In this question: The confusion about the proof of if function is continuous on a closed interval $[a,b]$,then it is integrable, the OP derives the following inequality:
$$\begin{equation} \ f(x')-\epsilon\leq\inf f(x)\leq f(x)\leq\sup f(x)\leq f(x')+\epsilon\end{equation}$$
Deriving this result, I am not getting less or equal inequality on outer parts of it. Instead, I am getting strict inequality:
Since $|f(x) - f(x')| < \varepsilon$, then $\sup f(x) - \inf f(x) < \varepsilon$, where $\sup f(x)$ and $\inf f(x)$ are taken over $[t_{i-1}, t_i]$.From it we have two inequalities:$$\begin{align*}\text{(i)}\quad & \sup f(x) < \inf f(x) + \varepsilon \\\text{(ii)}\quad & \inf f(x) > \sup f(x) - \varepsilon\end{align*}$$Also $\inf f(x) \leq f(x) \leq \sup f(x)$ by combining with $\text{(i)}$ and $\text{(ii)}$:$$\sup f(x) - \varepsilon < \inf f(x) \leq f(x) \leq \sup f(x) < \inf f(x) + \varepsilon$$It's clear that for any $x' \in [t_{i-1}, t_i]$:$$f(x') - \varepsilon \leq \sup f(x) - \varepsilon \qquad \text{and} \qquad \inf f(x) + \varepsilon \leq f(x') + \varepsilon$$
because $f(x') \leq \sup f(x)$ and $\inf f(x) \leq f(x')$
Therefore,$$f(x') - \varepsilon \leq \sup f(x) - \varepsilon < \inf f(x) \leq f(x) \leq \sup f(x) < \inf f(x) + \varepsilon \leq f(x') + \varepsilon$$or$$f(x') - \varepsilon < \inf f(x) \leq f(x) \leq \sup f(x) < f(x') + \varepsilon$$
As you can see the outer parts of my inequality are strict, unlike OP's inequality at the start of this question.
EDIT:I have been reminded in the comments that $a < b \implies a \leq b$. So my derivation is equivalent to OP's. Do note, however, that the converse is obviously not true, i.e. $a \leq b \not\!\!\!\implies a < b$