Monotone Functions and Countably Many Discontinuities

Monotonicity Restricts Discontinuities

A bounded monotone function can jump, but it cannot oscillate back and forth. Each interior jump occupies a nonempty interval of function values, and the intervals associated with different jumps do not overlap. Since every nonempty open interval contains a rational number, this gives a direct way to count the discontinuities. Combined with the Lebesgue criterion for Riemann integrability, the countability result also guarantees that every bounded monotone function on a compact interval is Riemann integrable.

Here, a function \(f:[a,b]\to\mathbb{R}\) is nondecreasing if \(f(x)\le f(y)\) whenever \(x\le y\). Nonincreasing means \(f(x)\ge f(y)\) whenever \(x\le y\). The function is bounded if there is a finite \(M\) such that \(|f(x)|\le M\) for every \(x\in[a,b]\). Continuity at an endpoint is understood relative to the interval: at \(a\), only values to the right are relevant, and at \(b\), only values to the left are relevant.

One-Sided Limits Detect Jumps

Consider first a bounded nondecreasing function. At each interior point \(x\in(a,b)\), define its left and right limits by

These quantities are finite because \(f\) is bounded. Monotonicity gives \(f(x-)\le f(x)\le f(x+)\). To see that the first quantity is the left limit, fix \(\varepsilon>0\). By the definition of supremum, some \(u<x\) satisfies \(f(u)>f(x-)-\varepsilon\). For \(u<t<x\), monotonicity then gives \(f(x-)-\varepsilon<f(u)\le f(t)\le f(x-)\). This is the left-limit condition. The right-limit condition follows in the same way from the infimum: choose \(v>x\) with \(f(v)<f(x+)+\varepsilon\); for \(x<t<v\), \(f(x+)\le f(t)\le f(v)<f(x+)+\varepsilon\).

It follows that \(f\) is continuous at \(x\) exactly when \(f(x-)=f(x)=f(x+)\). In particular, an interior discontinuity must satisfy \(f(x-)<f(x+)\). The open interval \((f(x-),f(x+))\) is therefore nonempty at every interior discontinuity.

Proof of Countability and Integrability

Proof. Let \(D\) be the set of interior discontinuities of \(f\). For each \(x\in D\), the open interval \((f(x-),f(x+))\) contains a rational number. If \(x<y\) are in \(D\), choose \(t\) with \(x<t<y\). The definitions of the one-sided limits and monotonicity give \(f(x+)\le f(t)\le f(y-)\). Thus the two open intervals \((f(x-),f(x+))\) and \((f(y-),f(y+))\) are disjoint. Fix an enumeration of the rational numbers and assign to each \(x\in D\) the first rational in that enumeration lying in \((f(x-),f(x+))\). Disjointness makes this assignment injective, so \(D\) is at most countable.

There are at most two further discontinuities, at \(a\) and \(b\). Indeed, the monotone one-sided limit exists at each endpoint, and a discontinuity there can only be a mismatch between that limit and the endpoint value. Adding at most two points to a countable set preserves countability.

A countable subset of \([a,b]\) has Lebesgue measure zero: if its points are listed as \(x_1,x_2,\ldots\), then for any \(\varepsilon>0\) cover \(x_n\) by an open interval of length \(\varepsilon/2^n\). The total length of this cover is at most \(\sum_{n=1}^{\infty}\varepsilon/2^n=\varepsilon\), so the set has outer measure zero. The Lebesgue criterion for Riemann integrability states that a bounded real-valued function on a compact interval is Riemann integrable if and only if its set of discontinuities has Lebesgue measure zero; this criterion was established in earlier material. Applying it to \(f\) proves Riemann integrability.

If \(f\) is nonincreasing, then \(-f\) is bounded and nondecreasing. It has the same discontinuities as \(f\), so the conclusion follows. If \(a=b\), there is only one point in the domain, the function is continuous in the relative topology, and its integral over the degenerate interval is zero. \(\square\)

The proof uses boundedness to ensure finite one-sided limits. It also explains why countability is stronger than needed for integrability: the criterion requires only that the discontinuity set have measure zero.

A Pitfall: A Jump Cannot Be Ignored by the Fundamental Theorem

For the single-jump function above, it is incorrect to apply the fundamental theorem of calculus across the jump and conclude that the integral equals \(f(1)-f(0)=2\). Although \(f'(x)=1\) away from \(x=1/2\), this function is discontinuous there, so that application of the theorem is not justified. The actual integral, computed on the two pieces, is \(1\), while \(f(1)-f(0)=2\). The jump contributes to the endpoint difference but is not an area of that size. Monotonicity guarantees Riemann integrability; it does not make a discontinuous function eligible for every calculus formula.

It is also important not to mistake “at most countably many” for “finitely many.” The rational-interval argument permits infinitely many jumps: their associated value intervals are disjoint, but their lengths can become arbitrarily small. Countability is enough here because every countable subset of a real interval has Lebesgue measure zero.