for $f:A \rightarrow \mathbb{R}$ and $x\in A$, the oscillation of $f$ at $x$ is $$\operatorname{Osc}_x(f):=\lim_{r \to 0^+} \text{diam}(f([x-r, x+r]\cap A))$$
where, diam$(S) := \sup \{|x-y|: x,y \in S\}$
Now, given that $f$ is an increasing function that has a jump discontinuity at $x$, how do I prove that $\operatorname{Osc}_x(f) = f(x^+) – f(x^-)$?
what I have tried is this:
since f is a jump discontinuity, hence $f(x^-) \neq f(x^+)$
since f is increasing, hence $f(x^-) < f(x^+)$ and $f(x) \in [f(x^-) , f(x^+)]$
now drawing a picture makes it obvious that $\operatorname{Osc}_x(f) = f(x^+) – f(x^-)$, but how do I rigorously prove that?
EDIT:-
A part of @EuYu 's answer wasn't quite clear to me, so I have added a clarification for that part. Could someone verify if this clarification is correct or not?
set $\epsilon_0>0$ such that $f(x_0^-) + \epsilon_0 < f(x_0^+) – \epsilon_0 $, this can be done since, $f(x_0^-) < f(x_0^+)$
Now, for any $\epsilon \in (0,\epsilon_0)$, there exists some $\delta > 0 $ such that
$$|f(x) – f(x_0^-)| < \epsilon,\qquad \text{and}\qquad |f(x)-f(x_0^+)|<\epsilon$$
for all $x \in (x_0-\delta,x_0)$ and $x \in (x_0, x_0 + \delta)$, respectively.
This implies that
$$\quad f(x_0^-) – \epsilon < f(x) < f(x_0^-) + \epsilon,\quad \forall x\in (x_0-\delta,x_0)$$ and
$$ f(x_0^+) – \epsilon <f(x) < f(x_0^+) + \epsilon,\quad \forall x\in (x_0,x_0+\delta)$$
and
$$ f(x_0^-) + \epsilon < f(x_0^+) – \epsilon $$
This implies that, $\forall \epsilon \in (0,\epsilon_0)$, there exists some $\delta > 0 $ such that
$$ f(x_0^-) – \epsilon < \sup_{x\in (x_0-\delta,x_0)}f(x) \le f(x_0^-) + \epsilon < f(x_0^+) – \epsilon < \sup_{x\in (x_0,x_0+\delta)}f(x) \leq f(x_0^+) + \epsilon,$$
This implies that $\forall \epsilon \in (0,\epsilon_0)$ , there exists some $\delta > 0 $ such that
$$ f(x_0^+) – \epsilon < \sup_{x\in (x_0-\delta,x_0+\delta)}f(x) \le f(x_0^+) + \epsilon $$
This implies that $\forall \epsilon>0$ , there exists some $\delta > 0 $ such that
$$ f(x_0^+) – \epsilon < \sup_{x\in (x_0-\delta,x_0+\delta)}f(x) \le f(x_0^+) + \epsilon $$
similarly $\forall \epsilon>0$, there exists some $\delta > 0 $ such that
$$ f(x_0^-) – \epsilon < \inf_{x\in(x_0-\delta,x_0+\delta)}f(x) \le f(x_0^-) + \epsilon$$
Best Answer
Let me denote the oscillation of the function $f$ on a set $S$ by $$\omega_f(S) = \sup_{x\in S}f(x) - \inf_{x\in S}f(x).$$ Suppose that $f$ has a jump discontinuity at $x_0$. You don't need the function to be increasing, just for the value of $f(x_0)$ to be between its one-sided limits $f(x_0^-)$ and $f(x_0^+)$. Without loss of generality, let's assume that $f(x_0^-) \le f(x_0) \le f(x_0^+)$.
Since the left and right limits exist by assumption, for any $\epsilon > 0$, there exists some $\delta > 0 $ such that $$|f(x) - f(x_0^-)| < \epsilon,\qquad \text{and}\qquad |f(x)-f(x_0^+)|<\epsilon$$ for all $x \in (x_0-\delta,x_0)$ and $x \in (x_0, x_0 + \delta)$, respectively. This implies that $$f(x_0^+) - \epsilon \le \sup_{x\in (x_0-\delta,x_0+\delta)}f(x) \le f(x_0^+) + \epsilon,$$and$$\quad f(x_0^-) - \epsilon \le \inf_{x\in(x_0-\delta,x_0+\delta)}f(x) \le f(x_0^-) + \epsilon.$$ Therefore the oscillation of $f$ on $(x_0-\delta,x_0+\delta)$ is bounded by $$f(x_0^+)-f(x_0^-) - 2\epsilon \le \omega_f((x_0-\delta,x_0+\delta)) \le f(x_0^+) - f(x_0^-) + 2\epsilon.$$ Since this holds for arbitrary $\epsilon > 0$, it follows that we must have $$\omega_f(x_0) = f(x_0^+) - f(x_0^-).$$