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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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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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Sources

Darboux integral (Wikipedia) (standard reference, not scraped)
J. Lebl, Basic Analysis I, The Riemann Integral (standard reference, not scraped)
J. Hunter, Chapter 11: The Riemann Integral (standard reference, not scraped)

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