Timeline for Demystifying the Caratheodory approach to measurability
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Jan 14, 2018 at 22:43 | comment | added | Akiva Weinberger | @FreeziiS A counterexample can be found otherwise on $(X,m^*)$ where $X$ has more than two elements and $m^*$ is $0$ on the empty set, $2$ on $X$, and $1$ on any other set. The inner and outer measures agree everywhere there, but no set other than $\emptyset$ and $X$ is measurable in the sense of Carathéodory (and $m^*$ fails to be a measure). | |
| Jan 14, 2018 at 22:41 | comment | added | Akiva Weinberger | How do you prove that claim, that is, that if $m^*$ was indeed from a countable additive measure then a subset of $X$ will be measurable in the sense of Caratheodory iff its outer and inner measures agree? | |
| Sep 13, 2015 at 23:47 | comment | added | Zardo | For the case $m^*(X) =\infty$, it is useful to note, that for every $A\subseteq X$ the statement "for every set $S$ we have $m^*(S)= m^*(A\cap S)+m^*(A^C\cap S)$" is equivalent to "for every set $S$ with $m^*(S)<\infty$ we have $m^*(A\cap S) = m^*(S) - m^*(A^C\cap S)$." The last expression is basically a "local" form of an inner measure. Requiring a finite outer measure for $S$ is no restriction since for sets of infinite outer measure the (first) equality always holds due to the sub-additivity of $m^*$. | |
| Sep 1, 2014 at 20:00 | comment | added | C-star-W-star | Is it still sufficient to require that inner and outer measure if the countably additive measure wasn't defined on an algebra? | |
| Nov 16, 2011 at 19:19 | vote | accept | Michael Greinecker | ||
| Jul 31, 2010 at 15:02 | history | answered | Mark | CC BY-SA 2.5 |