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