I have read this definition first time today. As far as I can understand it, it seems to me that difference between Riemann sums and a number L can be made small by changing norm of partition. Does the changing norm of partition change values of $L$ and $\varepsilon$? $\delta$ is dependent on $\varepsilon$, HOW? What does it mean that the set of all Riemann integrable functions on $[a,b]$ is $\mathcal R[a,b]$?
Thanks for help
On
If your function is continuous on the interval $[a,b]$ it is also uniformly continuous:
$$\forall, \epsilon > 0 \; \exists \delta>0: |x-y|< \delta \; x,y \in [a,b]\Rightarrow |f(x)-f(y)|<\epsilon $$
take the norm of your partitions($P_1,P_2$) to be less than $\delta$. Let $P_3$ refine both partitions $P_3 = P_1 \cup P_2$
$$S(f,P_1) - S(f,P_2) = \sum_{j=1}^N (f(x_{t_j})- f(y_{t_j}))(t_{j+1}-t_j) $$
Therefore \begin{align*} |S(f,P_1) - S(f,P_2)| &= \sum_{j=1}^N |(f(x_{t_j})- f(y_{t_j}))|(t_{j+1}-t_j) \\ &\leq \sum_{j=1}^N \epsilon (t_{j+1}-t_j) = \epsilon (b-a) \end{align*}
So as you reduce the norm of your partition, the distance to the limit of your sum gets smaller (at least an upper bound to this distance).
Just follow carefully on the order of quantifications. The function is Reimann integrable if there exists a number $L$ (and it remains fixed!) such that for any $\varepsilon >0$ (which you are free to choose as you like, but once you made that choice it remains fixed) there exists a $\delta_\varepsilon >0$ (which may depend on $\varepsilon $ in any way whatsoever (and typically it does not matter at all how it depends on it)) such that for any partition with mesh less than $\delta_\varepsilon $, the corresponding Riemann sum is within distance $\varepsilon $ from $L$. Finally, $\mathcal R[a,b]$ is just the name given to the set of all Riemann integrable functions on $[a,b]$.