We now turn to defining *integration* over domains in . We will take a very practical approach to integration at first. Rather than trying to make a general theory, we will define the integral as an *interated integral*. That is we will define the integral of a continuous function by integrating “one variable at a time:”

where is a *d-cell,*

(So a -cell in is a compact interval, a -cell in is a rectangle, and a -cell in is the Cartesian product of -cell in with a closed interval.)

A PDF version of this lecture is available here.

In more detail, if then for fixed the function

is continuous on . Thus is integrable and it makes sense to define:

Since is uniformly continuous (why?), it is an easy exercise to show that is continuous on the -cell . Iterating the procedure we obtain for each a function where with the following properties

- .
- for each .

Then *we define *

* *

Remark 1If we simply assume that is a Riemann integrable function of for each fixed then may not be a Riemann integrable function of . This is why we restrict our attention tocontinuous funcitons.Ultimately, using the Lebesgue theory of integration, it is possible to show that the iterated integral makes sense for all Lebesgue integral functions (which includes all Riemann integrable functions) and gives the same result as a more geometrically defined integral.

Recall that the *support of a function *is the smallest closed set containing all points at which does not vanish,

The set of all compactly supported continuous functions on is denoted . Given we have for sufficiently large by the Heine-Borel theorem. We define

To see that this definition makes sense we need the following

Theorem 2Let be a continuous function on . If and are -cells such that whenever then

Remark 3Recall that the symmetric difference of two sets and is

*Proof:* First suppose . So

where

Since vanishes whenever or it is an easy exercise to see that according to the above definitions .

Now for a general pair of -cells, is a -cell and so if vanishes on we have

** Permuation invariance of the integral. **

The above definition of the integral is very practical but it has one annoying defect: In principle the integral might depend on what order we choose to present the variables . Put differently we can ask whether *permuting* the coordinates changes the integral?

Definition 4Given a permutation we define the associatedpermutation map ofdenoted as follows

Remark 5Note that is a linear map. The matrix of consists of ‘s and ‘s, with exactly one in each row and in each column;

The question of whether the integral depends on the order of the variables can be phrased as asking whether, with the above definition,

Theorem 6Let be a -cell, let and let be a permutation map. Then (1) holds.

*Proof:* We first prove this result in a special case. Suppose is a *product function*, that is

so is a product of functions of a single variable. Then the left and right sides of (1) are both equal to

Since both sides of (1) are linear in the function , it follows that (1) holds whenever is a *linear combination of product functions*, i.e. for

with for and . Let us denote by the collection of all such linear combinations of product functions,

It is easy to see that is an *algebra*. The words “Stone-Weierstrass” should pop into your head now! Indeed the algebra is *nowhere vanishing* because the constant functions are in . Likewise it separates points since if then for some so the function

which is in satisfies while . So is dense in by Stone-Weierstrass!

Suppose we want to integrate in two different orders. Then for any we can find such that for all . But then

Since is arbitrary, we see that (1) holds

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