Alphabeta Math
DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-28
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Tagged partitions of [a,b], with a tag ξi in each subinterval, and the Riemann sum S(f,P,ξ)=∑if(ξi) Δi

Definition

Let a<b be reals and let P=(n,t) be a partition of [a,b], with subintervals Ii=[ti,ti+1] and lengths Δi=ti+1−ti for i<n (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).

A tagging of P is a sequence ξ:N→R (Sequences of reals: bounded, eventually, frequently, tails, subsequences) with

ξi  ∈  Iifor every i<n,ξk:=b  for k≥n,

the second clause being the same bookkeeping tail convention that 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 uses, so that ξ is a genuine sequence and no index above n is ever read. The pair (P,ξ) is a tagged partition of [a,b], and ξi is the tag of the i-th subinterval. The mesh of a tagged partition is the mesh ∥P∥ of its underlying partition.

Taggings exist, and no choice is involved in producing one. Setting ξi:=ti for i<n and ξk:=b for k≥n defines a tagging, since ti∈[ti,ti+1]=Ii (Intervals of R: the nine order-convex forms, nondegeneracy, and length). So every partition carries at least one tagging, exhibited by a formula. What is a selection is choosing a tag in each subinterval subject to a condition, as 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 does; there the family of choices is finite and the selection is a theorem of ZF.

For f:[a,b]→R and a tagged partition (P,ξ) the Riemann sum of f is

S(f,P,ξ)  :=  ∑i<nf(ξi) Δi,

the finite sum of Finite sums and finite products, by recursion, indexed by i∈N with i<n. It is a real number, being a finite sum of reals, and it is defined for every f, bounded or not: no supremum or infimum of f occurs in it.

A Riemann sum lies between the Darboux sums of the same partition

Suppose in addition that f is bounded (Lower bound, bounded below, bounded set), so that the Darboux sums of 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 are defined. Then for every tagging ξ of P,

L(f,P)  ≤  S(f,P,ξ)  ≤  U(f,P).

Indeed ξi∈Ii gives mi≤f(ξi)≤Mi (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), and multiplying by Δi>0 and summing over i<n preserves the two inequalities, by monotonicity of finite sums, clause 4 of Laws of finite sums and finite products, and the order axioms (Ordered field, Complete ordered field (least-upper-bound property)).

This one line is the whole of the easy half of 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: whatever the tags, a Riemann sum is trapped between the two Darboux sums, so control of U(f,P)−L(f,P) is control of every Riemann sum over P at once.

Remarks

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