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

Let $A\subseteq \mathbb{R}$ and let $f:A\to \mathbb{R}$. Write

$D:=\{x\in A:f\text{ is discontinuous at }x\},C:=A\setminus D$

(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). Then:

1. Pointwise exhaustion. $D=\{x\in A:{\omega }_f(x)>0\}$ (The oscillation ${\omega }_f(S)=\sup \{\mathrm{\mid }f(x)-f(y)\mathrm{\mid }:x,y\in S\}$ of $f$ on a set and the oscillation ${\omega }_f(c)={\inf }_{\delta >0}{\omega }_f(A\cap N_{\delta }(c))$ at a point, both taken in the extended reals), and $D$ is the union of the increasing sequence of superlevel sets $D={\bigcup }_{n\in \mathbb{N}}\{x\in A:{\omega }_f(x)\ge 1\mathrm{/}\iota (n+1)\}$ (The canonical natural $\iota (n)=n\cdot 1_F$ of a field), whose thresholds are $1,1\mathrm{/}2,1\mathrm{/}3,\dots$.
2. Descriptive form. There is an $F_{\sigma }$ set $F\subseteq \mathbb{R}$ and a $G_{\delta }$ set $V\subseteq \mathbb{R}$ ($F_{\sigma }$ and $G_{\delta }$ subsets of $\mathbb{R}$) with $D=A\cap F,C=A\cap V,V=\mathbb{R}\setminus F,$ and $F$ may be taken to be ${\bigcup }_{n\in \mathbb{N}}{G}_n$ with each $G_n$ a closed subset of $\mathbb{R}$ cutting down on $A$ to the $n$-th set of claim 1 (For every real $\epsilon >0$ the set $\{x\in A:{\omega }_f(x)\ge \epsilon \}$ is the intersection with $A$ of a closed subset of $\mathbb{R}$; in particular it is closed in $\mathbb{R}$ when $A=\mathbb{R}$).

In particular, when $A=\mathbb{R}$ the discontinuity set $D$ is an $F_{\sigma }$ subset of $\mathbb{R}$ and the continuity set $C$ is a $G_{\delta }$ subset, and claim 1 reads $D={\bigcup }_n\{x\in \mathbb{R}:{\omega }_f(x)\ge 1\mathrm{/}\iota (n+1)\}$.

Claim 1 is stated separately because it is what later arguments cite downstream. The exhaustion of $D$ by the superlevel sets $\{{\omega }_f\ge 1\mathrm{/}\iota (n+1)\}$ is exactly the form needed when a later proof establishes a property one threshold at a time, so the identity itself is recorded here and not only the descriptive conclusion of claim 2.

The statement is relative on purpose. For a general domain $A$ the sets $D$ and $C$ are subsets of $A$, and neither is $F_{\sigma }$ or $G_{\delta }$ in $\mathbb{R}$ in general; what the proof produces are two subsets of $\mathbb{R}$ that cut down to them. The absolute form is stated only for $A=\mathbb{R}$, which is the case used later by the realization and exact-continuity-set examples.

• The exhaustion is indexed from $0$ and the thresholds are $1\mathrm{/}\iota (n+1)$. $\mathbb{N}$ contains $0$, so the sequence of thresholds is $1,1\mathrm{/}2,1\mathrm{/}3,\dots$ and never $1\mathrm{/}\iota (0)$, which is not defined. Writing the union as ${\bigcup }_{n\ge 1}\{{\omega }_f\ge 1\mathrm{/}\iota (n)\}$ names the same family; the form above is used because a sequence in this library is a function on $\mathbb{N}$.

• The converse holds and is proved separately. Every $G_{\delta }$ subset of $\mathbb{R}$ is the continuity set of some function $\mathbb{R}\to \mathbb{R}$ (Every $G_{\delta }$ subset of $\mathbb{R}$ is the set of continuity points of some $f:\mathbb{R}\to \mathbb{R}$, so the $G_{\delta }$ sets are exactly the continuity sets), so the two classes coincide exactly. What the present theorem contributes is the direction that constrains: no function can have a continuity set that fails to be $G_{\delta }$, and No function $\mathbb{R}\to \mathbb{R}$ is continuous at every rational and discontinuous at every irrational, because $\mathbb{Q}$ is not $G_{\delta }$ spends that on $\mathbb{Q}$.