(33.2) Let be a nonempty bounded subset of . For fixed , let . Show that and .
Proof. Let . Then is an upper bound for as for all we have that so as .
Now, if for all (that is, is an upper bound for ), then for all so is an upper bound for . But then as is the least upper bound for , so . So .
Now notice that .
(33.6) Let be integrable on . Prove that, for any subset we have
Hint. For , we have .
Proof. The first part of the hint follows from the triangle inequality, as
The second part of the hint comes from how and for all .
On the other hand, we have that for any , properties of the supremum and infimum guarantee that there exists so that,
So we have that,
This holds for all , so we conclude that the desired inequality holds.
(33.5) Show that .
Proof. Using Theorems 33.4 and 33.5, together with the FTC I (for example), we have that (as )
(33.8) Let and be integrable functions on .
It is a fact (see Exercise 33.7) that if is integrable on , then so is . Prove that is integrable on . Hint. Use that .
Show that and are integrable on . You may use the results of Excercise 17.8 without proof.
Use that for integrable functions we have that , , , , are all integrable. Further applications of linearity of the integral prove (a), and together with prove the first part of (b). The second part then follows from and integrability of , etc.
(33.10) Let be the function,
Prove that is integrable on . Hint. See Excercise 33.11(c) and its solution in the textbook. (Why can we not apply the Dominated Convergence Theorem to prove that is integrable?)
Proof. Let . Notice that is continuous on each of and , and is hence integrable. This means that there exist partitions of and of so that
Notice that is a partition of and furthermore that
and,
(Why? The only terms in the two sums are those contributed by the interval but we know that the supremum and infimum of on any such interval are and respectively and that the width of this interval is ) So,
We can find any such for any , so we conclude that is integrable on .