“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.

1. 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.
2. 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?

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How general do you want $(Y,\mathscr{N})$ to be? If you allow $\mathscr{N}$ to be countably generated (which is what's used most often anyway), this is a rather easy exercise. –  t.b. Nov 14 '11 at 4:20
–  Byron Schmuland Nov 14 '11 at 4:41
@Byron: Your argument with the monotone class theorem was nice, too, but it only works if $\mu$ is finite, I think. I completely forgot about the other thread you linked to... –  t.b. Nov 14 '11 at 4:45
@t.b. I was considering arbitrary $\mathscr{N}$, but a reasonable hypothesis under which 1 and 2 are equivalent would be nice too! So by countably generated you mean there is some countable collection $G \subset 2^Y$ such that $\mathscr{N}$ is the smallest $\sigma$-algebra containing $G$? –  Mike Nov 14 '11 at 4:47
Yes, exactly. The argument is given by André Caldas in the thread Byron links to. A few minutes ago, Byron gave an argument using his favorite monotone class theorem, but unfortunately he deleted it. –  t.b. Nov 14 '11 at 4:50
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.