For $n \in \Bbb N$, can we extend the Lebesgue measure to some measure(or complete measure) on $\Bbb R^n$? Can we extent the Lebesgue measure to some measure on the power set of $\Bbb R^n$? (The answer is negative for $n=3$, by using Banach- Tarski s paradox) Many thanks for your help.
Asked
Active
Viewed 48 times
1
-
Such measure is possible only for $n=1$ and $n=2$. If $n\ge 3$ than the Banach-Tarski's theorem (paradox) provides a counter-example, it is not for the case $n=3$ only. – Mher Dec 03 '14 at 14:11
-
It depends on the axiom of choice. If you assume it, then even for $n=1$ there exist non Lebesgue measurable sets, making extending Lebesgue measure to power set of $\mathbb{R}^n$ impossible. – Milly Dec 03 '14 at 15:24
-
1Lebesgue measure may be extended, even to non-Lebesgue-measurable sets. – GEdgar Dec 03 '14 at 15:54
-
Banach-Tarski precludes a finitely-additive measure that is preserved by translations and rotations. I wonder if ali means something different? – GEdgar Dec 03 '14 at 15:59
-
(Mher)Why you say that such measure exists for $n=1$ or $n=2$. In my question a measuremeans that a countabily additive function on a sigma-algebra.\ – Arman Dec 31 '14 at 10:58
-
(GEdgar) The cllasical version of Banach-Tarski theorem, precludes a finitely-additive measure for $n=3$. But What happen when $n\neq 3$? For example can we extend the Lebesgue measure which is defined on Lebesgue measurable sets of $\Bbb,$ to some measure on the power set of $\Bbb R$? – Arman Dec 31 '14 at 11:06
-
((Whith many thanks for your answers.)) – Arman Dec 31 '14 at 11:07