# Exercises: Chapter 3, Section 2

1. Prove that is not of content 0 if for .

Suppose for are closed rectangles which form a cover for . By replacing the with , one can assume that for all . Let . Choose a partition which refines all of the partitions where Note that is a rectangle of the cover . Let be any rectangle in with non-empty interior. Since the intersection of any two rectangles of a partition is contained in their boundaries, if contains an interior point not in for some , then contains only boundary points of . So, if has non-empty interior, then is a subset of for some since the union of the is . The sum of the volumes of the rectangles of is the volume of , which is at most equal to the sum of the volumes of the . So is not of content 0 as it cannot be covered with rectangles of total area less than the volume of .

1. Show that an unbounded set cannot have content 0.

Suppose where are rectangles, say . Let where and . Then contains all the and hence also . But then is bounded, contrary to hypothesis.

2. Give an example of a closed set of measure 0 which does not have content 0.

The set of natural numbers is unbounded, and hence not of content 0 by part (a). On the other hand, it is of measure zero. Indeed, if , then the union of the open intervals for . contains all the natural numbers and the total volume of all the intervals is .

1. If C is a set of content 0, show that the boundary of also has content 0.

Suppose a finite set of open rectangles , . cover of and have total volume less than where . Let where . Then the union of the cover the boundary of and have total volume less than . So the boundary of is also of content 0.

2. Give an example of a bounded set of measure 0 such that the boundry of does not have measure 0.

The set of rational numbers in the interval is of measure 0 (cf Proof of Problem 3-9 (b)), but its boundary is not of measure 0 (by Theorem 3-6 and Problem 3-8).

2. Let be the set of Problem 1-18. If , show that the boundary of does not have measure 0.

The set closed and bounded, and hence compact. If it were also of measure 0, then it would be of content 0 by Theorem 3-6. But then there is a finite collection of open intervals which cover the set and have total volume less than . Since the set these open intervals together with the set of form an open cover of [0, 1], there is a finite subcover of . But then the sum of the lengths of the intervals in this finite subcover would be less than 1, contrary to Theorem 3-5.

3. Let be an increasing function. Show that is a set of measure 0.

Using the hint, we know by Problem 1-30 that the set of where if finite for every . Hence the set of discontinuities of is a countable union of finite sets, and hence has measure 0 by Theorem 3-4.

1. Show that the set of all rectangles where each and each are rational can be arranged into a sequence (i.e. form a countable set).

Since the set of rational numbers is countable, and cartesian products of countable sets are countable, so is the set of all -tuples of rational numbers. Since the set of these intervals is just a subset of this set, it must be countable too.

2. If is any set and is an open cover of , show that there is a sequence of members of which also cover .

Following the hint, for each in , there is a rectangle B of the type in part (a) such that has non-zero volume, contains and is contained in some in . In fact, we can even assume that is in the interior of the rectangle . In particular, the union of the interiors of the rectangles (where is allowed to range throughout ) is a cover of . By part (a), the set of these are countable, and hence so are the set of corresponding 's; this set of corresponding 's cover .