Lebesgue's Criterion for Riemann Integrability

Lebesgue's criterion is the standard bridge topic between elementary real analysis and measure theory in every rigorous analysis course, and it is the theorem that makes precise why the Riemann integral is inadequate for many purposes in modern mathematics. In functional analysis and Fourier analysis, it explains which functions arising as limits of continuous functions (e.g. pointwise limits, which can have wild discontinuity sets) remain Riemann integrable, informing when term-by-term integration of series is legitimate before invoking the more powerful Dominated Convergence Theorem from Lebesgue theory. In probability and statistics, indicator functions of measure-zero versus positive-measure sets govern which "boundary" or "singular" events can be safely ignored when computing expectations via Riemann-style integrals, and the language of "almost everywhere" properties traces directly to this dichotomy. In numerical analysis, quadrature rules (e.g. Simpson's rule, Newton–Cotes methods) implicitly rely on integrand regularity; the criterion tells the practitioner exactly how much discontinuity a function can tolerate before Riemann-based quadrature schemes lose their theoretical justification. In computer science and information theory, functions defined via case splits on Diophantine or number-theoretic conditions (analogues of Thomae's function) are analyzed via exactly this kind of discontinuity-set argument to establish integrability of complexity-theoretic or combinatorial generating functions. Finally, in mathematical logic and descriptive set theory, the criterion is a first, concrete example of how the "size" of a set (measure) rather than its cardinality governs analytic properties, a theme that recurs throughout the theory of Borel and analytic sets.

:::quiz[Which of the following bounded functions on $[0,1]$ is Riemann integrable by Lebesgue's criterion?]

• The indicator function of $\mathbb{Q}\cap[0,1]$
• The indicator function of a fat Cantor set of measure $2/3$
• Thomae's function (discontinuous exactly on the rationals) ✓
• The function equal to $1/x$ near $0$ and $0$ elsewhere (unbounded)

:::quiz[A bounded function on $[0,1]$ is discontinuous at every point of a nowhere-dense closed set of Lebesgue measure $0.4$. Is it Riemann integrable?]

• Yes, since nowhere-dense sets are always measure zero
• No, since the discontinuity set has positive measure ($0.4 \ne 0$) ✓
• Yes, since the set is closed
• Cannot be determined without knowing the function's specific values

Ask the tutor: Can you show me in full detail why the oscillation function $\omega_f$ is well-defined and why $D_\eta$ being closed doesn't require $f$ to be continuous anywhere at all?

The classical theorems already known — continuity implies integrability, monotonicity implies integrability, finitely many jumps do not obstruct integrability — all share a hidden feature: in each case the "bad set" (where $f$ fails to be continuous) is either empty, finite, or at worst countable. Riemann's integral is built from upper and lower Darboux sums, and what actually damages a Darsboux sum is not the number of discontinuities but the total width of intervals over which $f$'s values swing by a fixed nontrivial amount. A single jump discontinuity contributes an error to $U(f,P) - L(f,P)$ proportional to the width of the one subinterval containing it — negligible in the limit. Infinitely many jumps can still be negligible if they can be corralled into subintervals of arbitrarily small total length. That corralling condition is exactly measure zero. The right way to organize the discontinuity set is by how badly $f$ oscillates, not simply whether it is continuous. Define for each $\eta>0$ the set

Each $D_\eta$ turns out to be closed (a compactness-friendly set), and

since a point has $\omega_f(x)>0$ iff $\omega_f(x) \ge 1/n$ for some $n$. This decomposition is the technical heart of the theorem: it converts a statement about an arbitrary discontinuity set into a countable union of compact oscillation-level sets, each of which can be attacked with a finite subcover argument reminiscent of the Heine–Borel theorem. The picture below shows a bounded function with a large jump at one point, a countable sequence of shrinking jumps accumulating at another point, and continuity elsewhere — illustrating a discontinuity set that is countably infinite yet still measure zero, hence integrable by Lebesgue's criterion.

Example 1: A function with a countable, non-eventually-isolated discontinuity set. Let $f:[0,1]\to\mathbb{R}$ be defined by

This is Thomae's function. Its discontinuity set is $D_f = \mathbb{Q}\cap [0,1]$: one can show directly $f$ is continuous at every irrational and discontinuous at every rational (a standard fact, provable via the $\varepsilon$–$\delta$ definition using that only finitely many rationals in $[0,1]$ have denominator below any bound $N$). Since $\mathbb{Q}\cap[0,1]$ is countable, and every countable set has measure zero — enumerate it as $\{q_1,q_2,\dots\}$, cover $q_n$ by an interval of length $\varepsilon/2^{n+1}$, total length $<\varepsilon$ — Lebesgue's criterion gives immediately that $f$ is Riemann integrable, without any need to estimate Darboux sums directly. (One further computes $\int_0^1 f = 0$, but the criterion alone settles integrability.)

Example 2: A bounded function discontinuous on a "fat" set — non-integrable, showing sharpness. We build a fat Cantor set $C \subseteq [0,1]$: a Cantor-like set with positive Lebesgue measure, and a bounded function whose discontinuity set is exactly $C$, hence not integrable. Construction of $C$. Start with $[0,1]$. At stage $1$, remove the open middle interval of length $1/4$ (not $1/3$) from $[0,1]$, leaving two closed intervals of total length $1 - 1/4$. At stage $k$, from each of the $2^{k-1}$ remaining closed intervals, remove an open middle interval of length $1/4^{\,k+1}$ times... concretely, remove middle open intervals of total length $1/4^{k}$ at stage $k$ across all remaining pieces combined (there are standard variants; we specify total removed length at stage $k$ is $2^{k-1}\cdot \ell_k$ where $\ell_k = \dfrac{1}{4^k \cdot 2^{k-1}}$, chosen so total removed length across all stages is

Let $C = [0,1] \setminus \bigcup_{k} (\text{removed intervals})$. Then $C$ is closed (complement of a union of open intervals), nowhere dense (contains no interval, since at every stage every remaining piece is subdivided further), uncountable (it is a Cantor set, homeomorphic to the standard middle-thirds Cantor set via the tree of choices at each stage), and has measure

The function. Define

This is bounded ($0 \le f \le 1$). Its discontinuity set is exactly $C$: at any $x \in C$, since $C$ is nowhere dense, every neighborhood of $x$ contains points not in $C$ (where $f=0$) and (since $x\in C$ itself, or by density of $C$ in itself at non-isolated points — one checks $C$ has no isolated points by construction) also points of $C$ arbitrarily close, so $\Omega_f(B_\delta(x)) = 1$ for all $\delta$, giving $\omega_f(x) = 1 > 0$; at $x \notin C$, since $[0,1]\setminus C$ is open, a whole neighborhood of $x$ avoids $C$, so $f \equiv 0$ near $x$ and $f$ is continuous there. Since $D_f = C$ has measure $2/3 \ne 0$, Lebesgue's criterion says $f$ is not Riemann integrable — and indeed one can check directly that for every partition $P$, every subinterval of positive length meets both $C$ and its complement (since $C$ is nowhere dense but has positive measure, no subinterval can avoid it or be swallowed by it), so $U(f,P) = 1$ and $L(f,P) = 0$ for every partition, giving $U(f,P) - L(f,P) = 1$ always — confirming non-integrability directly, exactly as the criterion predicts.