Suppose , so is continuous on and it support,* *

* *is compact. Recall that we defined

with large enough that . Our main goal in these notes is to prove the change of variables formula:

Theorem 1Let be a one-to-one map such that for all . If has compact support with then

A PDF version of this lecture is available here.

** Open Mapping Theorem **

In proving the change of variables formula it will be useful to know that is in fact an *open set*. This is, in fact, a simple consequence of the inverse function theorem:

Thm 2 (Open Mapping Theorem)Let where is open and suppose for all . Then is open.

Remark 3In the applications below, will be one-to-one, however that is not required for the open mapping theorem.

*Proof:* Let . Suppose . Since is invertible, by the inverse function theorem, there are open sets such that , , and is a one-to-one map of onto . In particular we see that . Thus is open.

** Special cases of the Change of Variables Formula **

For some special types of maps the for which (1) is easily seen to hold. For example, if is a *translation *

* *where ** **is a fixed vector in then so . Since our definition of the integral implies

we see that (1) holds in this case.

Another important special case of(1) follows from the fact that the integral does not depend on the order in which we integrate the variables. Recall that a *permutation map* is a linear map such that for some permutation of .

Lemma 4Let be a permuation map then (1) holds.

*Proof:* Let . In a previous lecture we proved that

for any -cell. If then so with large enough we see that

Since is linear for each and so the proof will be complete once we show that . Thus the proof is completed by the following

Exercise 5Let be a permutation map. Prove that .

Another special case for which we know 1 is in 1D:

Lemma 6Suppose and for all where is open. If and , then is finite for each andwhere , the number of points mapped to by . In particular, if is one-to-one then the change of variables formula (1) holds for any with .

*Proof:* Since is an open subset of it is a countable or finite union of disjoint open intervals,

Note that . Furthermore, is one-to-one on and the image is an open interval for each (because for all ). Since is compact, there are finitely many intervals such that cover . Hence, without loss of generality we may assume that with and pairwise disjoint. Note that for each .

Because is compact it is closed. Thus for each there is a closed interval such that . Thus

Fix . Since on , there are two cases:

- for all . Then on and , so
- for all . Then on and so

In either case,

Since , this identity and (4) imply (3).

Remark 7If is an interval with , the integral is equal to . This integral is “unoriented” — it does not depend on the “orientation” of the interval and is positive whenever is. By contrast the usual notation refers to an “oriented” integral: if it is defined to be . In 1D this is rather trivial, but such distinctions will be important when we look at integration of differential forms.

The 1D change of variables formula rather directly shows a change of variables formula for a special class of maps on .

Definition 8A function with open in is called aprimitive mapif there is and such thatHere denotes the -th elementary basis vector and is the -th coordinate of . Thus a primitive map alters only one coordinate of .

Lemma 9Let be a one-to-one primitive map with non-singular derivative and let have support in . Then (1) holds.

*Proof:* First, suppose we have obtained (1) whenever is a primitive mapping that alters only the -th coordinate. Given a primitive mapping that alters only the -th coordinate with , let be the permutation map with the permutation that interchanges and but leaves all other indices unchanged. Note that is a primitive map that alters only the -th coordinate. Since

we see that

where we have used the permutation invariance of the integral and (1) for .

Thus we may assume without loss of generality that

What are the partial derivatives of ? Clearly, for we have and for . On the other hand

for each . It follows that

Since we see that for all .

Now fix and consider the following continuous functions of a single variable

and

Here is defined for any and has compact support (because does). However, is defined on the set

which is an open set in by virtue of the fact that is open in . Note that

and

Since and hence is one-to-one, we have

by the 1D change of variables formula (3). Integrating over the remaining variables yields (1).

** The primitive map factorization theorem, partitions of unity and the general change of variables formula **

The idea of the proof of Theorem 1 is to use what we know so far — that the identity holds for translations, permutations and primitive maps — to obtain the result for general maps . The key behind this proof is that *the class of maps for which (1) holds is closed under composition*:

Lemma 10Let and be one-to-one maps such that, for , for all and . If, for , (1) holds with for all with compact support in then (1) holds also with for all with compact support in .

*Proof:* Let and let have compact support in . By the chain rule

so we have

So the question is: *Can every one-to-one map with non-vanishing Jacobian be written as the composition of flips and primitive mappings? *The correct but limited answer is: “No.” A more useful observation is, “Not quite, but almost.” Namely, we can do this *locally.*

Thm 11 (Primitive map factorization theorem)Let with open in . If for some then there is an open set containing such that for , where and are translations, and , is a permutation map, and for each , is a one-to-one primitive map defined on an open set . Furthermore, the number of primitive maps is less than or equal to .

Remark 12It is implied by (5) thatthe domains satisfy and for .

Let us defer the proof of Theorem 11. Suppose we fix with throughout it’s domain . If we are lucky enough to have a continuous function with compact support in a neighborhood on which (5) holds then Lemma 10 immediately implies that (1) holds. However, what do we do if we are not so lucky? The answer is to use a *partition of unity*:

Thm 13 (Partition of Unity Theorem)Suppose is compact and is an open cover of . Then there exist functions such thatand, for each ,

- , and
- there is such that .

Remark 14We call the functions apartition of unity on subordinate to the open cover .We won’t need it here, but it is possible to take the functions for any or even in !

Partitions of unity are a very important tool throughout analysis for *localizing *proofs. An example of this is the following proof of the Change of Variables Theorem: *Proof:* .} Let and be as in the statement of the theorem and let . For each let be an open set on which the factorization (5) holds and such that . Let be a partition of unity on subordinate to the open cover — note that this *is *an open cover by the open mapping theorem. Then

However, since and the factorization (5) holds on we see by Lemma 10 that

Summing over , we see that 1 holds!

It remains to prove the existence of a partition of unity (Theorem 13) and the primitive map factorization theorem (Theorem 11). We start with the partition of unity: *Proof:* } For each there is such that . Since are open, then we can find radii such that

Since is an open cover of there are finitely many points such that

For each , I claim that we can find such that

- for all ,
- if ,
- if .

The functions can be defined recursively by taking and

for . Clearly . Suppose we have proved that

from which it follows that . Since we see by induction that (8) holds for each and thus, by (7), that for each . Finally for we have for some . Thus by (9)

from which it follows that .

Exercise 15Modify the definition (6) so as to produce a partion of unity with

** Proof of the Primitive Map Factorization Theorem **

It remains to prove Theorem 11: *Proof:* First suppose is a map as in the statement of the Theorem and let and . Then

Thus it suffices to prove (5) with but under the additional assumption that and . That is, it suffices to prove

Claim 16For each , wherethere is an open neigheborhood of such that for we have

where is a permutation map, for each , is a one-to-one primitive map defined on a neighborhood of and satisfying , and the number of primitive maps .

Let denote the zero map and for each let be the projection onto the first coordinates in ,

Note that . Now define for each the following class of maps

Clearly . We prove the claim by using backwards induction on to show that any can be factored as in (10) with the number of primitive maps . The base case is trivial since the only map in d is the identity , which is a permutation map. Likewise the case is trivial since any map is primitive.

Now suppose we have verified for some that (10) holds with primitive maps for any . Let , so

for some functions . Thus

Since we must have for some . Let be the permutation map that interchanges and , taking if , and let

So is primitive, and, because , is invertible. Thus by the inverse function theorem there is an open neighborhood of such that is a one-to-one map of onto a neighborhood of on which is continuosly differentiable. Consider the function . Clearly , and (by the chain rule), so . Furthermore

Thus for all and we see that . By the induction hypothesis, factors as

with a permutation map and primitive maps with . Thus for

where is a permutation map and the number of primitive maps satisfies