Is the completion of a saturated measure saturated? A measure $\mu$ on $(X,\mathcal{M})$ is saturated if every locally measurable set belongs to $\mathcal{M}$.  A set $E\subset X$ is locally measurable if $E\cap A\in \mathcal{M}$ whenever $A\in\mathcal{M}$ with $\mu(A)<\infty$.
It is easy to check that a $\sigma$-finite measure is saturated.  Thus a 
a related, much weaker, true statement is that the completion of a $\sigma$-finite measure is $\sigma$-finite, hence saturated.
EDIT: In the context of inducing an outer measure $\mu^*$ on $X$ from a measure $\mu$ on $(X,\mathcal M)$ (namely, by letting
$\mu^{*}(E)=\text{inf}\{\sum_1^\infty\mu(A_j): A_j\in\mathcal M, E\subset\bigcup_1^\infty A_j\}$), and then restricting it to the $\sigma$-algebra $\mathcal{M}^*$ of $\mu^*$-measurable sets to obtain a (complete) measure $\bar{\mu}$ (i.e., letting $\bar{\mu}=\mu^*|\mathcal{M^*})$, Michael Greinecker's counterexample has the following significance: In general, $\bar{\mu}$ is the saturation of the completion of $\mu$. Of course, if $\mu$ is $\sigma$-finite, then $\bar{\mu}$ is just the completion of $\mu$. But $\bar{\mu}$ may not just be the completion of $\mu$, even if $\mu$ is saturated. 
 A: No.
Let $\kappa$ be an uncountable index set and for each $k\in\kappa$, let $X_k$ be a disjoint copy of the set $\{0,1\}$, $\mathcal{M}_k=\{\emptyset,X_k\}$. We let $X=\bigcup_{k\in\kappa} X_k$ and $$\mathcal{M}=\{E\subseteq X:E\cap X_k\in\mathcal{M}_k\text{ for all }k\in\kappa\}.$$
Let $\mu$ be the measure on $(X,\mathcal{M})$ given by
$$\mu(A) = \begin{cases} \infty &\mbox{if } A\cap X_k\neq\emptyset\text{ for uncountably many }k.   \\ 
0 & \mbox{otherwise.}\end{cases}$$
It is easily seen that $(X,\mathcal{M},\mu)$ is a saturated measure space. We show that it has a completion that  is not saturated. 
Let $Z$ be the set that contains a copy of $0$ from each $X_k$. For each $k$, the null set consisting of the copy of $0$ and $1$ in $X_k$ is measurable and so the set containing a copy of $0$ in $X_k$ is in the completion. It follows that $Z$ is locally measurable in the completion. 
If $Z$ would be in the completion, we could write it as $Z=A\cup N$ with $A\in\mathcal{M}$ and $N$ being the subset of a null set. Now the only subset of $Z$ that lies in $\mathcal{M}$ is the empty set, since every set in $\mathcal{M}$ that contains some copy of $0$ in some $X_k$ it must also contain a copy of $1$ in that $X_k$. So we must have $Z=N$ and $Z$ must be the subset of a measure zero set in $\mathcal{M}$. But, again, if a set in $\mathcal{M}$ contains a copy of $0$ in some $X_k$ it must also contain a copy of $1$ in that $X_k$ and since $Z$ contains a copy of $0$ in every $X_k$ every superset of $Z$ in $\mathcal{M}$ must contain both a copy of $0$ and $1$ in each $X_k$. So the only superset of $Z$ in $\mathcal{M}$ is $X$, but $\mu(X)=\infty$. This contradiction shows that $Z$ is not in the completion, even though it is locally measurable in the completion.
