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Given a function $f:E\subset\mathbb R^n\longrightarrow\mathbb R$ and $x\in E$. The function $$\omega(f,E):=\sup\{|f(x)-f(y)|:x,y\in E\}$$ is called the oscillation of $f$ on $E$, and $$\omega(f,x):=\lim_{\delta\rightarrow0^+}\omega(f,U_\delta(x)\cap E)$$ is called the oscillation of $f$ at $x$.

Assume $E$ is compact and $\omega(f,x)\leq a\ \ \forall x\in E$. Prove that for all $\epsilon>0$, there exists a $\delta=\delta(\epsilon)>0$ such that: $$\omega(f,U_\delta(x)\cap E)\leq a+\epsilon\quad\forall x\in E$$

I think this question is similar to Lebesgue criterion for Riemann-integrability and Heine-Borel Theorem, but I am not sure of the context of that question or what I could take from there.

The furthest I got is arguing that for each $x\in E$ there is a $\delta_x>0$ such that for every $0<\delta\leq\delta_x$ we have $$\omega(f,U_\delta(x)\cap E)\leq \omega(f,x) +\epsilon\leq a+\epsilon\quad\forall x\in E$$ Since $E$ is compact, any open cover of $E$ has a finite subcover: $$\exists\ x_1,\cdots,x_k\in E(k\in\mathbb N^*,k<\infty): E\subseteq\bigcup_{i=1}^k\ U_{\delta_{x_i}}(x_i)$$ The task remains to connect these two observations.

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    $\begingroup$ I don't think the notation $U_\delta(x)$ is quite standard, I'm guessing it's a ball of radius or diameter $\delta$? $B$ would be more common than $U$ if so. Or maybe it's a "nice" set of small neighbourhoods? $\endgroup$ Commented Oct 6 at 16:07
  • $\begingroup$ It wasn't specified in the problem I was given, but I think it's safe to assume that $U_\delta(x) \equiv B(x,\delta)$. $\endgroup$ Commented Oct 6 at 16:25

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Let $\hat{\delta} > 0$ be a Lebesgue number of the cover $\{U_{\delta_{x_i}}(x_i)\}_{i\leq k}$, that is, if $diam \ A < \hat{\delta}$ then $A \subset U_{\delta_{x_i}}(x_i)$ for some $i$.

Given $x$, since $U_{\hat{\delta}/3}(x)\subset U_{\delta_{x_i}}(x_i)$ for some $i$ we have

$$osc(f,U_{\hat{\delta}/3}(x))\leq osc(f,U_{\delta_{x_i}}(x_i))\leq a+\epsilon.$$

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