Note:
This homework assignment is just an example, for practice with our end-of-semester material. It will not be collected or graded.
Exercises:
Let be integrable on . Prove that, for any subset we have
Hint. For , we have .
Note: This is the primary step in proving that \(|f|\) is integrable whenever \(f\) is.
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.
Show that .
Proof. Using Theorems 33.4 and 33.5, together with the FTC I (for example), we have that (as )
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 .
Let be a continuous function on and define
Show that is differentiable on and compute .
Let . Then is differentiable and by the FTC II as is continuous. Notice that . So by the chain rule, is differentiable and .
Let be the function,
Determine the function .
Sketch . Where is continuous?
Where is differentiable? Calculate at the points of differentiability.
By the FTC II, is continuous everywhere.
By the FTC II, is differentiable everywhere that is continuous, hence at all except possibly and .
We can prove that is not differentiable at and by showing that the left-hand and right-hand difference quotient limits disagree.