Infimum of Set of Oscillations on Set

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Lemma

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

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

Let $S_x$ be a set of real sets that contain (as an element) $x$.

Let:

$\map {\omega_f} x = \inf \set {\map {\omega_f} I: I \in S_x}$

where $\map {\omega_f} I$ denotes the oscillation of $f$ on the set $I$:

$\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$


Then:

$\map {\omega_f} x \in \R$ if and only if $\set {\map {\omega_f} I: I \in S_x}$ contains a real number.


Proof

Let:

$S = \set {\map {\omega_f} I: I \in S_x}$

We observe that:

$\inf S = \map {\omega_f} x$


Necessary Condition

Let $\inf S \in \R$.

We need to prove that $S$ contains a real number.


Note that $S$ is non-empty as the empty set does not admit an infimum (in $\R$).

Therefore, $S$ has at least one element.

Accordingly, there is an $I \in S_x$ such that $\map {\omega_f} I \in S$.


Let $I \in S_x$.

Therefore, $x \in I$.

From this follows by Oscillation on Set is an Extended Real Number that $\map {\omega_f} I$ is an extended real number.

Therefore $S$ is a set of extended real numbers as $S = \set {\map {\omega_f} I: I \in S_x}$.

Accordingly, $S$ contains a real number by Infimum of Subset of Extended Real Numbers is Arbitrarily Close as $\inf S \in \R$.

$\Box$


Sufficient Condition

Let $S$ contain a real number.

We need to prove that $\inf S \in \R$.


We have:

$S \cap \R$ is non-empty as $S$ contains a real number.


Let $I \in S_x$.

Therefore, $x \in I$.

From this follows by Oscillation on Set is an Extended Real Number that $\map {\omega_f} I \in \overline \R_{\ge 0}$.

Therefore:

$S$ is a subset of $\overline \R_{\ge 0}$ as $S = \set {\map {\omega_f} I: I \in S_x}$

Accordingly:

$S$ is bounded below.

From this follows that:

$S \cap \R$ is bounded below as $SR$ is a subset of $S$


We have:

$S \cap \R$ is bounded below
$S \cap \R$ is not empty

Therefore:

$\inf S \cap \R \in \R$ Continuum Property


We have:

$S$ is a set of extended real numbers as $S$ is a subset of $\overline \R_{\ge 0}$
$S$ is bounded below

Therefore:

$\inf S \in \R$ by Infimum of Real Subset as $\inf S \cap \R \in \R$

$\blacksquare$