Using a the definition of integral if on then there is a number such that for any positive there is a partition of then .
Now here is the trick. We can do it twice with
Now complete it.
Q: Prove that a bounded function is integrable on if and only if there exists a sequence of partitions satisfying
Here some of my (incomplete) thoughts:
Let . Since there exists a sequence of partitions satisfying , there exists an such that, for any . Thus, integrable on .
For this direction I am not sure how to show there exists a sequence of partitions. Since we are assuming is integrable, I know that there is a partion of such that for any positive .
Define and , where is the set of all possible partitions.
We are given is integrable on , so there exists a partition of such that . Since is integrable .
Let . Then, and .
So, we have a sequence of paritions such that, there exists an for which any implies the upper and lower sums converge to the same value.
I am not sure how to introduce the limiting variable, . I understand what is going on, but I am having trouble formalizing the idea of a sequence of partitions.
Sorry to bring up an old thread, but I have been away for a while and have not yeat figured out the proof.
I am unsure how to prove the existance of the sequence of partitions.
is integrable on , so for any we have that . Since is arbitrary it is certianly true that can be made less than for each . Therefore, there exists a sequenct of paritions such that , where , implies .
since is integrable. Thus, can be made less than half the prescribled as can . Since , we have that the lower and upper sums converge to the same limit. Thus,