] and define $n (x) : = \min {n \in N : x \in F_{n}}$ for $x \in E$ , which exists by [L2] since the set is a nonempty subset of $N$ ; define $h : [0, 1] \to R$ by $h (x) : = 2^{- n (x)}$ for $x \in E$ and $h (x) : = 0$ otherwise.

givenL1L2L5construct

2.1

$h$ is bounded with values in $[0, 1]$ : $0 < 2^{- n (x)} \leq 2^{0} = 1$ for $x \in E$ by [L5], and $h = 0$ off $E$ .

step 1.1L5

2.2

$h$ is discontinuous at every point of $E$ . Let $x \in E$ , so $h (x) = 2^{- n (x)} > 0$ by [L5], and let a real $ρ > 0$ be given. By [L9] the interval $(u, v) \subseteq N_{ρ} (x) \cap [0, 1]$ is nonempty with $u < v$ , hence is a nondegenerate interval, so by [L4] it is not contained in $E$ : there is $y \in (u, v)$ with $y \notin E$ , and then $h (y) = 0$ and $∣ h (x) - h (y) ∣ = 2^{- n (x)}$ . So the continuity condition fails at $x$ for $ε : = 2^{- n (x)}$ .

step 1.1L4L5L9

2.3

$h$ is continuous at every point of $[0, 1] ∖ E$ . Let $x \in [0, 1]$ with $x \notin E$ , so $h (x) = 0$ , and let a real $ε > 0$ be given. By [L7] fix a natural $N \geq 1$ with $2^{- N} \leq 1 / ι (N) < ε$ . For each $n \leq N$ one has $x \notin F_{n}$ , since $F_{n} \subseteq E$ , so [L3] supplies a real $ρ_{n} > 0$ with $N_{ρ_{n}} (x) \cap F_{n} = \emptyset$ ; put $ρ : = \min {ρ_{0}, \dots, ρ_{N}} > 0$ , which exists by [L6].

step 1.1L3L6L7L9choose

3.1

For $y \in [0, 1]$ with $∣ y - x ∣ < ρ$ : if $y \notin E$ then $h (y) = 0$ ; and if $y \in E$ then $y \notin F_{n}$ for every $n \leq N$ by step 2.3, so $n (y) > N$ and $h (y) = 2^{- n (y)} < 2^{- N} < ε$ by [L5] and step 2.3. In both cases $∣ h (y) - h (x) ∣ = h (y) < ε$ , so $h$ is continuous at $x$ .

step 1.1step 2.3L5L9

4.1

By steps 2.2 and 3.1 the set of discontinuities of $h$ in $[0, 1]$ is exactly $E$ , which has measure zero by hypothesis; $h$ is bounded by step 2.1 and $0 < 1$ , so [L8] gives that $h$ is Riemann integrable on $[0, 1]$ . The function $h$ constructed in step 1.1 therefore has all the stated properties.

step 2.1step 2.2step 3.1givenL8discharge-construct∎

Remarks

Where each hypothesis on $E$ is used. That $E$ is $F_{σ}$ is what makes the exhaustion $(F_{n})$ available and hence gives continuity off $E$ in step 3.1; that $E$ is null is used twice, once through [L4] to force discontinuity on $E$ , and once at the end through 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. Dropping either hypothesis breaks the example, and by For $f : A \to R$ the set of points of $A$ at which $f$ is discontinuous is the intersection with $A$ of an $F_{σ}$ subset of $R$ , and the set of points at which $f$ is continuous is the intersection with $A$ of a $G_{δ}$ subset; for $A = R$ the two sets are $F_{σ}$ and $G_{δ}$ outright and 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 neither can be dropped from the conclusion either.
The values $2^{- n}$ are a convenience. Any sequence of positive reals tending to $0$ would do in their place; what the proof needs is that the value at a point of $F_{n} ∖ (F_{0} \cup \dots \cup F_{n - 1})$ is small when $n$ is large, and that it is never $0$ on $E$ .
Two familiar instances. Taking $E$ at most countable recovers a function continuous exactly off a prescribed countable set, of which Thomae's function is the case $E = Q \cap [0, 1]$ in spirit though not in formula (Thomae's function is Riemann integrable on $[0, 1]$ with integral $0$ : it is continuous at every irrational, so its discontinuity set is countable, and every lower Darboux sum is $0$ ); taking $E$ to be the Cantor set recovers a function of the type of The indicator of the Cantor set is discontinuous exactly on the Cantor set, which is null, so it is Riemann integrable with integral $0$ even though it is discontinuous at uncountably many points, the Cantor set being closed, hence $F_{σ}$ , and null.
No analogue holds without nullity. The Smith-Volterra-Cantor set is closed, hence $F_{σ}$ , and it is the discontinuity set of its own indicator, which is not integrable (The indicator of the Smith-Volterra-Cantor set is discontinuous exactly on a nowhere dense set, and is not Riemann integrable, because that set does not have measure zero). So the $F_{σ}$ condition alone buys nothing.

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28Open item page →

For the Dirichlet function every uniform partition with rational tags gives Riemann sum $1$ , so the sums converge along that sequence of tagged partitions although the function is not integrable: the mesh condition of the Riemann definition quantifies over all tagged partitions and cannot be weakened to one sequence

Statement refuted

Refuted: that a bounded $f$ on $[a, b]$ is Riemann integrable with integral $I$ as soon as there is one sequence $(P_{N}, ξ^{N})$ of tagged partitions with $∥ P_{N} ∥ \to 0$ and $S (f, P_{N}, ξ^{N}) \to I$ (Tagged partitions of $[a, b]$ , with a tag $ξ_{i}$ in each subinterval, and the Riemann sum $S (f, P, ξ) = \sum_{i} f (ξ_{i}) Δ_{i}$ , 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 $\int_{a}^{b} f$ ).

The witness is again the Dirichlet function on $[0, 1]$ (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 $x$ ). Take $P_{N} : = U_{N}$ , the uniform partition into $N \geq 1$ parts, and tag each subinterval by its left endpoint $ξ_{i}^{N} : = ι (i) / ι (N)$ , a rational. Then

$∥ U_{N} ∥ = \frac{1}{ι (N)} ⟶ 0, S (1_{Q}, U_{N}, ξ^{N}) = 1 for every N \geq 1,$

so the Riemann sums converge, to $1$ ; and yet $1_{Q}$ is not Riemann integrable on $[0, 1]$ (The Dirichlet function on $[0, 1]$ has lower Darboux integral $0$ and upper Darboux integral $1$ , so it is bounded and not Riemann integrable).

What this shows about The Darboux and Riemann definitions agree: a bounded $f$ on $[a, b]$ is Darboux integrable with integral $I$ if and only if for every real $ε > 0$ there is a real $δ > 0$ such that $∣ S (f, P, ξ) - I ∣ < ε$ for every tagged partition of mesh below $δ$ . Condition 2 there quantifies over every tagged partition of mesh below $δ$ , tags included, and the quantifier cannot be replaced by the existence of one good sequence. Tagging the very same partition by irrationals instead gives Riemann sum $0$ , so for each $N$ the two taggings of $U_{N}$ give the two values $1$ and $0$ ; no real number is within $1 / 2$ of both.

Facts & Assumptions

Given: $g : [0, 1] \to R$ with $g (x) = 1$ for rational $x$ and $g (x) = 0$ for irrational $x$ ; for $N \geq 1$ the uniform partition $U_{N} = (N, t)$ of $[0, 1]$ with $t_{i} = ι (i) / ι (N)$ and $Δ_{i} = 1 / ι (N)$ ; and the tagging $ξ^{N}$ with $ξ_{i}^{N} : = t_{i}$ for $i < N$ .

[A1]

The refuted claim: if some sequence of tagged partitions of $[0, 1]$ has meshes tending to $0$ and Riemann sums tending to $I$ , then $g$ is Riemann integrable with integral $I$ .

[L1]

For $U_{N}$ : $t_{i} = ι (i) / ι (N) \in [0, 1]$ , $t_{i} < t_{i + 1}$ , $Δ_{i} = 1 / ι (N)$ , $\sum_{i < N} Δ_{i} = 1$ , and $∥ U_{N} ∥ = 1 / ι (N)$ (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).

[L2]

$ξ_{i}^{N} : = t_{i} \in I_{i}$ defines a tagging of $U_{N}$ , and $S (g, U_{N}, ξ^{N}) = \sum_{i < N} g (ξ_{i}^{N}) Δ_{i}$ (Tagged partitions of $[a, b]$ , with a tag $ξ_{i}$ in each subinterval, and the Riemann sum $S (f, P, ξ) = \sum_{i} f (ξ_{i}) Δ_{i}$ ).

[L3]

Each $t_{i} = ι (i) / ι (N)$ is rational, being a quotient of canonical naturals with $ι (N) > 0$ ; hence $g (t_{i}) = 1$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing, 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 $x$ ).

[L4]

Finite sums: scaling and $\sum_{i < N} 1 \cdot Δ_{i} = \sum_{i < N} Δ_{i}$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L5]

A constant sequence converges to its value; and for every real $η > 0$ there is $N \geq 1$ with $1 / ι (N) < η$ , so the sequence $N \mapsto 1 / ι (N)$ converges to $0$ , its terms being positive and decreasing below every positive bound (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences, For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Every complete ordered field is Archimedean, Canonical naturals are positive and strictly increasing, Basic properties of the absolute value).

[L6]

$g$ is not Riemann integrable on $[0, 1]$ ; its lower Darboux integral is $0$ and its upper Darboux integral is $1$ (The Dirichlet function on $[0, 1]$ has lower Darboux integral $0$ and upper Darboux integral $1$ , so it is bounded and not 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 $\int_{a}^{b} f$ ).

[L7]

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

[L8]

The Riemann condition of The Darboux and Riemann definitions agree: a bounded $f$ on $[a, b]$ is Darboux integrable with integral $I$ if and only if for every real $ε > 0$ there is a real $δ > 0$ such that $∣ S (f, P, ξ) - I ∣ < ε$ for every tagged partition of mesh below $δ$ requires, for each $ε > 0$ , a $δ > 0$ such that every tagged partition of mesh below $δ$ has its Riemann sum within $ε$ of the integral.

[L9]

Ordered-field arithmetic: the order is total and transitive, $ι (N) > 0$ for $N \geq 1$ , and $0 \neq 1$ (Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property)). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

Counterexample

technique · direct

1.1

For each $N \geq 1$ , $(U_{N}, ξ^{N})$ is a tagged partition of $[0, 1]$ of mesh $1 / ι (N)$ , by [L1] and [L2].

givenL1L2

1.2

By [L3] every tag $ξ_{i}^{N} = t_{i}$ is rational, so $g (ξ_{i}^{N}) = 1$ ; hence by [L2], [L4] and [L1], $S (g, U_{N}, ξ^{N}) = \sum_{i < N} 1 \cdot Δ_{i} = \sum_{i < N} Δ_{i} = 1$ .

givenL1L2L3L4

2.1

The sequence $N \mapsto ∥ U_{N} ∥ = 1 / ι (N)$ converges to $0$ and the sequence $N \mapsto S (g, U_{N}, ξ^{N})$ is constantly $1$ , hence converges to $1$ , by [L5].

step 1.1step 1.2L5

3.1

So the hypothesis of [A1] is met with $I : = 1$ ; but $g$ is not Riemann integrable on $[0, 1]$ by [L6]. [A1] is therefore refuted.

step 2.1A1L6

4.1

Moreover, for each $N \geq 1$ the same partition carries a tagging whose Riemann sum is $0$ : by [L7] each open interval $(t_{i}, t_{i + 1})$ contains an irrational, and choosing one in each of the $N$ subintervals is a finite selection, giving a tagging $ζ$ of $U_{N}$ with $g (ζ_{i}) = 0$ for every $i < N$ and hence $S (g, U_{N}, ζ) = 0$ by [L2] and [L4]. So for every real $δ > 0$ there are tagged partitions of mesh below $δ$ with Riemann sum $1$ and others with Riemann sum $0$ , and by [L9] no single real $I$ can satisfy the condition of [L8] at $ε : = 2^{- 1}$ .

step 1.1step 1.2L2L4L5L7L8L9∎

Remarks

What survives of the naive formulation. If $f$ is integrable then every sequence of tagged partitions with meshes tending to $0$ has Riemann sums tending to $\int_{a}^{b} f$ , by The Darboux and Riemann definitions agree: a bounded $f$ on $[a, b]$ is Darboux integrable with integral $I$ if and only if for every real $ε > 0$ there is a real $δ > 0$ such that $∣ S (f, P, ξ) - I ∣ < ε$ for every tagged partition of mesh below $δ$ ; the failure is only in the converse. So sequences of Riemann sums are a legitimate way to compute an integral once integrability is known, and no way at all to establish it.
The Darboux sums see the difference immediately. For every partition $P$ of $[0, 1]$ one has $L (g, P) = 0$ and $U (g, P) = 1$ (The Dirichlet function on $[0, 1]$ has lower Darboux integral $0$ and upper Darboux integral $1$ , so it is bounded and not Riemann integrable), and by Tagged partitions of $[a, b]$ , with a tag $ξ_{i}$ in each subinterval, and the Riemann sum $S (f, P, ξ) = \sum_{i} f (ξ_{i}) Δ_{i}$ every Riemann sum over $P$ lies between them. The rational tags realise the upper end and the irrational tags the lower end; the choice of tags moves the sum across the whole gap.
Choosing the irrational tags costs nothing. Step 4.1 selects one irrational in each of finitely many subintervals, which is a finite family; the same observation as in What this page costs in choice: Riemann's criterion, the Darboux-Riemann equivalence and integrability of a monotone function are theorems of ZF; integrability of a continuous function inherits the single use of countable choice inside Heine-Cantor; and only the forward half of the Lebesgue criterion spends countable choice, once, at the countable union of null sets applies, and no choice principle is used.

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28Open item page →

Thomae's function is nonnegative, Riemann integrable on $[0, 1]$ with integral $0$ , and nonzero at every rational, so a vanishing integral does not force a nonnegative integrand to vanish

Statement refuted

Refuted: that a nonnegative Riemann integrable function on $[a, b]$ with $\int_{a}^{b} f = 0$ vanishes identically (FALSE: a nonnegative Riemann integrable function on $[a, b]$ with $\int_{a}^{b} f = 0$ is identically zero, 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 $\int_{a}^{b} f$ ).

The witness is Thomae's function $t$ on $[0, 1]$ (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 $x$ ). It satisfies $0 \leq t \leq 1$ , it is Riemann integrable with $\int_{0}^{1} t = 0$ (Thomae's function is Riemann integrable on $[0, 1]$ with integral $0$ : it is continuous at every irrational, so its discontinuity set is countable, and every lower Darboux sum is $0$ ), and it is positive at every rational point of $[0, 1]$ — a dense set, and an infinite one. So the failure is not at a single stray point: the function is nonzero on a dense subset of the interval and the integral still vanishes.

The repaired statement asks for continuity. With $f$ continuous the conclusion is true, and the reason is exactly what fails here: a continuous function positive at one point is positive on a whole subinterval, whereas $t$ is positive only on a set that contains no interval (Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable). That repaired statement is not proved here, since the additivity of the integral over subintervals is not available at this point in the reading order.

Facts & Assumptions

Given: Thomae's function $t : [0, 1] \to R$ , 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 $x$ (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).

[A1]

The refuted claim: a nonnegative Riemann integrable function on a closed bounded interval with distinct endpoints whose integral is $0$ vanishes identically.

[L1]

$0 \leq t (x) \leq 1$ for every $x \in [0, 1]$ , and $t (x) = 1 / ι (q (x)) > 0$ at every 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).

[L2]

$t$ is Riemann integrable on $[0, 1]$ and $\int_{0}^{1} t = 0$ (Thomae's function is Riemann integrable on $[0, 1]$ with integral $0$ : it is continuous at every irrational, so its discontinuity set is countable, and every lower Darboux sum is $0$ , The lower and upper Darboux integrals of a bounded $f$ on $[a, b]$ as $\sup_{P} L (f, P)$ and $\inf_{P} U (f, P)$ , Darboux integrability as their equality, and the notation $\int_{a}^{b} f$ ).

[L3]

$t$ is discontinuous at every rational point and continuous at every irrational point, so its discontinuity set in $[0, 1]$ is $Q \cap [0, 1]$ (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 $t (c)$ ).

[L4]

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

[L5]

Ordered-field arithmetic: $0 < 2^{- 1} < 1$ , so $2^{- 1}$ lies in $[0, 1]$ and is rational; the order is total and transitive (Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property), Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

Counterexample

technique · direct

1.1

$t$ is nonnegative on $[0, 1]$ by [L1], and it is Riemann integrable there with $\int_{0}^{1} t = 0$ by [L2]; the interval $[0, 1]$ has $0 < 1$ by [L5].

givenL1L2L5

1.2

$t$ does not vanish identically: $2^{- 1}$ is a rational point of $[0, 1]$ by [L5], so $t (2^{- 1}) > 0$ by [L1].

givenL1L5

2.1

The hypotheses of [A1] hold for $t$ on $[0, 1]$ and its conclusion fails, so [A1] is refuted.

step 1.1step 1.2A1

3.1

The failure is dense, not isolated: by [L4] every nonempty open subinterval of $[0, 1]$ contains a rational, at which $t$ is positive by [L1]; so ${x \in [0, 1] : t (x) > 0}$ meets every subinterval of $[0, 1]$ with distinct endpoints. It is also exactly the set of discontinuities of $t$ , by [L3] and [L1].

step 1.2L1L3L4∎

Remarks

What the integral measures, and what it does not. $\int_{0}^{1} t = 0$ says that the upper Darboux sums can be made arbitrarily small, not that $t$ is small anywhere in particular. The set where $t$ is positive is $Q \cap [0, 1]$ , which is null (Every at most countable subset of $R$ 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 that is also exactly why $t$ is integrable at all.
Every upper Darboux sum is strictly positive. Each subinterval contains a rational, so $M_{i} > 0$ for every $i$ and $U (t, P) > 0$ for every partition $P$ ; the upper integral is nevertheless $0$ , an infimum of a set of positive numbers. Nothing is contradictory here, and it is the cleanest reminder that an infimum need not be attained.
The Dirichlet function is not available as a witness. It is also nonnegative and nonzero on a dense set, but it is not integrable at all (The Dirichlet function on $[0, 1]$ has lower Darboux integral $0$ and upper Darboux integral $1$ , so it is bounded and not Riemann integrable), so it cannot satisfy the hypotheses of the refuted claim. Thomae's function is the standard witness precisely because it repairs integrability while keeping the dense positive set.

The Riemann Integral: Definition and Integrability: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

∫01x2=1/3∫01​x2=1/3, computed from the Darboux definition with uniform partitions and the closed form ∑k<nk2=n(n−1)(2n−1)/6∑k<n​k2=n(n−1)(2n−1)/6

Example

Facts & Assumptions

Verification

Remarks

One refinement worked out for f(x)=x2f(x)=x2 on [0,1][0,1]: adding the point 1/21/2 to the trivial partition raises the lower sum from 00 to 1/81/8 and lowers the upper sum from 11 to 5/85/8

Example

Facts & Assumptions

Verification

Remarks

∫03⌊x⌋=3∫03​⌊x⌋=3: the floor function is nondecreasing, hence integrable, and the integral is computed from the uniform partitions

Example

Facts & Assumptions

Verification

Remarks

The indicator of the Cantor set is discontinuous exactly on the Cantor set, which is null, so it is Riemann integrable with integral 00 even though it is discontinuous at uncountably many points

Example

Facts & Assumptions

Verification

Remarks

Thomae's function is Riemann integrable on [0,1][0,1] with integral 00: it is continuous at every irrational, so its discontinuity set is countable, and every lower Darboux sum is 00

Example

Facts & Assumptions

Verification

Remarks

The Dirichlet function on [0,1][0,1] has lower Darboux integral 00 and upper Darboux integral 11, so it is bounded and not Riemann integrable

Statement refuted

Facts & Assumptions

Counterexample

Remarks

The indicator of the Smith-Volterra-Cantor set is discontinuous exactly on a nowhere dense set, and is not Riemann integrable, because that set does not have measure zero

Statement refuted

Facts & Assumptions

Counterexample

Remarks

For every FσFσ​ subset EE of [0,1][0,1] of measure zero there is a bounded Riemann integrable function on [0,1][0,1] whose set of discontinuities is exactly EE

Example

Facts & Assumptions

Verification

Remarks

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Thomae's function is nonnegative, Riemann integrable on [0,1][0,1] with integral 00, and nonzero at every rational, so a vanishing integral does not force a nonnegative integrand to vanish

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Sources

$\int_{0}^{1} x^{2} = 1 / 3$ , computed from the Darboux definition with uniform partitions and the closed form $\sum_{k < n} k^{2} = n (n - 1) (2 n - 1) / 6$

One refinement worked out for $f (x) = x^{2}$ on $[0, 1]$ : adding the point $1 / 2$ to the trivial partition raises the lower sum from $0$ to $1 / 8$ and lowers the upper sum from $1$ to $5 / 8$

$\int_{0}^{3} ⌊ x ⌋ = 3$ : the floor function is nondecreasing, hence integrable, and the integral is computed from the uniform partitions

The indicator of the Cantor set is discontinuous exactly on the Cantor set, which is null, so it is Riemann integrable with integral $0$ even though it is discontinuous at uncountably many points

Thomae's function is Riemann integrable on $[0, 1]$ with integral $0$ : it is continuous at every irrational, so its discontinuity set is countable, and every lower Darboux sum is $0$

The Dirichlet function on $[0, 1]$ has lower Darboux integral $0$ and upper Darboux integral $1$ , so it is bounded and not Riemann integrable

For every $F_{σ}$ subset $E$ of $[0, 1]$ of measure zero there is a bounded Riemann integrable function on $[0, 1]$ whose set of discontinuities is exactly $E$

Thomae's function is nonnegative, Riemann integrable on $[0, 1]$ with integral $0$ , and nonzero at every rational, so a vanishing integral does not force a nonnegative integrand to vanish