# Showing that all monotone functions are integrable

I am given the following proof:

Theorem. All monotone functions are integrable.

Proof. Without loss of generality, assume that $f$ is increasing on an interval $\left[a, b \right]$. Thus, $f(a) \le f(x) \le f(b)$, and $f$ is bounded on $\left[a, b \right]$. Given $\varepsilon >0$, there exists $k > 0$ such that

\begin{equation*} k \left[f(b) - f(a) \right] < \varepsilon. \end{equation*}

Let $P = \left\lbrace x_0, x_1, \dots, x_n \right\rbrace$ be a partition of $\left[a, b \right]$ such that $\Delta x_i \le k$ for all $i$. Since $f$ is increasing, it follows that

\begin{equation*} m_i = f(x_{i-1}) \ \text{and} \ M_i = f(x_i), \quad i = 1, 2, \dots, n. \end{equation*}

Where $m_i$ is the greatest lower bound of $f$ on $\left[ x_{i-1}, x_i \right]$, and $M_i$ is the least upper bound of $f$ on $\left[ x_{i-1}, x_i \right]$.

$U(f, P) - L(f, P) = \sum_{i=1}^n \left[ f(x_i) - f(x_{i-1}) \right] \Delta x_i$

$\le k \sum_{i=1}^n \left[ f(x_i) - f(x_{i-1}) \right] (*)$

$= k \left[f(b) - f(a) \right]$

$< \varepsilon.$

By Theorem 7.1.9 $f$ is integrable on $\left[ a, b \right]$. In the case that $f$ is monotone decreasing, we may use the same argument on $-f$.

I am just wondering if somebody could explain how we get from the line marked by (*) to the line after that.

## 2 Answers

\begin{align}\sum_{i=1}^n[f(x_i)-f(x_{i-1})] &= \sum_{i=1}^n f(x_i)-\sum_{i=1}^nf(x_{i-1}) \\&= \sum_{i=1}^n f(x_i)-\sum_{i=0}^{n-1}f(x_{i})\\&=f(x_n)-f(x_0)\end{align}

Just try to write down that sum.

You'll see that each term cancels out except the first and the last, leaving you with that expression.

By the way, it is false that if $f(x)$ is monotone on $[a, b)$ then $f(x)$ is bounded.. just take a function with an vertical asymptote.

• I would prefer to use $[a, b)$, because, to me, saying that $f$ is monotone on $[a, b]$ implies that $f(b)$ is defined and finite. – marty cohen Nov 23 '14 at 22:52
• @martycohen Nice point. Edited :) – Ant Nov 23 '14 at 22:53