Let $f,g:[a,b]\to\Bbb{R}$ integrable functions. Let's dot each partition $P$ in two different ways, choosing in all the interval $[t_{i-1},t_i]$ a point $\xi_i$ and a dot $\eta_i$. Show that: $$\lim_{|P|\to0} \sum_{i=1}^n f(\xi_i)g(\eta_i)(t_{i}-t_{i-1})=\int_a^b f(x)g(x)dx$$
I need something as accurate as possible, because i really need to understand this question and i'm stuck at the very beginning.
Let's first understand the question and to understand that we first need to write the definition of Riemann integrability.
Riemann Integrability: Let $f$ be defined and bounded on $[a, b]$. A set $P$ of the form $$P = \{x_{0}, x_{1}, x_{2}, \dots, x_{n}\}$$ is said to be a partition of $[a, b]$ if $$a = x_{0} < x_{1} < x_{2} < \dots < x_{n} = b$$ and the norm (or mesh) of $P$ (denoted by $|P|$) is defined by $|P| = \max_{i = 1}^{n}(x_{i} - x_{i-1})$. A sum of the form $$S(f, P) = \sum_{i = 1}^{n}f(\xi_{i})(x_{i} - x_{i - 1})$$ is called a Riemann sum for $f$ over partition $P$ where $\xi_{i}$ is any point in $[x_{i - 1}, x_{i}]$. The function $f$ is said to be Riemann integrable on $[a, b]$ if there is a number $I$ with the following property: For any given $\epsilon > 0$ there is a number $\delta > 0$ such that $$|S(f, P) - I| < \epsilon$$ whenever $|P| < \delta$. The number $I$ is called the Riemann integral of $f$ on $[a, b]$ and denoted by $\int_{a}^{b}f(x)\,dx$.
The condition mentioned in the above definition is stated symbolically as $$\lim_{|P| \to 0}S(f, P) = \lim_{|P| \to 0}\sum_{i = 1}^{n}f(\xi_{i})(x_{i} - x_{i - 1}) = \int_{a}^{b}f(x)\,dx$$ It is a standard result in the theory of Riemann integration that if $f, g$ are Riemann integrable over $[a, b]$ then their product $fg$ is also Riemann integrable over $[a, b]$. This means that $$\lim_{|P|\to 0}\sum_{i = 1}^{n}f(\xi_{i})g(\xi_{i})(x_{i} - x_{i - 1}) = \int_{a}^{b}f(x)g(x)\,dx$$ Your question asks to prove that in the above equation we can choose the points $\xi_{i}$ for $f$ and $g$ separately i.e. we can choose points $\xi_{i}$ for $f$ and points $\eta_{i}$ for $g$.
One can prove this by using the following criterion for Riemann integrability
Criterion for Riemann integrability: Let $f$ be defined and bounded on $[a, b]$. If $$P = \{x_{0}, x_{1}, x_{2}, \dots, x_{n}\}$$ then we define upper and lower Darboux sums $$U(f, P) = \sum_{i = 1}^{n}M_{i}(x_{i} - x_{i - 1}), L(f, P) = \sum_{i = 1}^{n}m_{i}(x_{i} - x_{i - 1})$$ where $$M_{i} = \sup\,\{f(x): x \in [x_{i - 1}, x_{i}]\},\,m_{i} = \inf\,\{f(x): x \in [x_{i - 1}, x_{i}]\}$$ The function $f$ is Riemann integrable on $[a, b]$ if and only if for every $\epsilon > 0$ there is a $\delta > 0$ such that $$U(f, P) - L(f, P) < \epsilon$$ for all partitions $P$ of $[a, b]$ with $|P| < \delta$.
Since $f$ is Riemann integrable on $[a, b]$ we have an $M$ such that $|f(x)| < M$ for all $x \in [a, b]$. Further note that $g$ is Riemann integrable and hence the above criterion applies. Therefore given any $\epsilon > 0$ there is a $\delta_{1} > 0$ such that $$U(g, P) - L(g, P) < \frac{\epsilon}{2M}$$ for all partitions $P$ with norm less than $\delta_{1}$.
Next let $I = \int_{a}^{b}f(x)g(x)\,dx$ and then we have a number $\delta_{2} > 0$ such that $$\left|\sum_{i = 1}^{n}f(\xi_{i})g(\xi_{i})(x_{i} - x_{i - 1}) - I\right| < \frac{\epsilon}{2}$$ whenever $|P| < \delta_{2}$. Now we can see that if $|P| < \delta = \min(\delta_{1}, \delta_{2})$ then we have \begin{align}A &= \left|\sum_{i = 1}^{n}f(\xi_{i})g(\eta_{i})(x_{i} - x_{i - 1}) - I\right|\notag\\ &\leq \left|\sum_{i = 1}^{n}f(\xi_{i})g(\eta_{i})(x_{i} - x_{i - 1}) - \sum_{i = 1}^{n}f(\xi_{i})g(\xi_{i})(x_{i} - x_{i - 1})\right|\notag\\ &\,\,\,\,\,\,\,\,+ \left|\sum_{i = 1}^{n}f(\xi_{i})g(\xi_{i})(x_{i} - x_{i - 1}) - I\right|\notag\\ &< \sum_{i = 1}^{n}|f(\xi_{i})||g(\eta_{i}) - g(\xi_{i})|(x_{i} - x_{i-1}) + \frac{\epsilon}{2}\notag\\ &< M\{U(g, P) - L(g, P)\} + \frac{\epsilon}{2}\notag\\ &< M\cdot\frac{\epsilon}{2M} + \frac{\epsilon}{2}\notag\\ &= \epsilon\notag \end{align} and this means that $$\lim_{|P|\to 0}\sum_{i = 1}^{n}f(\xi_{i})g(\eta_{i})(x_{i} - x_{i - 1}) = \int_{a}^{b}f(x)g(x)\,dx$$
Note: The same proof works using suitable modifications if the limit is taken via refinement of partitions instead of reducing their norms to $0$.
