Oscillation on Set is an Extended Real Number

Theorem
Let $f: D \to \R$ be a real function where $D \subseteq \R$.

Let $x$ be a point in $D$.

Let $I$ be a real set that contains (as an element) $x$.

Let $\map {\omega_f} I$ be the oscillation of $f$ on $I$:
 * $\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$

Then:
 * $\map {\omega_f} I \in \overline \R_{\ge 0}$

and:
 * $\map {\omega_f} I = \begin{cases}

\text{a positive real number} & \set {\size {\map f y - \map f z}: y, z \in I \cap D} \text{is bounded above} \\ \infty & \set {\size {\map f y - \map f z}: y, z \in I \cap D} \text{is not bounded above} \end{cases}$

Proof
We observe that $\size {\map f y - \map f z} = 0$ for $y = z = x$.

Therefore:
 * $0 \in \set {\size {\map f y - \map f z}: y, z \in I \cap D}$ as $x \in I \cap D$

Accordingly:
 * $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is non-empty

We have that $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is a real set as $\size {\map f y - \map f z} \in \R$ for every $y, z \in D$.

Every real number is less than $\infty$.

Therefore:
 * $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is bounded above in $\overline \R$

There are two cases: either $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is bounded above (in $\R$), or it is not.

First, assume that:
 * $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is bounded above (in $\R$)

We have that $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is bounded above and non-empty.

Therefore:
 * $\sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$ exists as a real number by the Continuum Property

We know that $\sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is greater than or equal to every element of $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$.

Also $0 \in \set {\size {\map f y - \map f z}: y, z \in I \cap D}$.

Therefore, $\sup \set {\size {\map f y - \map f z}: y, z \in I \cap D} \ge 0$.

We also have that $\sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$ exists as a real number.

Therefore, $\sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is a positive real number.

In other words:
 * $\map {\omega_f} I$ is a positive real number as $\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$

Next, assume that:
 * $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is not bounded above (in $\R$)

We have that $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is bounded above in $\overline \R$.

Therefore, $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ has $\infty$ as an upper bound.

A number that is less than $\infty$ is a real number or equal to $-\infty$.

No real number is an upper bound for $\set {\size {\map f y - \map f z} y, z \in I \cap D}$ as $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is not bounded above in $\R$.

$-\infty$ is not an upper bound for $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ because then every real number would be an upper bound for $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$.

Therefore, $\infty$ is the only upper bound for $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$.

Accordingly, $\infty$ is the least upper bound of $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$.

In other words:
 * $\map {\omega_f} I = \infty$ as $\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$

In either case, whether $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is bounded above or not:
 * $\map {\omega_f} I \in \overline \R_{\ge 0}$.