Riemann Integral Tagged Partition Independence Suppose the limit exists for some partitioning and sampling scheme.
How to prove the limit exists, and is the same limit, for all partitioning, sampling schemes?
I.e. The limit is independent of "tagged partition" choice.
(Not homework - just curious)
 A: Recall that if a function is Riemann integrable, then the upper and lower Riemann sums converge to the same value as the mesh of an arbitrary partition becomes arbitrarily small. Since these are upper and lower bounds the Riemann sum of every possible tagged partition over $[a,b]$, the squeeze theorem implies that they all must converge to $I$ as required.
A: You cannot fix both the partitioning scheme as well as sampling scheme. One of them has to be arbitrary. A simple example should suffice to convince you. Let $f(x) =0$ if $x$ is rational and $f(x) =1$ when $x$ is irrational and consider this function on interval $[0,1]$. If we fix the partitioning scheme to consider only uniform partitions (sub-intervals are all of equal length) and the tagging scheme is fixed to choose left-hand end-points of the sub-intervals of the partition then we can easily see that the Riemann sum is $0$ and therefore its limit is also $0$. But as can be easily seen the function is not Riemann integrable as every upper Darboux is $1$ and every lower Darboux sum is $0$.
On the other hand keeping one of schemes fixed and the other one arbitrary is sufficient to show that the function is Riemann integrable. But the proof of this is difficult (especially the part where sampling scheme is fixed). 
A: Hint: Think about refinements. 
A: Each partition consists of Intervals $I_j$ where $j \in \mathbb{N}$ for some finite amount of intervals. It suffices to prove that splitting one interval up into two intervals by adding a point makes the resulting Riemann integral more or equally precise.
