Let $(X,\mathscr{A},\mu)$ be a probability space and let $\{A\_1,\ldots,\}\subset\mathscr{A}$ be a countable family of sets with small measure: say $\mu(A\_i)\le\epsilon$. I am trying to show that one can find a countable (disjoint!) partition $\{B\_i\}$ of $X$ with the following property: Each $A\_i$ is covered by some $(B\_j)\_{j\in J}$ such that
$|J|$ is small (say, $1/\epsilon$) and $\mu(\cup\_{j\in J}B\_j)$ is not too large (say, $O(\epsilon)$ or even $O(\sqrt\epsilon)$).


We can assume that $\mathscr{A}$ is a Borel $\sigma$-algebra induced by some metric, if it helps.


Edit. It was pointed out by Fedja and others that the previous formulation, which required a condition like $\mu(B\_i)\le\epsilon^2$, has atomic counterexamples.



Covering families of sets by small-measure partitions
The answer is negative.
First we may always assume that there are only finitely many $B\_i$ -s: The sum of the measures of the $B\_i$-s converges so we may take the union of all but finitely many of them with measure of this union less that $\epsilon^2$ and replace this cofinite set of $B\_i$-s by their union.


Now let the $A\_i$ be independent sets of measure $\epsilon$ each (like in Andrey's comment) and $B\_i$ a finite cover as required. Each $A\_i$ is covered by a union of some $B\_j$-s whose measure (of the union) is of order $\epsilon$ (or whatever the required bound is- as long as it is of order smaller than 1). There are only finitely many such unions. The indicator functions of the $A\_i$-s tend weakly to the function which is constantly $\epsilon$. It follows that the measure of $A\_i$ intersection with each of the finite unions above tend to something of order the measure of the union times $\epsilon$ which is of order smaller that $\epsilon$ so it is impossible that all the $A\_i$ are covered by such unions.


