Originally posted by royalchicken
I don't understand the first and second sentences. The definition of the integral of g over (-infinity,infinity) is the limit as a --> infinity of the integral of g over any interval [-a,a], which is 'g(a)-'g(a) = 0. I see why the same does not go for h, because 2n grows faster than n, so I think I see why one can't just take limits as I did, but: ...[text shortened]... tions which tend to some limit as x-->+-infinity are zero, and look at other cases separately?
The definition you give of the integral is not the standard one. Usually, the integral over R is defined as the limit as m and n go to infinity of the integral over [-m,n], provided this limit exists.
From an analytical point-of-view it's a bit arbitrary to talk about the integral over R being the limit of the integral over [-n,n], as there's nothing special about 0, and an odd function g(x) is no more special than a translation of it, g(x-c), which won't be odd in general. Nevertheless, the limit you defined is given a name, the Cauchy principle value
of the integral, because it's useful some contexts, eg in conjunction with Cauchy's theorem, where the line -infinity to infinity is part of the limit of some loop in C you're integrating round, and the way you form the loop allows you to take the limit symmetrically about one of the axes.
'measurable' = Lebesgue measurable (which means that a Lebesgue integral can be defined), but then this won't mean much if you haven't done Lebesgue measure and integrals. Suffice it to say that the proof we were given that non-measurable functions to R even exist required Zorn's Lemma, an alternative formulation of the Axiom of Choice.