# Integrals

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.

## Integration Over a Rectangle

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 of for the partition are defined to be

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.

## Extending the Integral: Characteristic Functions

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.

## Extending the Integral: Partitions of Unity

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:

1. 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 .
2. 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 ).

1. If is another partition of unity subordinate to an admissible cover of , then is convergent and

2. If and are bounded and the set of discontinuities of is a set of measure zero, then is integrable in the extended sense.
3. 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 has

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 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: