# Converging of Riemann Sums with different partitions

I have seen the following following definition of Riemann integrals: $$\overline{\int_a^b }f(x)dx = \inf U(P,f) \underline{\int_a^b }f(x)dx = \sup L(P,f)$$

These are upper and lower Riemann integrals, respectively. Here $P$ is a partition of the interval $[a,b]$ into a finite number of subintervals. $U(P,f)$ is $\sum_i^nM_i\Delta x_i$ with $M_i = \sup f(x) ,(x_{i-1}\leq x \leq x_{i})$ and $L(P,f)$ is $\sum_i^nm_i\Delta x_i$ with $m_i = \inf f(x) ,(x_{i-1}\leq x \leq x_{i})$.

Riemann integral is defined if the upper and lower Riemann integrals are equal. The infimum and supremums are taken over all partitions on $[a,b]$ for both $U(P,f)$ and $L(P,f)$. My question is about the pattern of convergence for these upper and lower sums. Each partition $P$ has two parameters, the number of subintervals $n$ and the locations of partition points. How do these parameters affect the rate of converge in $U$ and $L$? It is expected that in general more subintervals mean a closer approximation to the real upper or lower integral but what about intermediate situations? For example, for a Riemann integrable function $f$ is it always $U(P_1,f) \leq U(P_2,f)$ when the number of subintervals in $P_1$ is larger than $P_2$, without any assumptions on the locations of the partition points? Can it be otherwise for some $P_1$ and $P_2$ with the same conditions ( $P_1$ with more subintervals than $P_2$)?

The fact that the definitions of the lower and upper integrals given on all partitions $P$, without any mention about the number and location of subintervals disturbed me. I wonder how these free parameters affect the rate of convergence?

• No, the number of points in the partitions is not enough to order the corresponding Darboux sums. Simple examples show this. Which ones have you tried? – Did Sep 6 '14 at 7:17
• Actually, I am used to the classic definition of the Riemann Integral: Just slicing $[a,b]$ into $n$ equal parts like $\dfrac{b-a}{n}\sum_{i=1}^{n} f(x_i)$ and taking $n$ to infinity. This definition with arbitrary subinterval lengths is somewhat new to me. What does define then the ordering of the Darboux sums, in general? – Ufuk Can Bicici Sep 6 '14 at 8:14
• ?? Well, the fact that one sum is greater than the other, of course. – Did Sep 6 '14 at 8:15
• I mean the number of partitions and locations of them, how do they affect sums, is there a strict rule in general? (Like larger $n$ means always a smaller upper sum, for example, for which you have said it does not in the first comment.) – Ufuk Can Bicici Sep 6 '14 at 8:25