I was given the following proof of a theorem, quick note about notation, the fact that $f$ is riemann integrable on $[a,b]$ we denote by $f\in\mathcal{R}([a,b])$
Theorem: Let $a,b\in\mathbb{R},a<b$, let $f$ be continuous on closed interval $[a,b]$, then $f\in\mathcal{R}([a,b])$
We use the following lemma:
Lemma: Let $a,b\in\mathbb{R},a<b$ and $f$ bounded on $[a,b]$. Then the following is equivallent: $$f\in\mathcal{R}([a,b])\tag{1}$$ $$\forall \epsilon >0 \exists \sigma_{[a,b]}:\overline{S}(f,\sigma)-\underline{S}(f,\sigma)<\epsilon$$
where $\sigma$ is a partition of $[a,b]$. Now, i understand the proof of the lemma and I'm not giving it here.
Proof. (of the theorem)
$f$ is continuous on $[a,b]$, so $f$ is bounded on $[a,b]$, also because $f$ is bounded on bounded interval, f is uniformly continuous. Let $\epsilon > 0$ be given, by assumption, we find $\delta>0$ s.t. for any pair $x,y\in[a,b]:|x-y|<\delta \Rightarrow |f(x)-f(y)|<\epsilon$. Choose partitioning $\sigma_{[a,b]}$ s.t. $\nu(\sigma)<\delta$. Then by uniform continuity for each $i\in\{1\ldots,n\}$ we have that $$\sup_{[x_{i-1},x_i]}f\leq \inf_{[x_{i-1},x_i]}f+\epsilon$$
thus $$\overline{S}(f,\sigma)-\underline{S}(f,\sigma)=\sum_{i=1}^n \bigg(\sup_{[x_{i-1},x_i]}f-\inf_{[x_{i-1},x_i]}f\bigg)(x_i-x_{i-1})\leq \sum_{i=1}^n \epsilon(x_i-x_{i-1})=\epsilon(b-a)$$
which by lemma shows that $f\in\mathcal{R}([a,b])$.
where $\nu(\sigma)$ is the norm of the partition, that is $\nu(\sigma)=\max\{x_{i}-x_{i-1}\mid i\in\{1,\ldots,n\}\}$
Now, the part I don't understand is why uniform continuity implies something like $\sup_{[x_{i-1},x_i]}f\leq \inf_{[x_{i-1},x_i]}f+\epsilon$ (bolded part). Would someone explain more thoroughly, what is going on? Do they mean that by continuity $f$ attains extreme values (suprema, and infima) on each of $\mathcal{I}:=[x_{i-1},x_i]$ and by $|f(x)-f(y)|<\epsilon$, this results in $\sup_\mathcal{I} f – \inf_\mathcal{I} f \leq \epsilon$ (now, why is there a $\leq$ instead of $<$ ?).
Best Answer
Suppose that $|s-r| <\delta$, and $|f(x)-f(y)| < \epsilon$ whenever $|x-y| <\delta$. Then, on the interval $[r,s]$ (in the domain of $f$), $$\epsilon \ge \sup_{x,y\in [r,s]} (f(x)-f(y)) =\sup_{x\in [r,s]}f(x) - \inf_{x\in [r,s]}f(y)$$ That equality is always true, even without continuity; the supremum minus the infimum is an upper bound for the variation, and we can approach arbitrarily close to each of them. What (uniform) continuity gets us is the inequality out front relating to $\epsilon$, as we can choose the mesh fine enough that the variation in any one subinterval is small.