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Finite measure of uncountably many disjoint sets implies at least one of them has measure zero?
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Given a probability space $(Omega,mathcalF, P).$ Suppose $mathcalA$is a subclass of $mathcalF$. Let $mathcalA_x,epsilon$ be the class of $mathcalA-$sets satisfying $xin A^circsubset Asubset B(x,epsilon)$ and let $partial mathcalA_x,epsilon$ be the class of their boundaries.
Claim: If $partialmathcalA_x,epsilon$ contains uncountably many disjoint sets, then at least one of them mush have $P-$ measure zero.
This is in section 2 from the book " convergence of probability measures(Billingsley)".
I think to proof this claim, we may need Borel-Cantelli lemma, but Borel-Cantelli lemma is for countably many sets.
Denote the metric on $Omega$ by $rho(x,y)$.
Since for any $AinmathcalA_x,epsilon$, $xin A^circsubset Asubset B(x,epsilon)$,
then $partial Asubset partial B(x,epsilon)subset y:rho(x,y)=epsilon$.
Therefore $cup_AinpartialmathcalA_x,epsilon Asubset y:rho(x,y)=epsilon$. Thus
$P(cup_AinpartialmathcalA_x,epsilon A) leq P(y:rho(x,y)=epsilon)<infty$
But then how can I get the conclusion?
Any help will be appreciated. Thanks
real-analysis probability elementary-set-theory
edited Jul 19 at 21:08
Mike Earnest
15.3k11644
asked Jul 19 at 20:31
MathProba
213
add a comment |Â
up vote
0
down vote
favorite
Given a probability space $(Omega,mathcalF, P).$ Suppose $mathcalA$is a subclass of $mathcalF$. Let $mathcalA_x,epsilon$ be the class of $mathcalA-$sets satisfying $xin A^circsubset Asubset B(x,epsilon)$ and let $partial mathcalA_x,epsilon$ be the class of their boundaries.
Claim: If $partialmathcalA_x,epsilon$ contains uncountably many disjoint sets, then at least one of them mush have $P-$ measure zero.
This is in section 2 from the book " convergence of probability measures(Billingsley)".
I think to proof this claim, we may need Borel-Cantelli lemma, but Borel-Cantelli lemma is for countably many sets.
Denote the metric on $Omega$ by $rho(x,y)$.
Since for any $AinmathcalA_x,epsilon$, $xin A^circsubset Asubset B(x,epsilon)$,
then $partial Asubset partial B(x,epsilon)subset y:rho(x,y)=epsilon$.
Therefore $cup_AinpartialmathcalA_x,epsilon Asubset y:rho(x,y)=epsilon$. Thus
$P(cup_AinpartialmathcalA_x,epsilon A) leq P(y:rho(x,y)=epsilon)<infty$
But then how can I get the conclusion?
Any help will be appreciated. Thanks
real-analysis probability elementary-set-theory
edited Jul 19 at 21:08
Mike Earnest
15.3k11644
asked Jul 19 at 20:31
MathProba
213
Yes, but the sets in the boundary of $mathcalA_x,epsilon$ may have a chance to be disjoint
– MathProba
Jul 19 at 21:07
Well, that was written when the $partial$ was missing from the Claim.
– user577471
Jul 19 at 21:21
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given a probability space $(Omega,mathcalF, P).$ Suppose $mathcalA$is a subclass of $mathcalF$. Let $mathcalA_x,epsilon$ be the class of $mathcalA-$sets satisfying $xin A^circsubset Asubset B(x,epsilon)$ and let $partial mathcalA_x,epsilon$ be the class of their boundaries.
Claim: If $partialmathcalA_x,epsilon$ contains uncountably many disjoint sets, then at least one of them mush have $P-$ measure zero.
This is in section 2 from the book " convergence of probability measures(Billingsley)".
I think to proof this claim, we may need Borel-Cantelli lemma, but Borel-Cantelli lemma is for countably many sets.
Denote the metric on $Omega$ by $rho(x,y)$.
Since for any $AinmathcalA_x,epsilon$, $xin A^circsubset Asubset B(x,epsilon)$,
then $partial Asubset partial B(x,epsilon)subset y:rho(x,y)=epsilon$.
Therefore $cup_AinpartialmathcalA_x,epsilon Asubset y:rho(x,y)=epsilon$. Thus
$P(cup_AinpartialmathcalA_x,epsilon A) leq P(y:rho(x,y)=epsilon)<infty$
But then how can I get the conclusion?
Any help will be appreciated. Thanks
real-analysis probability elementary-set-theory
edited Jul 19 at 21:08
Mike Earnest
15.3k11644
asked Jul 19 at 20:31
MathProba
213
Given a probability space $(Omega,mathcalF, P).$ Suppose $mathcalA$is a subclass of $mathcalF$. Let $mathcalA_x,epsilon$ be the class of $mathcalA-$sets satisfying $xin A^circsubset Asubset B(x,epsilon)$ and let $partial mathcalA_x,epsilon$ be the class of their boundaries.
Claim: If $partialmathcalA_x,epsilon$ contains uncountably many disjoint sets, then at least one of them mush have $P-$ measure zero.
This is in section 2 from the book " convergence of probability measures(Billingsley)".
I think to proof this claim, we may need Borel-Cantelli lemma, but Borel-Cantelli lemma is for countably many sets.
Denote the metric on $Omega$ by $rho(x,y)$.
Since for any $AinmathcalA_x,epsilon$, $xin A^circsubset Asubset B(x,epsilon)$,
then $partial Asubset partial B(x,epsilon)subset y:rho(x,y)=epsilon$.
Therefore $cup_AinpartialmathcalA_x,epsilon Asubset y:rho(x,y)=epsilon$. Thus
$P(cup_AinpartialmathcalA_x,epsilon A) leq P(y:rho(x,y)=epsilon)<infty$
But then how can I get the conclusion?
Any help will be appreciated. Thanks
real-analysis probability elementary-set-theory
edited Jul 19 at 21:08
Mike Earnest
15.3k11644
asked Jul 19 at 20:31
MathProba
213
edited Jul 19 at 21:08
Mike Earnest
15.3k11644
edited Jul 19 at 21:08
Mike Earnest
15.3k11644
edited Jul 19 at 21:08
Mike Earnest
15.3k11644
15.3k11644
asked Jul 19 at 20:31
MathProba
213
asked Jul 19 at 20:31
MathProba
213
asked Jul 19 at 20:31
MathProba
213
213
Yes, but the sets in the boundary of $mathcalA_x,epsilon$ may have a chance to be disjoint
– MathProba
Jul 19 at 21:07
Well, that was written when the $partial$ was missing from the Claim.
– user577471
Jul 19 at 21:21
add a comment |Â
Yes, but the sets in the boundary of $mathcalA_x,epsilon$ may have a chance to be disjoint
– MathProba
Jul 19 at 21:07
Well, that was written when the $partial$ was missing from the Claim.
– user577471
Jul 19 at 21:21
Yes, but the sets in the boundary of $mathcalA_x,epsilon$ may have a chance to be disjoint
– MathProba
Jul 19 at 21:07
Yes, but the sets in the boundary of $mathcalA_x,epsilon$ may have a chance to be disjoint
– MathProba
Jul 19 at 21:07
Well, that was written when the $partial$ was missing from the Claim.
– user577471
Jul 19 at 21:21
Well, that was written when the $partial$ was missing from the Claim.
– user577471
Jul 19 at 21:21
add a comment |Â
1 Answer
1
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oldest
votes
up vote
4
down vote
You don't need anything fancy like Borel-Cantelli. Just the axioms of measure theory. In fact, a more general result is true, in your same probability space:
For any uncountable collection $mathcal Asubset mathcal F$ of disjoint subsets of $Omega$, at most countably many elements of $mathcal A$ have nonzero measure.
Proof: For any $nin mathbb N$, let $mathcal A_n=Ain mathcal Amid P(A)>frac1n$. Note that $mathcal A_n$ has fewer than $n$ elements: if there were $n$ sets $E_1,E_2,dots,E_nin mathcal A_n$, then you would have
$$
P(E_1cup E_2cupdotscup E_n)=P(E_1)+dots+P(E_n)>frac1n+dots+frac1n=1,
$$
contradicting that $P$ is a probability measure. Now, let $mathcal A_0=Ain mathcal Amid P(A)>0$. Note that $mathcal A_0=bigcup_nge 1mathcal A_n$, so $mathcal A_0$ is a countable union of finite sets. This implies $mathcal A_0$ is countable, as claimed.
answered Jul 19 at 20:49
Mike Earnest
15.3k11644
This proof is so nice, thanks!
– MathProba
Jul 19 at 21:08
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
You don't need anything fancy like Borel-Cantelli. Just the axioms of measure theory. In fact, a more general result is true, in your same probability space:
For any uncountable collection $mathcal Asubset mathcal F$ of disjoint subsets of $Omega$, at most countably many elements of $mathcal A$ have nonzero measure.
Proof: For any $nin mathbb N$, let $mathcal A_n=Ain mathcal Amid P(A)>frac1n$. Note that $mathcal A_n$ has fewer than $n$ elements: if there were $n$ sets $E_1,E_2,dots,E_nin mathcal A_n$, then you would have
$$
P(E_1cup E_2cupdotscup E_n)=P(E_1)+dots+P(E_n)>frac1n+dots+frac1n=1,
$$
contradicting that $P$ is a probability measure. Now, let $mathcal A_0=Ain mathcal Amid P(A)>0$. Note that $mathcal A_0=bigcup_nge 1mathcal A_n$, so $mathcal A_0$ is a countable union of finite sets. This implies $mathcal A_0$ is countable, as claimed.
answered Jul 19 at 20:49
Mike Earnest
15.3k11644
This proof is so nice, thanks!
– MathProba
Jul 19 at 21:08
add a comment |Â
up vote
4
down vote
You don't need anything fancy like Borel-Cantelli. Just the axioms of measure theory. In fact, a more general result is true, in your same probability space:
For any uncountable collection $mathcal Asubset mathcal F$ of disjoint subsets of $Omega$, at most countably many elements of $mathcal A$ have nonzero measure.
Proof: For any $nin mathbb N$, let $mathcal A_n=Ain mathcal Amid P(A)>frac1n$. Note that $mathcal A_n$ has fewer than $n$ elements: if there were $n$ sets $E_1,E_2,dots,E_nin mathcal A_n$, then you would have
$$
P(E_1cup E_2cupdotscup E_n)=P(E_1)+dots+P(E_n)>frac1n+dots+frac1n=1,
$$
contradicting that $P$ is a probability measure. Now, let $mathcal A_0=Ain mathcal Amid P(A)>0$. Note that $mathcal A_0=bigcup_nge 1mathcal A_n$, so $mathcal A_0$ is a countable union of finite sets. This implies $mathcal A_0$ is countable, as claimed.
answered Jul 19 at 20:49
Mike Earnest
15.3k11644
This proof is so nice, thanks!
– MathProba
Jul 19 at 21:08
add a comment |Â
up vote
4
down vote
up vote
4
down vote
You don't need anything fancy like Borel-Cantelli. Just the axioms of measure theory. In fact, a more general result is true, in your same probability space:
For any uncountable collection $mathcal Asubset mathcal F$ of disjoint subsets of $Omega$, at most countably many elements of $mathcal A$ have nonzero measure.
Proof: For any $nin mathbb N$, let $mathcal A_n=Ain mathcal Amid P(A)>frac1n$. Note that $mathcal A_n$ has fewer than $n$ elements: if there were $n$ sets $E_1,E_2,dots,E_nin mathcal A_n$, then you would have
$$
P(E_1cup E_2cupdotscup E_n)=P(E_1)+dots+P(E_n)>frac1n+dots+frac1n=1,
$$
contradicting that $P$ is a probability measure. Now, let $mathcal A_0=Ain mathcal Amid P(A)>0$. Note that $mathcal A_0=bigcup_nge 1mathcal A_n$, so $mathcal A_0$ is a countable union of finite sets. This implies $mathcal A_0$ is countable, as claimed.
answered Jul 19 at 20:49
Mike Earnest
15.3k11644
You don't need anything fancy like Borel-Cantelli. Just the axioms of measure theory. In fact, a more general result is true, in your same probability space:
For any uncountable collection $mathcal Asubset mathcal F$ of disjoint subsets of $Omega$, at most countably many elements of $mathcal A$ have nonzero measure.
Proof: For any $nin mathbb N$, let $mathcal A_n=Ain mathcal Amid P(A)>frac1n$. Note that $mathcal A_n$ has fewer than $n$ elements: if there were $n$ sets $E_1,E_2,dots,E_nin mathcal A_n$, then you would have
$$
P(E_1cup E_2cupdotscup E_n)=P(E_1)+dots+P(E_n)>frac1n+dots+frac1n=1,
$$
contradicting that $P$ is a probability measure. Now, let $mathcal A_0=Ain mathcal Amid P(A)>0$. Note that $mathcal A_0=bigcup_nge 1mathcal A_n$, so $mathcal A_0$ is a countable union of finite sets. This implies $mathcal A_0$ is countable, as claimed.
answered Jul 19 at 20:49
Mike Earnest
15.3k11644
edited Jul 19 at 21:07
answered Jul 19 at 20:49
Mike Earnest
15.3k11644
answered Jul 19 at 20:49
Mike Earnest
15.3k11644
answered Jul 19 at 20:49
Mike Earnest
15.3k11644
15.3k11644
This proof is so nice, thanks!
– MathProba
Jul 19 at 21:08
add a comment |Â
This proof is so nice, thanks!
– MathProba
Jul 19 at 21:08
This proof is so nice, thanks!
– MathProba
Jul 19 at 21:08
This proof is so nice, thanks!
– MathProba
Jul 19 at 21:08
add a comment |Â
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Yes, but the sets in the boundary of $mathcalA_x,epsilon$ may have a chance to be disjoint
– MathProba
Jul 19 at 21:07
Well, that was written when the $partial$ was missing from the Claim.
– user577471
Jul 19 at 21:21