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.)
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 1 If 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 to continuous 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 2 Let be a continuous function on . If and are -cells such that whenever then
Remark 3 Recall that the symmetric difference of two sets and is
Proof: First suppose . So
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 4 Given a permutation we define the associated permutation map of denoted as follows
Remark 5 Note that is a linear map. The matrix of consists of ‘s and ‘s, with exactly one in each row and in each column;
for arbitrary -permutation?
Theorem 6 Let 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
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