7.1 Area Under a Monotonic Function

7.1   Theorem. Let $f$ be a monotonic function from the interval $[a,b]$ to $\mbox{{\bf R}}_{\geq 0}$. Let $\{P_n\}$ be a sequence of partitions of $[a,b]$ such that $\{\mu (P_n)\}\to 0$, and let

$$A_a^bf = \alpha \{ (x,y) \in \mbox{{\bf R}}^2: a \leq x \leq b \mbox{ and }0 \leq y \leq f(x) \}$$

Then

$$\{\alpha \Big(I_a^b(f,P_n)\Big)\}\to A_a^bf$$

and

$$\{\alpha\Big(O_a^b(f,P_n)\Big)\}\to A_a^b f.$$

(The notation here is the same as in theorem 5.40 and exercise 5.47.

Proof: We noted in theorem 5.40 and exercise 5.47 that

$$0 \leq \alpha\Big( O_a^b(f,P_n)\Big)-\alpha\Big( I_a^b(f,P_n)\Big)\leq \mu (P_n)\cdot\vert f(b)-f(a)\vert.$$ (7.2)

Since

$$\begin{eqnarray*} \lim \left \{ \mu (P_n)\cdot \vert f(b)-f(a) \vert\right\} & ... ...lim \{ \mu (P_n) \} \\ &=& \vert f(b) - f(a)\vert \cdot 0 = 0, \end{eqnarray*}$$

we conclude from the squeezing rule that

$$\lim \left\{ \alpha\Big( O_a^b(f,P_n)\Big)-\alpha\Big(I_a^b(f,P_n)\Big)\right\} =0.$$ (7.3)

We also have by (5.43) that

$$\alpha\Big( I_a^b(f,P_n)\Big)\leq A_a^b f\leq\alpha\Big( O_a^b(f,P_n)\Big),$$

so that

$$0 \leq A_a^b f - \alpha\Big( I_a^b(f,P_n)\Big) \leq \alpha\Big( O_a^b(f,P_n)\Big) - \alpha\Big( I_a^b(f,P_n)\Big).$$

By (7.3) and the squeezing rule

$$\lim \left\{ A_a^bf - \alpha\Big( I_a^b(f,P_n)\Big) \right\} = 0,$$

and hence

$$\lim \left\{ \alpha\Big( I_a^b(f,P_n)\Big) \right\} = A_a^bf.$$

Also,

$$\begin{eqnarray*} \lim \left\{ \alpha\Big( O_a^b(f,P_n)\Big) \right\} &=& \lim \... ...Big) \right\}\\ &=& A_a^bf + 0 = A_a^bf. \mbox{ $\diamondsuit$} \end{eqnarray*}$$

7.4   Definition (Riemann sum.) Let $P=\{x_0,x_1,\cdots ,x_n\}$ be a partition for an interval $[a,b]$. A sample for $P$ is a finite sequence $S=\{s_1,s_2,\cdots ,s_n\}$ of numbers such that $s_i\in[x_{i-1},x_i]$ for $1\leq i\leq n$. If $f$ is a function from $[a,b]$ to $\mbox{{\bf R}}$, and $P$ is a partition for $[a,b]$ and $S$ is a sample for $P$, we define

$$\sum (f,P,S)=\sum_{i=1}^n f(s_i)(x_i-x_{i-1})$$

and we call $\sum (f,P,S)$ a Riemann sum for $f,\;P$ and $S$. We will sometimes write $\sum([f(t)],P,S)$ instead of $\sum (f,P,S)$.

7.5   Example. If $f$ is an increasing function from $[a,b]$ to $\mbox{{\bf R}}_{\geq 0}$, and $P=\{x_0,\cdots ,x_n\}$ is a partition of $[a,b]$, and $S_l=\{x_0,\cdots ,x_{n-1}\}$, then

$$\sum (f,P,S_l)=\alpha(I_a^b(f,P)).$$

If $S_r=\{x_1,x_2,\cdots ,x_n\}$, then

$$\sum(f,P,S_r)=\alpha(O_a^b(f,P)).$$

If $${S_m=\{ {{x_0+x_1}\over 2},\cdots , {{x_{n-1}+x_n}\over 2}\}}$$ then

$$\sum (f,P,S_m)=\sum_{i=1}^n f\Big( {{x_{i-1}+x_i}\over 2}\Big)(x_i-x_{i-1})$$

is some number between $\alpha(I_a^b(f,P))$ and $\alpha(O_a^b(f,P))$.

7.6   Theorem (Area theorem for monotonic functions.) Let $f$ be a
monotonic function from the interval $[a,b]$ to $\mbox{{\bf R}}_{\geq 0}$. Then for every sequence $\{P_n\}$ of partitions of $[a,b]$ such that $\{\mu (P_n)\}\to 0$, and for every sequence $\{S_n\}$ where $S_n$ is a sample for $P_n$, we have

$$\{\sum (f,P_n,S_n)\}\to A_a^bf.$$

Proof: We will consider the case where $f$ is increasing. The case where $f$ is decreasing is similar.

For each partition $P_n=\{x_0,\cdots ,x_m\}$ and sample $S_n=\{s_1,\cdots ,s_m\}$, we have for $1\leq i\leq m$

$$\begin{eqnarray*} x_{i-1}\leq s_i\leq x_i &\mbox{$\Longrightarrow$}& f(x_{i-1})\... ...})(x_i-x_{i-1})\leq f(s_i)(x_i-x_{i-1})\leq f(x_i)(x_i-x_{i-1}). \end{eqnarray*}$$

Hence

$$\sum_{i=1}^mf(x_{i-1})(x_i-x_{i-1})\leq\sum_{i=1}^m f(s_i)(x_i-x_{i-1})\leq\sum_{i=1}^m f(x_i)(x_i-x_{i-1}),$$

i.e.,

$$\alpha\Big( I_a^b(f,P_n)\Big)\leq\sum (f,P_n,S_n)\leq\alpha\Big( O_a^b(f,P_n)\Big).$$

By theorem 7.1 we have

$$\{\alpha\Big(I_a^b(f,P_n)\Big)\}\to A_a^bf,$$

and

$$\{\alpha\Big( O_a^b(f,P_n)\Big)\}\to A_a^bf,$$

so by the squeezing rule,

$$\{\sum (f,P_n,S_n)\}\to A_a^bf.$$