with . The intervals , are called the

is called the

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 .

(5.32) |

We say that is

and we say that is

We say that is

A constant function on is both increasing and
decreasing on .

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

To see this, observe that since is increasing

so

and also

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

Now

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

Since for all , we have

for all , and hence

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

for every partition of . Since we can find partitions with smaller than any preassigned number, it follows that

(After we have discussed the notion of

Then

and

Also

If is any real number such that

then

If and , then

If and , then

If and , then

If and , then

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.

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.

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.

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

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

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]

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 .

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