If $(X,\mathcal{B},\mu)$ is a (non-necessarily complete) measure space, we can give two different notions of a property $P(x)$ that is true **almost-everywhere** :

(D1) There is a measurable set $A$ such that $\mu(X\backslash A)=0$ and such that for all $x\in A$, $P(x)$ is true.

(D2) The subset of all $x\in X$ such that $P(x)$ holds is measurable and its complement has zero measure.

(D1) seems the one commonly used, and is obviously weaker than (D2). Of course, (D1) and (D2) are equivalent when the measure is complete.

My question is : if we take (D2) as the definition of "almost-everywhere", do we "lose" anything fundamental in the theory of measure and integration ? It seems to me (but I'm not sure) that everything is still true, for instance, the integral of a positive measurable function is zero implies that the function is zero almost everywhere in the sense of (D2). The Fubini theorem seems again to hold with (D2) too. The completeness of $L^p$ spaces (with the equivalence relation associated to almost everywhere equality corresponding to (D2)) seems to hold again (but I'm not completely sure).

I ask this question because even if (D2) is a little more annoying to use when proving that something holds almost-everywhere (we have to check measurability of the set on which the property holds), in the end it gives statements that are more precise (because precisely mesurability is implied). Therefore, I am wondering if I'm not missing anything by using (D2) only.