Weighted sums and refinement

Try

An ordinary rectangle sum weights a function value by interval length. Now suppose an accumulating quantity α changes unevenly across the input interval. Weight the function value by the change in α instead. For the worked case f(x)=x and α(x)=x² on [0,2], the two equal-length intervals [0,1] and [1,2] have weights 1 and 3. Equal input widths do not imply equal weights.

Choose a tag, meaning any input in each closed cell. With midpoint tags 1/2 and 3/2, the weighted sum is (1/2)·1+(3/2)·3=5. Using the smallest and largest values on each cell gives an enclosure from 3 to 7. Predict whether halving both cells must increase every tagged sum. It need not. What refinement controls monotonically is the lower and upper enclosure, not every choice of tags.

The narrower enclosure [17/4,25/4] contains every tagged sum on this partition. Midpoint tags give 21/4. These numbers are sums, not an asserted exact integral.

Predict

We assume f is bounded and real-valued on a compact interval [a,b], with a<b. The real-valued function α is nondecreasing. Its cell increments are therefore nonnegative. Predict why this sign matters: multiplying a lower bound for f by a negative number would reverse the inequality. The theory developed here uses nonnegative increments throughout.

The total weight is W=α(b)−α(a), because all cell increments telescope. It is finite and nonnegative. If W=0, monotonicity forces α to be constant, and every cell weight is zero. Every bounded f then has all weighted sums equal to zero, regardless of its discontinuities. This boundary case should be checked before dividing by W in an error estimate.

Experiment

Reconstruct the table with paper arithmetic. Then choose the right endpoint as the tag in the first cell, the left endpoint in the second, and midpoints in the remaining two. Your sum must stay within the displayed enclosure. For a different experiment, retain f(x)=x but replace α(x)=x² by α(x)=2x. Every weight is now twice the cell length. Explain why this changes the contributions even though neither the domain nor the integrand changed.

Now insert a new partition point inside just one existing cell. Record the old weight and the two new weights. They add to the old weight. The new infima cannot be below the old infimum; the new suprema cannot be above the old supremum. These three facts are sufficient for the refinement proof, even if f has no minimum or maximum on a cell.

Finally try α(x)=−x on [0,1] with f(x)=x. The supposed lower contribution is 0 and the supposed upper contribution is −1. This is a diagnostic failure of the nonnegative-weight framework, not a negative integral that somehow satisfies the same lower/upper proof. Signed integrators need separate definitions and arguments.

Understand

A partition P consists of a=x₀<x₁<…<xₙ=b. On the closed cell Iᵢ=[xᵢ₋₁,xᵢ], let mᵢ=inf f(Iᵢ), Mᵢ=sup f(Iᵢ), and wᵢ=α(xᵢ)−α(xᵢ₋₁). Define L(f,α,P)=Σmᵢwᵢ and U(f,α,P)=ΣMᵢwᵢ. A tagged sum S(f,α,P,ξ)=Σf(ξᵢ)wᵢ permits any ξᵢ∈Iᵢ, including endpoints. Thus L≤S≤U. Adjacent cells share an endpoint in their extrema sets; this is intentional and will matter when α jumps.

If a point c splits [u,v], the two child weights are nonnegative and add to α(v)−α(u). Both child infima are at least the parent infimum; both child suprema are at most the parent supremum. Multiplying and adding proves that inserting c cannot decrease L or increase U. Repeated insertion gives, for a refinement R of P,

L(f,α,P) ≤ L(f,α,R) ≤ U(f,α,R) ≤ U(f,α,P).

For unrelated partitions P and Q, their finite union is a common refinement R. Hence L(P)≤L(R)≤U(R)≤U(Q). Suppressing f and α in this display changes only notation, not hypotheses.

Choose B with |f|≤B. Every sum lies between −BW and BW. Consequently the lower integral, defined as sup over all P of L(P), and the upper integral, defined as inf over all P of U(P), are finite and ordered. We call f integrable against α when they agree, and call their common value I.

They agree exactly when arbitrarily small Darboux gaps U(P)−L(P) exist. One direction follows because the difference between the upper and lower integrals is bounded by any such gap. Conversely, if both equal I, choose a lower sum greater than I−ε/2 and an upper sum less than I+ε/2. These choices may use different partitions, and neither extremum need be attained. Their common refinement has gap less than ε. This proves the criterion rather than assuming the two approximations already share a partition.

Here is the precise tagged-sum conclusion: for every ε>0 there is a partition P₀ such that every refinement of P₀ and every tag choice gives |S−I|<ε. Choose P₀ with gap below ε. Its lower and upper sums trap both I and every subsequent tagged sum. This is convergence directed by refinement: retaining specified partition points is part of the quantifier. It does not claim that every sufficiently small-mesh partition, whether or not it retains those points, has the same property.

The converse also holds. For a fixed partition, the supremum of its tagged sums equals U and the infimum equals L. To justify this without attained cell extrema, choose each tag with f-value within η/(1+W) of its cell supremum. The total deficit is at most ηW/(1+W), which is less than η. Use infima similarly. If all tags on P₀ lie within ε of a proposed value I, then L(P₀)≥I−ε and U(P₀)≤I+ε. Arbitrarily small tolerances give arbitrarily small gaps. The resulting integral equals I because it lies in all these shrinking enclosures.

Apply

Continuity gives a stronger useful existence result. For continuous f on [a,b], define its modulus ω(h) as the supremum of |f(x)−f(y)| over inputs at distance at most h. Compactness gives uniform continuity, so ω(h) tends to zero as h tends to zero. Every cell of a partition with mesh h has oscillation at most ω(h). Therefore

U(f,α,P)−L(f,α,P) ≤ ω(h)Σwᵢ = ω(h)W.

If W>0, choose h small enough to make this less than the requested error. If W=0, all sums are already zero. This proves every continuous f is integrable against every nondecreasing real α, even when α is discontinuous. It also supplies small-mesh control in this continuous case; it is not a proof of a general mesh equivalence for discontinuous integrands.

As a changed calculation, assume f is 3-Lipschitz and W=5. The gap is at most 15h. To guarantee a gap strictly below 1/10, choose mesh strictly below 1/150. Equality in the mesh target only guarantees a non-strict bound. Explain which quantities come from f, which from α, and which you can control by choosing the partition.

Compare this result with the zero-weight case using a rational indicator as f. The latter is integrable against constant α for a different reason: no cell receives weight. A zero weighted integral does not certify continuity or regularity of the integrand.

Continue

Refresh ordinary Darboux sums to compare length weights with increments of α. The proof of uniform continuity supplies the compact-domain input to the modulus estimate.

Keep the conventions visible in later jump examples: α is nondecreasing, increments use its actual endpoint values, extrema use closed cells, and tags may be any point in those cells. The next lesson tests why these endpoint conventions and the distinction between refinement and arbitrary small mesh cannot be omitted.