Mixing
measure of the union taken from a finite subcollection
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Let $c>0$ be a constant and $mathcal C$ be a collection of balls in $mathbb R^d$ such that $m(cup_Bin mathcal CB)>c$. I wonder if we can find finitely many $B_i$'s from $mathcal C$ such that $m(cup_iB_i)>c$. $m$ is the Lebesgue measure.
real-analysis measure-theory
asked Jul 26 at 23:22
Ti Wen
1,8021418
add a comment |Â
up vote
1
down vote
favorite
Let $c>0$ be a constant and $mathcal C$ be a collection of balls in $mathbb R^d$ such that $m(cup_Bin mathcal CB)>c$. I wonder if we can find finitely many $B_i$'s from $mathcal C$ such that $m(cup_iB_i)>c$. $m$ is the Lebesgue measure.
real-analysis measure-theory
asked Jul 26 at 23:22
Ti Wen
1,8021418
1
Are these necessarily open balls? If so, I think the answer is yes.
– Clayton
Jul 26 at 23:34
@Clayton Does it fail for non-open balls or arbitrary measurable sets?
– Ti Wen
Jul 26 at 23:46
1
Correct. If you allow closed balls, then each ball could be a singleton where the collection must be uncountable. In that case, finitely many will always have measure zero.
– Clayton
Jul 26 at 23:48
There seems to be a modern style of teaching Measure Theory that avoids mentioning inner measure. $S$ is Lebesgue measurable iff $m^o(S)=m^i(S)=m(S)$ where $m^o$ is Lebesgue outer measure, and the Lebesgue inner measure is $m^i(S)=sup m(T):T=overline T subset S.$... If $S$ is Lebesgue measurable and $c<m(S)<infty$ then $c<m(T)$ for some $compact $ $ Tsubset S.$
– DanielWainfleet
Jul 27 at 4:51
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $c>0$ be a constant and $mathcal C$ be a collection of balls in $mathbb R^d$ such that $m(cup_Bin mathcal CB)>c$. I wonder if we can find finitely many $B_i$'s from $mathcal C$ such that $m(cup_iB_i)>c$. $m$ is the Lebesgue measure.
real-analysis measure-theory
asked Jul 26 at 23:22
Ti Wen
1,8021418
Let $c>0$ be a constant and $mathcal C$ be a collection of balls in $mathbb R^d$ such that $m(cup_Bin mathcal CB)>c$. I wonder if we can find finitely many $B_i$'s from $mathcal C$ such that $m(cup_iB_i)>c$. $m$ is the Lebesgue measure.
real-analysis measure-theory
asked Jul 26 at 23:22
Ti Wen
1,8021418
asked Jul 26 at 23:22
Ti Wen
1,8021418
asked Jul 26 at 23:22
Ti Wen
1,8021418
asked Jul 26 at 23:22
Ti Wen
1,8021418
1,8021418
1
Are these necessarily open balls? If so, I think the answer is yes.
– Clayton
Jul 26 at 23:34
@Clayton Does it fail for non-open balls or arbitrary measurable sets?
– Ti Wen
Jul 26 at 23:46
1
Correct. If you allow closed balls, then each ball could be a singleton where the collection must be uncountable. In that case, finitely many will always have measure zero.
– Clayton
Jul 26 at 23:48
There seems to be a modern style of teaching Measure Theory that avoids mentioning inner measure. $S$ is Lebesgue measurable iff $m^o(S)=m^i(S)=m(S)$ where $m^o$ is Lebesgue outer measure, and the Lebesgue inner measure is $m^i(S)=sup m(T):T=overline T subset S.$... If $S$ is Lebesgue measurable and $c<m(S)<infty$ then $c<m(T)$ for some $compact $ $ Tsubset S.$
– DanielWainfleet
Jul 27 at 4:51
add a comment |Â
1
Are these necessarily open balls? If so, I think the answer is yes.
– Clayton
Jul 26 at 23:34
@Clayton Does it fail for non-open balls or arbitrary measurable sets?
– Ti Wen
Jul 26 at 23:46
1
Correct. If you allow closed balls, then each ball could be a singleton where the collection must be uncountable. In that case, finitely many will always have measure zero.
– Clayton
Jul 26 at 23:48
There seems to be a modern style of teaching Measure Theory that avoids mentioning inner measure. $S$ is Lebesgue measurable iff $m^o(S)=m^i(S)=m(S)$ where $m^o$ is Lebesgue outer measure, and the Lebesgue inner measure is $m^i(S)=sup m(T):T=overline T subset S.$... If $S$ is Lebesgue measurable and $c<m(S)<infty$ then $c<m(T)$ for some $compact $ $ Tsubset S.$
– DanielWainfleet
Jul 27 at 4:51
1
1
Are these necessarily open balls? If so, I think the answer is yes.
– Clayton
Jul 26 at 23:34
Are these necessarily open balls? If so, I think the answer is yes.
– Clayton
Jul 26 at 23:34
@Clayton Does it fail for non-open balls or arbitrary measurable sets?
– Ti Wen
Jul 26 at 23:46
@Clayton Does it fail for non-open balls or arbitrary measurable sets?
– Ti Wen
Jul 26 at 23:46
1
1
Correct. If you allow closed balls, then each ball could be a singleton where the collection must be uncountable. In that case, finitely many will always have measure zero.
– Clayton
Jul 26 at 23:48
Correct. If you allow closed balls, then each ball could be a singleton where the collection must be uncountable. In that case, finitely many will always have measure zero.
– Clayton
Jul 26 at 23:48
There seems to be a modern style of teaching Measure Theory that avoids mentioning inner measure. $S$ is Lebesgue measurable iff $m^o(S)=m^i(S)=m(S)$ where $m^o$ is Lebesgue outer measure, and the Lebesgue inner measure is $m^i(S)=sup m(T):T=overline T subset S.$... If $S$ is Lebesgue measurable and $c<m(S)<infty$ then $c<m(T)$ for some $compact $ $ Tsubset S.$
– DanielWainfleet
Jul 27 at 4:51
There seems to be a modern style of teaching Measure Theory that avoids mentioning inner measure. $S$ is Lebesgue measurable iff $m^o(S)=m^i(S)=m(S)$ where $m^o$ is Lebesgue outer measure, and the Lebesgue inner measure is $m^i(S)=sup m(T):T=overline T subset S.$... If $S$ is Lebesgue measurable and $c<m(S)<infty$ then $c<m(T)$ for some $compact $ $ Tsubset S.$
– DanielWainfleet
Jul 27 at 4:51
add a comment |Â
1 Answer
1
active
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votes
up vote
1
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accepted
Yes, if your balls are open balls. Since $mathbb R^d$ is second countable we can express any union of open balls as a countable union, say $cup_n=1^infty B_n$. Since $m(cup_n=1^infty B_n)=lim_nto infty m(cup_k=1^n B_k)$ we can find $n$ such that $m(cup_k=1^n B_k)>c$
answered Jul 26 at 23:35
Kavi Rama Murthy
20k2829
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Yes, if your balls are open balls. Since $mathbb R^d$ is second countable we can express any union of open balls as a countable union, say $cup_n=1^infty B_n$. Since $m(cup_n=1^infty B_n)=lim_nto infty m(cup_k=1^n B_k)$ we can find $n$ such that $m(cup_k=1^n B_k)>c$
answered Jul 26 at 23:35
Kavi Rama Murthy
20k2829
add a comment |Â
up vote
1
down vote
accepted
Yes, if your balls are open balls. Since $mathbb R^d$ is second countable we can express any union of open balls as a countable union, say $cup_n=1^infty B_n$. Since $m(cup_n=1^infty B_n)=lim_nto infty m(cup_k=1^n B_k)$ we can find $n$ such that $m(cup_k=1^n B_k)>c$
answered Jul 26 at 23:35
Kavi Rama Murthy
20k2829
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Yes, if your balls are open balls. Since $mathbb R^d$ is second countable we can express any union of open balls as a countable union, say $cup_n=1^infty B_n$. Since $m(cup_n=1^infty B_n)=lim_nto infty m(cup_k=1^n B_k)$ we can find $n$ such that $m(cup_k=1^n B_k)>c$
answered Jul 26 at 23:35
Kavi Rama Murthy
20k2829
Yes, if your balls are open balls. Since $mathbb R^d$ is second countable we can express any union of open balls as a countable union, say $cup_n=1^infty B_n$. Since $m(cup_n=1^infty B_n)=lim_nto infty m(cup_k=1^n B_k)$ we can find $n$ such that $m(cup_k=1^n B_k)>c$
answered Jul 26 at 23:35
Kavi Rama Murthy
20k2829
answered Jul 26 at 23:35
Kavi Rama Murthy
20k2829
answered Jul 26 at 23:35
Kavi Rama Murthy
20k2829
answered Jul 26 at 23:35
Kavi Rama Murthy
20k2829
20k2829
add a comment |Â
add a comment |Â
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1
Are these necessarily open balls? If so, I think the answer is yes.
– Clayton
Jul 26 at 23:34
@Clayton Does it fail for non-open balls or arbitrary measurable sets?
– Ti Wen
Jul 26 at 23:46
1
Correct. If you allow closed balls, then each ball could be a singleton where the collection must be uncountable. In that case, finitely many will always have measure zero.
– Clayton
Jul 26 at 23:48
There seems to be a modern style of teaching Measure Theory that avoids mentioning inner measure. $S$ is Lebesgue measurable iff $m^o(S)=m^i(S)=m(S)$ where $m^o$ is Lebesgue outer measure, and the Lebesgue inner measure is $m^i(S)=sup m(T):T=overline T subset S.$... If $S$ is Lebesgue measurable and $c<m(S)<infty$ then $c<m(T)$ for some $compact $ $ Tsubset S.$
– DanielWainfleet
Jul 27 at 4:51