# Exercises: Chapter 3, Section 4

1. Let be a set of content 0. Let be the set of all such that is not of content 0. Show that is a set of measure 0.

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.

2. Let be the union of all where is a rational number in written in lowest terms. Use to show that the word ``measure" in Problem 3-23 cannot be replaced with ``content".

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 .

3. Show by induction on that is not a set of measure 0 (or content 0) if for each .

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.

4. Let be integrable and non-negative, and let . Show that is Jordan measurable and has area .

One has and so by Fubini,

where is an upper bound on the image of .

5. If is continuous, show that

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 .

6. Use Fubini's Theorem to give an easy proof that if these are continuous.

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.

7. Use Fubini's Theorem to derive an expression for the volume of a set in obtained by revolving a Jordan measurable set in the -plane about the -axis.

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 .

8. Let be the set in Problem 1-17. Show that

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.

9. If and is continuous, define by

What is , for in the interior of ?

Let be in the interior of , fix . We have

by Fubini's Theorem.

10. Let be continuous and suppose is continuous. Define . Prove Leibnitz' Rule: .

Using the hint, we have . One has

11. If is continuous and is continuous, define .
1. Find and .

One has and where the second assertion used Problem 3-32.

2. If , find .

We have and so by the chain rule one has

12. Let be continuously differentiable and suppose . As in Problem 2-21, let

Show that .

One has

1. Let be a linear transformation of one of the following types:

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.

2. Prove that is the volume of for any linear transformation .

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.

13. (Cavalieri's principle). Let and be Jordan measurable subsets of . Let and define similarly. Suppose that each and are Jordan measurable and have the same area. Show that and have the same volume.

This is an immediate consequence of Fubini's Theorem since the inside integrals are equal.