Classification of Discontinuities and Monotone Functions

The classification of discontinuities is not an isolated curiosity; it is load-bearing machinery throughout analysis and its applications.

• Riemann integration theory. A bounded function on $[a,b]$ is Riemann integrable if and only if its set of discontinuities has Lebesgue measure zero (Lebesgue's criterion). Since every countable set has measure zero, the monotone function theorem immediately implies that every monotone function on a closed interval is Riemann integrable — a foundational result usually proved as an easy corollary of exactly this classification theorem.
• Distribution functions in probability. The cumulative distribution function (CDF) of any real random variable is nondecreasing. The theorem guarantees that a CDF has at most countably many jump discontinuities, which is precisely the set of "atoms" of the distribution — outcomes with positive probability mass. Continuous random variables are, by definition, those whose CDF has no jumps at all.
• Sturm–Liouville theory and generalized functions in physics. Step-like potentials and piecewise-defined material properties (e.g. in wave propagation through layered media) produce monotone or piecewise-monotone auxiliary functions whose jump sets must be tracked explicitly; the countability guarantee assures that numerical and analytic techniques only need to handle a discrete, listable set of interface points.
• Numerical analysis and root-finding. Bisection-type algorithms rely on the The Intermediate Value Theorem, which requires continuity; understanding exactly where a monotone function can fail continuity (only at isolated jump points) lets algorithms safely bracket roots away from the countable bad set.
• Cryptographic and combinatorial constructions. Monotone Boolean functions and monotone real-valued potential functions appear in the analysis of algorithms and in circuit complexity; discreteness (countability) of their discontinuity structure underlies arguments about worst-case behavior at threshold values.
• Real analysis pedagogy itself. The theorem is a favorite bridge result connecting Continuity of Polynomials, Rational, and Elementary Functions and One-Sided Limits to the more advanced machinery of measure theory, providing students their first taste of a "structure theorem": a global, countability-flavored conclusion extracted from a purely local, order-theoretic hypothesis.

:::quiz[The Dirichlet function $\mathbf{1}_{\mathbb{Q}}$ on $[0,1]$ shows that, without monotonicity, which conclusion of the theorem can fail?]

• Only the "countable" conclusion fails; discontinuities remain jumps
• Both conclusions fail: discontinuities are essential (not jumps) and uncountably many ✓
• The function remains continuous everywhere, so nothing fails
• Only the "jump" conclusion fails; the set remains countable

Ask the tutor: How does this classification change if we only require $f$ to be Riemann integrable rather than monotone — can its discontinuity set still be uncountable?

Picture the graph of $f$ near $c$ as two half-graphs, one drawn as $x \to c^-$ and one as $x \to c^+$. Continuity asks that both half-graphs approach the same height, and that this height matches the actual dot plotted at $f(c)$. A discontinuity is a failure of this matching, and the classification simply asks how the two half-graphs and the actual point disagree. If both half-graphs approach the same height but the dot is misplaced (or missing), you have a removable discontinuity — a single pinprick that could be patched by moving one point. If the two half-graphs approach different heights, no single repair works: there is a genuine step, a jump. If one of the half-graphs doesn't settle down at all — it oscillates forever, or shoots to infinity — the discontinuity is essential, and no simple picture of "two heights that disagree" even applies.

Now specialize to a monotone (say, nondecreasing) function on an interval. Because the function never turns around, the values it takes just to the left of $c$ are all bounded above by $f(c)$, and they only increase as $x$ climbs toward $c$. A bounded, monotone quantity must settle down — this is exactly the completeness of $\mathbb{R}$ at work, in the guise of "bounded monotone sequences converge." So the left-hand limit exists no matter what. The same argument, mirrored, gives the right-hand limit. The oscillating and infinite behaviors that generate essential discontinuities are therefore structurally impossible for a monotone function: it can only ever jump.

Example 1 (classifying discontinuities of an explicitly built function). Define $f : \mathbb{R} \to \mathbb{R}$ by

Classify the behavior of $f$ at $x=1$ and $x=3$. At $x=1$: Compute the one-sided limits using continuity of polynomials on each piece (see Continuity of Polynomials, Rational, and Elementary Functions):

Both one-sided limits exist and are equal, $L^- = L^+ = 2$. But $f(1) = 5 \neq 2$. This is exactly the removable case: the two half-graphs agree with each other but not with the plotted value. Redefining $f(1) := 2$ would make $f$ continuous at $1$; no such redefinition is possible at the other points below, which is the substantive difference between removable and jump singularities. At $x=3$: From the left,

From the right, we must check whether $\lim_{x\to 3^+} \frac{1}{x-3}$ exists as a finite number. As $x \to 3^+$, the denominator $x - 3 \to 0^+$ through positive values, so $\frac{1}{x-3} \to +\infty$; the limit does not exist as a real number. Since one of the one-sided limits fails to exist finitely, $x=3$ is an essential discontinuity of $f$ — no value of $f(3)$, and no amount of patching a single point, can repair this: the function is genuinely unbounded on every one-sided punctured neighborhood of $3$.

Example 2 (verifying the monotone-function theorem on a hand-built example with infinitely many jumps). Define $f : [0, 2] \to \mathbb{R}$ by

where $H$ is the Heaviside step function, $H(t) = 0$ for $t<0$ and $H(t)=1$ for $t \ge 0$. The jump points are $c_n = 1 - \tfrac1n$ for $n=1,2,3,\dots$, an increasing sequence converging to $1$, together they accumulate at $c_\infty = 1$ itself (check: is $f$ discontinuous at $1$ too? We check below.). Step 1: $f$ is nondecreasing. The term $x$ is strictly increasing, contributing rate $1$. Each term $\frac{1}{2^n}H(x - c_n)$ is nondecreasing (a jump up of size $\tfrac{1}{2^n}\ge0$, flat elsewhere). A sum of a strictly increasing function and nondecreasing functions is strictly increasing, in particular nondecreasing. So the hypothesis of the theorem is satisfied on $(0,2)$ (we work on the open interval to match the theorem; endpoint behavior is a one-sided variant). Step 2: locate discontinuities and check they are jumps. At each $c_n = 1 - \frac1n$ (for $n \ge 1$; note $c_1 = 0$ is the left endpoint, so consider $n\ge 2$ for genuinely interior points), the term $H(x-c_n)$ jumps from $0$ to $1$ exactly at $x = c_n$, while every other term in the (uniformly convergent, since $\sum 2^{-n}$ converges) series is continuous at $c_n$ (either identically $0$ nearby, since $c_m > c_n$ for $m>n$ keeps those terms at $0$ near $c_n$, or identically $1$ nearby, since $c_m < c_n$ for $m < n$ keeps those terms at $1$ near $c_n$, and there are only finitely many indices $m<n$, so no accumulation issue arises locally). Hence near $c_n$,

and the jump size is exactly $\frac{1}{2^n} > 0$: a genuine jump discontinuity, matching Part 2 of the theorem (never removable, never essential). Step 3: check $x=1$, the accumulation point of the $c_n$. For $x$ slightly less than $1$, say $x = 1-\eta$ with $\eta>0$ small, every $c_n$ with $n > 1/\eta$ satisfies $c_n > x$, so those terms are still $0$ at $x$; as $\eta \to 0^+$ (i.e. $x \to 1^-$), all the jump terms with $c_n < 1$ get absorbed, and

Here we used that the series $\sum 2^{-n}$ converges absolutely, so the limit of partial information as $x\to1^-$ is exactly the full infinite sum — this is where uniform convergence of the defining series is doing real work, guaranteeing $f(x) \to 1 + \sum_n 2^{-n}$ as $x \to 1^-$. Also $f(1) = 1 + \sum_n 2^{-n} = 2$ directly from the formula (every $H(1-c_n) = H(\tfrac1n) = 1$). And for $x>1$, $f(x) = x + 1$ exactly (all steps already taken), so $f(1^+) = 1+1 = 2$ as well. So $f$ is actually continuous at $x=1$, despite being the limit of infinitely many jump points! This confirms Part 3 of the theorem in a strong form: the discontinuity set is $\{c_2, c_3, c_4, \dots\}$, a countably infinite set (matching the theorem's guarantee of countability, realized here as genuinely infinite, but with the accumulation point itself remaining a point of continuity — nothing in the theorem forbids that, and this example shows why the countability conclusion is the correct, sharp one rather than "finite").