Let us call $\overline{\int_a^b}f(x)dx$ the Darboux upper integral of $f$ and $\underline{\int_a^b}f(x)dx$ the lower one.
Let us construct a partition of $[a,b]$ into $2^n$ intervals $[x_{k-1},x_k]$ defined by $x_k=a+k(b-a)/2^n$ and the corresponding Darboux sums$$\Delta_n=\frac{b-a}{2^n}\sum_{k=1}^{2^n}\sup_{x\in[x_{k-1},x_k]}f(x),\quad \delta_n=\frac{b-a}{2^n}\sum_{k=1}^{2^n}\inf_{x\in[x_{k-1},x_k]}f(x)$$
I see, 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$$\lim_n\Delta_n\geq\overline{\int_a^b}f(x)dx,\quad\quad\quad\lim_n\delta_n\leq \underline{\int_a^b}f(x)dx$$Moreover, in the case that, if the two Darboux integrals coincides, i.e. if $f$ is Riemann-Darboux integrable, I also see, by following standard techniques used to prove that $\overline{\int_a^b}f(x)dx=\underline{\int_a^b}f(x)dx$ if and only if $f$ is Cauchy integrable, that, in such a particular case the equality $\lim_n\Delta_n=\lim_n\delta_n=\int_a^bf(x)dx$, where $\int_a^bf(x)dx$ is the Riemann-Darboux, or Cauchy (it is the same), integral, holds.
I wonder whether $\lim_n\Delta_n=\overline{\int_a^b}f(x)dx$ and $\lim_n\delta_n=\underline{\int_a^b}f(x)dx$ hold in general, and how it can be proven.
I thank you all for any answer!
EDIT Mar 22'15: more general result here.