Definition D1
A partition of [a,b] is a finite set of points a = x₀ < x₁ < … < xₙ = b. Write Δxᵢ = xᵢ − xᵢ₋₁.
Proof.
(1)A partition that contains all the points of P together with additional points is called a refinement of P.convention
(2)P₁ ∪ P₂, formed by combining the points of two partitions P₁ and P₂, is a refinement of P₁ and a refinement of P₂.(1)
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Definition D2
For Iᵢ = [xᵢ₋₁, xᵢ], 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)Since mᵢ ≤ Mᵢ and Δxᵢ > 0, we have L(P) ≤ U(P) for every partition.(1), D1, A4, A2
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Definition D3
The upper integral is the infimum of U(P) over all partitions, and the lower integral is the supremum of L(P). When the two are equal, f is said to be integrable, and their common value is written ∫ₐᵇ f.
Proof.
(1)Since f is bounded, {U(P)} and {L(P)} are bounded, so the upper and lower integrals exist.assumption, D2, A3, A2
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Definition D4
f satisfies the Riemann criterion if for every ε > 0 there exists a partition P of [a,b] with U(P) − L(P) < ε.
Proof.
(1)This condition looks at a single partition. Neither the upper nor the lower integral appears.D2, convention
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Lemma L1
If X is nonempty and bounded and ε > 0, then there exists x ∈ X with sup X − ε < x and y ∈ X with y < inf X + ε.
Proof.
(1)Since X is bounded, sup X and inf X exist.assumption, A3
(2)Suppose x ≤ sup X − ε for every x ∈ X.assumption
(3)Then sup X − ε is an upper bound, and sup X is the least upper bound, so sup X ≤ sup X − ε.(1), (2), A5
(4)Subtracting sup X from both sides gives 0 ≤ −ε, that is, ε ≤ 0. This contradicts ε > 0.(3), A1, A2
(5)Hence there exists x ∈ X for which x ≤ sup X − ε fails, that is, sup X − ε < x.(4), A10
(6)The inf side is obtained in the same way using A6.(1), A6, A1, A2, A10
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Theorem T1
If f satisfies the Riemann criterion, then f is integrable on [a,b].
Proof.
(1)Let ε > 0 be given, and take a partition P with U(P) − L(P) < ε.assumption, D4
(2)The upper integral is the infimum of {U(P)}, so upper integral ≤ U(P), and lower integral ≥ L(P).D3, A4
(3)0 ≤ upper integral − lower integral ≤ U(P) − L(P) < εA8, (1), (2), A1, A2
(4)Since ε was arbitrary, upper integral − lower integral = 0.(3), A9
(5)The upper integral equals the lower integral, so f is integrable.(4), A1, D3
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Theorem T2
If f is integrable on [a,b], then f satisfies the Riemann criterion.
Proof.
(1)Let ε > 0 be given, and let I denote the common value of the upper and lower integrals.assumption, D3
(2)I is the infimum of {U(P)}, so there is a partition P₁ with U(P₁) < I + ε/2.(1), D3, L1, A1
(3)I is the supremum of {L(P)}, so there is a partition P₂ with L(P₂) > I − ε/2.(1), D3, L1, A1
(4)Set P = P₁ ∪ P₂. Then P is a refinement of P₁ and a refinement of P₂.D1
(5)By the refinement inequalities, U(P) ≤ U(P₁) and L(P) ≥ L(P₂).(4), D2, A7
(6)U(P) − L(P) ≤ U(P₁) − L(P₂) < (I + ε/2) − (I − ε/2) = ε(2), (3), (5), A1, A2
(7)Such a P exists for every ε > 0, so f satisfies the Riemann criterion.(6), D4
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Corollary C
For f bounded on [a,b], f is integrable if and only if f satisfies the Riemann criterion.
Proof.
(1)The Riemann criterion implies integrability.T1
(2)Integrability implies the Riemann criterion.T2
(3)Both directions hold, so the two statements are equivalent. Integrability is now decided by a single partition.(1), (2)
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