Graphics Reference
In-Depth Information
Having defined the integral, the next obvious task is to determine when it exists.
One can show that continuous functions are integrable, but there is a more general
condition. We shall not take the time to define the general concept of the measure
(“volume”) of a set. All we will need here is the following:
Definition.
A subset
X
of
R
n
has
measure zero
if, for every e>0, it can be covered
by a sequence of n-rectangles
A
i
so that
•
Â
()
<
vol
e
.
i
i
=
1
It is easy to see that every finite set has measure zero. But countable sets {
p
i
}
i=1
(like the rational numbers) also have measure zero because the ith one can be con-
tained in a small rectangle of volume e/2
i
and these volumes sum to e. The boundary
of “nice” sets, such as rectangles or closed disks, have measure zero. More generally,
“k-dimensional” subsets of
R
n
have measure zero when k < n. For example, Figure
4.22 shows why a segment is a set of measure zero in
R
2
. We can cover the segment
[(a,c),(b,d)] with n rectangles that have width (b - a)/n and height (d - c)/n. The total
area of these rectangles is
(
)
(
)
n
ba
n
--
=
dc
n
badc
n
-
-
and we can clearly make this area as small as desired by increasing n.
4.8.1. Theorem.
Let
A
be rectangle in
R
n
. A bounded function f :
A
Æ
R
is integrable
if and only if the set
{
a
Œ
A
| f is discontinuous at
a
}
has measure zero.
Proof.
See [Spiv65] or [Buck78].
Theorem 4.8.1 is a good result for functions defined on rectangles, but what if the
domain of a function is not a rectangle. Let
X
be a bounded subset of
R
n
and f :
X
Æ
R
. Define F :
R
n
Æ
R
by
y
(b,d)
d
b-a
n
d-c
n
c
(a,c)
Figure 4.22.
A segment is a set of measure zero
in the plane.
x
a
b
