Definition D1
For a partition P: a = x₀ < … < xₙ = b of [a,b], let Iᵢ = [xᵢ₋₁, xᵢ], Mᵢ = sup f(Iᵢ), mᵢ = inf f(Iᵢ), and set U(P) = Σ MᵢΔxᵢ, L(P) = Σ mᵢΔxᵢ.
Proof.
(1)This definition is carried over unchanged from Proof 46. f is assumed to be bounded.convention, assumption
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Definition D2
When the upper and lower integrals are equal, f is said to be integrable, and their common value is written ∫ₐᵇ f. This notation is defined only for a < b.
Proof.
(1)The case where the upper limit is smaller than the lower limit has no meaning yet. That gap is filled in D3.convention
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Definition D3
For a > b, set ∫ₐᵇ f = −∫ᵇₐ f by convention, and for a = b, set ∫ₐᵃ f = 0.
Proof.
(1)D2 was defined only for a < b. This is not a theorem but a convention extending the definition.D2, convention
(2)Being a convention, it cannot be asked whether it is true or false. What can be asked is what it preserves.convention
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Theorem T1
If a < b < c and f is integrable on [a,b] and on [b,c], then f is integrable on [a,c] and ∫ₐᵇ f + ∫ᵇᶜ f = ∫ₐᶜ f.
Proof.
(1)Let ε > 0 be given.assumption
(2)By the Riemann criterion there is a partition P₁ of [a,b] with U(P₁) − L(P₁) < ε/2 and a partition P₂ of [b,c] with U(P₂) − L(P₂) < ε/2.assumption, (1), A4, A1
(3)Joining P₁ and P₂ gives a partition P of [a,c] having b as one of its points.(2), D1
(4)The pieces are unchanged, so U(P) = U(P₁) + U(P₂) and L(P) = L(P₁) + L(P₂).(3), D1, A1
(5)U(P) − L(P) = (U(P₁) − L(P₁)) + (U(P₂) − L(P₂)) < ε(2), (4), A1, A2
(6)Since ε was arbitrary, by the Riemann criterion f is integrable on [a,c].(5), A4, D2
(7)L(P) ≤ ∫ₐᶜ f ≤ U(P).(6), A5
(8)L(P₁) ≤ ∫ₐᵇ f ≤ U(P₁) and L(P₂) ≤ ∫ᵇᶜ f ≤ U(P₂), so adding gives L(P) ≤ ∫ₐᵇ f + ∫ᵇᶜ f ≤ U(P).assumption, A5, (4), A2
(9)Both numbers lie between L(P) and U(P), so |∫ₐᶜ f − (∫ₐᵇ f + ∫ᵇᶜ f)| ≤ U(P) − L(P) < ε(7), (8), (5), A1, A2, A10
(10)Since ε was arbitrary, the difference is 0, and ∫ₐᶜ f = ∫ₐᵇ f + ∫ᵇᶜ f.(9), A3, A1
(11)This proof did not use the fundamental theorem. Proof 47 took additivity over intervals as an axiom, so using it would be circular.(10), D1
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Theorem T2
If f and g are continuous on [a,b] and α, β are constants, then ∫ₐᵇ (αf + βg) = α∫ₐᵇ f + β∫ₐᵇ g.
Proof.
(1)Since f and g are continuous, F(x) = ∫ₐˣ f and G(x) = ∫ₐˣ g are antiderivatives of f and g respectively.assumption, A6
(2)(αF + βG)′ = αf + βg.(1), A8
(3)αf + βg is continuous on [a,b].assumption, A9
(4)By the fundamental theorem, ∫ₐᵇ (αf + βg) = (αF + βG)(b) − (αF + βG)(a).(2), (3), A7
(5)Rearranging the right side gives α(F(b) − F(a)) + β(G(b) − G(a)).(4), A1
(6)Applying the fundamental theorem to F and G gives F(b) − F(a) = ∫ₐᵇ f, G(b) − G(a) = ∫ₐᵇ g.(1), assumption, A7
(7)Hence ∫ₐᵇ (αf + βg) = α∫ₐᵇ f + β∫ₐᵇ g.(5), (6), E
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Theorem T3
If f is continuous on an interval and a, b, c are points of that interval, then under the convention D3, regardless of order, ∫ₐᵇ f + ∫ᵇᶜ f = ∫ₐᶜ f.
Proof.
(1)Since f is continuous, f has an antiderivative F on that interval.assumption, A6
(2)If u < v, then by the fundamental theorem ∫ᵤᵛ f = F(v) − F(u).(1), A7
(3)If u = v, then by convention ∫ᵤᵘ f = 0 = F(u) − F(u).D3, A1
(4)If u > v, then by convention ∫ᵤᵛ f = −∫ᵥᵘ f = −(F(u) − F(v)) = F(v) − F(u).D3, (2), A1
(5)Hence regardless of order, ∫ᵤᵛ f = F(v) − F(u).(2), (3), (4)
(6)∫ₐᵇ f + ∫ᵇᶜ f = (F(b) − F(a)) + (F(c) − F(b)) = F(c) − F(a) = ∫ₐᶜ f(5), A1
(7)For a < b < c this equation is T1. The convention extends T1 to the remaining orders.(6), T1
(8)Had the convention been ∫ᵤᵛ f = ∫ᵥᵘ f for u > v, then (4) would become F(u) − F(v) and (6) would break. That is why the sign was attached.(4), (6), A1
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