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!

Thanks ahead

  • $\begingroup$ Isn’t this a corollary of the Vitali covering theorem? (en.m.wikipedia.org/wiki/Vitali_covering_lemma) $\endgroup$ – Anthony Quas Oct 9 at 15:14
  • $\begingroup$ @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. $\endgroup$ – BOS Oct 9 at 16:26
  • $\begingroup$ @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) $\endgroup$ – Denis T. Oct 9 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.

  • $\begingroup$ 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? $\endgroup$ – BOS Oct 9 at 13:08
  • $\begingroup$ I think that edit clarified things a bit. Both C and F are fixed at the beginning. $\endgroup$ – Denis T. Oct 9 at 17:18
  • $\begingroup$ I see, thank you for this clarification. I understand now that the answer to my question will depend on some regularity of the cover elements. Could you perhaps refer me to a source which covers (no pun intended) this issue? The Vitali-Lebesgue theorem is only relevant to the Lebesgue measure. Thanks again! $\endgroup$ – BOS Oct 9 at 19:03

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