Point oscillation and small interval covers

Try

On the interval [0,3], define f(1)=4 and f(x)=0 at every other point. A graph shows a single raised dot above an otherwise flat line. How much can nearby function values differ near x=1? No matter how closely you inspect, the neighborhood includes the raised center and other points with value zero. The largest possible difference remains 4. At x=2, a sufficiently small neighborhood misses the spike entirely and the difference becomes zero.

This measurement uses all values in a neighborhood, not their average. In particular, a single exceptional value matters. We call the resulting smallest local range the point oscillation. The term does not mean an average deviation, and does not require the function to repeatedly wiggle.

Use a second function g: its value is −1 for x below 2 and 2 for x at least 2. Before reading further, predict its local range at the jump, at the left endpoint, and at the right endpoint. Neighborhoods are always restricted to the given domain; there are no function values outside [0,3] to include.

Predict

The spike and step are different pictures but have the same kind of discontinuity certificate: a positive local range survives every reduction in neighborhood size. Changing the spike’s value from 4 to −4 leaves its oscillation 4. Changing the step’s value at its jump from 2 to 7 would increase that point’s oscillation to 8: the center itself belongs to every neighborhood.

An endpoint can also be discontinuous. If f(0) is changed to 5 while its nearby positive arguments retain value zero, its relative point oscillation at 0 becomes 5. One-sided neighborhoods do not automatically make endpoint discontinuities disappear.

Experiment

Draw the two worked functions on paper. Around each of the marked points 0,1,2,3, draw relative neighborhoods of radii 1/4 and 1/8. Record the supremum and infimum separately before subtracting. A supremum is a least upper bound; it need not be attained in general, even though the worked functions attain their local bounds.

Now construct a function that equals 3 at x=1, equals 1 at x=2, and equals zero elsewhere on [0,3]. List the points where its oscillation is at least 2, then those where it is at least 1/2. What happens at a point very close to 1 but different from it?

Check the changed two-spike construction

The threshold sets are respectively the singleton containing 1 and the two-point set containing 1 and 2. At any other fixed point, choose a radius smaller than its positive distance from both spikes. The neighborhood range is then zero. Being near a discontinuity is different from having positive oscillation at one’s own point.

For the two-point set, cover each point by an open interval of length ε/4. The total length is ε/2, less than the prescribed positive budget ε. Repeat this construction for smaller budgets: the points do not disappear, but their covering cost can be made arbitrarily small. This is the precise meaning of a null set used below.

Which spikes did the plot omit?

For a reduced rational p/q in [0,1], Thomae’s function T has value 1/q; at irrational inputs its value is 0. The finite function T_Q keeps only denominators q≤Q and is 0 everywhere else. Neither function connects these isolated points with lines.

Denominator cutoff Q24816 Equal partition intervals m48163264128 Reset

Predict the upper sum before revealing it. A spike on a partition boundary belongs to both neighboring closed cells. Every cell has infimum 0 because irrational points occur in every nonempty interval.

3 retained rational spikes. Circles are exact locations computed from reduced fractions; shaded rectangles are upper-sum areas for T_Q. The plot is a finite construction and does not determine the continuity or integral of T.

Reveal bounds and omitted-spike certificate

Inspect all exact cell memberships and rounded bounds

Both endpoints are included; heights and sums are rounded.

Closed cell	Retained rational inputs	Finite supremum	Full-function upper bound (rounded)
[0/8, 1/8]	0/1	1	1
[1/8, 2/8]	None	0	0.3333333
[2/8, 3/8]	None	0	0.3333333
[3/8, 4/8]	1/2	0.5	0.5
[4/8, 5/8]	1/2	0.5	0.5
[5/8, 6/8]	None	0	0.3333333
[6/8, 7/8]	None	0	0.3333333
[7/8, 8/8]	1/1	1	1

Start with Q=2 and m=8: explain why the spike at 1/2 contributes to two cells. Increase m to 128 with Q fixed, then increase Q. Which numerical changes support your prediction, and which infinite claims still require proof? Controls are temporary and award no completion.

In this explorer, the displayed rational spikes have only finitely many allowed denominators. They are not the full infinite Thomae function. Compare the finite-spike upper sum with the separately justified bound for omitted denominators. A boundary spike belongs to both adjacent closed partition cells, so do not discard it from either supremum. Increasing the finite cutoff adds known spikes; it does not turn a finite picture into a complete discontinuity proof.

Understand

Let f be bounded on a nondegenerate compact interval [a,b], meaning a and b are real and a is strictly smaller than b. For x in the interval and radius r>0, put V(x,r)=[a,b] intersected with the open interval from x−r to x+r. Define the local range as the supremum of f on V(x,r) minus its infimum there. Boundedness and the presence of x ensure both bounds are finite. Define ω(x), the point oscillation, as the infimum of these nonnegative ranges over all positive radii.

Continuity is equivalent to ω(x)=0. If f is continuous at x, for any ε>0 its values in a sufficiently small relative neighborhood differ from f(x) by less than ε/3. Its range is therefore at most 2ε/3, which is less than ε. This gives zero oscillation without assuming the bounds are attained. Conversely, if ω(x)=0, some neighborhood has range less than ε. Because f(x) and every nearby f(y) lie between its infimum and supremum, their difference in absolute value is less than ε. This is exactly relative continuity, including at the endpoints.

For a fixed positive threshold η, let E(η) be the set of points where ω(x)≥η. This set is compact. To prove it, first take a point outside it. There is r>0 such that the range on V(x,r) is less than η. Whenever the distance from y to x is less than r/2, the neighborhood V(y,r/2) lies inside V(x,r). Its range cannot be larger, so ω(y) is also less than η. The complement of E(η) is relatively open. Thus E(η) is relatively closed in the compact interval and hence compact. This argument is about all nearby values; checking a finite grid would not prove it.

The discontinuity set D is exactly the union of E(1/m) over positive integers m. A point is discontinuous exactly when its oscillation is positive; every positive number is at least 1/m for a sufficiently large integer m. Conversely, membership in any such threshold set forces positive oscillation. This expresses D as a countable union of compact sets, but does not yet prove that any of those sets is null.

Here a set is null if, for every ε>0, it can be covered by countably many open intervals whose lengths sum to less than ε. A countable set with listed points x₁,x₂,… is null: center an interval of length ε/2^(j+1) at x_j. The total is ε/2. For countably many null sets A_j, cover A_j with total length less than ε/2^(j+1). Enumerate all these intervals by traversing pairs of positive indices in diagonal order. Their total length is at most ε/2. This proves countable-union closure using a concrete error budget. Empty sets require no intervals.

Apply

Can the same reasoning show that every set is null, by writing it as a union of singleton points? No: the union theorem is countable. To see the obstruction concretely, prove that any finite interval cover of [a,b] has total length at least b−a. Partition [a,b] at all covering-interval endpoints lying inside it. Each positive-length partition cell is covered by at least one interval throughout its interior: an interval containing an interior point cannot end inside that cell. Assign each cell to one covering interval. The total assigned to an interval is no greater than its length; summing gives the lower bound b−a.

For a countable open cover of the compact interval, compactness supplies a finite subcover. Its total length is already at least b−a, and adding the other nonnegative lengths cannot reduce it. Therefore a cover with total length less than (b−a)/2 is impossible. A nondegenerate interval is not null, although each of its points individually is null. Compactness and the openness of the covering intervals are the bridge from the finite argument to countable covers.

In practice, do not confuse density with positive covering cost. The rational points in [a,b] are dense and countable, hence null. A set can meet every open subinterval while still admitting arbitrarily short total covers. The practice constructs a bounded function whose discontinuities are exactly such a dense countable set, and asks you to prove its threshold sets explicitly.

Continue

Retain three distinct objects: a finite list of sampled function values, the range over an entire relative neighborhood, and the infimum of those ranges as the neighborhood shrinks. Only the last object is point oscillation. Neither small observed differences nor small average deviations certify that it vanishes.

The next criterion uses the compact threshold sets and null-cover budgets to connect discontinuities with upper and lower sums. The present lesson supplies those ingredients; it has not assumed every discontinuity set is null. In a proof, state which set is being covered, the total length budget, and whether a finite subcover is needed. These details prevent a local picture from being mistaken for an assertion about all points.