Timeline for Codimension of Measurable Sets
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Feb 13, 2011 at 18:13 | vote | accept | Jim Belk | ||
| Feb 13, 2011 at 18:11 | comment | added | Jim Belk | It seems to be right now. Since each $S_J$ has outer measure $1$, and $[0,1] - S_J = S_{I-J}$, every $S_J$ is non-measurable as long as $J\ne\emptyset$ and $J\ne I$. Since $S_J + S_K = S_{J+K}$, it follows that $S_J$ and $S_K$ are distinct modulo $\mathcal{M}$ as long as $J\ne K$ and $J \ne I-K$. | |
| Feb 13, 2011 at 18:05 | history | edited | Simon Thomas | CC BY-SA 2.5 |
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| Feb 13, 2011 at 17:51 | history | edited | Simon Thomas | CC BY-SA 2.5 |
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| Feb 13, 2011 at 17:42 | comment | added | Thomas Kragh | @Simon: You didn't fix it you still have the two equivalent sets Jim described. You should rather look at all subsets $J \subset I$ such that $i\notin J$ for some fixed $i\in I$. | |
| Feb 13, 2011 at 17:39 | history | edited | Simon Thomas | CC BY-SA 2.5 |
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| Feb 13, 2011 at 17:21 | comment | added | Bill Johnson | Jim, the argument I suggest in my answer for forming the $2^c$ sets avoids that (probably very minor) problem but at the expense of knowing some infinite combinatorics. | |
| Feb 13, 2011 at 17:02 | comment | added | Jim Belk | Thanks! I didn't know the result of Lusin and Sierpinski that you mentioned, and it seems potentially very helpful. Unfortunately, the argument you gave isn't correct as written: it's not strictly true that the sets $S_J$, $J\subseteq I$ are distinct modulo $\mathcal{M}$. For example, $S_I = [0,1]$ is measurable, so $S_J$ and $S_{I-J}$ are equivalent modulo $\mathcal{M}$ for each $J$. Is there some reason that there should be $2^c$ different equivalence classes of $S_J$'s? | |
| Feb 13, 2011 at 13:31 | history | answered | Simon Thomas | CC BY-SA 2.5 |