Problem
Let $f,g: [a,b] \rightarrow \mathbb{R}$ integrable functions and $P$ a partition $t_0=a<...<t_i<...<t_n=b$
Even if we choose $\zeta_i,\eta_i \in [t_{i-1},t_i]$ with $\zeta_i \neq\eta_i$, prove that $$\lim \limits_{|P|\rightarrow 0} \sum \limits^n_{i=1}f(\zeta_i)g(\eta_i)(t_i-t_{i-1})=\int^{b}_{a}f(x)g(x)dx$$
Question
I know that the point we choose in $[t_{i-1},t_i]$ doen't change the value of $\int h(x)dx$ $\space$ for some $h$. But in this case we have two different functions evalueted in differents points.
In this case I think the way to solve this is argummenting that we could use a refinament $P'$ of $P$ such that $\zeta_i,\eta_i\in P'$ then we can choose arbitrary but equal points in these new intervals to calculate the integral with no problems.