# Approximating measure by partitions

Let $$(X,\mathcal{B},\mu)$$ be a non-atomic Borel probability space. We may assume that $$X\subseteq \mathbb{R}^d$$ is the open (or closed) unit ball, if it helps.

Let $$\mathcal{C}$$ be a countable collection of sets such that for $$\mu$$-a.e. $$x\in X$$, $$\exists \{c_n\}_{n\geq 1}\subset \mathcal{C}$$ s.t. $$\{x\}= \bigcap_{n\geq1}c_n$$, $$c_{n+1}\subseteq c_n$$ ($$n\geq1$$), and $$\mathrm{diam}(c_n)\xrightarrow[n\rightarrow\infty]{}0$$.

My question is the following:

Does there exist a positive constant $$\alpha>0$$, s.t. $$\forall r>0$$, $$\exists$$ a $$\textbf{disjoint}$$ sub-collection $$\mathcal{C}_{r}\subset \mathcal{C}$$ s.t. $$\forall c\in \mathcal{C}_{r}$$, $$\mathrm{diam}(c)\leq r$$, and $$\mu(\bigcup \mathcal{C}_{r})\geq \alpha$$ ?

Any tips, insights, or counter-examples would be greatly appreciated!

• Isn’t this a corollary of the Vitali covering theorem? (en.m.wikipedia.org/wiki/Vitali_covering_lemma) – Anthony Quas Oct 9 '19 at 15:14
• @AnthonyQuas unless I’m missing something, the use of the Vitali lemma is not trivial, since the cover I have is not necessarily by balls. I would be very happy to learn if you have a way to translate my problem to the Vitali lemma setup. – BOS Oct 9 '19 at 16:26
• @AnthonyQuas Vitali lemma is based on the fact that any point at distance less than $2r$ from a $r$-ball is covered by $3r$-ball with same center. It is clearly not true, even for convex coverings (consider several thin strips with common end) – Denis T. Oct 9 '19 at 17:34

No, such positive $$\alpha$$ may not exist.
Fix some fery fast growing function $$F$$. Let $$X$$ be a unit square, and choose $$\mathcal{C}$$ consisting of $$2^{-F(n)}$$-neigborhoods of $$2^{-n}$$ by $$2^{-n}$$ $$+$$-shapes with dyadic rational centers. They clearly satisfy your first condition, but there cannot be more than $$2^{2n}$$ disjoint crosses with diameter $$2^{-n}$$ on a unit square, and you can bound their area above by $$2 \times (2^{-F(n)} \times 2^{-n}) := S_n$$.
Function $$\sum_{n > N} 2^{2n}S_n$$ is an obvious upper bound for the area of any disjoint family with diameter of elements less than $$2^{-N}$$. So if this function has limit 0 ($$F = 2^{2^n}$$ will suffice) then $$\alpha$$ does not exist.
• thank you for this really insightful example. Although, I don’t believe it’s a counter example yet. $\alpha$ was not required to hold for all possible $\mathcal{C}$ simultaneously, but only for the one fixed $\mathcal{C}$ (i.e. one fixed $F(n)$). This example will give a constant which depends on the limit of the sum, given the fixed $F$ (i.e. given the fixed $\mathcal{C}$). Am I wrong? – BOS Oct 9 '19 at 13:08