Let $(X, \mathbb A, m)$ be a measurable space. Let $(Y, \mathbb B, v)$ a second measure space and $(X, \bar{\mathbb A}, \bar{m})$ , $(Y, \bar{\mathbb B}, \bar{v})$ other spaces, where $\bar{\mathbb A}$ denotes the completion of $\mathbb A$ with respect to $m$.
Now, if $f: (X, \mathbb A) \to (Y, \mathbb B)$ is a measurable function and $m(f^{-1} (B)) = 0$ for any $B \in \mathbb B$ such that $v(B) = 0$ : Why is $f: (X, \bar{\mathbb A}) \to (Y, \bar{\mathbb B})$ also measurable?
Suppose that $B\in\Bbb B$ is a null set, that is, that $v(B)=0$. Then the completion of $\Bbb B$ ensures that if $A\subset B$ then $\bar v(A)=0$. And because $f^{-1}(A)\subset f^{-1}(B)$ then we find that $\bar m(f^{-1}(A))=0$ because $m(f^{-1}(B))=0$ and now $\Bbb A$ is complete. Then $f$ is also measurable in the completion of the measure spaces.