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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:

  1. 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 $P\geq_1 P_c$, $|R(f,P)-I|<\epsilon$.
  2. 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?

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  • $\begingroup$ The orders which you have mentioned are ones found in literature. I wonder if any other sort of order on partitions has been studied. $\endgroup$ – Paramanand Singh Jun 28 '18 at 6:13
  • $\begingroup$ @ParamanandSingh I wonder whether if two orders produce equivalent definitions of the Riemann integral, that implies something about the relationship between the two orders, like a suborder relation or something else. $\endgroup$ – Keshav Srinivasan Jun 28 '18 at 6:16
  • $\begingroup$ The given two orders are "natural" as one comes directly from set inclusion and the other from the metric. I suppose something like the transitive hull of "is obtained by splitting each subinterval into sevenpieces" would work $\endgroup$ – Hagen von Eitzen Jun 28 '18 at 6:31
  • $\begingroup$ @HagenvonEitzen If there are multiple suborders of $\geq_1$ that would work, is there a minimal suborder of $\geq_1$ that would work? $\endgroup$ – Keshav Srinivasan Jun 28 '18 at 6:34
  • $\begingroup$ By the way the same two orders are not equivalent for Riemann Stieltjes integral and the order 1 is more general here. $\endgroup$ – Paramanand Singh Jun 28 '18 at 8:32

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