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.


Claim: $\lim_{n\to\infty}\delta_n=\underline{\int}_a^bf$.

Denote by $P$ a partition of $[a,b]$, by $L(f,P)$ the lower Darboux sum of $f$ or the partition $P$ and by $P_n$ the partition of $[a,b]$ in $2^n$ intervals of equal length (so that $\delta_n=L(f,P_n)$.) The sequence $L(f,P_n)$ is bounded and increasing. Let $I$ be its limit, and assume that $I<\underline{\int}_a^bf$. Then there exists a partition $P$ such that $I<L(f,P)\le\underline{\int}_a^bf$. Let $P_n^*$ be the partition formed with the points of $P_n$ and $P$. Let also $M$ be a bound of $|f|$ and $K$ the number of points in $P$. $L(f,P_n)$ and $L(f,P_n^*)$ differ on the intervals of $P_n$ that contain points of $P$. Then $$ 0<L(f,P_n^*)-L(f,P_n)\le2\,M\,K\,2^{-n}, $$ from where $$ L(f,P_n)\ge L(f,P_n^*)-2\,M\,K\,2^{-n}\ge L(f,P)-2\,M\,K\,2^{-n}. $$ Taking limits as $n\to\infty$ we get $I\ge L(f,P)$, a contradiction.

  • $\begingroup$ What a beautiful proof. Similarly, if $\lim_n\Delta_n>\overline{\int_a^b}$, then there would exist a partition such that $\lim_n\Delta_n>U(f,P)$ and $U(f,P_n)\leq U(f,P)+2MK2^{-n}$, which would lead to the analogous contradiction. I also notice that the all that has been said also hold for partitions into intervals of measure $k^{-n}$ for any $k\in\mathbb{N}\setminus\{0,1\}$. Very interesting. ¡Le agradezco de todo corazón, gentilísimo Profesor! $\endgroup$ – Self-teaching worker Oct 20 '14 at 15:46

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