Let $[a,b]$ be a closed interval in $\mathbb{R}$, and let $X$ be the set of tagged partitions of $[a,b]$. Now let’s define two partial orders over $X$. Let $P_1\geq_1 P_2$ if $P_1$ is a refinement of $P_2$ and let $P_1\geq_2 P_2$ if the mesh of $P_1$ is less than or equal to the mesh of $P_2$. ($\geq_2$ is a total order, but of course total orders are also partial orders.)
Using these two partial orders, we can make two equivalent definitions of the Riemann integral:
- We say that $\int_a^b f(x) dx = I$ if for every $\epsilon >0$ there exists a partition $P_c$ such that for all partitions $P\geq_1 P_c$, $|R(f,P)-I|<\epsilon$.
- We say that $\int_a^b f(x) dx = I$ if for every $\epsilon >0$ there exists a partition $P_c$ such that for all partitions $P\geq_2 P_c$, $|R(f,P)-I|<\epsilon$.
Where $R(f,P)$ is the Riemann sum of $f$ over $P$. But my question is, what other partial orders on the set of tagged partitions would produce an equivalent definition of the Riemann integral?
I noticed that $\geq_1$ is a “suborder” of $\geq_2$. That is, $P_1\geq_1 P_2$ implies $P_1\geq_2 P_2$, because a refinement of a partition will have a smaller mesh. Is there any suborder $\geq_0$ of $\geq_1$ that produces an equivalent definition of the Riemann integral?