Timeline for Demystifying the Caratheodory approach to measurability
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| Jan 15, 2018 at 9:07 | history | edited | coudy | CC BY-SA 3.0 |
clarification concerning regularity
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| Jan 15, 2018 at 9:02 | comment | added | coudy | @Weinberger I am assuming that the measure is regular: for all subset $A$ of $X$, there exists a measurable set $B$ containing $A$ with the same outer measure. Given an exterior measure $m^*$, we can always build a regular measure coinciding with $m^*$ on $m^*$-measurable sets, so it is not a big restriction. Let me edit my post. | |
| Jan 14, 2018 at 22:45 | comment | added | Akiva Weinberger | You say that, if $X$ is a space of finite measure, then $\mu^*(A)+\mu^*(X\setminus A)=\mu^*(X)$ iff $A$ satisfies Carathéodory's condition. This is not true: see here. | |
| Mar 28, 2017 at 12:24 | history | answered | coudy | CC BY-SA 3.0 |