Integrability of Monotone Functions

Monotone functions and their integrability underpin probability theory, where every cumulative distribution function $F(x) = P(X\le x)$ is nondecreasing, frequently has jump discontinuities (at atoms of the distribution), and yet integrals against $F$ (Riemann–Stieltjes integrals) are exactly what define expectations of random variables — the integrability theorem here is the base case making that machinery well-posed. In real analysis and measure theory, the countability of the discontinuity set of a monotone function (proved above) is the prototypical example motivating Lebesgue's Criterion for Riemann Integrability, and monotone functions are a standard testing ground for the concept of functions of bounded variation, since every function of bounded variation is a difference of two monotone functions. In numerical analysis, sorting algorithms produce monotone (sorted) sequences, and quantile functions (generalized inverses of CDFs) used throughout statistics and machine learning are monotone by construction, with integrability of these often invoked implicitly when computing quantile-based risk measures. In economics and optimization, monotone utility and cost functions are integrated routinely (e.g. computing consumer surplus as an area under a monotone demand curve), relying silently on exactly this theorem to guarantee the area is well defined even when the demand curve has discrete jumps from market frictions.

:::quiz[For a nondecreasing function $f$ on $[a,b]$, what is $\sup_{[x_{i-1},x_i]} f$ equal to, and why?]

• $f\left(\frac{x_{i-1}+x_i}{2}\right)$, by the mean value theorem
• $f(x_{i-1})$, since nondecreasing functions attain their max at the left endpoint
• $f(x_i)$, since $f$ is nondecreasing and $x_i$ is the rightmost point of the subinterval, and this value is actually attained ✓
• It need not exist unless $f$ is continuous

:::quiz[Why is boundedness of $f$ on $[a,b]$ guaranteed automatically once we know $f$ is monotone and $[a,b]$ is compact?]

• Because monotone functions are always continuous
• Because $f(a) \le f(x) \le f(b)$ (or the reverse) for all $x\in[a,b]$, trapping $f$ between two finite numbers ✓
• Because Riemann integrability was assumed in advance
• Because $[a,b]$ has measure zero

:::quiz[A monotone function on $[a,b]$ can have infinitely many points of discontinuity. What must nonetheless be true of that discontinuity set?]

• It must be finite
• It must be an interval
• It must be countable ✓
• It must have positive length

:::quiz[Which hypothesis, if dropped, breaks the theorem as shown by $g(x)=1/x$ on $(0,1]$?]

• Monotonicity of $g$
• The domain being a closed, bounded interval (compactness), which is needed to force boundedness ✓
• The existence of the Cauchy Integrability Criterion
• The countability of $\mathbb{Q}$

Ask the tutor: Can you show me why the same style of proof breaks down if $f$ is monotone but the interval is half-open, and how the notion of "measure zero" eventually repairs it?

Picture a nondecreasing function whose graph climbs, possibly with sudden vertical jumps, from $f(a)$ to $f(b)$. Split $[a,b]$ into $n$ equal subintervals of width $(b-a)/n$. Because $f$ is nondecreasing, on each subinterval the supremum is attained at the right endpoint and the infimum at the left endpoint — no need to hunt for where the max or min occurs, monotonicity hands it to you for free. The upper sum is the total area of rectangles that slightly overshoot the graph on the right of each subinterval; the lower sum uses rectangles that undershoot on the left. The difference $U(f,P)-L(f,P)$ is then a sum of little rectangles of width $(b-a)/n$ sitting exactly on top of each jump or rise of the function. When you stack all those little error-rectangles on top of one another (this is the clever trick — they all have the same width, so you can slide them together), their combined height is at most the total rise of the function, $f(b)-f(a)$. So the total error is at most $\frac{(b-a)}{n}\bigl(f(b)-f(a)\bigr)$, which visibly shrinks to $0$ as $n\to\infty$, regardless of how many discontinuities $f$ has or how erratically it jumps.

This corollary is exactly the phenomenon that later generalizes: a bounded function is Riemann integrable if and only if its set of discontinuities has Lebesgue measure zero, per Lebesgue's Criterion for Riemann Integrability. A countable set always has measure zero, so this theorem is a special case of that later, sharper result — but it is provable directly, with elementary tools, well before measure theory is available.

Example 1 (a genuinely discontinuous monotone function, non-symmetric jumps). Let $f:[0,4]\to\mathbb R$ be defined by

This $f$ is nondecreasing (check each case boundary: $0\le 2$, $2 \le 2.5$, $2.5 \le 7$) with jumps of size $2$, $0.5$, and $4.5$ at $x=1, 2.5, 3$ respectively — three unevenly spaced, unevenly sized jumps. By the theorem, $f$ is Riemann integrable on $[0,4]$. Apply the proof's construction with, say, $n=8$ equal subintervals of width $\Delta x = 4/8 = 0.5$, so $x_i = 0.5i$ for $i=0,\dots,8$: $0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4$. Evaluate $f$ at each node:

Upper sum uses $M_i = f(x_i)$ (right endpoint value, by the proof):

Lower sum uses $m_i = f(x_{i-1})$ (left endpoint value):

Oscillation: $U-L = 14.75-11.25 = 3.5$. The theorem's bound predicts

matching exactly — confirming the telescoping computation. To get oscillation below, say, $\varepsilon = 0.1$, the proof requires $n$ with $\frac{4\cdot 7}{n} < 0.1$, i.e. $n > 280$; any such $n$ works regardless of where the three jumps sit. Direct computation of the integral (splitting at the jump points, since $f$ is piecewise constant):

Indeed $L(f,P_8) = 11.25$ already equals the true integral for this coarse partition (because $P_8$ happens to include every jump point exactly), while $U(f,P_8)=14.75$ overshoots — illustrating how $U$ and $L$ squeeze the true value as predicted.

Example 2 (infinitely many discontinuities — one at every rational in the interval). Let $\{q_1, q_2, q_3, \dots\}$ enumerate $\mathbb{Q}\cap(0,3)$ (countable, since $\mathbb Q$ is countable). Define

i.e. $f(x)$ sums $2^{-n}$ over every index $n$ with $q_n<x$ (with $f(x)=0$ if no such $n$, and interpreting $f(0)=0$). This is a standard construction (a "jump function"): $f$ is nondecreasing, since increasing $x$ can only enlarge the set of indices $\{n : q_n < x\}$, hence can only add nonnegative terms to the sum. It is bounded, since

for every $x$ (geometric series). One can show $f$ is discontinuous at every single rational $q_n\in(0,3)$, with jump exactly $2^{-n}$ there, and continuous at every irrational point — so its discontinuity set is exactly the countably infinite set $\mathbb Q\cap(0,3)$, dense in $[0,3]$. By the theorem, $f$ is nonetheless Riemann integrable on $[0,3]$, despite being discontinuous at a dense set of points — discontinuous "almost everywhere you look" in the sense of density, though not in the sense of measure. Concretely, for any $\varepsilon>0$, choosing $n$ with $\frac{3(f(3)-f(0))}{n}<\varepsilon$, i.e. (since $f(0)=0$ and $f(3)\le 1$) any $n > 3/\varepsilon$ suffices, and the same equally-spaced partition $P_n$ from the proof forces $U(f,P_n)-L(f,P_n)<\varepsilon$ — the argument never needed to know where the (infinitely many) jumps were, only that their total size is bounded by $f(3)-f(0)\le 1$.