In order to simplify the presentation, we will develop integration using the Riemann (or Jordan) approach rather than the more general theory of Lebesgue integration.
Definition 1: (i) A partition of an interval
is a sequence
where
.
A partition of a rectangle
is an n-tuple
where
is a partition of
for every
. A second partition
of the same rectangle is
called a refinement of the first partition
if the set
is a subset
of the set
for each
.
(ii) If
for each
, then the partition P defines
the set of subrectangles of the partition
made up of all the
rectangles of the form
.
The volume
of the rectangle
is defined to be
(iii) If, in addition,
is a bounded function defined
on the rectangle
, then one defines for each subrectangle of
the functions:
Further, the lower and
The definitions have been constructed to make the following trivial, but
crucial Lemma true.
Lemma 1: With the notation as in Definition 1, every subrectangle
of
Definition 2: With the notation as in Definition 1, the lower
integral (respectively upper integral of
Proposition 1: A bounded function
Proof: The condition is clearly sufficient. On the other hand,
if
Let
Theorem 1: (Fubini's Theorem) Let
Then
In other words, the double integral can be calculated as either of
two iterated integrals:
Proof: This is mostly a matter of sorting out all the definitions.
If
One has:
For
But then:
and so, combining results, we get:
For the upper sums, one can argue analogously:
For
But then:
and so, combining results, we get:
Combining the results for upper and lower sums:
Since
and so
A similar argument shows that
and so one gets
Note: By interchanging the roles of
Exercise 1: State and prove a result which would make the following
definition legitimate and reasonable.
Definition 3: If
If
Definition 4: A subset
Exercise 2: Show that, if we replace open with closed
in Definition 4, the resulting definitions are equivalent.
Proposition 2: (i) The union of a countable collection of sets of
measure zero is also of measure zero.
(ii) A compact set is of measure zero if and only if it is of content zero.
(iii) A closed interval
Proof: (i) Suppose
(ii> Clearly, the condition is sufficent. On the other hand, if
(iii) If
Definition 5: Let
The oscillation
Proposition 3:(i) The bounded function
(ii) If
(iii) If
Proof: (i) Do this as an exercise.
(ii) If
Suppose
(iii) For each
Theorem 2: Let
Proof: Suppose
Now, suppose that the set of discontinuities of
Corollary 1: Let
Proof: Apply Theorem 2 given that the set of discontinuities of
Definition 6: A bounded set
Note: Although our extension was defined in a natural way, it
is not quite adequate. Indeed, we have seen that there are open
subsets of
The goal of this section is to define an integral for functions defined
on open sets. This is based on a technical result:
Theorem 3: (Partitions of Unity) Let
Definition 7: A set
Lemma 1: (i) If
(ii) If
Proof: This is just Exercises 1-22 and 2-26.
Proof: (of Theorem 3) Case 1: First, consider the case in which
Claim: There are compact sets
Proof: We make an inductive definition. Suppose, we have
already chosen
Then
By Lemma 1 (ii), there are
Now, choose a
Case 2: Now, let's prove the result in case
Case 3: Suppose now that
Definition 8: An open cover
Theorem 4: Let
Proof: (i) For
In particular, the right hand side converges. Absolute convergence implies
that we can re-arrange the series on the right as
A similar argument shows
that the same identities hold when the absolute values are removed.
(ii) Suppose
(iii) If
for the partition
are defined to be
is a union of finitely many subrectangles of
whose interiors are
pairwise disjoint. As a result, one has:
on
is defined
to be
(respectively
). If
,
then
is called integrable with integral
equal to this
common value. Another notation for the integral is
defined on
a rectangle
is integrable if and only if for every
,
there is a partition
such that
is integrable and
, then there are partitions
and
such that
be a refinement of both
and
. Then Lemma 1 implies that
is a partition of the desired type.
and
be closed rectangles and
be integrable. For
, let
be defined by
and let
and
be defined by:
and
are both integrable on
and
(respectively
) is a partition of
(respectively
), then
is a partition of
with subrectangles of the
form
where
is a subrectangle of
and
is a
subrectangle of
.
, one has
, and so
, one has
, and so
is integrable,
is integrable with integral equal to
is integrable with integral
,
as desired.
and
, one can show that
the order in which the integrals are iterated does not affect the result.
Extending the Integral: Characteristic Functions
, then its characteristic
function
is defined by
where
is a closed rectangle, then a bounded function
is said to be integrable with integral
provided that this last quantity is defined.
of
is said to be of
measure zero (respectively content zero) if for every
,
there is a countable infinite (respectively finite) sequence
of open rectangles which form a cover of
and such that
with
is not of measure zero.
is a countable sequence of
sets of measure zero and
. Then
has an open cover by
rectangles
such that
Then clearly the
are a cover of the union of
and the sum of
their volumes is less than
(why?). To enumerate the
, just
list them out in order of increasing
keeping those with the same value of
in order of increasing
.
is compact and of measure 0, with an open covering
of open
rectangles of total volume less than
. Then any finite subcover
satisfies the same condition. So
is also of content zero.
were of measure zero, it would be of content zero by
part (ii). Suppose
has a cover
by closed rectangles of
total length less than
. We can replace the
with their
intersections with
to show that the
can be assumed to be
subintervals of
. But then the endpoints of the
can be arranged
into increasing order to obtain a partition of
. Each
is
a union of certain of the subrectangles of this partition and so
So, if
, the interval
cannot be of content zero and so it is also not of measure zero.
be a bounded function and
.
Define the functions:
of
at
is defined to be
is continuous at
if
and only if
.
is closed,
is
bounded, and
, then
is closed.
is a bounded function defined on a closed
rectangle
and
for some
, then
there is a partition
of
such that
is not in
, then there is an open an open ball of radius
about
which is disjoint from
. Using this ball, one can check that
So
.
and
. Then
. Choose a
such that
. Then for all
such that
, we have
.
and so
. Thus the complement of
is open and
is
closed.
there is a closed rectangle with
in its interior
such that
Since
is compact, a
finite number of these are such that their interiors cover
. Choose for
any partition of
such that every subrectangle
of
is contained in one
of these finitely many
's. Then
for each
subrectangle
. It follows that
be a bounded function defined
on a closed rectangle
and let
Then
is integrable if and only if
is a set of measure zero.
is integrable and
. For
,
choose a partition
of
such that
.
Then the set of subrectangles
of
such that
has total volume less than
If
is a point where the oscillation of
is
at least
and if
is in the interior of one of the subrectangles
of
, then
. Since the boundaries of the
subrectangles of
is a set of measure zero, it follows that the set
of
with
oscillation at least
must be a set of content (and hence measure) zero.
Since the set of discontuities of the function
is just the union of the
countably many sets
for
, it follows
that the set of discontinuities of
is also of measure zero.
is of measure 0.
As before, let
be the set of
satisfying
Since
is compact by Proposition 3 and
is of
measure zero,
is of content zero. In particular, there is
a finite set of closed rectangles whose interiors cover
and such that their
total volume is at most
. First choose a partition
of
such
that every subrectangle of
is either completely contained with one of
the finitely many rectangles or else its interior is disjoint from all of these
rectangles. For each rectangle of the first type, one has
where
is an upper bound for |f(x)| for all
.
For each subrectangle S of the second type, one can use Proposition 3(iii)
to find a partition of the subrectangle such that the difference between
the upper and lower sums on this subrectangle is at most
.
Now replace P with a refinement of each of these partitions. Then we
have
.
By making
sufficiently small, we can make this sum as small as we like, and so
is integrable.
be a bounded set. Then
its characteristic function
is
integrable if and only if the boundary of
is of measure zero.
is precisely the boundary of
.
is called
Jordan measurable if its boundary is of measure zero. In this
case, the integral of its characteristic function is called
the content of
or its (n-dimensional) volume.
whose boundaries are not of measure zero. So even
constant functions on open sets might not be integrable. This is why we
need the next section.
Extending the Integral: Partitions of Unity
be an open cover
of a subset
There is a set
of
function
defined in an open set containing
and satifying:
, one has
and
where the sum is defined because
for every such
there is an open set
containing
where there
are but finitely many
such that
.
, there is a
such that
outside of some compact set contained in
.
of functions satisfying condition (1)
of Theorem 3 is called a
partition of unity of
. When
both conditions are satisfied, we call
a partition of unity of
subordinate to the cover
is a compact subset of an open set
. Then there is a compact subset
of
such that
is contained in the
interior of
.
is a compact subset of an open set
, then there is a
function
such that
on all of
and such
outside of a compact subset of
. In fact,
can
be chosen so it maps into
and such that
for all
.
is compact.
Replace
with a finite subcover of
, say
.
whose interiors cover
and such that
for every
such that the union of the their interiors and
the sets
contains
. Let
is a compact subset of
By Lemma 1 (i), there
is a compact set
contained in
and containing
in
its interior. This completes the induction.
functions
which are
positive on
and which are zero outside of some compact subset of
.
Then
for
in an open set
containing
. Define functions
by
function
such that
for
and such that
outside of a compact set contained in
. Then the functions
is the desired partition
of unity.
where
each
is compact and contained in the interior of
. To see
this, note that the compact set
has an open cover
So, our previous case shows that there is a partition of unity
for
subordinate to
Now
is a finite sum in some open set containing
and so we can take as our
partition of unity the set of all the
for all
for all
.
is an open set. This case follows
from the previous one by letting
be the set of
such that
and the distance from
to the boundary of
is at least
. The general case is now evident; one can simply replace the set
with the union of all the open subsets
in
.
of a set
is called admissible if each element of
is contained in
. Let
be a function and
be admissible for
(and so
is open). Then
is said to be
integrable in the extended sense with integral
provided that the terms of
the series are defined and
is convergent.
be a partition of unity subordinate to an
admissible cover
of an
open set
and
be a function integrable in the
extended sense (as defined using this this particular
).
is another partition of unity subordinate to an admissible
cover
of
, then
is convergent and
and
are bounded and the set of discontinuities of
is a set of measure zero, then
is integrable in the extended sense.
is Jordan-measurable and
is bounded, then the extended
definition of
agrees with the earlier definition of
.
,
except on
a compact set
dependent on
. For each
, there is an
open set containing
on which there are at most finitely many
which are non-zero at some point in the open set. These open sets cover
and so there is a finite subcover; hence there are only finitely many
whose restrictions to
are not identically zero. One has
for some closed rectangle
and that
for some
and all
. Then for any finite
subset
, one has
is Jordan-measurable and
, then there is
a compact Jordan-measurable set
such that
.
For any finite set
containing all the
which
are not identically zero on
, one has:
