For every $F_\sigma$ subset $E$ of $[0,1]$ of measure zero there is a bounded Riemann integrable function on $[0,1]$ whose set of discontinuities is exactly $E$

Let $E \subseteq [0,1]$ be an $F_\sigma$ subset of $\mathbb{R}$ ($F_\sigma$ and $G_\delta$ subsets of $\mathbb{R}$) of measure zero (Measure zero (a countable cover by intervals of total length below every $\varepsilon$) and content zero (a finite such cover)). Then there is a bounded function $h : [0,1] \to \mathbb{R}$, with values in $[0,1]$, that is Riemann integrable on $[0,1]$ and whose set of discontinuities is exactly $E$ (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$: the $\varepsilon$-$\delta$ 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).

The construction. Fix closed sets $F_0, F_1, F_2, \dots$ with $E = \bigcup_{n \in \mathbb{N}} F_n$ and put

$h(x) \;:=\; 2^{-n(x)} \ \text{ for } x \in E, \qquad h(x) \;:=\; 0 \ \text{ for } x \in [0,1]\setminus E,$

where $n(x) := \min\{\, n \in \mathbb{N} : x \in F_n \,\}$ is the least index of a closed set containing $x$ (The well-ordering principle). Nothing is selected: $n(x)$ is the least element of a set determined by $x$ and the fixed sequence $(F_n)$.

Why this is worth stating. Together with For $f : A \to \mathbb{R}$ the set of points of $A$ at which $f$ is discontinuous is the intersection with $A$ of an $F_\sigma$ subset of $\mathbb{R}$, and the set of points at which $f$ is continuous is the intersection with $A$ of a $G_\delta$ subset; for $A = \mathbb{R}$ the two sets are $F_\sigma$ and $G_\delta$ outright, which shows that a discontinuity set is always the trace of an $F_\sigma$ set, and with 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, which shows that an integrable function has a null discontinuity set, the example says that the two necessary conditions are also jointly sufficient: null and $F_\sigma$ is exactly what a discontinuity set of a Riemann integrable function on $[0,1]$ can be. The Cantor set and any at most countable subset of $[0,1]$ are instances.

Choice. The construction uses none; the only choice principle in the statement comes from the direction 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 used at the end, and that direction, "null $\Rightarrow$ integrable", is a theorem of ZF.

• Where each hypothesis on $E$ is used. That $E$ is $F_\sigma$ is what makes the exhaustion $(F_n)$ available and hence gives continuity off $E$ in step 3.1; that $E$ is null is used twice, once through [L4] to force discontinuity on $E$, and once at the end through 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. Dropping either hypothesis breaks the example, and by For $f : A \to \mathbb{R}$ the set of points of $A$ at which $f$ is discontinuous is the intersection with $A$ of an $F_\sigma$ subset of $\mathbb{R}$, and the set of points at which $f$ is continuous is the intersection with $A$ of a $G_\delta$ subset; for $A = \mathbb{R}$ the two sets are $F_\sigma$ and $G_\delta$ outright and 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 neither can be dropped from the conclusion either.

• The values $2^{-n}$ are a convenience. Any sequence of positive reals tending to $0$ would do in their place; what the proof needs is that the value at a point of $F_n \setminus (F_0 \cup \dots \cup F_{n-1})$ is small when $n$ is large, and that it is never $0$ on $E$.

• Two familiar instances. Taking $E$ at most countable recovers a function continuous exactly off a prescribed countable set, of which Thomae's function is the case $E = \mathbb{Q}\cap[0,1]$ in spirit though not in formula (Thomae's function is Riemann integrable on $[0,1]$ with integral $0$: it is continuous at every irrational, so its discontinuity set is countable, and every lower Darboux sum is $0$); taking $E$ to be the Cantor set recovers a function of the type of The indicator of the Cantor set is discontinuous exactly on the Cantor set, which is null, so it is Riemann integrable with integral $0$ even though it is discontinuous at uncountably many points, the Cantor set being closed, hence $F_\sigma$, and null.

• No analogue holds without nullity. The Smith-Volterra-Cantor set is closed, hence $F_\sigma$, and it is the discontinuity set of its own indicator, which is not integrable (The indicator of the Smith-Volterra-Cantor set is discontinuous exactly on a nowhere dense set, and is not Riemann integrable, because that set does not have measure zero). So the $F_\sigma$ condition alone buys nothing.