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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
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
|
(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
Then
|
(5.43) |
and
|
(5.44) |
Also
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.
5.49
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