It is usually claimed that the Riemann integral can be seen as the limit of a net, but this is not clear to me. I can see that if the Riemann exists, the limit of the net must exist and they must coincide. The converse is what I'm having problems with. The statement is equivalent to:
Let $f:[a,b]\to \mathbb R$, and suppose there is a $c\in \mathbb R$ such that that for any $\epsilon >0$ there is a partition $P$ such that any tagged partition $Q^*$ that refines $P$, we have that $$|\sum(f;Q^*)-c|<\epsilon$$ where $\sum(f;Q^*)$ is the Riemann sum of $f$ on $Q$. Then, for any $\epsilon >0$, there is a $\delta>0$ such that any tagged partition $Q^*$ with mesh $<\delta$ has $|\sum(f;Q^*)-c|<\epsilon$.
The obvious approach would be to pick $\delta=|P|$, with the $P$ given by the first condition, but it is not the case that $|Q|<|P|$ implies $\sum(f;Q^*)<\sum(f;P^*)$. It is false even if we take the upper Darboux sums instead.
By exploiting that $f$ is bounded it is however possible to see that the statement is true, at least for $\mathbb R$. The proof I have in mind does not work for infinite dimensional spaces since it uses the fact that for $\mathbb R$ the Darboux and Riemann integral coincide.
Is there a more direct approach that works for the general case too?
