# 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.

\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}
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