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 $A=\mathbb{R}$

Let $A\subseteq \mathbb{R}$, let $f:A\to \mathbb{R}$ and let $\epsilon \in \mathbb{R}$ with $\epsilon >0$. Put

$E_{\epsilon }:=\{x\in A:{\omega }_f(x)\ge \epsilon \}$

(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). Then there is a closed $G\subseteq \mathbb{R}$ (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen) with

$E_{\epsilon }=A\cap G\mathrm{.}$

In particular, if $A=\mathbb{R}$ then $E_{\epsilon }$ is itself a closed subset of $\mathbb{R}$.

The set $G$ is produced explicitly and does not depend on any choice: it is the complement of

$U:=\{y\in \mathbb{R}:{\omega }_f(A\cap N_{\rho }(y))<\epsilon \text{ for some real }\rho >0\},$

which the proof shows to be open. Note that $y$ ranges over all of $\mathbb{R}$ here and not only over $A$; the expression ${\omega }_f(A\cap N_{\rho }(y))$ is the oscillation of $f$ on a subset of $A$ and makes sense for every real $y$, taking the value $\sup \mathrm{\varnothing }=-\mathrm{\infty }$ when $A\cap N_{\rho }(y)=\mathrm{\varnothing }$ (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, Every subset of $\bar{\mathbb{R}}$ has a least upper bound and a greatest lower bound in $\bar{\mathbb{R}}$, agreeing with the real supremum and infimum on nonempty sets bounded in $\mathbb{R}$).

• Why the strict inequality is on the open side. The set $U$ is defined by a strict inequality and an existential quantifier over $\rho$, which is what makes it open; its complement is then closed, and the superlevel set $\{{\omega }_f\ge \epsilon \}$ is what remains of it inside $A$. Defining $E_{\epsilon }$ with a strict inequality instead, as $\{{\omega }_f>\epsilon \}$, would not give a closed set in general, and the exhaustion of 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 is arranged so that only the non-strict form is ever needed.

• The relative form is the honest one. For general $A$ the set $E_{\epsilon }$ is a subset of $A$ and there is no reason for it to be closed in $\mathbb{R}$: taking $A=(0,1)$ and $f$ the restriction of a function with oscillation $\ge \epsilon$ everywhere gives $E_{\epsilon }=(0,1)$, which is not closed. What is always true is the displayed identity $E_{\epsilon }=A\cap G$, and that is what the theorems downstream use.