# How do i prove that "If $f$ is riemann-stieltjes integrable with respect to $\alpha$, then $f,\alpha$ have no common discontinuity?

Let $\alpha$ be a monotonically increasing function on $[a,b]$ and $f\in\mathscr{R}(\alpha)$.

I googled it, but i couldn't find a text using relatively easy concepts to prove this. (For example, I don't think dual space concept is necessary to prove this.)

I've seen this theorem quite frequently on this website and it seems it's a very important theorem, so i want to prove it. (I cannot believe why this theorem is not in Rudin's PMA)

I may be missing something, but with $f=1_{[1,2]}$ and $\alpha = 1_{(1,2]}$, it would seem that on $[0,2]$ $f$ is Riemann Stieltjes integrable wrt $\alpha$. They have a common discontinuity at $1$, but with the partition $\pi = (0,1,2)$ I get $\int f d \alpha = 0$. –  copper.hat Dec 21 '12 at 8:20
@copper.hat I think you are right. I think hypothesis should be rephrased as '$\alpha,f$ are discontinuituous at the same side' and proof for this is relatively easy to me. Thank you! –  Katlus Dec 24 '12 at 2:31