I have some trouble with the Riemann integral, specifically, the definition of it in an article on wikipedia.
We say that the Riemann integral of $f$ equals s if the following condition holds:
For a given $\varepsilon>0$, there exists $\delta$ such that for any tagged partition $x_0,\cdots,x_n$ and $t_0,\cdots,t_{n-1}$ whose mesh is less than $\delta$, we have $$\left|\sum_{i=0}^{n-1}f(t_i)(x_{i+1}-x_i)-s\right|<\varepsilon.$$
Unfortunately, this definition is very difficult to use. It would help to develop an equivalent definition of the Riemann integral which is easier to work with. We develop this definition now, with a proof of equivalence following.(?) Our new definition says that the Riemann integral of f equals s if the following condition holds:
For all $\varepsilon>0$, there exists a tagged partition $x_0,\cdots,x_n$ and $t_0,\cdots,t_{n-1}$ such that for any refinement $y_0,\cdots,y_m$ and $s_0,\cdots,s_{m-1}$ of $x_0,\cdots,x_n$ and $t_0,\cdots,t_{n-1}$, we have $$\left|\sum_{i=0}^{m-1}f(s_i)(y_{i+1}-y_i)-s\right|<\varepsilon.$$
- It is easy to show that the first definition implies the second.
- To show that the second definition implies the first, it is easiest to use the Darboux integral. First one shows that the second definition is equivalent to the definition of the Darboux integral; for this I want you to give me some hints. However, I know that it is also easy to show that a Darboux integrable function satisfies the first definition.
The original article says that the second definition is equivalent to the definition of the Darboux integral; for this see the article on Darboux integration. Well, I couldn't find anything useful information.