A bounded function on $[a,b]$ whose set of discontinuities is at most countable is Riemann integrable

Let $a<b$ be reals and let $f:[a,b]\to \mathbb{R}$ be bounded (Lower bound, bounded below, bounded set). If the set

$D=\{x\in [a,b]:f\text{ is discontinuous at }x\}$

(Continuity of $f:A\to \mathbb{R}$ at a point of $A$ and on $A$: the $\epsilon$-$\delta$ condition, its agreement with ${\lim }_{x\to c}f(x)=f(c)$ at a limit point, and continuity at an isolated point, Discontinuity of $f$ at a point of its domain, and its classification: removable discontinuity, jump discontinuity and essential discontinuity, equivalently Rudin's discontinuities of the first and of the second kind) is at most countable (Finite, countably infinite, countable, uncountable), then $f$ is Riemann integrable on $[a,b]$ (The lower and upper Darboux integrals of a bounded $f$ on $[a,b]$ as ${\sup }_{P}L(f,P)$ and ${\inf }_{P}U(f,P)$, Darboux integrability as their equality, and the notation ${\int }_a^{b}f$).

No choice principle is used. Only the implication "$D$ null $\Rightarrow$ $f$ integrable" of Lebesgue's criterion for Riemann integrability: a bounded $f$ on $[a,b]$ is Riemann integrable if and only if its set of discontinuities has measure zero is invoked, and that implication is a theorem of ZF; Every at most countable subset of $\mathbb{R}$ has measure zero is choice-free as well, since a listing of $D$ is a single object and the cover is a formula in the index.

The converse fails badly: the indicator of the Cantor set is discontinuous at uncountably many points and is integrable, because the Cantor set is null (The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points). What is true in both directions is Lebesgue's criterion for Riemann integrability: a bounded $f$ on $[a,b]$ is Riemann integrable if and only if its set of discontinuities has measure zero itself.

• The two classical instances. A function with finitely many discontinuities is covered (A bounded function on $[a,b]$ that is continuous except at finitely many points is Riemann integrable proves it again by an elementary argument that costs no Lebesgue criterion), and so is a monotone function, whose discontinuity set is at most countable by Froda's theorem (Froda's theorem: the set of discontinuities of a monotone function on an interval is at most countable, the injection into $\mathbb{N}$ being built from one fixed enumeration of the rationals by least index, so no choice principle is used). The direct proof of A monotone function on $[a,b]$ is Riemann integrable: for the uniform partition into $N$ parts the upper minus lower sum telescopes to $\mathrm{\mid }f(b)-f(a)\mathrm{\mid }(b-a)\mathrm{/}\iota (N)$ is nevertheless kept, because it is quantitative and elementary.

• Thomae's function is the standard witness that this corollary has content. It is discontinuous at every rational and continuous at every irrational (The Dirichlet function is continuous at no point of $\mathbb{R}$, and Thomae's function is continuous at every irrational and at no rational, so its set of continuity points is exactly the set of irrationals and its oscillation at $c$ equals $t(c)$), so its discontinuity set is countable and it is integrable, while its Dirichlet counterpart, discontinuous everywhere, is not (FALSE: every bounded function on $[a,b]$ is Riemann integrable).

• Countability is sufficient and not necessary, and it is not even the right invariant. The Cantor set is uncountable and null, while the Smith-Volterra-Cantor set is uncountable and not null (The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero); the first is a discontinuity set of an integrable function and the second is not. Cardinality decides nothing here; measure does.