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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.

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  • $\begingroup$ If $B_i$ are disjoint and cover $A_i$'s, then they have to be a refinement of ring generated by $A_i$'s, right? $\endgroup$ – erz May 10 '20 at 22:31
  • $\begingroup$ Yes, I think that’s right. $\endgroup$ – Aryeh Kontorovich May 11 '20 at 4:55
  • $\begingroup$ but then isn't $A_n=[\frac{1}{n},\frac{1}{n}+\varepsilon]$ a counterexample? Among $B_i$ you will have $(\frac{1}{n+1},\frac{1}{n}]$, as well as $(\frac{1}{n+1}+\varepsilon,\frac{1}{n}+\varepsilon]$, and so you won't be able to cover $A_j$'s with a finite number of $B_j$, or am i misreading the question? $\endgroup$ – erz May 11 '20 at 5:50
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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.

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Consider $X = [0, 1]^{\aleph_0}$ with cylindrical sigma algebra and product measure (of Lebesgue ones). Let $A_i = [0,1]\times\ldots \times [0, \varepsilon]_i \times [0,1]\times\ldots$ (cylinder with measure epsilon). What are $B_i$? It seems it is not possible.

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Let $X=[0,1]$, let $\epsilon\gt0$, and let $\{A_i:i\in\mathbb N\}$ be the set of all $A\subseteq[0,1]$ such that $A$ is a finite union of rational intervals and $\mu(A)\lt\epsilon$. Let $\{B_i:i\in\mathbb N\}$ be any countable partition of $[0,1]$. Given any $n\in\mathbb N$, we can find $A_i$ which has nonempty intersection with each of the sets $B_1,B_2,\dots,B_n$, whence $\{B_j:j\in J\}$ covers $A_i$ only if $J\supseteq\{1,2,\dots,n\}$. Therefore $|J|$ may be required to be arbitrarily large. Moreover, if the sets $B_j$ are measurable, $\mu(\bigcup_{j\in J}B_j)$ may be required to be arbitrarily close to $1$.

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