I'm studying the Darboux definition of integrability, which I completely explained here. There's an exercise that asks me to prove that the Darboux Integrability is equivalent to Riemann Integrability, but this Riemann integral is defined as the following:
It first defines a 'pointed' partition (I don't know how to say it in ensligh) by the following: a 'pointed' partition $[a,b]$ is a pair $P^*=(E,ξ)$, where $P=\{t_0, t_1, \cdots, t_n\}$ is a partition of $[a,b]$ and $ξ = (ξ_1, \cdots, ξ_n)$ is a list of $n$ chosen numbers such that $t_{i-1}\le ξ_i\le t_i$ for each $i=1,\cdots ,n$.
Now, the Riemann Integral is defined as:
$$\sum(f,P^*) = \sum_{i=1}^n f(ξ_i)(t_i-t_{i-1})$$
(I didn't understand the notation for the left hand side of the equation, by the way)
Finally, I'm asked to prove the following: given $f,g:[a,b]\to \mathbb{R}$ integrable functions, for the entire partition $P=\{t_0, \cdots, t_n\}$ of $[a,b]$ let $P^* = (P,ξ)$ and $P^{\#} = (P, η)$ be pointed partitions of $P$, then:
$$\lim_{|P|\to 0}\sum f(ξ_i)g(η_i)(t_i-t_{i-1}) = \int_a^b f(x)g(x) \ dx$$
I guess here I need to prove that the Riemann Integral of the product of two functions if the darboux integral of $f(x)g(x)$, but it seems too obvious, I just need to verify that $f,g$ are integrable, then their product is too, isn't it? I'm pretty sure this should be a hard question. Is there another interpretation that I'm missing?