Timeline for Zero measurability of zero-sets of polynomials
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Apr 9, 2020 at 6:22 | comment | added | diadochos | Another proof by mathematical induction can be found in this article: Okamoto, M. (1973). Distinctness of the Eigenvalues of a Quadratic form in a Multivariate Sample. The Annals of Statistics, 1(4), 763–765. doi.org/10.1214/aos/1176342472 | |
| Dec 8, 2016 at 16:25 | comment | added | passerby51 | @MohanRamachandran, Oh... I see. Thanks. | |
| Dec 8, 2016 at 6:35 | comment | added | Mohan Ramachandran | @passerby51 The set B will be empty when p is monic in x . | |
| Dec 8, 2016 at 5:13 | comment | added | passerby51 | @MohanRamachandran, thanks, but which aspect it simplifies? Not sure if I can see what you mean. | |
| Dec 8, 2016 at 5:09 | comment | added | passerby51 | @NateEldredge, thanks, yes you are right. I forgot that Fubini in fact says something about general measurable sets and not just products. | |
| Dec 7, 2016 at 23:57 | comment | added | Mohan Ramachandran | By a linear change of variables you can ensure your polynomial is monic in the x variable. This will simplify your argument. | |
| Dec 7, 2016 at 22:54 | comment | added | Nate Eldredge | Your argument works fine. If $p(x,y) = p_k(y) x^k + \dots + p_1(y) x + p_0(y)$, then $B$ is the intersection of the zero sets of the $p_i$. Some of the $p_i$ could be the zero polynomial, but by assumption, at least one is not. So its zero set has measure zero; thus so does $B$. Also, your "disintegration theorem" is just Fubini again; to make it easier to see, try writing your measure expressions as the integral of an indicator function. | |
| Dec 7, 2016 at 20:44 | answer | added | Misha | timeline score: 5 | |
| Dec 7, 2016 at 20:42 | comment | added | passerby51 | @Misha, Could you elaborate a bit? | |
| Dec 7, 2016 at 20:29 | comment | added | Misha | Follows from existence of a triangulation with smooth open simplices. | |
| Dec 7, 2016 at 20:18 | comment | added | passerby51 | @roy smith, Great! Thanks. Would that imply the real case? It doesn't seem to be a direct implication, the dimension of the Lebesgue measures are different for the real and complex cases. | |
| Dec 7, 2016 at 20:05 | comment | added | roy smith | Gunning and Rossi prove, on p.9, the zero locus of a holomorphic function has Lebesgue measure zero, if you want the complex case. | |
| Dec 7, 2016 at 20:00 | history | edited | passerby51 | CC BY-SA 3.0 |
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| Dec 7, 2016 at 19:49 | history | edited | passerby51 | CC BY-SA 3.0 |
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| Dec 7, 2016 at 19:33 | history | asked | passerby51 | CC BY-SA 3.0 |