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# 5.3 Monotonic Functions

5.27   Definition (Partition.) Let be real numbers with . A partition of the interval is a finite sequence of points

with . The intervals , are called the subintervals of the partition , and is the subinterval of for . The largest of the numbers is called the mesh of the partition , and is denoted by . The partition

is called the regular partition of into equal subintervals.

5.28   Example. Let

Then is a partition of into subintervals and .

The regular partition of into subintervals is .

If is the regular partition of into equal subintervals, then .

5.29   Exercise. Find a partition of into five subintervals, such that , or explain why no such partition exists.

5.30   Exercise. Find a partition of into five subintervals, such that , or explain why no such partition exists.

5.31   Definition (Monotonic function.) Let be an interval in , and let be a function. We say that is increasing on if
 (5.32)

and we say that is decreasing on if

We say that is strictly increasing on if

and we say that is strictly decreasing on if

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

A constant function on is both increasing and decreasing on .

5.33   Notation () Let be a function from the interval to the non-negative numbers. We will write

i.e., is the set of points under the graph of over the interval .

Let be an increasing function from the interval to the non-negative numbers. Let be a partition of and let

Then

 (5.34)

To see this, observe that since is increasing

so

and also

By equation (5.34) and monotonicity of area, we have
 (5.35)

Now
 (5.36)

Now let be the mesh of (cf. definition 5.27) so that

Since for all , we have

for all , and hence
 (5.37)

Now

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

Now suppose that is any real number that satisfies

We will show that We have

It follows from (5.35) that

Thus
 (5.38)

for every partition of . Since we can find partitions with smaller than any preassigned number, it follows that
 (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 be an increasing function from the interval to , and let be a partition of . Let

 (5.41) (5.42)

Then
 (5.43)

and
 (5.44)

Also
 (5.45) (5.46)

If is any real number such that

then

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 to denote areas inside and to denote sets containing . Draw a picture corresponding to the above figure for a decreasing function.

5.48   Definition (Right triangle ) Let and be non-zero real numbers, and let . We define the triangle to be the set of points between the line segment and the -axis. By example 4.8, we know that the equation of the line through and is . Hence we have:

If and , then

If and , then

If and , then

If and , then

5.49   Remark. We know from Euclidean geometry that
 (5.50)

I would like to show that this relation follows from our assumptions about area. If , and are the reflections and rotation defined in definition 4.9, then we can show without difficulty that for and

so by invariance of area under symmetry,

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

and thus our formula holds when is negative and is positive.

5.51   Theorem. Let and be non-zero real numbers, and let be the set defined in definition 5.48. Then

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

Let . It appears from the figure, and is straightforward to show, that

By translation invariance of area,

We have

and

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

i.e.,

Thus our theorem will follow if we can show that the segment is a zero-area set. We will prove this as the next theorem.

5.52   Theorem. Let be a point in . Then

Proof: If or , then 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 and are non-zero. Since any segment can be rotated or reflected to a segment where is in the first quadrant, we may further assume that and are both positive. Let be a positive integer, and for let

Then
 (5.53)

since

For each we have

Also the sets and are almost disjoint whenever and . (If and differ by more than 1, then and are disjoint, and if and differ by 1, then consists of a single point.) By additivity for almost-disjoint sets (assumption 5.25), it follows that

By (5.53) and monotonicity of area we have
 (5.54)

In order to conclude from this that We now make use of the Archimedean property of the real numbers, which says that for any real number there is a positive integer with . We know , since all areas are non-negative. Suppose (in order to get a contradiction) that is positive. Then by the Archimedean property, there is a positive integer such that . This implies that , and this contradicts (5.54). Hence is not positive, and we conclude that .

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 and be points in . Show that segment is a zero are set. (Use theorem 5.52. Do not reprove theorem 5.52).

5.56   Entertainment (Area of a triangle) Let , and be three points in , and let be the triangle with vertices , and . Let

Then the box is an almost-disjoint union of and three triangles which are translates of triangles of the form . Since you know how to find the area of a box and of a triangle , you can find the area of .

Using this remark show that for the triangles pictured below, , and .

Then choose another triangle with vertices and , where the coordinates of the points are related in a way different from the ways shown for and , and calculate the area of . You should find that

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

Next: 5.4 Logarithms. Up: 5. Area Previous: 5.2 Further Assumptions About   Index
Ray Mayer 2007-09-07