Show 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

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.

Answer 1

Since \begin{align*} f(\xi_{i})g(\eta_{i})=f(\xi_{i})g(\xi_{i})+f(\xi_{i})(g(\eta_{i})-g(\xi_{i})), \end{align*} it suffices to show that \begin{align*} \sum_{i=1}^{n}f(\xi_{i})(g(\eta_{i})-g(\xi_{i}))(t_{i}-t_{i-1})\rightarrow 0. \end{align*} But $f$ being Riemann integrable, it is bounded, say, $|f|\leq M$, then \begin{align*} &\left|\sum_{i=1}^{n}f(\xi_{i})(g(\eta_{i})-g(\xi_{i}))(t_{i}-t_{i-1})\right|\\ &\leq M\sum_{i=1}^{n}|g(\eta_{i})-g(\xi_{i})|(t-{i}-t_{i-1})\\ &\leq M\sum_{i=1}^{n}\left(\sup_{I_{i}}g-\inf_{I_{i}}g\right)(t_{i}-t_{i-1}), \end{align*} which can be controlled by arbitrarily small by Cauchy criterion of Riemann integrability.