Try
On [0,1], let f be zero below 1/3 and equal to 3 from 1/3 onward. Use the increasing weight function α(x)=x+2·1[x≥3/4]. An interval receives its ordinary width plus an extra weight 2 if it carries the jump at 3/4. Predict the weighted integral before calculating.
The ordinary-width contribution is 3·(2/3)=2. Near the weight jump, f is constantly 3, so the extra contribution is 2·3=6. The total is 8. The integrand and integrator both have discontinuities, but at different locations. A rule banning every discontinuous integrand would miss this valid example.
The distinction we need is local: does a small interval near a discontinuity of f carry a small amount of α variation? Ordinary width alone is insufficient when an integrator can place a fixed jump in arbitrarily short intervals.
Predict
Recall the two definitions from the preceding lessons. Weighted Darboux integrability requires partitions with arbitrarily small upper-minus-lower gaps and controls all refinements of suitable partitions. The stronger mesh assertion controls every tagged partition whose largest cell is sufficiently short, without requiring selected partition points to be present.
They cannot be equated without proof. For α=1 on [1/2,1] and zero below, take f=1 strictly above 1/2 and zero elsewhere. A partition containing 1/2 puts the only positive weight on the cell ending there, where f is zero. Its weighted gap is zero. But a cell straddling 1/2 allows tags giving either zero or one, however short the cell becomes. Refinement existence holds while the mesh assertion fails.
Predict why this obstruction disappears for continuous f. Every tag in a shrinking cell carrying a jump then produces nearly the same function value. The jump need not get smaller; the variation of the integrand across it does.
Experiment
Use f(x)=1+x and α(x)=x³ on [0,2]. Its integrand is 1-Lipschitz: a horizontal change of at most h produces a vertical change of at most h. Every cell of a partition with mesh h therefore has oscillation at most h. Its weighted Darboux gap is at most h times the total weight, which telescopes to α(2)−α(0)=8.
| Maximum cell width | Guaranteed weighted gap bound | What it covers |
|---|
| 1/4 | 2 | Every partition and every tag choice within this width |
| 1/20 | 2/5 | The same universal class with smaller cells |
| 1/100 | 2/25 | A bound, not the measured error of one sum |
Choose a width that guarantees error below 1/50; h=1/500 works because 8h=2/125. The next lesson computes the integral as 20. You do not need that value to establish the error bound: both the integral and every tagged sum lie between the corresponding lower and upper sums.
Understand
Let f be bounded on [a,b], with a strictly smaller than b, and α nondecreasing. Write V=α(b)−α(a). For a partition P with closed cells I, its weighted gap is the sum of osc(f;I) times Δα(I). All weights are nonnegative and their sum is V.
Continuous f. Uniform continuity implies that, for any positive tolerance t, all sufficiently short cells have oscillation less than t. One may use a smaller continuity tolerance first so taking suprema does not lose the strict target. The gap is then at most tV. If V=0, monotonicity forces every weight to vanish and all sums are zero. Otherwise let t shrink to obtain weighted Darboux integrability. The same bound holds for every sufficiently fine partition. Each tagged sum and the Darboux integral lie between that partition’s bounds, proving mesh convergence directly. If f is L-Lipschitz, the explicit error bound is LV times the mesh.
Finitely many discontinuities. Suppose all discontinuities lie in a finite set D and α is continuous at each such point, relative to [a,b]. Choose B≥1 with |f|≤B. For a desired gap η>0, surround each c in D by an inner relative open neighborhood of radius r_c and an outer relative closed interval of radius 2r_c. Choose these radii so the sum of the outer intervals’ α increments is less than η/(4B). This is possible by continuity of α at each c. At an endpoint, it uses a one-sided increment, including the endpoint value of α.
Remove the inner neighborhoods. The remaining set K is compact, and f restricted to K is continuous, hence uniformly continuous. Choose a mesh smaller than all r_c and sufficiently small that every cell contained in K has oscillation less than η/[2(V+1)]. If K is empty there are no such cells; if D is empty use uniform continuity on the whole domain instead.
A bad cell is one meeting an inner neighborhood. The mesh restriction puts it entirely inside that neighborhood’s outer interval. Assign each bad cell to one such outer interval. Partition cells have disjoint interiors; monotonicity implies their increments, when summed inside an outer interval, are at most that interval’s increment. Endpoint overlaps do not double the increment: consecutive differences telescope. Thus the total bad-cell weight is less than η/(4B), and its gap contribution is less than η/2 because each oscillation is at most 2B.
Every other cell lies in K. Its contribution is controlled by the small good-set oscillation, with total at most ηV/[2(V+1)], which is less than η/2. The full gap is less than η. This argument controls all partitions below the chosen mesh, so it proves both weighted Darboux existence and mesh convergence. It does not assume the integrator is continuous away from D.
Apply
Test the endpoint hypothesis explicitly. Let f(0)=7 and f(x)=2 for positive x on [0,1]. For α(x)=x², the first-cell extra contribution is bounded by 5x₁² and disappears as the cell shrinks; the integral is 2. For β(0)=0 and β(x)=1 for positive x, the first cell always has weight 1. Tags at zero give sum 7 and positive tags give sum 2. Its weighted gap stays 5. Relative continuity at an endpoint is essential, not decorative terminology.
The finite-discontinuity theorem is sufficient, not necessary. An integrator constant on a region assigns zero weight to cells entirely inside it; even complicated discontinuities there can be harmless. The practice supplies a rational-indicator example supported strictly inside such a flat region and asks for a direct fine-partition proof.
Continue
When auditing existence, identify where the integrand varies, where the integrator accumulates variation, and which definition is claimed. Length control and increment control coincide for α(x)=x, but not for an arbitrary increasing α. Separate a convenient refining sequence from a proof about every fine tagged partition.
The next lesson identifies a smooth situation where increments can be replaced by derivative-weighted lengths. A jump contribution is already a warning that this reduction requires hypotheses: an almost-everywhere derivative does not record every change in an integrator.
