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May 13, 2010 at 3:11 comment added Victor Protsak How about this: A function is $Riemann{\ }integrable{\ }\iff$ the sup of lower $Darboux{\ }sums$ is equal to the inf of the upper Darboux sums $\iff$ it is bounded and the $Lebesgue{\ }measure$ of the set of discontinuity points is 0.
May 13, 2010 at 2:41 comment added Qiaochu Yuan Part of the confusion might be that textbooks such as Rudin develop the theory of the Riemann integral by using the Darboux integral. I also remember being told that Rudin goes on to assume that properties proven for the Darboux integral hold for the Riemann integral (and maybe even the Cauchy integral) without proving that this works. Sneaky.
May 13, 2010 at 2:18 comment added VA. en.wikipedia.org/wiki/Darboux_integral and en.wikipedia.org/wiki/Riemann_integral. So Darboux sums use only min and max on each interval $[x_i,x_{i+1}]$, whereas Riemann sums use $f(t_i)$ for some $t_i\in [x_i,x_{i+1}]$.
May 12, 2010 at 10:30 comment added Andrea Ferretti Well, I don't know Darboux integral, and sadly I don't remember the source where I read this, so I have no way to check.
May 12, 2010 at 3:19 comment added VA. Are you sure that Darboux did not invent Darboux integral, and Riemann actually did invent Riemann integral? The definitions are different, although equivalent.
May 11, 2010 at 9:45 comment added Andrea Ferretti Say you want to integrate a fuction f on [0,1]. Then you define its integral, if it exists, as the limit of $\sum_{i = 0}^{n-1} f(i/n)/n$. Of course this does not make much sense unless f is continous. Still it is the most naif definition of integral one can come with and sometimes it works (and replacing the limit over n with the limit over the net of partitions of [0,1] almost gives you the right definition of Riemann's integral).
May 11, 2010 at 2:24 comment added Qiaochu Yuan What is the "Cauchy integral"? I tried searching for it, but I only get results about the Cauchy integral theorem or formula...
May 10, 2010 at 19:18 history answered Andrea Ferretti CC BY-SA 2.5