Try
On [0,2], use the signal f(x)=1+x and increasing weight α(x)=x³. Equal-width cells near 2 carry more weight than equal-width cells near 0 because α grows faster there. Predict whether ordinary area under f alone captures the weighted total.
It does not: the ordinary integral is 4, whereas the weighted integral will be 20. A derivative explains the changed weighting only after a proof links increments to a continuous density. Here the density is α′(x)=3x². Thus the candidate ordinary integral is the integral of (1+x)3x², equal to [x³+3x⁴/4] from 0 to 2, or 20.
The point of the theorem is not merely to produce a useful answer. It establishes that this ordinary integral equals the limit of all weighted tagged sums under specified smoothness conditions.
Predict
On one cell [u,v], the mean value theorem supplies an intermediate point ξ with α(v)−α(u)=α′(ξ)(v−u). But the tagged sum uses f(t) at a chosen tag t, and ξ need not equal t. Predict how to control the error in replacing α′(ξ) by α′(t).
Both points lie in the same short cell. Uniform continuity of α′ therefore makes their derivative values close, independently of the cell and the tag. A bounded f and a fixed total interval length convert that local error into a global estimate.
Now predict what would go wrong for α(x)=x+4·1[x≥2/3]. Its derivative equals 1 away from the jump, but a fixed increment 4 remains in one arbitrarily short cell. The continuous-derivative hypothesis fails precisely where a length density would miss the extra contribution.
Experiment
For the cubic weight, compare two cells of equal width 1/2. Calculate the weight directly from endpoint cubes, then compare with the midpoint derivative approximation.
| Cell | Exact α increment | Midpoint derivative times width |
|---|
| [0,1/2] | 1/8 | 3/32 |
| [3/2,2] | 37/8 | 147/32 |
Both approximations differ from the exact increment by 1/32. The mean value theorem promises some correct intermediate derivative, not necessarily the midpoint derivative. Smaller cells reduce the accumulated comparison error; a quadrature rule’s behavior should not be confused with an exact identity on each cell.
For a changed smooth signed weight, let β(x)=x−x² on [0,1]. It rises and then falls. With f(x)=x, the weighted total is the integral of x(1−2x), equal to −1/6. Although β(1)−β(0)=0, the integral is not zero. Explain how contributions with opposite signs can cancel for constant f but fail to cancel for a varying f.
Understand
Smooth nonnegative weights. Assume f is real and continuous and α is C¹ and nondecreasing on [a,b]. C¹ means the derivative is continuous up to the endpoints with the appropriate one-sided interpretation. On every partition cell, the mean value theorem gives Δα_i=α′(ξ_i)Δx_i. Compare its weighted sum with the ordinary tagged sum for fα′, using the same tags t_i. Their difference in absolute value is at most
||f||∞ times the sum of |α′(ξ_i)−α′(t_i)|Δx_i.
Let ω(h) be the largest variation of α′ between inputs at distance at most h. Uniform continuity gives ω(h) tending to zero. Since ξ_i and t_i lie in one cell, the total difference is bounded by ||f||∞(b−a)ω(mesh). The ordinary tagged sums converge to the integral of the continuous product fα′. The weighted sums converge by the preceding continuous-integrand theorem. Their limits agree:
∫f dα = ∫f(x)α′(x) dx.
Define signed bounded-variation weights before using them. For a real function β, its variation on an interval is the supremum, over finite partitions, of the sum of absolute increments. Assume this supremum is finite on [a,b]. Set v(t)=Var on [a,t] of β and v(a)=0. For s strictly below t, take a partition of [a,s] whose variation sum exceeds v(s)−ε and append t. This proves v(t)≥v(s)−ε+|β(t)−β(s)|. Letting ε decrease to zero yields
v(t)−v(s) ≥ |β(t)−β(s)|.
Consequently p(t)=[v(t)+β(t)−β(a)]/2 and q(t)=[v(t)−β(t)+β(a)]/2 are nondecreasing: each increment is half the sum of a variation increment and a signed increment whose magnitude is no larger. They satisfy β=β(a)+p−q. This is a proved monotone decomposition, not an assumed ability to subtract arbitrary positive-weight arguments.
For continuous f, define ∫f dβ as ∫f dp minus ∫f dq. Each monotone integral exists under both convergence conventions by the preceding lesson. For every tagged partition their sum difference is exactly Σf(tᵢ)Δβᵢ, because the constant β(a) contributes zero increments. Thus these signed tagged sums converge as mesh tends to zero. Any other monotone decomposition gives exactly the same sums and therefore the same limit: the definition is independent of the decomposition.
Each sum has absolute value at most ||f||∞ times Σ|Δβᵢ|, which is at most ||f||∞Var(β). Passing to the limit proves the signed variation bound. A bound using only |β(b)−β(a)| can fail because positive and negative variation cancel. For β=x−x² on [0,1], the function rises from 0 to 1/4 and falls back to 0, so its variation is 1/2 although its endpoint increment is zero.
Every C¹ integrator has bounded variation: the fundamental theorem bounds each absolute increment by the integral of |β′| over its cell, and summing gives a finite bound. The earlier mean-value comparison also works for signed derivatives. Hence its derivative-density integral is the same signed tagged limit just constructed. The changed example’s value −1/6 respects the variation bound and disproves the cancelling endpoint bound.
Integration by parts from an exact identity. Suppose both f and β are continuous and of bounded variation. On every cell, the endpoint product difference equals
f(x_i)[β(x_i)−β(x_(i−1))] + β(x_(i−1))[f(x_i)−f(x_(i−1))].
Verify the identity by expanding and cancelling the two middle products. Sum over cells: the product differences telescope to f(b)β(b)−f(a)β(a). The two sums on the right use opposite endpoint tags. The first converges to I_β(f) and the second to I_f(β), both justified by the bounded-variation construction with continuous integrands. Therefore
∫f dβ + ∫β df = f(b)β(b)−f(a)β(a).
No derivative or monotonicity is needed for this version. Each function is continuous as an integrand and has bounded variation as an integrator, so both required limits exist. Conversely, this proof has not established a formula for discontinuous bounded-variation functions with shared jumps. Such an extension needs its own convention and jump terms; the present hypotheses should not be silently removed.
Apply
For f(x)=x² and β(x)=sin x on [0,1], in radians, the companion integral is ∫sin x d(x²)=2∫x sin x dx=−2cos 1+2sin 1. The endpoint product is sin 1. Subtracting gives ∫x² d(sin x)=2cos 1−sin 1. Add the two answers to check the boundary term, rather than treating it as optional bookkeeping.
Return to the nonsmooth integrator α=x+4·1[x≥2/3] and f=x². The width contribution is 1/3, and the jump contribution is 4(2/3)²=16/9. Thus the weighted integral is 19/9. Integrating f against the derivative 1 where it exists gives only 1/3. The omitted 16/9 is a real jump contribution, not a numerical rounding error.
A Lebesgue–Stieltjes measure can represent such a jump as an atom, once a one-sided normalization and endpoint convention are specified. That theory is separate from the tagged-sum construction. Agreement in suitable continuous-integrand cases does not justify applying a measure integral’s existence or null-set rules to every Riemann–Stieltjes problem. The existing measure lesson develops a different route through absolute continuity and a density theorem.
Continue
For a new weighted calculation, first state the integrator class. A continuous derivative supplies the proven density reduction. A jump requires a separate increment contribution. A signed bounded-variation integrator with continuous integrand uses total variation for bounds rather than a cancelling endpoint increment. Integration by parts requires both integrals to have been defined under sufficient hypotheses before passing to the limit.
The practice asks you to reconstruct the mean-value estimate, derive the telescoping identity, audit a zero-total-increment mistake, and calculate a mixed smooth-plus-jump example. Successful symbolic manipulation is only one part of the evidence; the existence and convention checks explain why the symbols represent the intended limit.
