From exercise 2.8 of "Measure theory and probability theory (pag. 73)" by Krishna and Soumendra:
Let $f:\mathbb{R} \rightarrow \mathbb{R}$. Let $\bar{f}(x)=\inf_{\delta \gt 0}\sup_{\left|y-x\right| \lt \delta} f(y)$.
Show that, for $t \in \mathbb{R}$, $\{x: \bar{f}(x) < t\}$ is open and, therefore, $\bar{f}(x)$ is Borel-measurable.
Here I can see that $\bar{f}(x)=\omega_f(x) + f(x)$ where $\omega_f(x)$ is the oscillation of $f$ at $x$. To better explain, $h_x(\delta)=\sup\{f(y): y \in I_{\delta}(x)\}$ is a non decreasing function which sends $\mathbb{R}^{+}$ in $\left[f(x), +\infty\right]$. Thus its single sided limits exist (allowing $+\infty$) and, in particular, we can write $\bar{f}: \mathbb{R} \rightarrow \mathbb{R}\cup \{+\infty\}$ as
$\bar{f}(x)=\lim_{\delta \rightarrow0^+}h_x(\delta) \geq f(x)$
equality holds if and only if $f$ is continuous at $x$.
Still I can't figure out how to use this to solve the exercise. Any help would be appreciated, thank you.
Best Answer
$\omega_f(x)$ is not useful. Directly follow the given hint.
Let $\overline f(x) <t$. Let $s \in (f(x),t)$. There exist $\delta >0$ such that $f(y) <s$ whenever $|x-y| <\delta$. Let $|x-u|<\frac {\delta} 2$. For any $z$ with $|z-u| <\frac {\delta} 2$ we have $f(z) <s$ because $|z-x| \leq |z-u|+|u-x| <\frac {\delta} 2+\frac {\delta} 2=\delta$. Hence $\overline f(u)\leq s <t$. We have proved that $\overline f(x) <t$ implies $\overline f(u) <t$ for all $z$ in some interval around $x$. Hence $\{x: \overline f(x) <t\}$ is an open set.