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

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Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$

Statement

Let $a < b$ be reals and let $f, g : [a, b] \to R$ be integrable (The lower and upper Darboux integrals of a bounded $f$ on $[a, b]$ as $\sup_{P} L (f, P)$ and $\inf_{P} U (f, P)$ , Darboux integrability as their equality, and the notation $\int_{a}^{b} f$ ). Then:

$f + g$ is integrable on $[a, b]$ and $\int_{a}^{b} (f + g) = \int_{a}^{b} f + \int_{a}^{b} g$ ;
for every real $λ$ , $λ f$ is integrable on $[a, b]$ and $\int_{a}^{b} λ f = λ \int_{a}^{b} f$ ;
consequently, for all reals $λ, μ$ the function $λ f + μ g$ is integrable and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g;$
the same identity holds with oriented limits: if $u, v \in [a, b]$ and $f, g$ are integrable between $u$ and $v$ (The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ ), then $\int_{u}^{v} (λ f + μ g) = λ \int_{u}^{v} f + μ \int_{u}^{v} g$ .

Linearity of the integral is not linearity of the Darboux sums, and the proof of claim 1 has to squeeze rather than compute. On a subinterval the inequality $\sup (f + g) \leq \sup f + \sup g$ can be strict — take $f (x) = x$ and $g (x) = - x$ on $[0, 1]$ , where the left side is $0$ and the right side is $1$ — so $U (f + g, P)$ is in general strictly below $U (f, P) + U (g, P)$ and no identity between upper sums is available. Claim 2, by contrast, is an identity at the level of the sums, with the roles of $\sup$ and $\inf$ exchanged when $λ < 0$ .

Facts & Assumptions

Given: Reals $a < b$ , integrable $f, g : [a, b] \to R$ , reals $λ, μ$ , and a real $ε > 0$ .

[L1]

Riemann's criterion: a bounded $h$ on $[a, b]$ is integrable if and only if for every real $η > 0$ there is a partition $P$ with $U (h, P) - L (h, P) < η$ (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]

For every partition $P$ and bounded $h$ : $L (h, P) \leq \underline{\int_{a}^{b}} h \leq \overline{\int_{a}^{b}} h \leq U (h, P)$ , and $h$ is integrable exactly when the two integrals agree, their common value being $\int_{a}^{b} h$ ; the lower integral is $\sup_{P} L (h, P)$ and the upper is $\inf_{P} U (h, P)$ (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 $\int_{a}^{b} f$ , Suprema and infima are unique).

[L3]

$L (h, P) = \sum_{i < n} m_{i} (h) Δ_{i}$ and $U (h, P) = \sum_{i < n} M_{i} (h) Δ_{i}$ , where $m_{i} (h) = \inf h [I_{i}]$ and $M_{i} (h) = \sup h [I_{i}]$ over the subintervals $I_{i}$ of $P = (n, t)$ , with $Δ_{i} > 0$ ; an integrable function is bounded, and a sum of two bounded functions and a scalar multiple of a bounded function are bounded (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 $U (f, P) = \sum_{i} M_{i} Δ_{i}$ , 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, Lower bound, bounded below, bounded set).

[L4]

If $P^{'}$ refines $P$ then $L (h, P) \leq L (h, P^{'}) \leq U (h, P^{'}) \leq U (h, P)$ ; the common refinement $P_{1} \lor P_{2}$ refines both (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 $2 M (n^{'} - n) ∥ P ∥$ , 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).

[L5]

Finite sums are additive and homogeneous: $\sum_{i < n} (x_{i} + y_{i}) = \sum_{i < n} x_{i} + \sum_{i < n} y_{i}$ and $\sum_{i < n} λ x_{i} = λ \sum_{i < n} x_{i}$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products, clauses 1 and 2).

[L6]

A supremum is the least upper bound and an infimum the greatest lower bound; both exist for a nonempty bounded set and are unique (Complete ordered field (least-upper-bound property), Greatest lower bound (infimum), Suprema and infima are unique).

[L7]

Reflection: writing $- S = {- s : s \in S}$ , a real $u$ is an upper bound of $S$ exactly when $- u$ is a lower bound of $- S$ , and conversely; hence $\sup (- S) = - \inf S$ and $\inf (- S) = - \sup S$ for nonempty bounded $S$ , by [L6] (Reflection through zero exchanges upper and lower bounds).

[L8]

The constant function $0$ is integrable with $\int_{a}^{b} 0 = 0$ (If $m \leq f \leq M$ on $[a, b]$ then $m (b - a) \leq L (f, P) \leq \underline{\int_{a}^{b}} f \leq \overline{\int_{a}^{b}} f \leq U (f, P) \leq M (b - a)$ for every partition $P$ ; in particular every constant function is integrable, with $\int_{a}^{b} c = c (b - a)$ ).

[L9]

Ordered-field arithmetic: adding a constant and multiplying by a positive quantity preserve an inequality, the order is total and transitive, and a real $x$ with $∣ x ∣ < η$ for every real $η > 0$ is $0$ (Ordered field, Complete ordered field (least-upper-bound property)). These order facts are used in their nonstrict form as well, obtained by adjoining the case of equality.

[L10]

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

Proof

technique · direct

1.1givenL3

$f$ , $g$ , $f + g$ and $λ f$ are bounded on $[a, b]$ , so all their Darboux sums and integrals are defined.

1.2L3L6

For every partition $P = (n, t)$ and every $i < n$ : $f (x) + g (x) \leq M_{i} (f) + M_{i} (g)$ for $x \in I_{i}$ , so $M_{i} (f) + M_{i} (g)$ is an upper bound of $(f + g) [I_{i}]$ and $M_{i} (f + g) \leq M_{i} (f) + M_{i} (g)$ by [L6]; dually $m_{i} (f + g) \geq m_{i} (f) + m_{i} (g)$ .

1.3givenL1L4choose

Fix partitions $P_{1}$ and $P_{2}$ with $U (f, P_{1}) - L (f, P_{1}) < ε \cdot 2^{- 1}$ and $U (g, P_{2}) - L (g, P_{2}) < ε \cdot 2^{- 1}$ , and put $P : = P_{1} \lor P_{2}$ .

1.4L8

Claim 2, the case $λ = 0$ . Then $λ f$ is the constant function $0$ , integrable with integral $0 = 0 \cdot \int_{a}^{b} f$ .

1.5L3L6L9

Claim 2, the case $λ > 0$ . For every partition $P$ and every $i < n$ , $λ M_{i} (f)$ is an upper bound of $(λ f) [I_{i}]$ , and any upper bound $w$ of $(λ f) [I_{i}]$ gives the upper bound $w / λ$ of $f [I_{i}]$ , whence $M_{i} (f) \leq w / λ$ and $λ M_{i} (f) \leq w$ ; so $M_{i} (λ f) = λ M_{i} (f)$ by [L6], and dually $m_{i} (λ f) = λ m_{i} (f)$ .

1.6L3L7

Claim 2, the case $λ = - 1$ . For every $P$ and $i < n$ , $(- f) [I_{i}] = - (f [I_{i}])$ , so $M_{i} (- f) = - m_{i} (f)$ and $m_{i} (- f) = - M_{i} (f)$ by [L7].

2.1step 1.3L4

By [L4], $U (f, P) - L (f, P) < ε \cdot 2^{- 1}$ and $U (g, P) - L (g, P) < ε \cdot 2^{- 1}$ .

2.2step 1.2L3L5L9

Summing the inequalities of step 1.2 over $i < n$ against the positive weights $Δ_{i}$ and using [L5] gives $L (f, P) + L (g, P) \leq L (f + g, P) \leq U (f + g, P) \leq U (f, P) + U (g, P)$ .

2.3step 1.5L1L3L5L9

With step 1.5 and [L5], $U (λ f, P) = λ U (f, P)$ and $L (λ f, P) = λ L (f, P)$ for $λ > 0$ ; hence $U (λ f, P) - L (λ f, P) = λ (U (f, P) - L (f, P))$ , which [L1] makes smaller than any prescribed positive number by choosing $P$ suitably, so $λ f$ is integrable.

2.4step 1.6L1L2L5L7

With step 1.6 and [L5], $U (- f, P) = - L (f, P)$ and $L (- f, P) = - U (f, P)$ , so $U (- f, P) - L (- f, P) = U (f, P) - L (f, P)$ and $- f$ is integrable by [L1]; and by [L7] applied to the sets of Darboux sums, $\overline{\int_{a}^{b}} (- f) = - \underline{\int_{a}^{b}} f$ and $\underline{\int_{a}^{b}} (- f) = - \overline{\int_{a}^{b}} f$ , so $\int_{a}^{b} (- f) = - \int_{a}^{b} f$ .

3.1step 1.1step 2.1step 2.2L1L9

Hence $U (f + g, P) - L (f + g, P) \leq (U (f, P) - L (f, P)) + (U (g, P) - L (g, P)) < ε$ , so $f + g$ is integrable by [L1], $ε > 0$ having been arbitrary.

3.2step 1.5step 2.3L2L6L9

Moreover the set of lower sums of $λ f$ is $λ$ times the set of lower sums of $f$ , and a supremum scales by a positive factor, by the argument of step 1.5 applied to that set; so $\underline{\int_{a}^{b}} λ f = λ \underline{\int_{a}^{b}} f$ , and likewise for the upper integrals, giving $\int_{a}^{b} λ f = λ \int_{a}^{b} f$ .

4.1step 2.1step 2.2step 3.1L2

Both $\int_{a}^{b} (f + g)$ and $\int_{a}^{b} f + \int_{a}^{b} g$ lie in the interval from $L (f, P) + L (g, P)$ to $U (f, P) + U (g, P)$ : the first by [L2] and step 2.2, the second by [L2] applied to $f$ and to $g$ separately.

4.2step 2.3step 3.2step 2.4L9

Claim 2 for $λ < 0$ . Then $- λ > 0$ and $λ f = (- λ) (- f)$ , so steps 2.3, 2.4 and 3.2 give integrability and the required identities and $\int_{a}^{b} λ f = (- λ) \int_{a}^{b} (- f) = (- λ) (- \int_{a}^{b} f) = λ \int_{a}^{b} f$ .

5.1step 2.1step 4.1L9

That interval has length less than $ε$ by step 2.1, so $∣ \int_{a}^{b} (f + g) - \int_{a}^{b} f - \int_{a}^{b} g ∣ < ε$ ; as $ε > 0$ was arbitrary the difference is $0$ , which is claim 1.

5.2step 1.4step 2.3step 3.2step 4.2L9

Claim 2 is now proved in all three cases $λ = 0$ , $λ > 0$ and $λ < 0$ , which are exhaustive by trichotomy.

6.1step 5.1step 5.2

Claim 3. By claim 2 the functions $λ f$ and $μ g$ are integrable with integrals $λ \int_{a}^{b} f$ and $μ \int_{a}^{b} g$ , and by claim 1 their sum is integrable with the sum of those integrals.

7.1step 6.1L10∎

Claim 4. If $u < v$ then $u, v \in [a, b]$ and claim 3 applies verbatim on $[u, v]$ ; if $u = v$ both sides are $0$ by [L10]; and if $u > v$ then applying the case $u < v$ to the pair $(v, u)$ and multiplying by $- 1$ gives the identity, by [L10].

Remarks

Why claim 1 cannot be an identity of Darboux sums. The example in the statement shows $\sup (f + g) < \sup f + \sup g$ is possible on a single subinterval, so $U (f + g, P) = U (f, P) + U (g, P)$ is false in general. What survives is the pair of inequalities of step 1.2, and they are enough because the gap between them is squeezed to $0$ by 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 two scalar cases really are different. For $λ > 0$ the extreme values scale; for $λ < 0$ they are exchanged, because multiplying by a negative reverses the order (Reflection through zero exchanges upper and lower bounds). Merging the cases and writing $U (λ f, P) = λ U (f, P)$ for all $λ$ would be false at $λ = - 1$ , where the correct identity is $U (- f, P) = - L (f, P)$ .
The set of integrable functions on $[a, b]$ is closed under the operations named here and under more. Products, absolute values and the lattice operations are also integrable, but none of them is obtained from linearity alone: the proofs of If $f, g$ are integrable on $[a, b]$ then so are $∣ f ∣$ , $f^{2}$ , $f g$ , $\max (f, g)$ and $\min (f, g)$ , and $∣ \int_{a}^{b} f ∣ \leq \int_{a}^{b} ∣ f ∣$ all pass through If $f$ is integrable on $[a, b]$ with values in $[m, M]$ and $φ$ is continuous on $[m, M]$ , then $φ \circ f$ is integrable, with linearity used only to recombine the pieces.

Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Integrable functions on [a,b][a,b] form a set closed under sums and scalar multiples, and ∫ab(λf+μg)=λ∫abf+μ∫abg∫ab​(λf+μg)=λ∫ab​f+μ∫ab​g

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$