A weighted sum can concentrate its entire weight at one jump. On [0,1], define α(x)=0 below c=1/2 and α(x)=1 at and above c. For a partition cell [u,v], its weight is α(v)−α(u), not its horizontal width. Exactly one cell has weight 1; every other cell has weight zero.
Now let f(x)=0 for x≤c and f(x)=1 for x>c. In indicator notation, α=1_[c,1] and f=1_(c,1]. The different brackets are deliberate. If the partition contains c, the weighted cell ends at c and f is zero throughout that closed cell. Both weighted Darboux sums are zero. Refining the partition preserves c and preserves this calculation.
Predict what happens if a very fine partition instead straddles c. Its weighted cell contains both sides, so tags can produce different answers. Making cells small is not the same instruction as retaining a specified partition point.
Predict
Use m=4 in the explorer. Its aligned partition contains c; its straddling partition contains consecutive points 3/8 and 5/8 around c. Both partitions have mesh 1/4, but their weighted cells differ.
Exact worked values for α=1_[1/2,1] and f=1_(1/2,1]
Partition
Cell with weight 1
Lower / upper sum
Left / right tag sum
Aligned, m=4
[1/4,1/2]
0 / 0
0 / 0
Straddling, m=4
[3/8,5/8]
0 / 1
0 / 1
The refinement integral is 0: there is a partition with zero gap and every refinement of it gives sum zero. But sequences of arbitrarily fine straddling partitions admit sums 0 or 1. Therefore convergence for every sufficiently small mesh and every tag choice fails. These conclusions are compatible because their quantifiers differ.
Experiment
Does a small cell carry a small weight?
Fix c=1/2. The integrator α(x) is 0 before c and 1 from c onward. Its increment is 1 on exactly one cell, however narrow that cell becomes. The weight is an increment of α, not the cell’s width.
Predict the sum before revealing it. Aligned nodes are j/m. Straddling nodes are 0, the odd multiples of 1/(2m), and 1. Both have maximum cell width 1/m; their treatment of the jump differs.
Shaded cell: [0.25, 0.5], width 0.25, weight 1. Circle: its selected tag, approximately 0.5. Every other cell contributes zero because its weight is zero.
Inspect every cell, weight and tagged contribution
Closed-cell extrema and integrator increments; decimals rounded.
Cell
Width
Weight Δα
Tag
f(tag)
Contribution
[0, 0.25]
0.25
0
0.25
0
0
[0.25, 0.5]
0.25
1
0.5
0
0
[0.5, 0.75]
0.25
0
0.75
1
0
[0.75, 1]
0.25
0
1
1
0
Compare left and right tags on straddling cells at m=4 and m=64. Then switch to the continuous integrand. Finally include c as a node and change only f(c). Explain which definition of convergence each observation tests. Controls are temporary and award no completion.
Choose the strict step integrand, compare aligned and straddling partitions, then change m while keeping the tag rule fixed. Identify the unique positive increment before reading the tagged sum. The highlighted weight represents an integrator increment; it is not the geometric area of the narrow cell.
Next change only the integrand’s value at c, making it 1 there. The new integrand g=1_[c,1] has both values 0 and 1 on the weighted cell even when c is a partition node. Predict why no refinement can reduce its weighted gap. Finally choose the continuous identity integrand. Its weighted cell values approach 1/2 from either partition family, and every permitted tag approaches the same value. Continuity repairs this stronger mesh convergence in a way that choosing one convenient tag cannot.
The explorer is a finite exact example. The next arguments establish the claims for all partitions, not just its two displayed families.
Understand
Keep the integration convention visible
For bounded f and nondecreasing α on [a,b], use the closed cells of a partition P. Write L(P) and U(P) for the infimum and supremum sums weighted by the nonnegative increments of α. The definition in the preceding lesson is equality of the lower and upper weighted Darboux integrals, equivalently the existence of partitions with arbitrarily small gap U(P)−L(P).
This is equivalent to the following tagged statement: for every η>0 there is a partition P₀ such that every refinement of P₀, with every tag choice, gives a sum within η of the integral. Indeed, refinement increases the lower sum and decreases the upper sum, trapping all tagged sums in the original small bracket. Conversely, at a fixed finite partition one can approximate every cell supremum or infimum by tags. Thus trapping all its tagged sums near I traps both extrema there and makes the Darboux gap arbitrarily small. Unattained suprema do not cause a problem; finite approximations suffice.
If all sufficiently small meshes give the same tagged limit, choose a fine P₀; all refinements remain fine. So that stronger condition implies the Darboux/refinement one. The strict-step example proves that the converse fails under these definitions. Sources sometimes choose the stronger condition as their definition of a Stieltjes integral; always check the convention before transferring a theorem.
Two point values, two different outcomes
For the strict-step f, a partition containing c has its only positive increment on a cell ending at c. The integrand vanishes on that entire cell, yielding integral zero. For any h>0 small enough, a partition can instead contain a straddling cell [c−h,c+h] and subdivide the remaining pieces to mesh at most 2h. A tag below c gives sum zero and a tag above c gives sum one. Letting h decrease proves the failure of the all-small-mesh condition.
For g=1_[c,1], take any partition. Its unique positive increment is on the first cell whose right endpoint reaches c. Its left endpoint is strictly below c, and its right endpoint is at least c. Thus its g-infimum is 0 and its g-supremum is 1, whether or not c is a node. The gap is exactly 1 for every partition, so no weighted Darboux integral exists. A special tag cannot replace the required control of all tags. Nor can a length-zero point modification be dismissed when it coincides with a jump in the integrator.
A continuous integrand sees the full interior jump
Suppose α is constant with value A to the left of an interior c, with value A+J to the right, and α(c)=A+θJ, where J≥0 and 0≤θ≤1. If c is a partition node, the two neighboring cells receive weights θJ and (1−θ)J. Otherwise a single straddling cell receives weight J. All other increments vanish.
Every tag in a positively weighted cell is within the mesh of c. For continuous f, if that mesh is small enough to make |f(t)−f(c)|<ε/J, the difference between the sum and Jf(c) is less than ε. Handle J=0 directly: all sums are zero. Consequently every small-mesh tagged sum approaches Jf(c), independently of θ. The intermediate integrator value redistributes the weight between two cells but cannot change its total. For example, a jump of 4 at c=3/2 contributes 4f(3/2); with f(x)=x² this equals 9.
Endpoints use only increments inside the domain
For a pure initial jump, the weight is α(a+)−α(a) and continuous f contributes f(a) times that difference. For a pure final jump, the weight is α(b)−α(b−), contributing f(b) times that difference. The first and last cells show this directly as their tags approach the relevant endpoint. There is no outside-domain value to insert.
In particular α=1_[a,b] is identically 1 on its domain. It has no increments and contributes zero. In contrast α=1_(a,b] has an initial jump of 1 and contributes f(a). Calling both descriptions “a step at a” without specifying the endpoint value would lose essential information.
Apply
Work through the six tasks with the convention written at the top of your page. One changes the strict-step weight to 3 and places it at 3/5. Another splits a jump of 5 between the two cells at 2, and then adds separate endpoint contributions. Each calculation should follow from the actual differences α(v)−α(u), not from a remembered mass diagram with unspecified brackets.
For finitely many jumps and a continuous integrand, take a mesh smaller than the separation of distinct jump locations. Each weighted cell can then be assigned to its unique location, with adjacent cells sharing a location’s mass if it is a partition node. If V=α(b)−α(a) and the integrand’s variation over distance h is at most ω(h), the tagged error from the finite jump formula is at most Vω(h). For an L-Lipschitz integrand this is VLh. This bound concerns discretization under the exact stated function rules; it does not quantify measurement or floating-point error.
Continue
Separate three questions in your proof record: whether small weighted Darboux gaps exist, whether all tags on refinements of a chosen partition are controlled, and whether all sufficiently small meshes are controlled. The first two are equivalent here; the third needs an additional argument and can fail.
For continuous integrands and finitely many pure jumps, that additional argument has now been supplied. The following integration-by-parts lesson will require its own hypotheses and endpoint bookkeeping. Do not silently replace the current convention with a measure-theoretic one or assume arbitrary shared discontinuities behave like continuous values.
Apply · explain · revise
Practice and reasoning
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Define tagged convergence along refinements as follows: for every η>0 there is a partition P₀ such that every refinement P of P₀, with every choice of tags, has sum within η of I. Prove this is equivalent to weighted Darboux integrability with integral I. Show that convergence of all tagged sums whenever the mesh is sufficiently small implies it, but explain why a refinement requirement is not itself a mesh requirement.
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On [0,1] set c=3/5, α(x)=3·1_[c,1](x), and f(x)=1_(c,1](x). Prove weighted Darboux integrability and find the integral. Construct partitions with mesh tending to zero and two choices of tags whose sums are respectively 0 and 3. Explain why this does not contradict the integral under the stated convention.
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Keep c and α from the previous task but change f to g=1_[c,1]. Prove every weighted Darboux gap is exactly 3, whether or not the partition contains c. Determine whether assigning the jump point to a convenient tag can establish an integral. Explain why “the values differ at only one point” is insufficient here.
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On [0,3] let α(x)=−1 for x<2, α(2)=1, and α(x)=4 for x>2. For any continuous f prove its weighted integral is 5f(2), independently of the chosen value of α(2) anywhere in [−1,4]. Compute it for f(x)=x². Include a proof for all-small-mesh tagged sums, not just a partition whose node is 2.
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On [0,2], let α(0)=−2, α(x)=1 for 0<x<1, α(1)=2, α(x)=6 for 1<x<2, and α(2)=8. Compute the weighted integral of f(x)=2x+1 by proving the endpoint and interior contributions. Compare the integrators 1_[0,2] and 1_(0,2] on this same domain.
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Let α be a nondecreasing function on [a,b] that is constant between finitely many interior points c₁<…<c_s, with arbitrary intermediate values at those points consistent with monotonicity, and possibly jumps at a and b relative to the domain. Prove for continuous f the finite sum rule: f(a)[α(a+)−α(a)]+Σ f(c_j)[α(c_j+)−α(c_j−)]+f(b)[α(b)−α(b−)]. Include the case of no interior jumps, zero weights, and a numerical error bound from a known modulus of continuity.
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Reflect and continue
Before marking this lesson complete, explain one key idea without the worked solution and identify a condition that matters. Marking complete records participation, not independent mastery.
