# $\lim_n \frac{b-a}{n}\sum_{k=1}^{n}\sup_{x\in[x_{k-1},x_k]}f$, $\lim_n \frac{b-a}{n}\sum_{k=1}^{n}\inf_{x\in[x_{k-1},x_k]}f$ and Darboux integrals

Let us use the notation $\overline{\int_a^b}f(x)dx$ for the Darboux upper integral of $f$ and $\underline{\int_a^b}f(x)dx$ for the lower one.

Let us construct a partition of $[a,b]$ into $n$ intervals $[x_{k-1},x_k]$ defined by $x_k=a+k(b-a)/n$ and les us consider the corresponding Darboux sums$$\Delta_n=\frac{b-a}{n}\sum_{k=1}^{n}\sup_{x\in[x_{k-1},x_k]}f(x),\quad \delta_n=\frac{b-a}{n}\sum_{k=1}^{n}\inf_{x\in[x_{k-1},x_k]}f(x).$$

It is clear, by taking the definitions of $\sup$ and $\inf$, and the fact that such partitions are subsets of all partitions of $[a,b]$ into countably many closed intervals, into account, that $\lim_n\Delta_n\geq\overline{\int_a^b}f(x)dx$ and $\lim_n\delta_n\leq \underline{\int_a^b}f(x)dx$.

I wonder whether, analogously to what happens for partitions into $m^n$ intervals with $m\ge 2$, $m\in\mathbb{N}$, it is true that $\lim_n\Delta_n=\overline{\int_a^b}f(x)dx$ and $\lim_n\delta_n= \underline{\int_a^b}f(x)dx$. I have not been able to prove it and suspect that it is not true, but I would like to ask whether the equalities $\lim_n\Delta_n=\overline{\int_a^b}f(x)dx$ and $\lim_n\delta_n= \underline{\int_a^b}f(x)dx$ are true and, if they are, how it can be proved. I thank anybody for any answer!

Restate cleverly Theorem 7 (a) removing the integrability assumption and substituting $\,\int_a^b f\,$ with $\,\underline{\int_a^b}f$ and $\,\overline{\int_a^b}f$ respectively. On p. 271, $\,I\,$ becomes $\,\overline{\int_a^b}f$.
• ...and by taking into account that, with the notation of lemma 3, $L(f,P')\le L(f,P)+N(M-m)\delta$, we analogously get that $\underline{\int_a^b}f\ge L(f,P_n)\ge L(f,P_n'-\frac{1}{n})\ge L(f,Q_n)-\frac{1}{n}>\underline{\int_a^b}f-\frac{2}{n}$. I $\infty$-ly thank you! Mar 22, 2015 at 11:30