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 $f(x)=f(y+z)$.