Try
A rising graph can jump without having a prescribed value at the jump. On [0,2], define f(x)=0 before x=1, f(1)=1, and f(x)=3 after x=1. Approach 1 from each side. The left values approach 0 and the right values approach 3, while the value at 1 is neither side limit. This is still a nondecreasing function: moving right never lowers the output.
Draw its two horizontal pieces with open circles at (1,0) and (1,3), then add the filled point (1,1). Predict what changes if that filled point moves to height 2. The jump size remains 3. A jump measures the difference between side limits, not the distance from one side to the chosen point value.
| Input | f(x) | What this tells us |
|---|
| 0.9 | 0 | Left approach |
| 0.99 | 0 | Same left value |
| 1 | 1 | Actual point value |
| 1.01 | 3 | Right approach |
| 1.1 | 3 | Same right value |
The table illustrates the specified formula. Five samples alone would not prove what an unspecified function does between them.
Predict
Suppose now that f is any nondecreasing real-valued function on [a,b], with a strictly less than b. Its values are trapped between f(a) and f(b). Predict whether boundedness alone guarantees side limits. It does not: the rational indicator alternates between 0 and 1 arbitrarily close to every point. Order is the additional mechanism that traps nearby values near a supremum or infimum.
For an interior point x, predict the order of three quantities: the left limit, f(x), and the right limit. They must appear in that order. Consequently a discontinuity cannot be repaired by moving f(x) if the two side limits differ. Conversely, if the side limits coincide, monotonicity already forces f(x) to have the same value.
Experiment
Construct a second nondecreasing graph on [0,3]. Use endpoint values f(0)=−1 and f(3)=5; on the open intervals (0,1), (1,2), (2,3), use heights 0, 2, 4. Set f(1)=1 and f(2)=3. Mark every filled and open endpoint before checking the following diagnosis.
There are two interior jumps, each of size 2, and two relative endpoint discontinuities, each with discrepancy 1. At 0 there is only a right approach inside the domain; at 3 there is only a left approach. There is no need to extend the function beyond [0,3]. Changing f(0) to 0 repairs that endpoint, while changing f(1) to any value in [0,2] preserves monotonicity but cannot remove its jump.
Now try to place ten jumps of size at least 1/4 in a graph whose total increase is 7/3. Their combined increase would be at least 10/4, exceeding 7/3. This finite budget will give a second countability proof, rather than merely a numerical observation.
Understand
Define f(x−) as the supremum of f(t) for a≤t<x, and f(x+) as the infimum of f(t) for x<t≤b. These sets are nonempty and bounded when x is interior. To prove that the first quantity is a limit, let L=f(x−) and ε>0. The supremum definition supplies t₀<x with f(t₀)>L−ε. For every t between t₀ and x, monotonicity gives L−ε<f(t)≤L. This is the required one-sided limit argument, without assuming L is attained. For U=f(x+), choose t₁>x with f(t₁)<U+ε. Then x<t<t₁ implies U≤f(t)<U+ε.
Thus f(x−)≤f(x)≤f(x+). Interior continuity is equivalent to equality of all three. At a, define only the existing right limit by the corresponding infimum; continuity means it equals f(a). At b use the left supremum and compare it to f(b). These are relative-domain conventions, not two-sided definitions with invented outside values.
At an interior discontinuity x, the open value interval J(x)=(f(x−),f(x+)) has positive length. If x<y, choose t strictly between them. The definitions and monotonicity imply f(x+)≤f(t)≤f(y−), so J(x) and J(y) are disjoint. Fix an enumeration of the rational numbers. Assign to each discontinuity the first rational inside J(x). Density guarantees a witness; disjointness makes the assignment injective. There are at most countably many interior discontinuities, and the endpoints add at most two.
Alternatively, for any finite ordered collection of jump points, their jump sizes sum to at most f(b)−f(a): the intervening increases are nonnegative. For each positive integer n, there can therefore be only finitely many jumps of size at least 1/n. Every positive jump lies in one of these finite sets. This proves countability without pretending that all jumps have a common positive size.
Apply
Can infinitely many jumps accumulate at a continuous point? Consider F(x)=Σ from n=1 to infinity of 2^(−n) times the indicator of [1/n,1], on [0,1]. The indicator is 1 when x belongs to that interval and 0 otherwise. Each term is nondecreasing; the entire sum stays between 0 and 1.
The tail after N is at most 2^(−N), uniformly for all x. Near a point different from the first N thresholds, their finite sum is constant. Consequently F is continuous at every point outside the set of thresholds, including 0: choose a small neighborhood excluding the first N thresholds, then make the tail small. At 1/n for n≥2, isolate that threshold among the first N terms. All other finite terms are locally constant, and the tail tends uniformly to zero; the jump is exactly 2^(−n). At 1, the left limit is 1/2 while F(1)=1, so this is an endpoint discontinuity.
Explain why this construction does not contradict countability. Then change the weights to 3^(−n). The same uniform-tail reasoning works, with tail 1/(2·3^N), interior jump 3^(−n), and endpoint discrepancy 1/3. A finite plot cannot prove either the continuity at 0 or the existence of infinitely many jumps; the tail estimate does both jobs.
Continue
Return to suprema and the Archimedean property if the approximate-extremum step needs practice. The existing monotone integrability proof bounds Darboux gaps directly by mesh size times total increase. It does not need the discontinuities to be listed first.
The next question is what replaces monotone order for a general bounded function. Local oscillation measures the spread of nearby values. The later criterion relates its discontinuity set to Riemann integrability; it must not be inferred from countability alone for functions outside today’s hypotheses.
