Try
A function can have discontinuities densely packed into an interval and still have integral zero. Define Thomae’s function T on [0,1]. If x=p/q is rational in lowest terms, with q a positive integer, set T(x)=1/q. At irrational x, set T(x)=0. In particular T(0)=T(1)=1 because their reduced denominators are 1. The reduced-form requirement matters: writing 1/2 as 100/200 must not change its height.
Plot the following points as separate spikes, not as a joined polyline. Along the baseline, place several irrational inputs with value 0. Predict whether a dense set of spikes necessarily creates a positive integral.
| Input x | Reduced denominator | T(x) |
|---|
| 0 | 1 | 1 |
| 1/4 | 4 | 1/4 |
| 1/3 | 3 | 1/3 |
| 1/2 | 2 | 1/2 |
| 2/3 | 3 | 1/3 |
| 3/4 | 4 | 1/4 |
| 1 | 1 | 1 |
| √2/2 | Irrational | 0 |
There are infinitely many rational spikes in every interval, but only finitely many spikes above any fixed positive height. That distinction drives the proofs.
Predict
At a rational point p/q, irrational inputs approach while their outputs stay 0. The positive value 1/q prevents continuity. At an irrational input x, however, nearby rational spikes can be forced to have large denominators and therefore small heights. Predict which of these statements needs a finite-set argument: “the rationals are dense” or “all nearby spike heights are small.” Density alone does not control heights.
Also compare T with the rational indicator D, which equals 1 at every rational and 0 at every irrational. Both differ from the zero function only on a countable set. Nevertheless one is Riemann integrable and the other is not. The location of changed values and the resulting discontinuity set are different kinds of information.
Experiment
List all reduced rationals in [0,1] with denominator at most 3: 0, 1/3, 1/2, 2/3, 1. Away from these five points, T is at most 1/4. Choose disjoint short neighborhoods around them, with irrational interior endpoints and combined length less than 1/20. Include their endpoints in a partition. On the neighborhood cells the supremum is at most 1; on all other cells it is at most 1/4. Thus the upper sum is at most 1/20+1/4=3/10. This is an upper bound, not the integral’s exact value.
To obtain an arbitrarily small bound, increase the denominator cutoff as well as shrinking the exceptional neighborhoods. Changing only their widths while retaining cutoff 3 leaves the crude 1/4 remainder. Every lower sum is zero because every nondegenerate cell contains an irrational input.
For a nearby action, use denominator cutoff 9 and total exceptional length less than 1/100. The same estimate gives an upper sum below 11/100. Explain why this finite certificate still does not say there are only finitely many discontinuities.
Understand
Fix an irrational x₀ and ε>0. Choose an integer Q with 1/(Q+1)<ε. There are finitely many reduced rationals in [0,1] with denominator at most Q: for each q≤Q, the numerator has only q+1 possibilities. Since x₀ is not one of them, choose a positive relative neighborhood avoiding all of them. Throughout that neighborhood, T is either 0 or at most 1/(Q+1). Hence |T(x)−T(x₀)|<ε. This proves continuity at every irrational point. At each rational point, an irrational sequence approaching within the domain proves discontinuity, including the two endpoints. The exact discontinuity set is therefore the rationals in [0,1].
For the direct integral proof, choose Q so that 1/(Q+1)<ε/2. Enclose the finite high-spike set in disjoint small relative closed intervals with total length less than ε/2. Their interior boundaries can be irrational; the endpoint spikes 0 and 1 use one-sided neighborhoods. A partition containing these boundaries has no high spike in its remaining cells, including their endpoints. The bad cells contribute less than ε/2 because T≤1; the remaining cells contribute less than ε/2 because their total length is at most 1 and their suprema are at most 1/(Q+1). Thus U(T,P)<ε while L(T,P)=0. The Darboux criterion proves that T is Riemann integrable and its integral is 0. This proof does not assume the discontinuity criterion it illustrates.
Now suppose f and g are both bounded Riemann-integrable functions on [a,b] and agree outside a null set E. Here “null” means coverable by intervals of arbitrarily small total length. Such a set cannot contain a nondegenerate interval, so its complement is dense. By integral linearity, h=f−g is integrable. Each partition cell contains a zero of h, so every lower sum is at most 0 and every upper sum is at least 0. Their common integral must therefore be 0. Consequently the integrals of f and g are equal.
Both integrability assumptions are essential. Changing zero into the rational indicator only changes a countable null set, yet every cell then has supremum 1 and infimum 0. Its Darboux gap is always the length of the whole interval. It is not Riemann integrable. The statement that null modifications preserve a Lebesgue integral is a different theorem with a different integrability framework.
Apply
Replace each rational spike height by 1/√q. The finite-high-spike argument still works: outside denominators at most Q, the heights are at most 1/√(Q+1), which tends to zero. The exact continuity set remains the irrationals and the integral remains 0. If instead every rational spike has height 1, that decay disappears and the rational-indicator counterexample returns. Explain the role of decay rather than simply repeating that the exceptional set is countable.
Finally, suppose a finite data table contains only zeros. The zero function fits it. So does the rational indicator modified to zero at those finitely many sampled inputs. The latter still takes both 0 and 1 in every nondegenerate interval and is not integrable. These observations cannot certify integrability without a function specification or additional hypotheses.
Equality of integrals also has no converse about exceptional sets: x and 1−x on [0,1] both integrate to 1/2, but agree only at x=1/2. Their difference is nonzero on intervals of positive length.
Continue
Reuse partitions and Darboux sums for the upper-minus-lower criterion, and integral properties for the linearity step. The present lesson adds dense-exception proofs rather than replacing those foundations.
Keep three claims separate in your written work: where two functions differ, where each is discontinuous, and whether each is integrable. The bounded discontinuity criterion can connect the last two; equality outside a null set alone cannot supply the missing Riemann-integrability assumption.
