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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.

Powder
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  • This notation $ |P| \rightarrow 0 $ means mesh$(P) \rightarrow 0$ – Powder Nov 19 '19 at 21:39
  • Fair enough. That seems like a confusing notation, as $|P|$ is already defined as the cardinality of a set, but my google search did show that it is used to represent $\operatorname{mesh}(P)$, as well. – SlipEternal Nov 19 '19 at 22:24
  • I agree with you! I used $|P|$ because is how my book define. – Powder Nov 19 '19 at 22:26
  • I could see $\lVert P \rVert = \operatorname{mesh}(P)$, as that seems an appropriate norm for a partition. But, it is what it is. :) – SlipEternal Nov 19 '19 at 22:29

1 Answers1

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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.

user284331
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