Let $X$ and $Y$ be sets and let $\mathscr{M}$ (resp. $\mathscr{N}$) be a $\sigma$-algebra of subsets of $X$ (resp. $Y$). Let $f:X \to Y$ be some function.
Even without any measures lying around, we can still make sense of what it means for $f$ to be measurable (ie. elements of $\mathscr{N}$ pull back to elements of $\mathscr{M}$).
Now let us specify some not necessarily complete (countably additive) measure $\mu:\mathscr{M} \to [0,\infty]$. I will call a set $N \subset X$ $\mu$-null if there exists N' \in \mathscr{M} such that N \subset N' and \mu(N') = 0. Suppose I tell you that $f$ is "$\mu$-measurable". It seems to me there are two sensible ways to interpret this.
- I take the completion of the measure space $(X,\mathscr{M},\mu)$ and require that $f$ be measurable after replacing $\mathscr{M}$ with the resulting, possibly larger, $\sigma$-algebra. This is equivalent to requiring that, for all $B \in \mathscr{N}$, $f^{-1}(B) = A \cup N$ where $A \in \mathscr{M}$ and $N$ is $\mu$-null.
- I require that there exist some measurable function $g:X \to Y$ such that $f=g$, $\mu$-almost-everywhere (ie $\{x \in X: f(x) \neq g(x)\}$ is $\mu$-null).
It isn't too hard to see that 2 implies 1. I sort of suspect the converse fails, but I can't think of a counterexample. Thoughts?