Timeline for Meagre sets with measure zero
Current License: CC BY-SA 4.0
11 events
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| Jan 21, 2021 at 18:10 | comment | added | Nate Eldredge | @GeraldEdgar: As noted in the post linked by Christian Remling, for every Borel probability measure on a metric space, there is a meager set of full measure. With Wiener measure, a nice example is to take the Holder space $A = C^{0,\alpha}[0,1] \subset C[0,1]$ for $\alpha < 1/2$. It's a standard fact that Brownian motion is a.s. $\alpha$-Holder-continuous for $\alpha < 1/2$, so $\mu(A) = 1$, but $A$ is easily seen to be $\sigma$-compact (by Arzela-Ascoli) and therefore meager. | |
| Jan 21, 2021 at 18:06 | comment | added | Gerald Edgar | (ii) I think this is likely false as well. Can we prove (or disprove) this for the classical Wiener measure on $C[0,1]$, for example? | |
| Jan 21, 2021 at 17:58 | comment | added | Gerald Edgar | (i) You did not require completeness, you could easily have $X$ first category in itself. For example, the subspace of normed space $l^2$ consisting of sequences that are zero except for a finite number of coordinates. A normed space is certainly locally convex and metrizable. | |
| Jan 21, 2021 at 17:41 | comment | added | Nik Weaver | @ChristianRemling ah, nice! | |
| Jan 21, 2021 at 16:57 | comment | added | Christian Remling | @NikWeaver: This was answered here, the answer is no: mathoverflow.net/questions/342798/… | |
| Jan 21, 2021 at 16:50 | comment | added | Nik Weaver | Perhaps a more interesting question is: is there a Borel probability measure on $\mathbb{R}$ with respect to which every meager set is null? | |
| Jan 21, 2021 at 16:28 | comment | added | Fedor Petrov | (i) sure you can, this is rather tautological. But sometimes a measure with greater supports (intersecting $A$) may also work. | |
| Jan 21, 2021 at 16:28 | history | edited | Manolis D | CC BY-SA 4.0 |
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| Jan 21, 2021 at 16:26 | comment | added | Manolis D | @Fedor: (i) By this logic, I guess I can basically take any measure whose support lies outside $A$. (ii) Ah, right. I keep thinking about the infinite-dimensional $X$ so don't see the obvious. Thanks for pointing that out. | |
| Jan 21, 2021 at 15:33 | comment | added | Fedor Petrov | (i) say, a $\delta$-measure concentrated in a point outside $A$. (ii) if $X=\mathbb{R}$, a Gaussian measure is equivalent to Lebesgue measure and it does not always work. | |
| Jan 21, 2021 at 15:28 | history | asked | Manolis D | CC BY-SA 4.0 |