I'm reading a proof that a function is jump continuous if and only if there is a sequence of staircase functions that converges uniformly to it.
Here $I=[\alpha,\beta] \subseteq \mathbb R$. The author said that
Now we select a refinement $\mathfrak{Z}_{1}=\left(\xi_{0}, \ldots, \xi_{k}\right)$ of $\mathfrak{Z}_{0}$ with $$ \|f(s)-f(t)\|<1 / n \text { for } s, t \in\left(\xi_{j-1}, \xi_{j}\right) \text { and } j=1, \ldots, k $$
Could you please verify if my understanding of how to define $\mathfrak{Z}_{1}$ is correct?
From the definition of jump continuous functions, we have $\lim_{s \to x^-} f(s) =a$ and $\lim_{s \to x^+} f(s)=b$ exist. This means there exist $\alpha(x),\beta(x)$ such that $\alpha(x) \le s < x$ implies $|f(s)-a| < 1/(2n)$ and that $x < s \le \beta(x)$ implies $|f(s)-b| < 1/(2n)$.
As such, $|f(s)-f(t)| \le |f(s)-a|+|f(t)-a| < 1/(2n)+1/(2n) = 1/n$ for all $s,t \in [\alpha(x),x)$. Similarly, $|f(s)-f(t)| \le |f(s)-b|+|f(t)-b| < 1/(2n)+1/(2n) = 1/n$ for all $s,t \in (x,\beta(x)]$.
Then we define a refinement $$\mathfrak{Z}_{1} = \{x_j \mid 0 \le j \le m\} \cup \{\alpha(x_j) \mid 0 \le j \le m\} \cup \{\beta(x_j) \mid 0 \le j \le m\}$$