# Covering families of sets by small-measure partitions

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.

• 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? – erz May 10 '20 at 22:31
• Yes, I think that’s right. – Aryeh Kontorovich May 11 '20 at 4:55
• 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? – erz May 11 '20 at 5:50

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.
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.
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 $$1$$.