net convergence implies bounded? I just saw the following theorem:

Theorem Let $\alpha:[a,b] \to \mathbb{R}$ be a mapping. If the Riemann-Stieltjes integral $$I(f) := \int_a^b f(t) \, d\alpha(t)$$ exists for all continuous functions $f:[a,b] \to \mathbb{R}$, then $\alpha$ is of bounded variation.

in this answer.
but I'm confused by this step in the proof:

Since, by assumption, $I^{\Pi}(f) \to I(f)$ as $|\Pi| \to 0$ for all $f \in C[a,b]$, we have
  $$\sup_{\Pi} |I^{\Pi}(f)| \leq c_f < \infty$$

Does he use the fact that if $a_n$ converge, then $\sup_n a_n<\infty$ ? But I have heard from someone that the convergence of Riemann sums is a kind of "net convergence", does this convergence have the same property of ordinary convergence?

You can take limits along one sequence of partitions with the norms (i.e. the maximum of the lengths of subintervals) tending to zero. There is no need to use nets here since you are not proving the existence of limits of Riemann Steiltje sums.
