Timeline for Covering measure one sets by closed null sets
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 15, 2016 at 19:48 | vote | accept | Piotr Szewczak | ||
| Dec 13, 2016 at 19:26 | comment | added | Ashutosh | No worries, the comment only came a minute before your answer. | |
| Dec 13, 2016 at 19:22 | comment | added | Will Brian | @Ashutosh: Sorry, I didn't see your comments until I posted my answer. It looks like essentially the same idea. | |
| Dec 13, 2016 at 19:20 | answer | added | Will Brian | timeline score: 3 | |
| Dec 13, 2016 at 19:16 | comment | added | Ashutosh | Note that $f^{-1}$ preserves the measures of subintervals of $I$ hence it also preserves the measures of all Borel subsets of $I$. So if $\cal{F}$ is a family of compact measure zero subsets of $K$ that cover $K$, then $\{f[A]: A \in \cal{F}\}$ is a family of compact measure zero sets that cover $I$. So Q1 has a positive answer. | |
| Dec 13, 2016 at 19:16 | comment | added | Ashutosh | Here's another try: Suppose $K \subseteq [0, 1]$ is a compact positive measure and for every open set $U$, either $U \cap K$ is empty or has positive measure. Let $f: K \to I = [0, \mu(K)]$ be defined by $f(x) = \mu([0, x] \cap K)$. Then $f$ is a continuous surjection which is strictly increasing except on a countable set. | |
| Dec 13, 2016 at 18:20 | comment | added | Ashutosh | You are right. I take it back. | |
| Dec 13, 2016 at 18:14 | comment | added | Piotr Szewczak | @Ashutosh: Could you explain it? For a meager set of positive measure, its rational translates cannot cover everything. | |
| Dec 13, 2016 at 17:31 | comment | added | Goldstern | @PiotrSzewczak I assumed that you have checked it. I thought it is worth mentioning, because other people might use the results or references there - either to find an answer, or perhaps to ask interesting related questions. | |
| Dec 13, 2016 at 17:22 | comment | added | Ashutosh | For Q1, use the fact if a family of sets can cover a set of positive measure, then their rational translates can cover everything. | |
| Dec 13, 2016 at 17:13 | comment | added | Piotr Szewczak | @Goldstern: Thanks, we already consulted this reference, and failed to find an answer there. | |
| Dec 13, 2016 at 17:12 | history | edited | Piotr Szewczak | CC BY-SA 3.0 |
added 242 characters in body
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| Dec 13, 2016 at 16:46 | comment | added | Goldstern | The ideal generated by closed measure zero sets is discussed in Bartoszynski-Judah, section 2.6. | |
| Dec 13, 2016 at 16:42 | review | First posts | |||
| Dec 13, 2016 at 16:48 | |||||
| Dec 13, 2016 at 16:39 | history | asked | Piotr Szewczak | CC BY-SA 3.0 |