Timeline for Support of a regular measure Reg
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 17, 2018 at 11:14 | vote | accept | Tanmoy Paul | ||
| Nov 10, 2018 at 14:18 | comment | added | Taro Tokyo | Still false.<br> $ K=[0,1]^{\mathbb{N}}$ with the product topology <br> $ \mathbb{0}= (0,0,...) \in K$<br> $\mu(A) = \mathbb{\delta}_{\mathbb{0}}$<br> $S=K\setminus \{\mathbb{0}\}$<br> $f := 1$<br> Because $\{\mathbb{0}\}$ is not a Baire set (while it is compact), the only Baire set contained in $S^{c}$ is $\emptyset$, and $$1 = \int_{K}fd\mu \ne 0= \int_{S} fd\mu $$ | |
| Nov 10, 2018 at 12:34 | comment | added | Tanmoy Paul | According to my notations is it possible to conclude that, $\int_Sf(t)d\mu (t)=\int_Kf(t)d\mu(t)$ if $S$ is Borel and $f\in C(K)$. Here remember that $\mu(E)=0$ if $E\subseteq K\setminus S$ is a Baire set. | |
| Nov 5, 2018 at 14:08 | vote | accept | Tanmoy Paul | ||
| Nov 10, 2018 at 14:16 | |||||
| S Nov 4, 2018 at 7:42 | history | suggested | Robert Furber | CC BY-SA 4.0 |
Changed Lebesgue to the correct spelling, added grammar fixes to make 6 characters
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| Nov 4, 2018 at 7:25 | review | Suggested edits | |||
| S Nov 4, 2018 at 7:42 | |||||
| Nov 4, 2018 at 6:19 | history | answered | Taro Tokyo | CC BY-SA 4.0 |