S.O.S. Mathematics CyberBoard

[SexyBack]

If you have a continuous function f then it is Stieltjes integrable w.r.t. g(x)=x which means the integral has value I and given any e there's a partition s.t. any refinement P satisfies |S(P;f) - I| < e. I'm trying to prove that you get the same value for I when you do the calc 1 style definition where you use the endpoints of evenly spaced intervals as your partition and take the limit, something like and then take the limit as n goes to infinity. I was going the route of trying to show that I can refine the partition so that it has the same sample points as the calc 1 summation, but I can't fix the integrator, it's always 1/n in the calc 1 def. but since the Stieltjes partition is given just by saying f is integrable I can't guarentee that there's a refinement with evenly spaced partition points. So I'm stuck, anyone know what else I could try?

[SexyBack]

Ok, I've narrowed the problem to this. Let e > 0, since f is integrable there's a partition P s.t. if Q is any refinement then |S(P;f) - integral| < e. I need to choose an N s.t. if n > N then P can be refined into Q and specific points z_i can be chosen s.t. . My problem is I need to be able to take a partition of [0,1] P = (x1, x2, ..., xa) and refine it into Q = (x1, x2, ..., xm) s.t.|n(x_i - x_i-1) - 1| < e for all i (where n is already chosen). Intuitively this means adding points to the partition so that you even out the spacing. But I'm not sure how to show that this is possible.

[Opalg]