1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.