5.3 Monotonic Functions

5.27   Definition (Partition.) Let $a,b$ be real numbers with $a\leq b$. A partition $P$ of the interval $[a,b]$ is a finite sequence of points

$$P=\{x_0,x_1,\cdots ,x_n\}$$

with $a=x_0\leq x_1\leq x_2\cdots \leq x_n=b$. The intervals $[x_0,x_1],[x_1,x_2],\cdots$, $[x_{n-1},x_n]$ are called the subintervals of the partition $P$, and $[x_{j-1},x_j]$ is the $j^{th}$ subinterval of $P$ for $1\leq j\leq n$. The largest of the numbers $x_j-x_{j-1}$ is called the mesh of the partition $P$, and is denoted by $\mu (P)$. The partition

$$\{ a,\; a+{{(b-a)}\over n},\; a+{{2(b-a)}\over n},\cdots , a+{{n(b-a)}\over n}=b\}$$

is called the regular partition of $[a,b]$ into $n$ equal subintervals.

5.28   Example. Let

$$P=\{ 0,{1\over {16}},{1\over 8},{1\over 4},{1\over 2},1\}$$

Then $P$ is a partition of $[0,1]$ into $5$ subintervals and $${\mu (P)=1-{1\over 2}={1\over 2}}$$.

The regular partition of $[1,2]$ into $5$ subintervals is $${\{1,{6\over 5},{7\over 5},{8\over 5},{9\over 5},2\}}$$.

If $Q_n$ is the regular partition of $[a,b]$ into $n$ equal subintervals, then $\mu (Q_n)={{b-a}\over n}$.

5.29   Exercise. Find a partition $P$ of $[0,1]$ into five subintervals, such that $\mu(P) = {4\over 5}$, or explain why no such partition exists.

5.30   Exercise. Find a partition $Q$ of $[0,1]$ into five subintervals, such that $\mu(Q) = {1\over 10}$, or explain why no such partition exists.

5.31   Definition (Monotonic function.) Let $J$ be an interval in $\mbox{{\bf R}}$, and let $f\colon J\to\mbox{{\bf R}}$ be a function. We say that $f$ is increasing on $J$ if

$$\mbox{ for all }x,y \mbox{ in } J \Big[(x\leq y)\mbox{$\hspac... ...rightarrow\hspace{1ex}$}(f(x) \leq f(y))\Big]\mbox{{\nonumber}}$$ (5.32)

and we say that $f$ is decreasing on $J$ if

$$\mbox{ for all }x,y \mbox{ in } J \Big[(x\leq y)\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}(f(x)\geq f(y))\Big].$$

We say that $f$ is strictly increasing on $J$ if

$$\mbox{ for all }x,y \mbox{ in } J \Big[(x < y) \mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}(f(x) < f(y))\Big]$$

and we say that $f$ is strictly decreasing on $J$ if

$$\mbox{ for all }x,y \mbox{ in } J \Big[(x<y) \mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}(f(x) > f(y))\Big].$$

We say that $f$ is monotonic on J if $f$ is either increasing on $J$ or decreasing on $J$, and we say that $f$ is strictly monotonic on $J$ if $f$ is either strictly increasing or strictly decreasing on $J$.

A constant function on $J$ is both increasing and decreasing on $J$.

5.33   Notation ($S_a^bf$) Let $f$ be a function from the interval $[a,b]$ to the non-negative numbers. We will write

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

i.e., $S_a^bf$ is the set of points under the graph of $f$ over the interval $[a,b]$.

Let $f$ be an increasing function from the interval $[a,b]$ to the non-negative numbers. Let $P=\{x_0,\cdots ,x_n\}$ be a partition of $[a,b]$ and let

$$\begin{eqnarray*} I_a^b (f,P)&=& \bigcup_{i=1}^n B(x_{i-1},x_i\colon 0,f(x_{i-1}))\\ O_a^b (f,P)&=& \bigcup_{i=1}^n B(x_{i-1},x_i\colon 0,f(x_i)). \end{eqnarray*}$$

Then

$$I_a^b (f,P)\subset S_a^bf\subset O_a^b(f,P).$$ (5.34)

To see this, observe that since $f$ is increasing

$$x_{i-1}\leq x\leq x_i\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}f(x_{i-1})\leq f(x)\leq f(x_i),$$

so

$$\begin{eqnarray*} (x,y) \in I_a^b(f,P) &\mbox{$\Longrightarrow$}& (x,y) \in B(x_... ...mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}(x,y) \in S_a^bf. \end{eqnarray*}$$

and also

$$\begin{eqnarray*} (x,y)\in S_a^b f &\mbox{$\Longrightarrow$}& x_{i-1}\leq x\leq ... ... for some } i\\ &\mbox{$\Longrightarrow$}& (x,y)\in O_a^b(f,P). \end{eqnarray*}$$

By equation (5.34) and monotonicity of area, we have

$$\alpha \Big( I_a^b(f,P)\Big)\leq\alpha (S_a^b f)\leq \alpha \Big( O_a^b(f,P)\Big).$$ (5.35)

Now

$${\alpha \Big(O_a^b(f,P)\Big)-\alpha \Big( I_a^b (f,P)\Big)}\mbox{{}}$$

$$= \sum_{i=1}^n (x_i-x_{i-1})f(x_i)-\sum_{i=1}^n (x_i-x_{i-1})f(x_{i-1})$$

$$= \sum_{i=1}^n (x_i-x_{i-1})\Big(f(x_i)-f(x_{i-1})\Big).$$ (5.36)

Now let $\mu (P)$ be the mesh of $P$ (cf. definition 5.27) so that

$$0\leq x_i-x_{i-1}\leq\mu (P) \mbox{ for } 1\leq i\leq n.$$

Since $f(x_i)-f(x_{i-1})\geq 0$ for all $i$, we have

$$(x_i-x_{i-1})\Big(f(x_i)-f(x_{i-1})\Big)\leq\mu (P)\Big( f(x_i)-f(x_{i-1})\Big)$$

for all $i$, and hence

$$\sum_{i=1}^n(x_i-x_{i-1})\Big(f(x_i)-f(x_{i-1})\Big) \leq \sum_{i=1}^n\mu (P)\Big(f(x_i)-f(x_{i-1})\Big)$$

$$= \mu (P)\sum_{i=1}^n\Big( f(x_i)-f(x_{i-1})\Big).$$ (5.37)

Now

$$\begin{eqnarray*} \sum_{i=1}^n\Big(f(x_i)-f(x_{i-1})\Big)&=&\Big(f(x_n)-f(x_{n-1... ...f(x_1)-f(x_0)\Big)\\ &=&f(x_n)-f(x_0)=f(b)-f(a).\phantom{\Big)} \end{eqnarray*}$$

so by equations (5.37) and (5.36), we have

$$\alpha\Big(O_a^b(f,P)\Big)-\alpha\Big(I_a^b(f,P)\Big)\leq\mu(P)\Big(f(b)-f(a)\Big).$$

Now suppose that $A$ is any real number that satisfies

$$\alpha(I_a^b(f,P)) \leq A \leq \alpha(O_a^b(f,P)) \mbox{ for every partition $P$ of $[a,b]$.}$$

We will show that $A = \alpha(S_a^b f).$ We have

$$-\alpha(O_a^b(f,P)) \leq -A \leq -\alpha(I_a^b(f,P)).$$

It follows from (5.35) that

$$\alpha(I_a^b(f,P))-\alpha(O_a^b(f,P)) \leq \alpha(S_a^b f) - A \leq \alpha(O_a^b(f,P))-\alpha(I_a^b(f,P)).$$

Thus

$$-\mu(P)(f(b) - f(a)) \leq \alpha(S_a^b f) - A \leq \mu(P)(f(b) - f(a))$$ (5.38)

for every partition $P$ of $[a,b]$. Since we can find partitions $P$ with $\mu (P)$ smaller than any preassigned number, it follows that

$$A = \alpha(S_a^bf).$$ (5.39)

(After we have discussed the notion of limit, we will come back and reconsider how (5.39) follows from (5.38). For the present, I will just say that the implication is intuitively clear.) We have now proved the following theorem:

5.40   Theorem. Let $f$ be an increasing function from the interval $[a,b]$ to $\mbox{{\bf R}}_{\geq 0}$, and let $P=\{x_0,x_1,\cdots ,x_n\}$ be a partition of $[a,b]$. Let

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

$$I_a^b(f,P) = \bigcup_{i=1}^n B\Big(x_{i-1},x_i\colon 0,f(x_{i-1})\Big),$$ (5.41)

$$O_a^b(f,P) = \bigcup_{i=1}^n B\Big(x_{i-1},x_i\colon 0,f(x_i)\Big),$$ (5.42)

$$A_a^b f = \alpha(S_a^bf).$$

Then

$$\alpha\Big( I_a^b(f,P)\Big)\leq A_a^b f\leq\alpha\Big( O_a^b(f,P)\Big)$$ (5.43)

and

$$\alpha\Big(O_a^b(f,P)\Big)-\alpha\Big(I_a^b(f,P)\Big)\leq\mu(P)\Big(f(b)-f(a)\Big).$$ (5.44)

Also

$$\alpha\Big( I_a^b(f,P)\Big) = \sum_{i=1}^nf(x_{i-1})(x_i-x_{i-1})$$ (5.45)

$$\alpha\Big( O_a^b(f,P)\Big) = \sum_{i=1}^nf(x_i)(x_i-x_{i-1}).$$ (5.46)

If $A$ is any real number such that

$$\alpha(I_a^b(f,P)) \leq A \leq \alpha(O_a^b(f,P)) \mbox{ for every partition $P$ of $[a,b]$},$$

then

$$A = A_a^b(f).$$

The following picture illustrates the previous theorem.

5.47   Exercise. A version of theorem 5.40 for decreasing functions is also valid. To get this version you should replace the word ``increasing" by ``decreasing" and change lines (5.41), (5.42), (5.44), (5.45) and (5.46). Write down the proper versions of the altered lines. As usual, use $I$ to denote areas inside $S_a^bf$ and $O$ to denote sets containing $S_a^bf$. Draw a picture corresponding to the above figure for a decreasing function.

5.48   Definition (Right triangle $T_{\mathbf{{\bf c}}}$) Let $a$ and $b$ be non-zero real numbers, and let $\mathbf{{\bf c}}=(a,b)$. We define the triangle $T_{\mathbf{{\bf c}}} = T_{(a,b)}$ to be the set of points between the line segment $[{\bf0}\mathbf{{\bf c}}]$ and the $x$-axis. By example 4.8, we know that the equation of the line through ${\bf0}$ and $\mathbf{{\bf c}}$ is $y= {b\over a}x$. Hence we have:

If $a>0$ and $b>0$, then $T_{(a,b)} = \{(x,y): 0 \leq x \leq a \mbox{ and }0 \leq y \leq {b\over a}x \}$

If $a>0$ and $b<0$, then $T_{(a,b)} = \{(x,y): 0 \leq x \leq a \mbox{ and }{b\over a}x \leq y \leq 0\}$

If $a<0$ and $b<0$, then $T_{(a,b)} = \{(x,y): a \leq x \leq 0 \mbox{ and }{b\over a}x \leq y \leq 0\}$

If $a<0$ and $b>0$, then $T_{(a,b)} = \{(x,y): a \leq x \leq 0 \mbox{ and }0 \leq y \leq {b\over a}x \}$

5.49   Remark. We know from Euclidean geometry that

$$\alpha(T_{(a,b)}) = {1\over 2} \vert a\vert\vert b\vert.$$ (5.50)

I would like to show that this relation follows from our assumptions about area. If $H$, $V$ and $R_\pi$ are the reflections and rotation defined in definition 4.9, then we can show without difficulty that for $a>0$ and $b>0$

$$\begin{eqnarray*} T_{(-a,b)} &=& H(T{(a,b)}),\\ T_{(a,-b)} &=& V(T_{(a,b)}), \mbox{ and }\\ T_{(-a,-b)} &=& R_\pi(T_{(a,b)}) \end{eqnarray*}$$

so by invariance of area under symmetry,

$$\alpha(T_{(a,b)}) = \alpha(T_{(-a,b)}) =\alpha(T_{(a,-b)}) = \alpha(T_{(-a,-b)})$$

when $a$ and $b$ are positive. It follows that if we prove formula (5.50) when $a$ and $b$ are positive, then the formula holds in all cases. For example if $a$ and $b$ are positive, and we know that (5.50) holds when $a$ and $b$ are positive, we get

$$\alpha(T_{(-a,b)}) =\alpha(T_{(a,b)}) = {1\over 2}\vert a\vert\vert b\vert = {1\over 2}\vert-a\vert\vert b\vert,$$

and thus our formula holds when $a$ is negative and $b$ is positive.

5.51   Theorem. Let $a$ and $b$ be non-zero real numbers, and let $T_{(a,b)}$ be the set defined in definition 5.48. Then

$$\alpha(T_{(a,b)}) = {1\over 2} \vert a\vert\vert b\vert.$$

Proof: By the previous remark, if is sufficient to prove the theorem for the case when $a$ and $b$ are positive. So suppose that $a$ and $b$ are positive.

Let $E = (a,b) + R_\pi(T_{(a,b)})$. It appears from the figure, and is straightforward to show, that

$$E = \{(x,y):0\leq x\leq a \mbox{ and }{b\over a}x \leq y \leq b\}.$$

By translation invariance of area,

$$\alpha(E) = \alpha(R_\pi(T_{(a,b)})) = \alpha(T_{(a,b)}).$$

We have

$$E \cup T_{(a,b)} = B(0,a:0,b),$$

and

$$E \cap T_{(a,b)} = [{\bf0}\mathbf{{\bf c}}] \mbox{ where } \mathbf{{\bf c}}= (a,b).$$

By the addition rule for area (assumption 5.14) we have

$$\begin{eqnarray*} ab &=& \alpha(B(0,a:0,b))\\ &=& \alpha( E \cup T_{(a,b)} ) ... ...)})\\ &=& 2\alpha(T_{(a,b)}) -\alpha([{\bf0}\mathbf{{\bf c}}]), \end{eqnarray*}$$

i.e.,

$$\alpha(T_{(a,b)}) = {1\over 2}ab +{1\over 2}\alpha([{\bf0}\mathbf{{\bf c}}]).$$

Thus our theorem will follow if we can show that the segment $[{\bf0}\mathbf{{\bf c}}]$ is a zero-area set. We will prove this as the next theorem.

5.52   Theorem. Let $\mathbf{{\bf c}}=(a,b)$ be a point in $\mbox{{\bf R}}^2$. Then

$$\alpha([\bf0\mathbf{{\bf c}}]) = 0.$$

Proof: If $a=0$ or $b=0$, then ${[{\bf0}\mathbf{{\bf c}}]}$ is a box with width equal to zero, or height equal to zero, so the theorem holds in this case. Hence we only need to consider the case where $a$ and $b$ are non-zero. Since any segment $[{\bf0}\mathbf{{\bf c}}]$ can be rotated or reflected to a segment $[{\bf0}\mbox{{\bf q}}]$ where $\mbox{{\bf q}}$ is in the first quadrant, we may further assume that $a$ and $b$ are both positive. Let $n$ be a positive integer, and for $1\leq j\leq n$ let

$$B^n_j = B\left({a(j-1)\over n},{aj\over n}: {b(j-1)\over n},{bj\over n}\right).$$

Then

$${[}{\bf0}\mathbf{{\bf c}}] \subset \bigcup_{j=1}^n B^n_j,$$ (5.53)

since

$$\begin{eqnarray*} \mbox{{\bf x}}\in [{\bf0}\mathbf{{\bf c}}] &\mbox{$\Longrighta... ...box{{\bf x}}\in B^n_j \mbox{ for some $j$ with } 1\leq j\leq n. \end{eqnarray*}$$

For each $j$ we have

$$\alpha(B^n_j) = {a\over n}\cdot{b\over n} = {ab\over n^2}.$$

Also the sets $B^n_j$ and $B^n_k$ are almost disjoint whenever $1 \leq j,k \leq n$ and $j\neq k$. (If $j$ and $k$ differ by more than 1, then $B^n_j$ and $B^n_k$ are disjoint, and if $j$ and $k$ differ by 1, then $B^n_j\cap B^n_k$ consists of a single point.) By additivity for almost-disjoint sets (assumption 5.25), it follows that

$$\alpha(\bigcup_{j=1}^n B^n_j) = \sum_{j=1}^n\alpha(B^n_j) =\sum_{j=1}^n {ab\over n^2} = {nab\over n^2} ={ab\over n}.$$

By (5.53) and monotonicity of area we have

$$\alpha([0\mathbf{{\bf c}}]) \leq \alpha(\bigcup_{j=1}^n B^n_j) = {ab\over n} \mbox{ for every positive integer } n.$$ (5.54)

In order to conclude from this that $\alpha([{\bf0}\mathbf{{\bf c}}]) = 0$ We now make use of the Archimedean property of the real numbers, which says that for any real number $x$ there is a positive integer $n$ with $n> x$. We know $\alpha([{\bf0}\mathbf{{\bf c}}])\geq 0$, since all areas are non-negative. Suppose (in order to get a contradiction) that $\alpha([{\bf0}\mathbf{{\bf c}}])$ is positive. Then by the Archimedean property, there is a positive integer $N$ such that $N > {ab\over \alpha([{\bf0}\mathbf{{\bf c}}])}$. This implies that $\alpha([{\bf0}\mathbf{{\bf c}}]) > {ab\over N}$, and this contradicts (5.54). Hence $\alpha([{\bf0}\mathbf{{\bf c}}])$ is not positive, and we conclude that $\alpha([{\bf0}\mathbf{{\bf c}}]) = 0$. $\diamondsuit$

Archimedes' statement of the Archimedean property differs from our statement. He assumes that

Further, of unequal lines, unequal surfaces, and unequal solids, the greater exceeds the less by such a magnitude as, when added to itself, can be made to exceed any assigned magnitude among those which are comparable with [it and with] one another.[2, page 4]

5.55   Exercise. Let $\mbox{{\bf a}}$ and $\mbox{{\bf b}}$ be points in $\mbox{{\bf R}}^2$. Show that segment $[\mbox{{\bf a}}\mbox{{\bf b}}]$ is a zero are set. (Use theorem 5.52. Do not reprove theorem 5.52).

5.56   Entertainment (Area of a triangle) Let $\mbox{{\bf x}}_1 = (x_1,y_1)$, $\mbox{{\bf x}}_2 = (x_2,y_2)$ and $\mbox{{\bf x}}_3 = (x_3,y_3)$ be three points in $\mbox{{\bf R}}^2$, and let $T$ be the triangle with vertices $\mbox{{\bf x}}_1$, $\mbox{{\bf x}}_2$ and $\mbox{{\bf x}}_3$. Let

$$\begin{eqnarray*} x_s &=& \mbox{smallest of $x_1$, $x_2$ and $x_3$}\\ x_l &=& ... ...and $y_3$}\\ y_l &=& \mbox{largest of $y_1$, $y_2$ and $y_3$}. \end{eqnarray*}$$

Then the box $B(x_s,x_l:y_s,y_l)$ is an almost-disjoint union of $T$ and three triangles which are translates of triangles of the form $T_{\mathbf{{\bf c}}}$. Since you know how to find the area of a box and of a triangle $T_{\mathbf{{\bf c}}}$, you can find the area of $T$.

Using this remark show that for the triangles pictured below, $\alpha(T^1) = {1\over 2} (a_1b_2- a_2 b_1)$, and $\alpha(T^2) ={1\over 2} (a_2b_1-a_1b_2)$.

Then choose another triangle $T^3$ with vertices ${\bf0},\mbox{{\bf a}}$ and $\mbox{{\bf b}}$, where the coordinates of the points are related in a way different from the ways shown for $T^1$ and $T^2$, and calculate the area of $T^3$. You should find that

$$\alpha(T^3) = {1\over 2} \vert a_1b_2-a_2b_1\vert$$

in all cases. Notice that if some coordinate is zero, the formula agrees with theorem 5.51.