For every real $\varepsilon > 0$ the set $\{\,x \in A : \omega_f(x) \ge \varepsilon\,\}$ 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 $\varepsilon \in \mathbb{R}$ with $\varepsilon > 0$. Put

$E_\varepsilon \;:=\; \{\, x \in A : \omega_f(x) \ge \varepsilon \,\}$

(The oscillation $\omega_f(S) = \sup\{\,|f(x) - f(y)| : x, y \in S\,\}$ 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

$E_\varepsilon \;=\; A \cap G .$

In particular, if $A = \mathbb{R}$ then $E_\varepsilon$ 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

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

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 $\sup \varnothing = -\infty$ when $A \cap N_\rho(y) = \varnothing$ (The oscillation $\omega_f(S) = \sup\{\,|f(x) - f(y)| : x, y \in S\,\}$ 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 $\overline{\mathbb{R}}$ has a least upper bound and a greatest lower bound in $\overline{\mathbb{R}}$, 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 $\{\omega_f \ge \varepsilon\}$ is what remains of it inside $A$. Defining $E_\varepsilon$ with a strict inequality instead, as $\{\omega_f > \varepsilon\}$, 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 $E_\varepsilon$ 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 $\ge \varepsilon$ everywhere gives $E_\varepsilon = (0,1)$, which is not closed. What is always true is the displayed identity $E_\varepsilon = A \cap G$, and that is what the theorems downstream use.