Check proof. Riemann sum for product of two functions. 
Let $f,g:[a,b] \longrightarrow \mathbb{R}$ integrable functions. Prove that
  $$\displaystyle \lim_{|P|\to 0}\sum f(\lambda_{i})g({\eta_{i}})(t_{i} - t_{i-1}) = \int\limits_{a}^{b}f(x)g(x)\mathrm{d}x.$$

$\textbf{My idea:}$ Writting $f(\lambda_{i})g(\eta_{i}) = f(\lambda_{i})g(\lambda_{i}) + f(\lambda_{i})(g(\eta_{i}) - g(\lambda_{i}))$, then
$$\sum\limits_{1}^{n}f(\lambda_{i})g(\eta_{i})(t_{i} - t_{i-1}) = \sum\limits_{1}^{n}f(\lambda_{i})g(\lambda_{i})(t_{i} - t_{i-1}) + \sum\limits_{1}^{n}f(\lambda_{i})(g(\eta_{i}) - g(\lambda_{i}))(t_{i} - t_{i-1}).$$
Since $M = \sup\limits_{[a,b]}f$ and $g$ is a integrable function
$$\lim_{|P|\to 0}\sum\limits_{1}^{n}f(\lambda_{i})(g(\eta_{i}) - g(\lambda_{i}))(t_{i} - t_{i-1}) = 0.$$
The result follows.
Is a correct ideia?
!!! $P^{\ast} = (P, \lambda)$, $P^{\sharp} = (P, \eta)$ !!!
 A: Presumably you have shown that the product $fg$ is Riemann integrable and are focusing on showing that the points at which $f$ and $g$ are evaluated in the Riemann sum need not be the same.
Following your steps this reduces to showing that
$$\lim_{|P|\to 0}\sum_{i=1}^{n}f(\lambda_{i})(g(\eta_{i}) - g(\lambda_{i}))(t_{i} - t_{i-1}) = 0.
$$
Note that
$$\left|\sum_{i=1}^{n}f(\lambda_{i})(g(\eta_{i}) - g(\lambda_{i}))(t_{i} - t_{i-1})\right| \leqslant \sum_{i=1}^{n}|f(\lambda_{i})||g(\eta_{i}) - g(\lambda_{i})|(t_{i} - t_{i-1}) \\ \leqslant M\sum_{i=1}^{n}(M_i(g) - m_i(g))(t_{i} - t_{i-1}),  $$
where $M_i(g) = \sup_{t \in [t_{i-1},t_i]} g(t)$ and $m_i(g) = \inf_{t \in [t_{i-1},t_i]} g(t).$
Choose points $z_i$ and $z'_i$ such that for $\epsilon > 0$
$$g(z_i) \geqslant M_i(g) - \epsilon/(2M(b-a)), \\ g(z'_i) \leqslant m_i(g) + \epsilon/(2M(b-a)). $$
Then we have
$$\left|\sum_{i=1}^{n}f(\lambda_{i})(g(\eta_{i}) - g(\lambda_{i}))(t_{i} - t_{i-1})\right| \leqslant M \left(\sum_{i=1}^n g(z_i)(t_i - t_{i-1}) - \sum_{i=1}^n g(z_i')(t_i - t_{i-1}) \right) + \epsilon \\ \leqslant M \left|\sum_{i=1}^n g(z_i)(t_i - t_{i-1}) - \int_a^bg\right|+  M\left|\sum_{i=1}^n g(z_i')(t_i - t_{i-1}) - \int_a^bg\right| + \epsilon. $$
The limit of the RHS as $\|P\| \to 0$ is $\epsilon$.  Since $\epsilon$ can arbitrarily close to $0$, the limit of the LHS is $0$.  
