=== Lemma === {{:Oscillation at Point (Infimum) equals Oscillation at Point (Limit)/Lemma}} {{qed|lemma}} === Necessary Condition === Let $\map {\omega_f} x \in \R$. We need to prove: :$\map {\omega^L_f} x \in \R$ :$\map {\omega^L_f} x = \map {\omega_f} x$ where: :$\map {\omega^L_f} x = \displaystyle \lim_{h \mathop \to 0^+} \map {\omega_f} {\openint {x - h} {x + h} }$ :$\map {\omega_f} {\openint {x - h} {x + h} } = \sup \set {\size {\map f y - \map f z}: y, z \in \openint {x - h} {x + h} \cap D}$ :$\map {\omega_f} x = \inf \set {\map {\omega_f} I: I \in N_x}$ :$\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$ Let $\epsilon \in \R_{>0}$. Then an $I \in N_x$ exists such that: :$\map {\omega_f} I - \map {\omega_f} x < \epsilon$ by Infimum of Set of Oscillations on Set is Arbitrarily Close Let $I$ be such an element of $N_x$. We observe in particular that $\map {\omega_f} I \in \R$. A neighborhood in $N_x$ contains an open subset that contains the point $x$. So, $I$ contains such an open subset as $I \in N_x$. Therefore, a $\delta \in \R_{>0}$ exists such that $\openint {x - \delta} {x + \delta}$ is a subset of $I$. Let $h$ be a real number that satisfies: $0 < h < \delta$. We observe that $\openint {x - h} {x + h} \subset I$. Also, $\openint {x - h} {x + h} \in N_x$. We have: :$I \in N_x$ :$\openint {x - h} {x + h} \in N_x$ :$\openint {x - h} {x + h} \subset I$ :$\map {\omega_f} I \in \R$ from which follows by Oscillation on Subset: :$\map {\omega_f} {\openint {x - h} {x + h} } \in \R$ :$\map {\omega_f} {\openint {x - h} {x + h} } \le \map {\omega_f} I$ We have that: :$\map {\omega_f} {\openint {x - h} {x + h} } \in \set {\map {\omega_f} {I'}: I' \in N_x}$ as $\openint {x - h} {x + h} \in N_x$. Also, $\map {\omega_f} x$ is a lower bound for $\set {\map {\omega_f} {I'}: I' \in N_x}$ by the definition of $\map {\omega_f} x$. Therefore: :$\map {\omega_f} {\openint {x - h} {x + h} } \ge \map {\omega_f} x$ We find: {{begin-eqn}} {{eqn | l = \map {\omega_f} x \le \map {\omega_f} {\openint {x - h} {x + h} }       | o = \le       | r = \map {\omega_f} I }} {{eqn | ll= \leadsto       | l = 0 \le \map {\omega_f} {\openint {x - h} {x + h} } - \map {\omega_f} x       | o = \le       | r = \map {\omega_f} I - \map {\omega_f} x       | c =  }} {{eqn | ll= \leadsto       | l = 0 \le \map {\omega_f} {\openint {x - h} {x + h} } - \map {\omega_f} x       | o = \le       | r = \map {\omega_f} I - \map {\omega_f} x < \epsilon       | c = as $\map {\omega_f} I - \map {\omega_f} x < \epsilon$ is true }} {{eqn | ll= \leadsto       | l = 0 \le \map {\omega_f} {\openint {x - h} {x + h} } - \map {\omega_f} x       | o = <       | r = \epsilon       | c =  }} {{eqn | ll= \leadsto       | l = \size {\map {\omega_f} {\openint {x - h} {x + h} } - \map {\omega_f} x}       | o = <       | r = \epsilon       | c =  }} {{end-eqn}} which means that $\displaystyle \lim_{h \mathop \to 0^+} \map {\omega_f} {\openint {x - h} {x + h} }$ exists and equals $\map {\omega_f} x$ by the definition of limit. In other words, $\map {\omega^L_f} x \in \R$ and $\map {\omega^L_f} x = \map {\omega_f} x$. {{qed|lemma}} === Sufficient Condition === Let $\map {\omega^L_f} x \in \R$. We need to prove: :$\map {\omega_f} x \in \R$ :$\map {\omega_f} x = \map {\omega^L_f} x$ where: :$\map {\omega_f} x = \inf \set {\map {\omega_f} I: I \in N_x}$ :$\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$ :$\map {\omega^L_f} x = \displaystyle \lim_{h \mathop \to 0^+} \map {\omega_f} {\openint {x - h} {x + h} }$ We have: :$\displaystyle \lim_{h \mathop \to 0^+} \map{\omega_f} {\openint {x - h} {x + h} } \in \R$ as $\map {\omega^L_f} x \in \R$ Therefore, $\map {\omega_f} {\openint {x - h} {x + h} } \in \R$ for a small enough $h$ in $\R_{>0}$ by the definition of limit. Let $h$ be such a real number. We observe that $\openint {x - h} {x + h}$ is a neighborhood in $N_x$. We have: :$\openint {x - h} {x + h} \in N_x$ :$\map {\omega_f} {\openint {x - h} {x + h} } \in \R$ Accordingly: :$\map {\omega_f} x \in \R$ by Infimum of Set of Oscillations on Set $\map {\omega_f} x = \map {\omega^L_f} x$ follows by Lemma. This finishes the proof of the theorem. {{qed}} Category:Real Analysis 9fanzs4g1wiajx7jhvn2b5rk2aq0wir