TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$

Statement

Let $a < c < b$ be reals and let $f : [a, b] \to R$ be bounded (Lower bound, bounded below, bounded set). Then:

$f$ is integrable on $[a, b]$ (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 $\int_{a}^{b} f$ ) if and only if its restrictions to $[a, c]$ and to $[c, b]$ are integrable;
and in that case $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f .$
Oriented form. Let $α < β$ be reals, let $f : [α, β] \to R$ be integrable, and let $u, v, w \in [α, β]$ be arbitrary. Then, with the convention of The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ , $\int_{u}^{v} f + \int_{v}^{w} f = \int_{u}^{w} f .$

Claim 3 is where The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ earns its place: it holds for every arrangement of the three points, including the degenerate ones, and it is the form used everywhere below.

Facts & Assumptions

Given: Reals $a < c < b$ and a bounded $f : [a, b] \to R$ ; and, for claim 3, reals $α < β$ , an integrable $f : [α, β] \to R$ and points $u, v, w \in [α, β]$ . Let a real $ε > 0$ be given.

[L1]

Riemann's criterion on any closed bounded interval with distinct endpoints (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 $U (f, P) - L (f, P) < ε$ ).

[L2]

A function integrable on $[p, q]$ is integrable on every $[p^{'}, q^{'}] \subseteq [p, q]$ with $p^{'} < q^{'}$ (A function integrable on $[a, b]$ is integrable on every closed subinterval).

[L3]

For a partition $R = (n, t)$ and bounded $h$ : $L (h, R) = \sum_{i < n} m_{i} Δ_{i}$ , $U (h, R) = \sum_{i < n} M_{i} Δ_{i}$ , and $L (h, R) \leq \int_{‾} h \leq \int^{‾} h \leq U (h, R)$ , the integral being the common value of the two 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 partition of $[p, q]$ is a pair $(n, t)$ with $n \geq 1$ , $t_{0} = p$ , $t_{i} < t_{i + 1}$ for $i < n$ and $t_{k} = q$ for $k \geq n$ ; its subintervals are $[t_{i}, t_{i + 1}]$ for $i < n$ (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).

[L5]

Finite sums split at an intermediate index, with $\sum_{k = p}^{q - 1} x_{k} = \sum_{j < q - p} x_{p + j}$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products, clause 3).

[L6]

With oriented limits, $\int_{v}^{u} h = - \int_{u}^{v} h$ and $\int_{u}^{u} h = 0$ (The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ ).

[L7]

Ordered-field arithmetic: adding a constant preserves an inequality, the order is total and transitive, and a real of absolute value below every positive real is $0$ (Ordered field, Complete ordered field (least-upper-bound property)).

Proof

technique · direct

1.1

Claim 1, forward. If $f$ is integrable on $[a, b]$ then, since $a < c$ and $c < b$ , [L2] gives integrability on $[a, c]$ and on $[c, b]$ .

1.2

The splice. Let $P_{1} = (n_{1}, t^{1})$ be a partition of $[a, c]$ and $P_{2} = (n_{2}, t^{2})$ one of $[c, b]$ . Define $P : = (n_{1} + n_{2}, t)$ by $t_{i} : = t_{i}^{1}$ for $i \leq n_{1}$ , $t_{n_{1} + j} : = t_{j}^{2}$ for $j \leq n_{2}$ , and $t_{k} : = b$ for $k \geq n_{1} + n_{2}$ . The two prescriptions agree at $i = n_{1}$ , where $t_{n_{1}}^{1} = c = t_{0}^{2}$ ; and $t_{0} = a$ , $t_{n_{1} + n_{2}} = t_{n_{2}}^{2} = b$ , with $t_{i} < t_{i + 1}$ for every $i < n_{1} + n_{2}$ . So $P$ is a partition of $[a, b]$ .

L4construct

1.3

Claim 1, converse. Suppose $f$ is integrable on $[a, c]$ and on $[c, b]$ , and use [L1] on each to fix $P_{1}$ with $U (f, P_{1}) - L (f, P_{1}) < ε \cdot 2^{- 1}$ and $P_{2}$ with $U (f, P_{2}) - L (f, P_{2}) < ε \cdot 2^{- 1}$ .

L1choose

2.1

The first $n_{1}$ subintervals of $P$ are those of $P_{1}$ and the last $n_{2}$ are those of $P_{2}$ , with the matching lengths, so by [L3] and the splitting law [L5], $L (f, P) = L (f, P_{1}) + L (f, P_{2})$ and $U (f, P) = U (f, P_{1}) + U (f, P_{2})$ .

step 1.2L3L4L5

3.1

For the splice $P$ of those two, step 2.1 gives $U (f, P) - L (f, P) < ε$ ; as $ε > 0$ was arbitrary and $f$ is bounded, [L1] makes $f$ integrable on $[a, b]$ .

step 1.2step 2.1step 1.3L1L7

4.1

Claim 2. With $P_{1}$ , $P_{2}$ and $P$ as above, [L3] puts $\int_{a}^{b} f$ between $L (f, P)$ and $U (f, P)$ , that is between $L (f, P_{1}) + L (f, P_{2})$ and $U (f, P_{1}) + U (f, P_{2})$ by step 2.1; and [L3] applied on $[a, c]$ and on $[c, b]$ puts $\int_{a}^{c} f + \int_{c}^{b} f$ between the same two numbers.

step 2.1step 1.3step 3.1L3

5.1

Those two numbers differ by less than $ε$ by step 1.3, so $∣ \int_{a}^{b} f - \int_{a}^{c} f - \int_{c}^{b} f ∣ < ε$ ; as $ε > 0$ was arbitrary the difference is $0$ , which is claim 2.

step 1.3step 4.1L7

6.1

Claim 3, first the sorted case. Let $x \leq y$ in $[α, β]$ . Then $\int_{α}^{y} f = \int_{α}^{x} f + \int_{x}^{y} f$ . Indeed if $α < x < y$ this is claim 2 applied on $[α, y]$ , where $f$ is integrable by [L2]; if $x = α$ the middle term is $0$ by [L6] and the identity is trivial; and if $x = y$ the last term is $0$ by [L6] and the identity is again trivial.

step 5.1L2L6

7.1

Put $F (x) : = \int_{α}^{x} f$ for $x \in [α, β]$ , which is defined by [L2] and [L6]. Then $\int_{x}^{y} f = F (y) - F (x)$ for all $x, y \in [α, β]$ : for $x < y$ this is step 6.1 rearranged; for $x = y$ both sides are $0$ by [L6]; and for $x > y$ the case already proved gives $\int_{y}^{x} f = F (x) - F (y)$ , and [L6] negates both sides.

step 6.1L2L6construct

8.1

Claim 3. For arbitrary $u, v, w \in [α, β]$ , step 7.1 gives $\int_{u}^{v} f + \int_{v}^{w} f = (F (v) - F (u)) + (F (w) - F (v)) = F (w) - F (u) = \int_{u}^{w} f$ .

step 7.1algebra∎

Remarks

The oriented form is not proved by listing six orderings. Step 7.1 shows that the oriented integral between two points is a difference of values of one function of one variable, after which claim 3 is the cancellation $(F (v) - F (u)) + (F (w) - F (v)) = F (w) - F (u)$ , valid however the three points are arranged and however many of them coincide. The case analysis is confined to the two lines of step 7.1, and no appeal to symmetry is made anywhere.
The function $F$ of step 7.1 is the integral function, and it is given its own item, The integral function $F (x) : = \int_{a}^{x} f$ of an integrable $f$ , because the rest of the page is about it. Nothing there re-proves step 7.1; it cites this theorem.
Boundedness on the whole of $[a, b]$ is a hypothesis of claim 1 in both directions. Boundedness on $[a, c]$ and on $[c, b]$ separately does give boundedness on the union, so the converse could be stated with the hypothesis split; it is stated globally because 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 $U (f, P) = \sum_{i} M_{i} Δ_{i}$ needs it globally to make $U (f, P)$ meaningful for a partition of $[a, b]$ .

Depends on

Used by

Dependency tree · two levels

39 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Riemann integral (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 6 (standard reference, not scraped)
J. Lebl, Basic Analysis I, Properties of the Riemann integral (standard reference, not scraped)
Encyclopedia of Mathematics, Integral calculus (standard reference, not scraped)

For a<c<ba<c<b: ff is integrable on [a,b][a,b] if and only if it is integrable on [a,c][a,c] and on [c,b][c,b], and then ∫abf=∫acf+∫cbf∫ab​f=∫ac​f+∫cb​f; with the oriented form for arbitrary a,b,ca,b,c