"measurable with respect to completion" vs "equals a measurable function almost everywhere"? 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?
 A: Here's a counterexample with an $\aleph_1$-generated $\sigma$-algebra on the target space.  This is really a fact about any set of cardinality $\aleph_1$, but to make notation a bit easier we work with $X = Y = \omega_1 \times \{0,1\}$ (where as usual $\omega_1$ is the first uncountable ordinal).  On both sides we use the measure that assigns measure $0$ to countable sets and measure $1$ to cocountable sets (and those are the only sorts of sets we'll end up measuring).  We start by equipping $Y$ with the countable/cocountable $\sigma$-algebra on $\omega_1 \times \{0,1\}$, and we equip $X$ with the somewhat coarser $\sigma$-algebra $\mathcal{M}$ consisting of sets of the form $A \times \{0,1\}$, where $A \subseteq \omega_1$ is countable or cocountable.  Finally, our function $f$ is the identity.
Certainly we have condition 1, since the completion of $\mathcal{M}$ with respect to the measure is the entire countable/cocountable $\sigma$-algebra.  On the other hand, we don't have condition 2.  Towards a contradiction, suppose there is some $\mathcal{M}$-measurable function $g$ agreeing with the identity off of a null (thus countable) set $C \subseteq \omega_1 \times \{0,1\}$.  Choose $\alpha \in \omega_1$ such that $(\alpha, 0)$ is not in $C \cup g(C)$.  Then the singleton $S=\{(\alpha,0)\}$ is measurable in $Y$, but $g^{-1}(S) = \{(\alpha,0)\}$ is not $\mathcal{M}$-measurable, a contradiction.
