I have a question about the choice of the bound for uniform continuity (why is it $\frac{\epsilon}{2(b-a)}$) in Spivak's proof of integrability of continuous functions on $[a, b]$.
Here is his proof:
Notice, first, that $f$ is bounded on $[a,b]$, because it is continuous on $[a, b]$. To prove that $f$ is integrable on $[a, b]$, we want to use Theorem 2, and show that for every $\epsilon > 0$ there is a partition $P$ of $[a, b]$ such that $$ U(f, P) - L(f, P) < \epsilon. $$ Now we know, by Theorem 1 of the Appendix to Chapter 8, that $f$ is uniformly continuous on $[a, b]$. So there is some $\delta > 0$ such that for all $x$ and $y$ in $[a, b]$, $$ \text{if } |x - y| < \delta, \text{ then } |f(x) - f(y)| < \frac{\epsilon}{2(b-a)}. $$ The trick is simply to choose a partition $P = {t_0, \ldots, t_n}$ such that each $|t_i - t_{i-1}| < \delta$. Then for each $i$ we have $$ |f(x) - f(y)| < \frac{\epsilon}{2(b-a)} \qquad \text{for all } x, y \text{ in } [t_{i-1}, t_i], $$ and it follows easily that $$ M_i - m_i \leq \frac{\epsilon}{2(b-a)} < \frac{\epsilon}{b-a}. $$ Since this is true for all $i$, we then have $$ U(f, P) - L(f, P) = \sum_{i=1}^n (M_i - m_i)(t_i - t_{i-1}) \ < \frac{\epsilon}{b-a} \sum_{i=1}^n t_i - t_{i-1} \ = \frac{\epsilon}{b-a} \cdot b - a \ = \epsilon, $$ which is what we wanted. $\blacksquare$
It's not clear to me why he picks $\frac{\epsilon}{2(b-a)}$, since it seems he is not using this fact.
The only reason I can see is the fact that $M_i - m_i \leq \frac{\epsilon}{2(b-a)}$ is not a strict inequality, which allows us to then state it is strictly less than $\frac{\epsilon}{b-a}$
And, in fact, I am not sure why $M_i - m_i \leq \frac{\epsilon}{2(b-a)}$ instead of $M_i - m_i < \frac{\epsilon}{2(b-a)}$ since we have that $|f(x) - f(y)| < \frac{\epsilon}{2(b-a)}$ which would even further make his choice of the bound peculiar.
