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ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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A step function integrated by additivity over subintervals, and the same value from the definition

Example

Let f:[0,2]→R be

f(x)  :=  {10≤x<1,31≤x≤2.

Then f is integrable on [0,2] and

∫02f  =  1+3  =  4.

Two routes are carried out below. The first splits at the jump with For a<c<b: f is integrable on [a,b] if and only if it is integrable on [a,c] and on [c,b], and then ∫abf=∫acf+∫cbf; with the oriented form for arbitrary a,b,c and evaluates each piece with Changing an integrable function at finitely many points changes neither its integrability nor its integral and the constant case of If m≤f≤M on [a,b] then m(b−a)≤L(f,P)≤∫ab‾f≤∫ab‾f≤U(f,P)≤M(b−a) for every partition P; in particular every constant function is integrable, with ∫abc=c(b−a). The second computes lower and upper Darboux sums directly on a three-point partition and squeezes.

Why this is the smallest interesting example. The jump sits at the interior point 1, which is a partition point of both routes, and a partition point lies in two subintervals (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). That is exactly the case singled out in the proof of Changing an integrable function at finitely many points changes neither its integrability nor its integral, and it is why the value of f at the single point 1 is irrelevant to both the integrability and the value while the upper sums still see it.

Facts & Assumptions

Given: The function f above on [0,2], and a real η with 0<η<1.

[L3]

Changing an integrable function at finitely many points changes neither its integrability nor its integral (Changing an integrable function at finitely many points changes neither its integrability nor its integral).

[L6]

Ordered-field arithmetic: the order is total and transitive, and a real that is ≤4+2η for every η in (0,1) is ≤4 (Ordered field, Complete ordered field (least-upper-bound property), Intervals of R: the nine order-convex forms, nondegeneracy, and length).

Verification

technique · direct
1.1

f is nondecreasing on [0,2]: for x≤y the only possibility not giving f(x)=f(y) is x<1≤y, where f(x)=1<3=f(y). Hence f is bounded and integrable on [0,2], and likewise on [0,1] and on [1,2], by [L1].

givenL1
1.2

Second route. Let Pη be the partition of [0,2] with point set {0, 1−η, 1, 2}, so n=3 with Δ0=1−η, Δ1=η and Δ2=1.

givenL5construct
2.1

First route, the piece [0,1]. The constant function 1 on [0,1] agrees with f at every point of [0,1] except x=1, a single point; so by [L3] and [L4], ∫01f=∫011=1⋅(1−0)=1.

step 1.1L3L4
2.2

The piece [1,2]. There f is the constant 3, so ∫12f=3⋅(2−1)=3 by [L4].

step 1.1L4
2.3

On I0=[0,1−η], f is constantly 1, so m0=M0=1. On I1=[1−η,1], f takes the value 1 at every point below 1 and 3 at 1, so m1=1 and M1=3 by [L5]. On I2=[1,2], f is constantly 3, so m2=M2=3.

step 1.2givenL5
3.1

By [L2] with c:=1, ∫02f=∫01f+∫12f=1+3=4.

step 1.1step 2.1step 2.2L2
3.2

Hence L(f,Pη)=1(1−η)+1⋅η+3⋅1=4 and U(f,Pη)=1(1−η)+3η+3⋅1=4+2η, by [L5].

step 2.3L5
4.1

By [L4] the number ∫02f lies between L(f,Pη)=4 and U(f,Pη)=4+2η for every η∈(0,1), so 4≤∫02f≤4 by [L6]; the second route agrees with step 3.1.

step 1.1step 3.1step 3.2L4L6∎

Remarks

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources