Yes, it does exist ( while you need to correct the codomain of your function) and it equals 0 . I think the Lebesgue sum will show you this, as , in Lebesgue theory we define ## a \cdot \inft=0 ## for finite ##a##. The convergence criterion for Lebesgue integrals show it exists, and it equals ##0##. Only one of your partitions will contain a non-zero value. The problem is this contradicts the fact that the integral is supposed to equal ##1##.I reply to both.
I know Dirac Delta is better treated as a distribution (I've studied them), but my question is related to what I wrote.
Has it any meaning to write a function the way I wrote it? If it has, in which sense should I consider the write "f(0) = +oo"? Then, is it Lebesgue measureable? WWGD answered to this writing that the integral is 0, because that "function" is different than 0 just in a set which has Lebesgue measure = 0. Since I'm not an expert of Lebesgue thery, my question is if that is possibile even for values of x where f = +oo. I only have seen definitions of "Lebesgue measureable function f in (a, b) " when f is bound inside the interval (a, b) (can be infinite at the limit for x->a or x->b).