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Let $f: [a,b] \longrightarrow \mathbb{R}$ be an increasing function. If $x_1, \cdots, x_n \in [a,b]$ are distinct, show that $\sum_{i=1}^{i=n} o(f,x_i) < f(b) - f(a).$

I would like to know if my proof is correct and, if that's the case, an opinion if my proof writting is fine and an advice about how to write this better. Thanks in advance!

My attempt:

Let $P := \{ [t_{i-1},t_i] \ ; i = 1, \cdots, n, \ t_0 = a, t_n = b \ and \ x_i \in (t_{i-1},t_i) \ for \ each \ i \}$ a partition of $[a,b]$. The definition of oscillation of $f$ at $x$ given by Spivak is $o(f,x) := \lim_{\delta \rightarrow 0} [\sup f(B(x,\delta)) - \inf f(B(x,\delta))]$, then $[\lim \inf f((x_i - \delta, x_i + \delta)), \ \lim \sup f((x_i - \delta, x_i + \delta))] \subset [t_{i-1},t_i]$ when $\delta \rightarrow 0$ and $o(f,x_i) = \lim \sup f((x_i - \delta, x_i + \delta)) - \lim \inf f((x_i - \delta, x_i + \delta)) < f(t_i) - f(t_{i-1})$ for $i = 1, \cdots, n$. It's clear now that $\sum_{i=1}^{i=n} o(f,x_i) < \sum_{i=1}^{i=n} f(t_i) - f(t_{i-1}) = f(b) - f(a). \square$

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