Let $I\subset \mathbb{R}$ be an interval. By definition $\text{diam}f(I)=\sup_{x, x'\in I}|f(x)-f(x')|$, $\text{osc}(f, I)=\sup_{x\in I}f(x)-\inf_{x\in I}f(x)$ and $\text{osc}(f, x_0)=\lim_{\epsilon\to 0}\text{osc}(f, B_{\epsilon}(x_0))$.
Is it true that $$\text{diam}f(I)=\text{osc}(f, I)$$ and $$\text{osc}(f, x_0)=\inf\{\text{osc}(f, B_{\delta}(x_0)):\delta>0\}=\limsup_{x\to x_0}f(x)-\liminf_{x\to x_0}f(x)$$ ? I think so but I want to be sure.
On the real line diameter of any set $A\subseteq \mathbb R$ can be expressed in terms of its $\inf$ and $\sup$, namely $\operatorname{diam}A = \sup A-\inf A$. (At least of every nonempty set; I don't want to deal with the diameter of empty set right now.) In particular, $\operatorname{diam} f(I)=\operatorname{osc}(f, I)$.
The equality $\lim_{\delta\to 0^+}\operatorname{osc}(f, B_{\delta}(x_0)) =\inf\{\operatorname{osc}(f, B_{\delta}(x_0)):\delta>0\}$ follows from the fact that oscillation is a nondecreasing function of $\delta$.
The equality $ \operatorname{osc}(f, x_0) = \limsup_{x\to x_0}f(x)-\liminf_{x\to x_0}f(x)$ is also correct, but with the caveat: on the right we might have $\infty-\infty$ or $(-\infty)-(-\infty)$, neither of which is particularly meaningful. In both those cases $f$ is unbounded in every neighborhood of $x_0$, which makes the left side $\infty$.
