Try
A bounded graph may jump and still have a well-defined area. On [0,1], let f(x)=0 for x<2/5 and f(x)=2 for x≥2/5. Draw the two horizontal pieces, marking the value at the jump. Predict the upper-minus-lower rectangle area for a partition that places a boundary exactly at 2/5.
A boundary alone does not remove the jump from every closed partition cell. The left cell includes the value 2 at its right endpoint. We must control its width as well.
Same function, progressively smaller cell containing both heights| Partition points | Cell with oscillation 2 | Upper minus lower sum |
|---|
| 0, 2/5, 1 | [0, 2/5] | 4/5 |
|---|
| 0, 3/10, 2/5, 1 | [3/10, 2/5] | 1/5 |
|---|
| 0, 39/100, 2/5, 1 | [39/100, 2/5] | 1/50 |
|---|
In each row the lower sum is 6/5; the upper sum exceeds it by the displayed gap. Replacing 39/100 by 2/5−h, with 0<h<2/5, gives gap 2h. Hence the jump does not prevent integrability, and the integral is 6/5. The rectangle mechanism, not a visual guess, explains why.
Predict
Which feature should matter for a bounded function: the number of discontinuities, their total interval-cover length, or the height of the largest jump? A single large jump can be confined to arbitrarily short cells. Even some uncountable discontinuity sets can be covered by intervals of arbitrarily small total length.
Call a set null when, for every δ>0, open intervals cover it with summed lengths less than δ. This is a length condition, not a count of points. A finite set is null by putting a tiny interval around each point. A countable union of null sets is null: give its mth set a cover budget δ/2^(m+1) and combine the covers. No nondegenerate interval is null; the preceding lesson proves the interval-cover length inequality.
Predict the precise theorem: a bounded real function on a compact interval is Riemann integrable if and only if its discontinuity set is null. Boundedness and the compact interval are hypotheses, not conclusions. We prove the statement for a<b; a one-point interval has zero Darboux sums and relative continuity automatically.
Experiment
For a bounded function f, write osc(f;I)=sup over I of f minus inf over I of f. A partition cell contributes its oscillation times its width to the Darboux gap. Imagine marking each cell red when its oscillation is at least a chosen positive ε. Make a two-column ledger: total width of marked cells, and the contribution of all unmarked cells.
If |f|≤B, the marked-cell contribution is at most 2B times their total width. The unmarked-cell contribution is at most ε(b−a). Thus there are two independent controls: short total width where large oscillations remain, and small oscillation everywhere else. The proof below constructs a partition that provides both.
As a changed paper action, replace the worked jump by height 5 beginning at 3/4. Construct a partition whose Darboux gap is less than 1/100. One choice is 0, 3/4−1/1000, 3/4, 1; its gap is 5/1000=1/200. Record the strict error target and the chosen width rather than saying only “take a fine partition.”
Understand
The point oscillation ω_f(x) is the infimum, over radii r>0, of osc(f;[a,b] intersected with (x−r,x+r)). The preceding lesson establishes two facts: ω_f(x)=0 exactly at points of relative continuity, and each positive threshold set K_ε consisting of points with ω_f(x)≥ε is compact. Its compactness is essential in the reverse direction below.
The Darboux criterion says bounded f is integrable exactly when partitions can make U(f,P)−L(f,P) arbitrarily small.
Integrability forces a null discontinuity set
Fix ε>0 and a requested cover budget δ>0. Integrability supplies a partition with Darboux gap less than εδ/2. Mark the cells whose oscillation is at least ε. Because their contributions to the gap are at least ε times their lengths, their summed lengths are less than δ/2.
Every point of K_ε that is not a partition endpoint lies in the interior of a marked cell. Indeed, a sufficiently small relative neighborhood lies within its containing cell, so its point oscillation is at most that cell’s oscillation. A threshold point therefore forces that cell to have oscillation at least ε.
Enlarge the finitely many marked closed cells to open intervals, adding less than δ/4 in total length. Separately cover all partition endpoints with open intervals of total length less than δ/4. This covers K_ε within budget δ, proving it is null. The separate endpoint step matters: a boundary point can have large point oscillation because of values across the other adjacent cell; assigning it to one arbitrarily chosen cell can lose that information.
A point is discontinuous exactly when its oscillation is positive, so the discontinuity set D is the union, over positive integers m, of K_(1/m). Both inclusions follow directly: any positive number exceeds some 1/m, and membership in such a threshold implies positive oscillation. The countable null-cover budget proves D is null. We have not claimed that D is countable.
A null discontinuity set gives small Darboux gaps
Now suppose D is null. Choose B≥1 with |f|≤B, write ℓ=b−a, and let η>0 be a desired gap bound. Set ε=η/(2ℓ) and δ=η/(4B). Since K_ε is a compact subset of D, an open interval cover with total length less than δ has a finite subcover of K_ε. Let O be the union of these finitely many intervals; its total union length is no greater than their summed lengths.
For each x outside K_ε, the definition ω_f(x)<ε supplies a relative open neighborhood V_x with oscillation less than ε. Together O and these neighborhoods cover [a,b]. Extract a finite subcover. If K_ε is empty, simply use the neighborhoods and omit O.
A Lebesgue number λ>0 for this finite open cover guarantees that every subset of [a,b] with diameter less than λ is contained in one cover member. Here is a compactness justification. Around each point choose a relative ball of radius 2r lying in some cover member. Finitely many corresponding balls of radius r cover the interval. Set λ to the smallest of their positive radii. Any set of diameter less than λ, after choosing one point in it, lies in the doubled ball associated with a smaller ball containing that point. It therefore lies in the required cover member.
Choose a partition with every cell length less than λ. If a cell has oscillation at least ε, it cannot lie in a V_x and must lie wholly in O. These cells have disjoint interiors. Their total length is at most the length of O, hence less than δ; this elementary inequality follows by subdividing at all interval endpoints and adding the covered segment lengths. Overlapping cover intervals cannot increase the union length beyond the sum of their lengths.
The Darboux gap is consequently at most εℓ plus 2B times the marked-cell total. It is strictly less than η/2+2Bδ=η. Since η was arbitrary, the Darboux criterion proves integrability. This completes both directions, not merely a heuristic that discontinuities have “small area.”
Apply
Use the six written tasks to reconstruct each proof with its length budgets visible. The closed-null-set task is deliberately not restricted to countably many exceptional points. The weighted rational/irrational example requires proving that its discontinuity set contains an interval, rather than assuming that every uncountable set has positive length.
The endpoint task changes only two values of the continuous function x on [0,1]. Those endpoints become genuine discontinuities under relative continuity, yet their finite set is null and the integral stays 1/2. Compute the cell extrema to see how shrinking widths remove their contribution. Conversely, changing values on a null set can create discontinuities elsewhere through density; the criterion asks about the discontinuity set of the resulting function, not merely the set where a formula was edited.
Do not use one unsuccessful partition sequence as a general proof of nonintegrability. Either show that every partition retains a positive gap, or identify a non-null discontinuity set and apply the proved criterion. The practice makes this distinction explicit.
Continue
For a new bounded example, identify its point oscillation before invoking the theorem. Record the exceptional set, an actual null cover or a proof that no such cover exists, and how the Darboux conclusion follows. If the function is unbounded, this bounded Riemann criterion is unavailable; an improper integral is a different definition and needs its own endpoint analysis.
The next exceptional-function lesson supplies further tests of the boundary between dense sets, null sets and discontinuity sets. Keep the cover budget, finite subcover and mesh choice as separate steps: compactness provides a finite geometric control that a list of sample points cannot replace.
