Are there any direct ways of proving that if is real-valued and Riemann integrable on , then is bounded?
Because the usual proof would be by contraposition:
Outline of "Usual Proof"
1. There exists a sequence in that converges to , such that for every , .
2. Let be a partition of . Then the set of Riemann sums of corresponding to is an unbounded set of real numbers.
3. Thus cannot be Riemann integrable on .
Suppose we have a refinement then .
Now here comes the messy part.
By using right hand sums with respect to we change one number in the sum we get .
Use the number for the bound.
I hope you can at least see how it might work.
EDIT: Indirect proof.