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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28
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The Darboux and Riemann definitions agree: a bounded f on [a,b] is Darboux integrable with integral I if and only if for every real ε>0 there is a real δ>0 such that ∣S(f,P,ξ)−I∣<ε for every tagged partition of mesh below δ

Statement

Let a<b be reals, let f:[a,b]→R be bounded (Lower bound, bounded below, bounded set) and let I∈R. The following are equivalent.

  1. (Darboux) f is Darboux integrable on [a,b] with ∫abf=I (The lower and upper Darboux integrals of a bounded f on [a,b] as sup⁡PL(f,P) and inf⁡PU(f,P), Darboux integrability as their equality, and the notation ∫abf).
  2. (Riemann) For every real ε>0 there is a real δ>0 such that ∣ S(f,P,ξ)−I ∣  <  ε for every tagged partition (P,ξ) of [a,b] with ∥P∥<δ (Tagged partitions of [a,b], with a tag ξi in each subinterval, and the Riemann sum S(f,P,ξ)=∑if(ξi) Δi, Partition of [a,b] as a finite strictly increasing list a=t0<t1<⋯<tn=b, its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions).

The quantifier over tagged partitions is universal, and that is the content. Condition 2 constrains every tagged partition of small mesh at once, tags included; it is not a statement about one sequence of tagged partitions, and it cannot be weakened to one. The companion page of this pair exhibits a non-integrable function whose Riemann sums are constant along such a sequence.

Boundedness is a hypothesis of both conditions as stated here. Condition 1 presupposes it, since the Darboux sums of an unbounded function do not exist (For bounded f on [a,b] and a partition P: the infimum mi and supremum Mi of f on the i-th subinterval, and the lower and upper Darboux sums L(f,P)=∑imiΔi and U(f,P)=∑iMiΔi); condition 2 makes sense for unbounded f as well, and in fact implies boundedness, but that implication is not proved here and is not used: every application on this page starts from a bounded f.

Facts & Assumptions

Given: Reals a<b, a bounded f:[a,b]→R, a real M≥0 with ∣f(x)∣≤M for every x∈[a,b], and a real I. Put M+:=M+1, so M+>0 and ∣f(x)∣≤M+ for every x.

[L1]

For a partition P=(n,t) of [a,b]: n≥1, the subintervals Ii=[ti,ti+1] are nonempty, Δi=ti+1−ti>0, ∑i<nΔi=b−a, and ∥P∥=max⁡{Δi:i<n}. The uniform partition UN into N≥1 parts has ∥UN∥=(b−a)/ι(N). The common refinement P∨P0 refines both, and nP∨P0≤nP+nP0−1 (Partition of [a,b] as a finite strictly increasing list a=t0<t1<⋯<tn=b, its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions, Intervals of R: the nine order-convex forms, nondegeneracy, and length).

[L2]

L(f,P)=∑i<nmiΔi and U(f,P)=∑i<nMiΔi with mi=inf⁡f[Ii] and Mi=sup⁡f[Ii]; L(f,P)≤∫ab‾f≤∫ab‾f≤U(f,P); f is integrable exactly when the two integrals coincide, and then ∫abf is their common value (For bounded f on [a,b] and a partition P: the infimum mi and supremum Mi of f on the i-th subinterval, and the lower and upper Darboux sums L(f,P)=∑imiΔi and U(f,P)=∑iMiΔi, The lower and upper Darboux integrals of a bounded f on [a,b] as sup⁡PL(f,P) and inf⁡PU(f,P), Darboux integrability as their equality, and the notation ∫abf).

[L3]

S(f,P,ξ)=∑i<nf(ξi)Δi for a tagging ξ of P, and L(f,P)≤S(f,P,ξ)≤U(f,P) when f is bounded (Tagged partitions of [a,b], with a tag ξi in each subinterval, and the Riemann sum S(f,P,ξ)=∑if(ξi) Δi).

[L4]

Riemann's criterion: a bounded f is integrable if and only if for every real ε>0 there is a partition P with U(f,P)−L(f,P)<ε (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).

[L5]

If P′ refines P then L(f,P)≤L(f,P′), U(f,P′)≤U(f,P), and moreover U(f,P)−U(f,P′)≤2M+ ι(n′−n) ∥P∥ and L(f,P′)−L(f,P)≤2M+ ι(n′−n) ∥P∥ (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)≤L(f,P′)≤U(f,P′)≤U(f,P) when P′ refines P, and L(f,P)≤U(f,Q) for arbitrary partitions P and Q; moreover the two changes are at most 2M(n′−n)∥P∥).

[L6]

ε-characterisations: if u=sup⁡S with S nonempty then for every real η>0 there is s∈S with s>u−η; dually for the infimum (Epsilon characterisation of the supremum, Epsilon characterisation of the infimum).

[L7]

A family of nonempty sets indexed by a natural number n has a choice function, and this is a theorem of ZF; the family used below is indexed by i<n, which is exactly that listed form. Every natural-number-indexed list of nonempty sets has a choice function on its family of values states it in that form and expressly declines to identify it with "every finite family of nonempty sets has a choice function", no definition of finiteness being available where it is proved (Every natural-number-indexed list of nonempty sets has a choice function on its family of values, Choice function).

[L8]

For every real η>0 there is a natural N≥1 with 1/ι(N)<η; ι is nonnegative, additive and nondecreasing on N, and ι(N)>0 for N≥1 (For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε, Every complete ordered field is Archimedean, The canonical natural ι(n)=n⋅1F of a field, Canonical naturals are positive and strictly increasing).

[L9]

Finite sums: additivity, scaling, monotonicity in the terms (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L10]

Ordered-field arithmetic and the absolute value: adding a constant and multiplying by a positive quantity preserve an inequality; the order is total and transitive; ∣x∣<c exactly when −c<x<c for c>0 (Basic properties of the absolute value, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, The multiplicative identity is positive, Ordered field, Complete ordered field (least-upper-bound property)). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

Proof

technique · direct
1.1

Condition 2 implies condition 1. Assume condition 2 and let a real ε>0 be given. Fix δ>0 as in condition 2 for this ε, and put θ:=ε/(b−a)>0 by [L10].

givenL10choose
1.2

Condition 1 implies condition 2; this half of the proof is steps 1.2, 2.2, 2.3, 3.3, 4.2, 5.2 and 6.2, and its symbols are its own. Assume f is integrable with ∫abf=I and let a real η>0 be given. By [L4] fix a partition P0=(n0,t0) with U(f,P0)−L(f,P0)<η⋅2−1.

givenL4L10choose
2.1

A partition of mesh below δ exists: by [L8] fix N≥1 with 1/ι(N)<δ/(b−a) and take P:=UN, so ∥P∥=(b−a)/ι(N)<δ by [L1] and [L10]. Write P=(n,t).

step 1.1L1L8L10choose
2.2

By [L2] and integrability, L(f,P0)≤∫ab‾f=I=∫ab‾f≤U(f,P0). Hence U(f,P0)−I≤U(f,P0)−L(f,P0)<η⋅2−1 and I−L(f,P0)≤U(f,P0)−L(f,P0)<η⋅2−1, that is U(f,P0)<I+η⋅2−1 and L(f,P0)>I−η⋅2−1.

step 1.2L2L10
2.3

Put δ0:=η⋅(8 M+ ι(n0))−1, a positive real since M+>0 and ι(n0)>0 by [L8] and n0≥1 by [L1].

step 1.2L1L8L10construct
3.1

For each i<n the set Xi:={ x∈Ii:f(x)>Mi−θ } is nonempty by [L6], since Mi=sup⁡f[Ii] and f[Ii] is nonempty by [L1]. By [L7] the finite family {Xi:i<n} has a choice function g; put ξi:=g(Xi) for i<n and ξk:=b for k≥n, a tagging of P.

step 2.1L1L2L6L7choose
3.2

Likewise the sets Yi:={ x∈Ii:f(x)<mi+θ } are nonempty by [L6], and [L7] supplies a tagging ζ of P with ζi∈Yi for i<n.

step 2.1L1L2L6L7choose
3.3

Let (Q,υ) be any tagged partition of [a,b] with ∥Q∥<δ0, and write Q=(nQ,u) and R:=Q∨P0, with R=(nR,r). By [L1], R refines both Q and P0, and nR−nQ≤n0−1, so ι(nR−nQ)≤ι(n0) by [L8].

step 2.3L1L8given
4.1

S(f,P,ξ)≥U(f,P)−ε: by step 3.1, f(ξi)≥Mi−θ for i<n, so multiplying by Δi>0 and summing gives S(f,P,ξ)≥∑i<n(Mi−θ)Δi=U(f,P)−θ∑i<nΔi=U(f,P)−θ(b−a)=U(f,P)−ε, by [L9], [L1] and [L3]. Symmetrically S(f,P,ζ)≤L(f,P)+ε.

step 3.1step 3.2L1L3L9L10
4.2

By [L5] applied to the refinement R of Q, U(f,Q)−U(f,R)≤2M+ι(nR−nQ)∥Q∥≤2M+ι(n0)δ0=η⋅4−1, and likewise L(f,R)−L(f,Q)≤η⋅4−1.

step 2.3step 3.3L5L8L10
5.1

By condition 2 both ∣S(f,P,ξ)−I∣<ε and ∣S(f,P,ζ)−I∣<ε, since ∥P∥<δ. Hence U(f,P)≤S(f,P,ξ)+ε<I+2ε and L(f,P)≥S(f,P,ζ)−ε>I−2ε, by step 4.1 and [L10].

step 1.1step 2.1step 4.1L10
5.2

By [L5] applied to the refinement R of P0, U(f,R)≤U(f,P0) and L(f,R)≥L(f,P0). Combining with step 4.2 and step 2.2: U(f,Q)≤U(f,R)+η⋅4−1≤U(f,P0)+η⋅4−1<I+η⋅2−1+η⋅4−1, and symmetrically L(f,Q)>I−η⋅2−1−η⋅4−1.

step 2.2step 3.3step 4.2L5L10
6.1

By [L2], ∫ab‾f≤U(f,P)<I+2ε and ∫ab‾f≥L(f,P)>I−2ε, and since ∫ab‾f≤∫ab‾f by [L2], both integrals lie strictly between I−2ε and I+2ε; in particular ∣∫ab‾f−I∣≤2ε and ∣∫ab‾f−I∣≤2ε.

step 5.1L2L10
6.2

By [L3], L(f,Q)≤S(f,Q,υ)≤U(f,Q), so step 5.2 gives I−η⋅2−1−η⋅4−1<S(f,Q,υ)<I+η⋅2−1+η⋅4−1, whence ∣S(f,Q,υ)−I∣<η⋅2−1+η⋅4−1<η by [L10]. Since (Q,υ) was an arbitrary tagged partition of mesh below δ0, condition 2 holds with this δ0.

step 5.2L3L10
7.1

Step 6.1 holds for every real ε>0. If ∫ab‾f≠I, taking ε:=∣∫ab‾f−I∣⋅4−1>0 would give ∣∫ab‾f−I∣≤∣∫ab‾f−I∣⋅2−1, which is false for a positive quantity; so ∫ab‾f=I, and the same argument gives ∫ab‾f=I. Hence f is integrable with ∫abf=I by [L2], which is condition 1.

step 6.1L2L10
8.1

Steps 1.1, 2.1, 3.1, 3.2, 4.1, 5.1, 6.1 and 7.1 prove that condition 2 implies condition 1; steps 1.2, 2.2, 2.3, 3.3, 4.2, 5.2 and 6.2 prove the converse. The two halves share no symbol, the first working with ε,δ,P,ξ,ζ,θ and the second with η,δ0,P0,Q,υ,R, and together they give the stated equivalence.

step 7.1step 6.2∎

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