# Exercises: Chapter 3, Section 1

1. Let be defined by

Show that is integrable and that .

Apply Theorem 3-3 to the partition where . For this partition, .

2. Let be integrable and let except at finitely many points. Show that is integrable and .

For any , there is a partition of in which every subrectangle has volume less than . In fact, if you partition by dividing each side into equal sized subintervals and , then the volume of each subrectangle is precisely which is less than as soon as . Furthermore, if is any partition, then any common refinement of this partition and has the same property.

If and is a partition of , then any point is an element of at most of the subrectangles of . The intuitive iddea of the proof is that the worst case is when the point is in a `corner'; the real proof is of course an induction on m.

Let and be a partition as in Theorem 3-3 applied to and . Let be a refinement of such that every subrectangle of has volume less than where , is the number of points where and have values which differ, and (resp. ) are upper (resp. lower) bounds for the values for all . Then the hypotheses of Theorem 3-3 are satisfied by and , and so is integrable.

In fact, and where is any upper bound for the volume of the subrectangles of , because the terms of the sum can differ only on those subrectangles which contain at least one of the points where and differ. Taking differences gives

3. Let be integrable.
1. For any partition of and any subrectangle of , show that and and therefore and .

For each , one has and since greatest lower bounds are lower bounds. Adding these inequalities shows that is a lower bound for , and so it is at most equal to the greatest lower bound of these values. A similar argument shows the result for . Since , , and are just positively weighted sums of the , , and the result for can be obtained by summing (with weights) the inequalities for the . A similar argument shows the result for .

2. Show that is integrable and .

Let (resp. ) be a partition as in Theorem 3-3 applied to (resp. ) and . Let be a common refinement of and . Then by part (a) and Lemma 3-1,

. By Theorem 3-3, is integrable.

Further

By the squeeze principle, one concludes that .

3. For any constant , show that .

We will show the result in the case where ; the other case being proved in a similar manner. Let be a partition as in Theorem 3-3 applied to and . Since and for each subrectangle of , we have

By Theorem 3-3, applied to and , the function is integrable; by the squeeze principle, its integral is .

4. Let and be a partition of . Show that is integrable if and only if for each subrectangle the function , which consists of restricted to , is integrble, and that in this case .

Suppose that is integrable and . Let be a partition of as in Theorem 3-3 applied to and . Let be a common refinement of and . Then there is a partition of whose subrectangles are precisely the subrectangles of which are contained in . Then . By Theorem 3-3, it follows that is integrable.

Suppose that all the are integrable where is any subrectangle of . Let be a partition as in Theorem 3-3 applied to and where is the number of rectangles in . Let be the partition of A obtained by taking the union of all the subsequences defining the partitions of the (for each dimension). Then there are refinements of the whose rectangles are the set of all subrectangles of which are contained in . One has

By Theorem 3-3, the function is integrable, and, by the squeeze principle, it has the desired value.

5. Let be integrable and suppose . Show that .

By Problem 3-3, the function is integrable and . Using the trivial partition in which is the only rectangle, we have since . This proves the result.

6. If is integrable, show that is integrable and .

Consider the function . For any rectangle contained in , we have and . If , then . On the other hand, if , then . Let be a partition as in Theorem 3-3 applied to and . Then this implies that

So, is integrable by Theorem 3-3.

Similarly, one can show that is integrable. But then by Problem 3-3, it follows that is integrable. But then, so if integrable. Further, since , Problem 3-5 implies that . Since by Problem 3-3 (c), it follows that .

7. Let be defined by

Show that is integrable and .

Let . Choose a positive integer so that . Let be any partition of such that every point with lies in a rectangle of of height (in the direction) at most . Since there are at most such pairs , such a exists and the total volume of all the rectangles containing points of this type is at most . Since , the contribution to from these rectangles is also at most . For the remaining rectangles , the value of and their total volume is, of course, no larger than 1; so their contribution to is at most . It follows that . By Theorem 3-3, is integrable and the squeeze principle implies that its integral is 0.