TITLE: On the definition of "almost-everywhere" for non-complete measure spaces
QUESTION [8 upvotes]: 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.

REPLY [3 votes]: There are relatively concrete examples of a Borel set $E$ in the plane $\mathbb R^2$ such that the projection
$$
p_1(E) = \{ x \in \mathbb R | \exists y, (x,y) \in E\}
$$
is not a Borel set.  (But it is a Lebesgue set.)  LINK
So, as far as Borel sets in the line are concerned, condition $p_1(E)$ is a non-measurable condition.  There is a Borel set $Q \supset p_1(E)$ such that $Q \setminus p_1(E)$ is a null set.  And then:  
$\qquad$for almost all $x \in Q$ there exists $y$ such that $(x,y) \in E$  
is true in sense (D1) but not in sense (D2).