(32.2) Let the function be defined as,
Compute the upper and lower Darboux integrals for on the interval .
Hint. Here’s how you can show that : For any partition , if , explain why
Is integrable on the interval ?
For any interval , by density of the irrationals, we have that . So, for any partition of , we have that , so we conclude that .
Consider now the partition . We have, by density of the rationals and the definition of that
So, , and as is the infimum over all , we have proven that .
To prove the other inequality, let be any partition of . Observe that
This inequality holds for all , so for all partitions of . From this we notice that as it is a greatest lower bound. Together now, we conclude that .
No, as in part (a) we have proved that .
Let be a bounded function on . Suppose that is integrable. Does it follow that is also integrable? If so, prove it. If not, provide a counterexample.
No. Let for and for irrational . Then is integrable but is not by a calculation similar to Example 22.2.
(32.6) Let be a bounded function on . Suppose there exist sequences and of upper and lower Darboux sums for such that . Show is integrable on and that .
Proof. Notice first that the sequences and are defined by two sequences of partitions and for which and . In particular, we have that as they are upper and lower Darboux sums.
Let . Then there exists so that for we know that . Notice that
.
Take any and let . We have just proved that is a partition for which . In other words, we have proved that is integrable.
Notice that for all , so . In particular, when , we see that . So ; in particular, converges. As , we conclude furthermore that .
(32.7) Let be integrable on and suppose is a function on so that except perhaps at finitely many in .
Show that is integrable on and that . Hint: First reduce to the case where for all .
Proof. Suppose and that at almost all of , except for when we have . Suppose first that .
Let . If , consider the partition . If , consider the partition . Otherwise, consider the partition . In all cases, observe that and that . In particular,
So is integrable and .
Suppose instead now that ; but by considering the function , we have just proved that , and we may use linearity of the integral to realize that in this case as well.
Now let be any function integrable on , and suppose that except at some point . Considering the function which is 0 at all points except for , realize that we have shown above that , and by linearity of the integral,
We have now shown completely the case in which differs by integrable at precisely one point. But what about finitely many points? We may use induction! Suppose that at all but points , and take as an inductive hypothesis that . Then notice that if at all but the points we have that actually at all but one point . But we have shown above that
So, we have proven the case for any finite number of differing points.