Integrability through oscillation control

Try

The step function equal to zero before x=1/3 and one from x=1/3 onward on [0,1] is discontinuous, yet integrable. Most partition intervals see a constant function; only intervals touching the jump can contribute to the upper-minus-lower gap. Enclosing the jump in a short interval makes that uncertainty small. This shows why continuity is a sufficient condition for integrability, not a necessary one.

Predict

Predict why monotonicity also helps. For an increasing function on a closed interval, the extrema on each subinterval occur at its endpoints, even when there are jumps. The oscillations are endpoint increments. Adding those increments telescopes to f(b)−f(a), so fine partition widths control the total sum gap. The proof does not need to count or locate every discontinuity.

Experiment

Compare two mechanisms.

Function class	Source of oscillation control	Resulting gap estimate
Continuous on [a,b]	Uniform continuity	Small oscillation × total length
Increasing on [a,b]	Telescoping increments	mesh(P) × (f(b)−f(a))
One-point modification	Small exceptional intervals	Bound × exceptional length

For the step example, place partition points just before and just after 1/3. Calculate the gap and shrink the enclosing interval. The function’s value exactly at the jump can affect finite sums but not the eventual integral.

Understand

Prove continuous functions on [a,b] are integrable. Compactness first gives boundedness, required for Darboux sums, and uniform continuity. Given ε>0 and a<b, choose δ so inputs less than δ apart have output difference less than ε/(2(b−a)). Choose a partition with mesh below δ. On each subinterval, supremum minus infimum is at most ε/(2(b−a)), because every pair of values differs by less than that bound. Summing yields U−L≤ε/2<ε. The factor two handles the possible nonattainment of suprema without turning a strict pointwise inequality into an unjustified strict supremum inequality.

For an increasing function, U−L=Σ(f(x_i)−f(x_(i−1)))Δx_i≤mesh(P)(f(b)−f(a)). If the endpoint difference is positive, choose the mesh below ε divided by that difference; if it is zero, the function is constant. Decreasing functions follow by negation.

Apply

Repair “integrable implies continuous except at finitely many points.” The statement is false; integrable functions can have infinitely many discontinuities. This unit does not prove the full measure-theoretic characterization. A bounded function that differs from zero only at isolated points 1/n, with values tending to zero there, gives a useful example to analyze by separating a finite initial collection from a uniformly small tail.

Changing an integrable function at one point preserves its integral. Isolate that point in partition intervals of arbitrarily small total length. A common bound on old and new values controls their possible contribution there, while outside those intervals the functions agree. Refining a small-gap partition proves integrability and makes the integral difference arbitrarily small.

When equal integrals force equal functions

Suppose two continuous functions satisfy f≤g on a nondegenerate interval [a,b], where a<b. If their integrals agree, must their graphs agree everywhere? Compare a narrow continuous tent with a one-point spike before answering.

Can a tiny area hide a high peak?

Case Continuous tent, height stays oneContinuous tent, common slope bound twoOne-point spike, continuity failsSigned line, nonnegativity fails Scale n 24816 Peak position Interior: 1/2Endpoint: 0

Predict whether increasing n changes the area, maximum, or both. Scale and position affect the tent cases; the signed line is fixed. The spike uses position only. All quantities are dimensionless.

Integral: 0.25; maximum: 1. Continuous: yes; nonnegative: yes. Lipschitz constant: 4.

The gold rectangle has height 0.5, width 0.125, and area 0.0625. This is a proved lower bound, not the whole tent area. A one-sided rectangle also works at the endpoint.

The maximum remains one as the area shrinks; the slope bound grows with n.

Reset

Exact piecewise-linear areas, not sample estimates. Temporary controls do not establish mastery or test an infinite sequence.

Set h=g−f. It is continuous, nonnegative and has integral zero. If h(c)=η>0 at some c, continuity gives a relative neighborhood of c on which h≥η/2. Because a<b, that neighborhood contains a closed interval of positive length ℓ, even if c is an endpoint. Integral monotonicity and nonnegativity outside it imply ∫h≥ηℓ/2>0, a contradiction. Therefore h=0 everywhere, and f=g. The monotonicity used here follows directly from the ordered lower and upper sums; the next lesson develops these integral properties further.

This completes the continuity-sensitive equality argument in MIT 18.100B problem set 9, question 1. It also identifies three essential boundaries. A nonnegative one-point spike has zero integral without vanishing, so integrability cannot replace continuity. The continuous function 2x−1 on [0,1] has zero integral by signed cancellation, so the order hypothesis cannot disappear. On a degenerate interval [a,a], all integrals are zero regardless of the functions’ values.

Small integral is a different claim

For n≥2, the continuous tent hₙ(x)=max(1−n|x−1/2|,0) on [0,1] has maximum one and integral 1/n. Thus integrals tending to zero do not force uniform convergence to zero, even when every function is continuous and nonnegative. At the endpoint, max(1−nx,0) instead has integral 1/(2n). Continuity creates some positive-width neighborhood for each fixed positive peak, but those widths need not be common across a sequence.

A common Lipschitz bound supplies the missing uniform control. Suppose h≥0 on [0,1] and |h(x)−h(y)|≤L|x−y| with L>0. Compactness supplies a maximizer c with height H. One side of c has length at least 1/2. Along that side, h≥H/2 for distance at most H/(2L). Hence

∫₀¹h ≥ (H/2) min(H/(2L), 1/2).

This is a lower bound, not a sharp equality claim. If H≤L it becomes H²/(4L); if H>L it becomes H/4. If L=0, h is constant and ∫h=H. For a sequence with one common L, any subsequence of heights at least ε would force areas at least (ε/2) min(ε/(2L),1/2), contradicting areas tending to zero. Therefore common-Lipschitz control plus nonnegative vanishing integrals does imply uniform convergence to zero.

In the second tent mode, hₙ(x)=max(1/n−2|x−1/2|,0): its common Lipschitz constant is two, maximum is 1/n, and area is 1/(2n²). The display shows the one-sided lower rectangle separately from the full tent. Explain why replacing the growing slopes of the first family with a common slope changes the conclusion.

Continue

Optional depth considers the bounded function that is one at every 1/n and zero elsewhere. Its exceptional points accumulate at zero, but all can be enclosed by a short interval near zero plus finitely many tiny intervals. The same small-total-length strategy proves integrability with integral zero. Work through the finite choices explicitly rather than calling a countable set automatically negligible without a theorem.

Save a continuous-function proof and a monotone-function proof side by side, identifying where compactness or telescoping enters. Next, integral properties and the fundamental theorem will connect these order-based constructions to derivatives and practical evaluation.