# Why are monotone functions Riemann integrable on a closed interval?

Monotone functions are continuous except countably many points. If function is Riemann integrable it has only a finite number of discontinuity points. So how monotone functions are Riemann integrable on closed interval always?

• "If function is Riemann integrable it has only a finite number of discontinuity points." That's not true; consider the function $f(x)=\begin{cases}1/2&\text{if$1/2<x\le1$,}\\1/4&\text{if$1/4<x\le1/2$,}\\1/8&\text{if$1/8<x\le1/4$,}\\&\vdots\\0&\text{if$0=x$}\end{cases}$ on the interval $[0,1]$. – Rahul Sep 5 '14 at 20:31
• Using the measure of the set of discontinuities is fairly high tech. One can prove that monotone functions are Riemann integrable using the definition. – Ayman Hourieh Sep 5 '14 at 20:35

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 $\delta = \max_j (x_j - x_{j-1})$. 