All:
Say $f$ is a measurable (integrable, actually) function over the Lebesgue-measurable set $S$, with $m(S)>0$.
Now, since $m(S)>0$, there exists a non-measurable subset S' of $S$, and we can then write:
S=S'\cup (S\setminus S').
How would we then go about dealing with this (sorry, I don't know how to Tex an integral)
\int_S f\,d\mu=\int_{S'} f\,d\mu+ \int_{S\setminus S'}f\,d\mu? (given that S' and S\setminus S' are clearly disjoint)
Doesn't this imply that the integral over the non-measurable subset S' can be defined?
It also seems , using inner- and outer- measure, that if S' is non-measurable, i.e. $m^*
So I'm confused here. Thanks for any comments.
Edit: what confuses me here is this:
We start with a set equality $A=B$ (given as S=S'\cup (S-S'), so that $A=S$, B=S-S', from which we cannot conclude:
$\int_A f=\int_B f$ , it is as if we had $x=y+z$ , but we cannot then conclude, for any decomposition of $x$, that $f(x)=f(y+z)$.