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
is a union of finitely many subrectangles of
whose interiors are
pairwise disjoint. As a result, one has:

**Definition 2:** With the notation as in Definition 1, the **lower
integral** (respectively **upper integral** of
on
is defined
to be
(respectively
). If
,
then
is called **integrable** with integral
equal to this
common value. Another notation for the integral is

**Proposition 1:** A bounded function
defined on
a rectangle
is integrable if and only if for every
,
there is a partition
such that

**Proof:** The condition is clearly sufficient. On the other hand,
if
is integrable and
, then there are partitions
and
such that

Let be a refinement of both and . Then Lemma 1 implies that is a partition of the desired type.

**Theorem 1:** (Fubini's Theorem) Let
and
be closed rectangles and
be integrable. For
, let
be defined by
and let
and
be defined by:

Then and are both integrable on and

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
(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:

For , one has , and so

But then:

and so, combining results, we get:

For the upper sums, one can argue analogously:

For , one has , and so

But then:

and so, combining results, we get:

Combining the results for upper and lower sums:

Since is integrable,

and so is integrable with integral equal to

A similar argument shows that

and so one gets is integrable with integral , as desired.

**Note:** By interchanging the roles of
and
, one can show that
the order in which the integrals are iterated does not affect the result.

**Exercise 1:** State and prove a result which would make the following
definition legitimate and reasonable.

**Definition 3:** If
, then its **characteristic
function**
is defined by

If where is a closed rectangle, then a bounded function is said to be integrable with integral provided that this last quantity is defined.

**Definition 4:** A subset
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

**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 with is not of measure zero.

**Proof:** (i) Suppose
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
.

(ii> Clearly, the condition is sufficent. On the other hand, if 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.

(iii) If 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.

**Definition 5:** Let
be a bounded function and
.
Define the functions:

The **oscillation**
of
at
is defined to be

**Proposition 3:**(i) The bounded function
is continuous at
if
and only if
.

(ii) If is closed, is bounded, and , then is closed.

(iii) If is a bounded function defined on a closed rectangle and for some , then there is a partition of such that

**Proof:** (i) Do this as an exercise.

(ii) If 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 .

Suppose and . Then . Choose a such that . Then for all such that , we have . and so . Thus the complement of is open and is closed.

(iii) For each 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

**Theorem 2:** Let
be a bounded function defined
on a closed rectangle
and let
Then
is integrable if and only if
is a set of measure zero.

**Proof:** Suppose
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.

Now, suppose that the set of discontinuities of 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.

**Corollary 1:** Let
be a bounded set. Then
its characteristic function
is
integrable if and only if the boundary of
is of measure zero.

**Proof:** Apply Theorem 2 given that the set of discontinuities of
is precisely the boundary of
.

**Definition 6:** A bounded set
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**.

**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
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.

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
be an open cover
of a subset
There is a set
of
function
defined in an open set containing
and satifying:

- For each , 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 .
- For each , there is a such that outside of some compact set contained in .

**Definition 7:** A set
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

**Lemma 1:** (i) If
is a compact subset of an open set
. Then there is a compact subset
of
such that
is contained in the
interior of
.

(ii) If 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 .

**Proof:** This is just Exercises 1-22 and 2-26.

**Proof:** (of Theorem 3) **Case 1:** First, consider the case in which
is compact.
Replace
with a finite subcover of
, say
.

**Claim:** There are compact sets
whose interiors cover
and such that
for every

**Proof:** We make an inductive definition. Suppose, we have
already chosen
such that the union of the their interiors and
the sets
contains
. Let

Then 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.

By Lemma 1 (ii), there are 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

Now, choose a function such that for and such that outside of a compact set contained in . Then the functions is the desired partition of unity.

**Case 2:** Now, let's prove the result in case
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
.

**Case 3:** Suppose now that
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
.

**Definition 8:** An open cover
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.

**Theorem 4:** Let
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
).

- If
is another partition of unity subordinate to an admissible
cover
of
, then
is convergent and
- If and are bounded and the set of discontinuities of is a set of measure zero, then is integrable in the extended sense.
- If
is Jordan-measurable and
is bounded, then the extended
definition of
agrees with the earlier definition of
.
**Proof:**(i) For , 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 hasIn 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 for some closed rectangle and that for some and all . Then for any finite subset , one has

(iii) If 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: