TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck 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 $\int_a^b(\lambda f+\mu g) = \lambda\int_a^b f + \mu\int_a^b g$

Statement

Let $a < b$ be reals and let $f, g : [a,b] \to \mathbb{R}$ 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 $\lambda$ , $\lambda f$ is integrable on $[a,b]$ and $\int_a^b \lambda f = \lambda \int_a^b f$ ;
consequently, for all reals $\lambda, \mu$ the function $\lambda f + \mu g$ is integrable and $\int_a^b(\lambda f + \mu g) \;=\; \lambda\int_a^b f \;+\; \mu\int_a^b g ;$
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 $\int_u^v(\lambda f + \mu g) = \lambda\int_u^v f + \mu\int_u^v g$ .

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 $\sup(f+g) \le \sup f + \sup g$ 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 $\lambda < 0$ .

Facts & Assumptions

Given: Reals $a < b$ , integrable $f, g : [a,b] \to \mathbb{R}$ , reals $\lambda, \mu$ , and a real $\varepsilon > 0$ .

[L1]

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

[L2]

For every partition $P$ and bounded $h$ : $L(h,P) \le \underline{\int_a^b} h \le \overline{\int_a^b} h \le U(h,P)$ , 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]

$L(h,P) = \sum_{i<n} m_i(h)\Delta_i$ and $U(h,P) = \sum_{i<n}M_i(h)\Delta_i$ , 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 $\Delta_i > 0$ ; 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 $L(f,P) = \sum_i m_i \Delta_i$ and $U(f,P) = \sum_i M_i \Delta_i$ , 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 $L(h,P) \le L(h,P') \le U(h,P') \le U(h,P)$ ; the common refinement $P_1 \vee P_2$ 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: $L(f,P) \le L(f,P') \le U(f,P') \le U(f,P)$ when $P'$ refines $P$ , and $L(f,P) \le U(f,Q)$ for arbitrary partitions $P$ and $Q$ ; moreover the two changes are at most $2M(n' - n)\|P\|$ , 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 $\sum_{i<n}\lambda x_i = \lambda\sum_{i<n}x_i$ (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 $m \le f \le M$ on $[a,b]$ then $m(b-a) \le L(f,P) \le \underline{\int_a^b} f \le \overline{\int_a^b} f \le U(f,P) \le M(b-a)$ 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 $|x| < \eta$ for every real $\eta > 0$ 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

1.1

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

givenL3

1.2

For every partition $P = (n,t)$ and every $i < n$ : $f(x) + g(x) \le M_i(f) + M_i(g)$ for $x \in I_i$ , so $M_i(f) + M_i(g)$ is an upper bound of $(f+g)[I_i]$ and $M_i(f+g) \le M_i(f) + M_i(g)$ by [L6]; dually $m_i(f+g) \ge m_i(f) + m_i(g)$ .

L3L6

1.3

Fix partitions $P_1$ and $P_2$ with $U(f,P_1) - L(f,P_1) < \varepsilon \cdot 2^{-1}$ and $U(g,P_2) - L(g,P_2) < \varepsilon \cdot 2^{-1}$ , and put $P := P_1 \vee P_2$ .

givenL1L4choose

1.4

Claim 2, the case $\lambda = 0$ . Then $\lambda f$ is the constant function $0$ , integrable with integral $0 = 0 \cdot \int_a^b f$ .

1.5

Claim 2, the case $\lambda > 0$ . For every partition $P$ and every $i < n$ , $\lambda M_i(f)$ is an upper bound of $(\lambda f)[I_i]$ , and any upper bound $w$ of $(\lambda f)[I_i]$ gives the upper bound $w/\lambda$ of $f[I_i]$ , whence $M_i(f) \le w/\lambda$ and $\lambda M_i(f) \le w$ ; so $M_i(\lambda f) = \lambda M_i(f)$ by [L6], and dually $m_i(\lambda f) = \lambda m_i(f)$ .

L3L6L9

1.6

Claim 2, the case $\lambda = -1$ . For every $P$ and $i < n$ , $(-f)[I_i] = -\bigl(f[I_i]\bigr)$ , so $M_i(-f) = -m_i(f)$ and $m_i(-f) = -M_i(f)$ by [L7].

L3L7

2.1

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

step 1.3L4

2.2

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

step 1.2L3L5L9

2.3

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

step 1.5L1L3L5L9

2.4

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 $\int_a^b(-f) = -\int_a^b f$ .

step 1.6L1L2L5L7

3.1

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

step 1.1step 2.1step 2.2L1L9

3.2

Moreover the set of lower sums of $\lambda f$ is $\lambda$ 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 $\underline{\int_a^b}\lambda f = \lambda \underline{\int_a^b} f$ , and likewise for the upper integrals, giving $\int_a^b \lambda f = \lambda\int_a^b f$ .

step 1.5step 2.3L2L6L9

4.1

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.

step 2.1step 2.2step 3.1L2

4.2

Claim 2 for $\lambda < 0$ . Then $-\lambda > 0$ and $\lambda f = (-\lambda)(-f)$ , so steps 2.3, 2.4 and 3.2 give integrability and the required identities and $\int_a^b \lambda f = (-\lambda)\int_a^b(-f) = (-\lambda)\bigl(-\int_a^b f\bigr) = \lambda\int_a^b f$ .

step 2.3step 3.2step 2.4L9

5.1

That interval has length less than $\varepsilon$ by step 2.1, so $\bigl|\int_a^b(f+g) - \int_a^b f - \int_a^b g\bigr| < \varepsilon$ ; as $\varepsilon > 0$ was arbitrary the difference is $0$ , which is claim 1.

step 2.1step 4.1L9

5.2

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

step 1.4step 2.3step 3.2step 4.2L9

6.1

Claim 3. By claim 2 the functions $\lambda f$ and $\mu g$ are integrable with integrals $\lambda\int_a^b f$ and $\mu\int_a^b g$ , and by claim 1 their sum is integrable with the sum of those integrals.

step 5.1step 5.2

7.1

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].

step 6.1L10∎

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 $\varepsilon > 0$ there is a partition $P$ with $U(f,P) - L(f,P) < \varepsilon$ .
The two scalar cases really are different. For $\lambda > 0$ the extreme values scale; for $\lambda < 0$ they are exchanged, because multiplying by a negative reverses the order (Reflection through zero exchanges upper and lower bounds). Merging the cases and writing $U(\lambda f, P) = \lambda U(f,P)$ for all $\lambda$ would be false at $\lambda = -1$ , 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 $\lvert f\rvert$ , $f^{2}$ , $fg$ , $\max(f,g)$ and $\min(f,g)$ , and $\bigl\lvert\int_a^b f\bigr\rvert \le \int_a^b\lvert f\rvert$ all pass through If $f$ is integrable on $[a,b]$ with values in $[m,M]$ and $\varphi$ is continuous on $[m,M]$ , then $\varphi \circ f$ is integrable, with linearity used only to recombine the pieces.

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Sources

Riemann integral (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 6 (standard reference, not scraped)
Carnegie Mellon 21-269, Riemann integration notes (standard reference, not scraped)
Encyclopedia of Mathematics, Integral calculus (standard reference, not scraped)

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