The set of where is not continuous is contained in the union of the sets where and are not continuous. These last two sets are of measure 0 by Theorem 3-8; so theee first set is also of measure 0. But then is integrable by Theorem 3-8.
If has content 0, then it is bounded by Problem 3-9 (a); so it is a subset of an closed rectangle . Since has content 0, one has for some open rectangles the sum of whose volumes can be made as small as desired. But then the boundary of is contained in the closure of , which is contained in the union of the closures of the (since this union is closed). But then the boundary of must be of content 0, and so is Jordan measurable by Theorem 3-9. Further, by Problem 3-5, one has which can be made as small as desired; so .
Let be the set of rational numbers in . Then the boundary of is , which is not of measure 0. So does not exist by Theorem 3-9.
Using the hint, let be a partition of where is a closed rectangle containing . Then let be a rectangle of of positive volume. Then is not of measure 0 by Problem 3-8, and so . But then there is a point of outside of ; so . Since this is true of all , one has . Since this holds for all partitions of , it follows that if the integral exists.
Following the hint, let be a positive integer and . Let . Let be a partition of such that . Then if is a rectangle of which intersects , we have . So . By replacing the closed rectangles with slightly larger open rectangles, one gets an open rectangular cover of with sets, the sum of whose volumes is at most . So has content 0. Now apply Theorem 3-4 to conclude that has measure 0.
The set of where is not continuous is which is not of measure 0. If the set where is not continuous is not of measure 0, then is not integrable by Theorem 3-8. On the other hand, if it is of measure 0, then taking the union of this set with the set of measure 0 consisting of the points where and differ gives a set of measure 0 which contains the set of points where is not continuous. So this set is also of measure 0, which is a contradiction.
This is an immediate consequence of Problem 3-12 and Theorem 3-8.
Suppose is Jordan measurable. Then its boundary is of content 0 by Theorem 3-9. Let and choose a finite set for of open rectangles the sum of whose volumes is less than and such that the form a cover of the boundary of . Let be a partition of such that every subrectangle of is either contained within each or does not intersect it. This satisfies the condition in the statement of the problem.
Suppose for every , there is a partition as in the statement. Then by replacing the rectangles with slightly larger ones, one can obtain the same result except now one will have in place of and the will be open rectangles. This shows that the boundary of is of content 0; hence is Jordan measurable by Theorem 3-9.
Let be a closed rectangle containing . Apply Problem 3-21 with as the Jordan measurable set. Let be the partition as in Problem 3-21. Define . Then and clearly is Jordan measurable by Theorem 3-9. Further .