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I am presented this definition of the Riemann Integral:

i) A Riemann sum of $f$ with resepct to a partition $P = {x_0,...,x_n}$ of $[a,b]$ generated by samples $t_j\in[x_{j-1},_j]$ is a sum

$S(f,P,t_j):=\sum_{j=1}^nf(t_j)\Delta x_j$

ii) The Riemann sums of $f$ are said to converge to $I(f)$ as $||P||\rightarrow0$ iff given $\epsilon$ there is a partition $P_\epsilon$ of $[a,b]$ such that

$P={x_0,...,x_n}\supseteq P_\epsilon$ implies $|S(f,P,t_j)-I(f)|<\epsilon$ for all choices of $t_j \in[x_{j-1},x_j], j = 1,2,...,n$

Edited out>For all $\epsilon>0$, there exists a tagged partition $y_0,y_1...y_{m-1}$ and $r_0, r_1...r_{m-1}$ such that for every tagged partition $x_0,x_1...x_{n-1}$ and $t_0, t_1...t_{n-1}$ which is a refinement of $y_0,y_1...y_{m-1}$ and $r_0, r_1...r_{m-1}$, we have $|\sum_{i=0}^{n-1}f(t_i)(x_{i+1}-x_i)-s|<\epsilon$, where $s:=\int_a^b f(x) dx$

My question is, what is the purpose of specifying that the inequality be fulfilled for all refinements rather than just for the original partition? Is it a redundant condition?

When I compare with this with the definition of the (equivalent) Darboux integral: For every $\epsilon>0$ there exist a partition $P$ such that the difference between the Upper Integral and Lower Integral is $<\epsilon$.

I note that in the definition of the riemann integral, it is emphasized that the partition has to be chosen such that not only its Riemann sum is close to s, but all refinements of that partition has to fulfil that requirement. This is not so in the definition of the Darboux Integral.

In fact I am led to believe that if there exist a partition $P_y$ that gives us $|\sum_{i=0}^{n-1}f(r_i)(y_{i+1}-y_i)-s|<\epsilon$, then all refinements of $P_y$ (call it $P_x$) will make $|\sum_{i=0}^{n-1}f(t_i)(x_{i+1}-x_i)-s|<\epsilon$ as well. My reasoning is that since any choice of tags(say choose sampling points to be the point that gives us the supremum) must make $|\sum_{i=0}^{n-1}f(r_i)(y_{i+1}-y_i)-s|<\epsilon$, any subdivision of a particular interval necessarily gives us a equals or smaller riemann sums.

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  • $\begingroup$ If you consider the function that is one on the rationals and zero on the irrationals, there always exists a partition as fine as you want where the riemann sum is zero by just making all the $r_n$ irrational. But you can't find one that works for all refinements because that would have to include ones with rationals and irrationals. Does that make sense to you? If so I'll elaborate and post it as an answer. $\endgroup$ – Gregory Grant Jan 5 '16 at 15:50
  • $\begingroup$ You see the refinements do make the intervals $(y_{i+1}-y_i)$ smaller but you cannot make an inequality to control the $f(r_i)$'s where you evaluate the function. So your argument that one partition implies all refinements fails there. $\endgroup$ – Gregory Grant Jan 5 '16 at 16:02
  • $\begingroup$ Sorry, forgot to add in that in the definition from Wade the choice of tags are made arbitrary; I'll quote him verbatim this time to avoid further confusion, see edit. As seen, the definition states for all choices of t, which means that the example you gave does not fulfil the hypothesis for any single partition due to the density of irrationals as you mentioned $\endgroup$ – Zhanfeng Lim Jan 6 '16 at 2:58

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