1)Let I be an integral, how we can know that it existe in sens of Lebesgue and not in sens of Riemann?
2)an integral of Lebesgue, has a relation with the derivation or no? i.e if F'=f then integral of f in sens of Lebesgue is F?
how to integrate a function Lebesgue?
SPIT! forum acting up again so I'll try AGAIN
(2) first: If the Riemann or Darboux [R or D] integral exists then all three integrals [Lebesgue (L), Riemann, and Darboux] exist and are the same. Thus, in this case, if F'=f, then then L integral exists and is equal to D and R integral which is F. However, note that formally, the L integral is initially considered only on some set, i.e. is a definite integral, as, initially, the R and D integrals are. So that the L integral between a and b of F' would be f(b)-f(a) as it is for the R & D integrals.
For (1) the classic example of where the L integral exists and the R and D integrals don't is the 'belongs to' function, i.e. the indicator function, on [0,1] for the rationals. That is, let
f(x) = 0 if x is not rational; \(\displaystyle 0\, \le\, x\, \le\, 1\)
and
f(x) = 1 if x is rational; \(\displaystyle 0\, \le\, x\, \le\, 1\)
Then, as explained on
http://en.wikipedia.org/wiki/Lebesgue_integration
since the rationals are countable on [0,1], the L integral 'from 0 to 1' of f is zero whereas neither the R nor D integral exists.