Proof that monotone functions are Riemann integrable

Proof

We compute the Darboux sums for a partition into equal pieces and use Riemann's criterion.

1. Let $a < b$ and let $f \colon [a, b] \to \mathbb{R}$ be monotone. First assume that $f$ is increasing, that is, $x \le y$ implies $f(x) \le f(y)$.
2. Then $f(a) \le f(x) \le f(b)$ for all $x \in [a, b]$. So $f$ is bounded, and its Darboux sums are defined.
3. Let $\varepsilon > 0$. By the Archimedean property, take $n \ge 1$ with $n \varepsilon > (b - a)(f(b) - f(a))$. Let $P$ have the points $x_i = a + i (b - a)/n$ for $i = 0, \dots, n$.
4. On $[x_{i-1}, x_i]$ we have $f(x_{i-1}) \le f(x) \le f(x_i)$, and both bounds are values of $f$ there. So $M_i = f(x_i)$ and $m_i = f(x_{i-1})$.
5. Each subinterval has length $(b - a)/n$. So the sum telescopes: $$U(f, P) - L(f, P) = \frac{b - a}{n} \sum_{i=1}^{n} \big(f(x_i) - f(x_{i-1})\big) = \frac{b - a}{n} \big(f(b) - f(a)\big) < \varepsilon.$$
6. Such a partition exists for each $\varepsilon > 0$. By Riemann's criterion, $f$ is Riemann integrable.
7. Now assume that $f$ is decreasing. Then $f(b) \le f(x) \le f(a)$, so $f$ is bounded. Let $\varepsilon > 0$. Take $n \ge 1$ with $n \varepsilon > (b - a)(f(a) - f(b))$, and let $P$ be the partition into $n$ equal pieces as in step 3.
8. Now $M_i = f(x_{i-1})$ and $m_i = f(x_i)$. As in step 5, $U(f, P) - L(f, P) = \frac{b - a}{n} (f(a) - f(b)) < \varepsilon$. As in step 6, $f$ is Riemann integrable.