We define the oscillation of a set $A$ to be $\omega_f(A)=|\sup\limits_{x\in A}f(x)-\sup\limits_{y \in A}f(y)|$.
We define the oscillation of a point $x \in A$ to be $\omega_f(x)=\inf\{\omega_f(x-\epsilon, x+\epsilon)\cap A) : \epsilon > 0\}$.
Proposition: Given $\epsilon > 0$ suppose $\omega_f(x) < \epsilon$ for each $x \in [a,b]$. Then show that there is a $\delta > 0$ such that for every closed interval $I \subseteq [a,b]$ with $l(I) < \delta$ we have $\omega_f(I) < \epsilon$.
To begin, I want to make sure I understand what I am reading. Is this saying that all the $x$ in some closed interval are uniformly oscillating (or oscillated)? So because of this uniform oscillation, I can pick any $I=[x_i, x_{i+1}]\subseteq [a,b]$ with $|x_{i+1}-x_i|<\delta$ and have $\omega_f(I)<\epsilon$. I'm not sure how to prove this one because I'm not sure how to pick endpoints for $I$.
Proposition: Show that $\{x \in [a,b] : \omega_f(x) \geq \delta \}$ is a closed set for each $\delta > 0$.
I think I was able to prove this as follows: Let $U= \{x \in [a,b] : \omega_f(x) \geq \delta \}$. So I want to show $U^c=\{x \in [a,b] : \omega_f(x) < \delta \}$ is open. Let $x_0 \in U^c$. Since $\omega_f(x_0) < \delta$ there must be an $r>0$ such that $\omega_f(B(x_0,r))<\delta$. So take $y \in B(x_0,r)$. Then there must exist $r'>0$ such that $y \in B(y, r')\subset B(x_0,r)$ and $\omega_f(y) \leq \omega_f(B(y, r')) \leq \omega_f(B(x_0,r))< \delta$. This shows that $y$ is a member of $U^c$ and since $y$ was chosen arbitrarily, we see that $U^c$ is open. Hence $U$ is closesed.