Why are monotone functions Riemann integrable on a closed interval?
Suppose $f$ is nondecreasing. For any partition $a = x_0 < x_1 \ldots < x_n = b$ of your interval $[a,b]$, any Riemann sum is between the left Riemann sum $L = \sum_{j=1}^n f(x_{j-1})(x_j - x_{j-1})$ and the right Riemann sum $R = \sum_{j=1}^n f(x_{j})(x_j - x_{j-1})$. The difference between them is at most $(f(b) - f(a)) \delta$ where $\delta = \max_j (x_j - x_{j-1})$.
Proof without words:
