Timeline for In practice, how is the Lebesgue measure usually generalized?
Current License: CC BY-SA 4.0
14 events
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| May 19, 2021 at 12:20 | comment | added | Tom Leinster | In response to your general question: there is a definition of the uniform measure on a compact metric space (Section 9 of arxiv.org/abs/1908.11184). This gives Haar measure for homogeneous spaces, Lebesgue measure for compact subsets of $\mathbb{R}^n$, and counting measure for finite spaces. | |
| May 19, 2021 at 10:14 | history | edited | gmvh | CC BY-SA 4.0 |
Capitalisation of proper names
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| May 18, 2021 at 7:43 | answer | added | Kostya_I | timeline score: 2 | |
| May 18, 2021 at 3:58 | comment | added | Andreas Blass | I think measures on spaces like the space of lines in the plane are studied in integral geometry. For example, there is a theorem (Crofton's formula) saying that the length of any (nice) curve $C$ in the plane equals (up to some normalization factor) the integral over the space of lines of the number of intersections of the line with $C$. | |
| May 18, 2021 at 3:55 | comment | added | Andreas Blass | @RobertFurber The lines in the projective plane (over $\mathbb R$) form a compact space, identified with the Grassmannian of $2$-planes in $\mathbb R^3$. The space of affine lines in $\mathbb R^2$ is the space of projective lines minus a single element (the line at infinity), which shouldn't affect the measure. | |
| May 18, 2021 at 1:35 | answer | added | ldrinehart | timeline score: 6 | |
| May 17, 2021 at 22:18 | comment | added | Anthony Quas | In the specific case of lines in $\mathbb R^2$, the set of lines that do not pass through 0 is in a natural bijection with $SL_2(\mathbb R)/G$ where $G$ is the subgroup of matrices of the form $\begin{pmatrix}1&0\\t&1\end{pmatrix}$. | |
| May 17, 2021 at 21:44 | comment | added | Anthony Quas | I think the answer is that Haar measure is the natural extension of the concept to (locally compact) groups. You can obtain spaces such as the space of lines as quotients of groups (I.e. quotients by a non-normal subgroup) and you can push forward Haar measure to the quotient in some cases. | |
| May 17, 2021 at 17:14 | comment | added | exfret | Looking more deeply at the Betrand's paradox wikipedia page, there is discussion that it hasn't yet been fully resolved, so this seems to be a deep question. I wonder why that is, and if this happens often in practice. | |
| May 17, 2021 at 15:56 | comment | added | Alessandro Codenotti | @RobertFurber good point, I should learn how to read. In a very general setting I know that Charatonik and others looked at measures defined on the Vietoris hyperspace of closed sets of a space with a nice measure, but I expect the "lines in the plane" case to be considerably nicer even though I don't see an answer off the top of my head | |
| May 17, 2021 at 15:46 | comment | added | Robert Furber | @AlessandroCodenotti Since exfret discussing Bertrand's paradox, I think the space of lines must be lines not required to pass through the origin, which is not the Grassmannian (and noncompact). After all, the Grassmannian of lines through the origin for $\mathbb{R}^2$ is just the circle (with its usual rotationally invariant measure). | |
| May 17, 2021 at 15:31 | comment | added | Alessandro Codenotti | Grassmanians have a natural "Haar" measure by thinking about them as quotients of $O(n)$. See the "associated measures" section en.m.wikipedia.org/wiki/Grassmannian | |
| May 17, 2021 at 14:10 | review | Close votes | |||
| May 22, 2021 at 3:06 | |||||
| May 17, 2021 at 13:51 | history | asked | exfret | CC BY-SA 4.0 |