Completion with respect a collection of measures. Given a set of measures $\mathcal{M}$ on a measurable space $(\Omega,\mathscr{F})$, the $\mathcal{M}$-completion of $\mathscr{F}$ is defined as
$$ \overline{\mathscr{F}}^\mathcal{M}:=\bigcap_{\mu\in\mathcal{M}}\mathscr{F}^\mu,$$
where $\mathscr{F}^\mu$ is the completion of $\mathscr{F}$ with respect to the measure $\mu$. Let $\mathscr{N}_\mathcal{M}$ denote the set of all $\mathcal{M}$-null sets, that is, the collection of all sets $A\subset\Omega$ such that $\mu^*(A)=0$ for all $\mu\in\mathcal{M}$.
It is clear that $$\sigma(\mathscr{F},\mathscr{N}_\mathcal{M})\subset\overline{\mathscr{F}}^\mathcal{M}.$$
 The question is whether both are actually the same. I think not, but can't think of a counter example at the moment.
The relevance of the question has to do with the notion of sufficient $\sigma$--algebra and minimal sufficient $\sigma$--algebras in statistics.

The answer is no.
Example.  Let $\Omega=[0,1]$.
$$\mathcal{F}=\{Y\subset\Omega : Y \subseteq [0,1]\, \& \,Y \,\text{is countable or} \, [0,1]\setminus Y\,\text{is countable}\}$$
Let $\mathcal{M}=(\delta_x: x \in [0,1]\}$ be the family of Dirac probability measures on $(\Omega,\cal{F})$. Then $\cal{N}_M=\{ \emptyset\}$. Hence $\sigma(\cal{F},\cal{N}_M)=\sigma(\cal{F},\{ \emptyset\})=\cal{F}$.
On the other hand,  $\overline{\cal{F}}^{\mathcal{M}}=\cal{P}([0,1])$, because $\cal{F}^{\delta_x}=\cal{P}([0,1])$ for each $x \in [0,1]$, where $\cal{P}([0,1])$ denotes the power set of the set $[0,1]$. Clearly, $\cal{P}([0,1])\neq \cal{F}$.
