Definition D1
A partition of [a,b] is a finite set of points a = x₀ < x₁ < … < xₙ = b. Write Δxᵢ = xᵢ − xᵢ₋₁ and ‖P‖ = max Δxᵢ.
Proof.
(1)Each piece is Iᵢ = [xᵢ₋₁, xᵢ], and Δxᵢ > 0.convention
(2)A partition containing all the points of P and more is called a refinement of P.convention
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Definition D2
Let Mᵢ = sup f(Iᵢ), mᵢ = inf f(Iᵢ), and set U(P) = Σ MᵢΔxᵢ, L(P) = Σ mᵢΔxᵢ.
Proof.
(1)Since f is bounded on [a,b], Mᵢ and mᵢ exist on each piece.assumption, A3
(2)U(P) is called the upper sum and L(P) the lower sum. Neither is called an area.convention
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Definition D3
The upper integral is the infimum of U(P) over all partitions, the lower integral the supremum of L(P). When they coincide, f is called integrable, and the common value is written ∫ₐᵇ f.
Proof.
(1)Let m = inf f([a,b]), M = sup f([a,b]). Then m ≤ mᵢ ≤ Mᵢ ≤ M on every piece.assumption, A3, A4, A5, A6, A2
(2)Hence for every partition, m(b−a) ≤ L(P) ≤ M(b−a) and m(b−a) ≤ U(P) ≤ M(b−a).(1), D1, D2, A1, A2
(3)Both sets are bounded, so the upper and lower integrals exist.(2), A3
(4)Area comes out of this definition, not the other way around.convention
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Lemma L1
Choosing a point ξᵢ ∈ Iᵢ in each piece and setting R(P,ξ) = Σ f(ξᵢ)Δxᵢ, we have L(P) ≤ R(P,ξ) ≤ U(P).
Proof.
(1)Since ξᵢ ∈ Iᵢ, we have mᵢ ≤ f(ξᵢ) ≤ Mᵢ.D2, A4
(2)Since Δxᵢ > 0, mᵢΔxᵢ ≤ f(ξᵢ)Δxᵢ ≤ MᵢΔxᵢ.(1), D1, A2
(3)Adding the n inequalities gives L(P) ≤ R(P,ξ) ≤ U(P).(2), A2, A8, D2
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Lemma L2
If P′ is a refinement of P, then L(P) ≤ L(P′) ≤ U(P′) ≤ U(P).
Proof.
(1)First consider adding a single point y inside a piece Iᵢ of P.D1, assumption
(2)Since [xᵢ₋₁, y] ⊆ Iᵢ, every value of f on it is at most Mᵢ.(1), D2, A4
(3)Hence the sup on [xᵢ₋₁, y] is at most Mᵢ. The same holds for [y, xᵢ].(2), A5
(4)Mᵢ Δxᵢ = Mᵢ(y − xᵢ₋₁) + Mᵢ(xᵢ − y)A1
(5)Substituting (3) into the two terms of (4), the sum of the two new terms is at most MᵢΔxᵢ.(3), (4), A2
(6)The other pieces are unchanged, so U(P′) ≤ U(P).(5), D2, A2
(7)Applying A6 to the infima in the same way gives L(P) ≤ L(P′).(1), A4, A6, A1, A2
(8)A refinement is obtained by adding points finitely many times, so (6)(7) carry through.(6), (7), D1, A8
(9)Since mᵢ ≤ Mᵢ, L ≤ U for every partition.D2, A4, A2
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Lemma L3
For any two partitions P, Q, L(P) ≤ U(Q); hence lower integral ≤ upper integral.
Proof.
(1)The partition P ∪ Q, combining all points of P and Q, is a refinement of both P and Q.D1
(2)L(P) ≤ L(P∪Q) ≤ U(P∪Q) ≤ U(Q)(1), L2
(3)Fixing Q, U(Q) is an upper bound of all L(P), so lower integral ≤ U(Q).(2), A5, D3
(4)This holds for every Q, so lower integral ≤ upper integral.(3), A6, D3
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Lemma L4
If for every ε > 0 there is a partition P with U(P) − L(P) ≤ ε, then f is integrable.
Proof.
(1)Let ε > 0 be given and choose such a P.assumption
(2)Upper integral ≤ U(P) and lower integral ≥ L(P).D3, A4
(3)0 ≤ upper integral − lower integral ≤ U(P) − L(P) ≤ εL3, (1), (2), A1, A2
(4)ε was arbitrary, so upper integral − lower integral = 0.(3), A7
(5)Upper integral = lower integral, so f is integrable.(4), A1, D3
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Theorem T
If f is continuous on [a,b], then f is integrable on [a,b].
Proof.
(1)Let ε > 0 be given.assumption
(2)Set η = ε / (b − a). Then η > 0.assumption, A1, A2
(3)There is δ > 0 such that |f(s) − f(t)| < η for all s, t with |s − t| < δ.assumption, (2), A10
(4)Choose n with (b − a)/n < δ and divide [a,b] into n equal parts; then every Δxᵢ < δ.(3), A9, D1
(5)For s, t in one piece, |s − t| ≤ Δxᵢ < δ, so f(s) < f(t) + η.(3), (4), A2
(6)Fixing t and taking the sup over s gives Mᵢ ≤ f(t) + η.(5), A5, D2
(7)For every t, f(t) ≥ Mᵢ − η, so mᵢ ≥ Mᵢ − η.(6), A6, A1, D2
(8)Therefore Mᵢ − mᵢ ≤ η.(7), A1, A2
(9)U(P) − L(P) = Σ(Mᵢ − mᵢ)Δxᵢ ≤ η Σ Δxᵢ = η(b − a) = ε(8), (2), D2, A1, A2, A8
(10)Such a P exists for every ε, so f is integrable.(9), L4
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Corollary C
If f is continuous on [a,b], then merely making ‖P‖ small enough brings the Riemann sum R(P,ξ) as close to ∫ₐᵇ f as desired, however the ξ are chosen.
Proof.
(1)Given ε > 0, take the δ provided by (3) of T. This δ does not depend on the partition.T, A10
(2)For any partition with ‖P‖ < δ, lines (5)–(9) of T hold verbatim, so U(P) − L(P) ≤ ε.(1), T
(3)Since f is integrable, L(P) ≤ ∫ₐᵇ f ≤ U(P).T, D3, A4
(4)Also L(P) ≤ R(P,ξ) ≤ U(P).L1
(5)Both numbers lie between L(P) and U(P), so |R(P,ξ) − ∫ₐᵇ f| ≤ U(P) − L(P) ≤ ε.(2), (3), (4), A1, A2
(6)Therefore ∫ₐᵇ f is the limit of the Riemann sums as ‖P‖ → 0.(5), D3
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