Following the hint, is integrable with by Problem 3-15 and Fubini's Theorem. We have . Now is equivalent to the condition that either or . Both of these having integral 0 implies by Problem 3-18 that the sets where their integrand is non-zero are of measure 0, and so is also of measure 0.
The set is the set of rational numbers in which is of measure 0, but not of content 0, because the integral of its characteristic function does not exist. To see that the set has content 0, let . Let be such that . Then the set can be covered by the rectangles and for each in lowest terms with , the rectangle where . The sum of the areas of these rectangles is less than .
This follows from Problem 3-8 and Theorem 3-6, but that is not an induction.
Fubini's Theorem and induction on show that and so does not have content 0, and hence is not of measure 0.
One has and so by Fubini,
where is an upper bound on the image of .
where the upper bounds need to be determined.
By Fubini, the left hand iterated integral is just where
Applying Fubini again, shows that this integral is equal to .
Following the hint, if is not zero for some point , then we may assume (by replacing with if necessary that it is positive at . But then continuity implies that it is positive on a rectangle containing . But then its integral over is also positive.
On the other hand, using Fubini on gives:
Similarly, one has
Subtracting gives: which is a contradiction.
To avoid overlap, it is convenient to keep the set in the positive half plane. To do this, let be the original Jordan measurable set in the -plane, and replace it with . Theorem 3-9 can be used to show that is Jordan measurable if is.
The problem appears to be premature since we really want to be able to do a change of variables to cylindrical coordinates. Assuming that we know how to do that, the result becomes .
but that does not exist.
The problem has a typo in it; the author should not have switched the order of the arguments of as that trivializes the assertion.
The iterated integrals are zero because the inside integral is the zero function. The last integral cannot exist by Theorem 3-9 and Problem 1-17.
What is , for in the interior of ?
Let be in the interior of , fix . We have
by Fubini's Theorem.
Using the hint, we have . One has
One has and where the second assertion used Problem 3-32.
We have and so by the chain rule one has
Show that .
One has
If is a rectangle, show that the volume of is .
In the three cases, is , 1, and 1 respectively. If the original rectangle , then is
in the first case, is a cylinder with a parallelogram base in the second case, and is the same rectangle except that the intervals in the and places are swapped in the third case. In the second case, the parallelogram base is in the and directions and has corners . So the volumes do not change in the second and third case and get multiplied by in the first case. This shows the result.
If is non-singular, then it is a composition of linear transformations of the types in part (a) of the problem. Since is multiplicative, the result follows in this case.
If is singular, then is a proper subspace of and is a compact set in this proper subspace. In particular, is contained in a hyperplane. By choosing the coordinate properly, the hyperplane is the image of a linear transformation from into made up of a composition of maps of the first two types. This shows that the compact portion of the hyperplane is of volume 0. Since the determinant is also 0, this shows the result in this case too.
This is an immediate consequence of Fubini's Theorem since the inside integrals are equal.