$sigma{A_i cap B_j}=sigma{{A_i} cup {B_j}}=sigma{sigma{A_i} cup sigma{B_j}}$
$begingroup$
Let ${A_1,A_2,...,A_m}$ and ${B_1,B_2,...,B_n}$ be two partitions of $Omega$. I am searching a proof of the following facts:
a) ${A_i cap B_j}$ is a partition of $Omega$.
b) $sigma{A_i cap B_j}=sigma{{A_i} cup {B_j}}=sigma{sigma{A_i} cup sigma{B_j}}$
I already managed to proof a) and the inclusions $sigma{A_i cap B_j}subseteqsigma{{A_i} cup {B_j}}$,$sigma{{A_i} cup {B_j}}subseteqsigma{sigma{A_i} cup sigma{B_j}}$,
But I have problems to prove:
$sigma{sigma{A_i} cup sigma{B_j}}subseteqsigma{A_i cap B_j}.$
Note: here, $sigma { A_i }$ refers to the $sigma$-algebra generated by ${ A_i }$ in $Omega$; ditto for $sigma{ B_j }$, $sigma { A_i cap B_j }$, etc.
Please, someone help me with this.
probability probability-theory measure-theory
edited Dec 6 '18 at 7:09
Brahadeesh
6,46942363
asked Dec 5 '18 at 19:09
Israel BarquínIsrael Barquín
276
$endgroup$
add a comment |
$begingroup$
Let ${A_1,A_2,...,A_m}$ and ${B_1,B_2,...,B_n}$ be two partitions of $Omega$. I am searching a proof of the following facts:
a) ${A_i cap B_j}$ is a partition of $Omega$.
b) $sigma{A_i cap B_j}=sigma{{A_i} cup {B_j}}=sigma{sigma{A_i} cup sigma{B_j}}$
I already managed to proof a) and the inclusions $sigma{A_i cap B_j}subseteqsigma{{A_i} cup {B_j}}$,$sigma{{A_i} cup {B_j}}subseteqsigma{sigma{A_i} cup sigma{B_j}}$,
But I have problems to prove:
$sigma{sigma{A_i} cup sigma{B_j}}subseteqsigma{A_i cap B_j}.$
Note: here, $sigma { A_i }$ refers to the $sigma$-algebra generated by ${ A_i }$ in $Omega$; ditto for $sigma{ B_j }$, $sigma { A_i cap B_j }$, etc.
Please, someone help me with this.
probability probability-theory measure-theory
edited Dec 6 '18 at 7:09
Brahadeesh
6,46942363
asked Dec 5 '18 at 19:09
Israel BarquínIsrael Barquín
276
$endgroup$
$begingroup$
Hi and welcome to Math.SE. I don't know what you are meaning with $sigma$. Could you provide a little more context?
$endgroup$
– Daniele Tampieri
Dec 5 '18 at 19:16
1
$begingroup$
@DanieleTampieri: Sigma algebra.
$endgroup$
– Alex R.
Dec 5 '18 at 19:17
add a comment |
$begingroup$
Let ${A_1,A_2,...,A_m}$ and ${B_1,B_2,...,B_n}$ be two partitions of $Omega$. I am searching a proof of the following facts:
a) ${A_i cap B_j}$ is a partition of $Omega$.
b) $sigma{A_i cap B_j}=sigma{{A_i} cup {B_j}}=sigma{sigma{A_i} cup sigma{B_j}}$
I already managed to proof a) and the inclusions $sigma{A_i cap B_j}subseteqsigma{{A_i} cup {B_j}}$,$sigma{{A_i} cup {B_j}}subseteqsigma{sigma{A_i} cup sigma{B_j}}$,
But I have problems to prove:
$sigma{sigma{A_i} cup sigma{B_j}}subseteqsigma{A_i cap B_j}.$
Note: here, $sigma { A_i }$ refers to the $sigma$-algebra generated by ${ A_i }$ in $Omega$; ditto for $sigma{ B_j }$, $sigma { A_i cap B_j }$, etc.
Please, someone help me with this.
probability probability-theory measure-theory
edited Dec 6 '18 at 7:09
Brahadeesh
6,46942363
asked Dec 5 '18 at 19:09
Israel BarquínIsrael Barquín
276
$endgroup$
Let ${A_1,A_2,...,A_m}$ and ${B_1,B_2,...,B_n}$ be two partitions of $Omega$. I am searching a proof of the following facts:
a) ${A_i cap B_j}$ is a partition of $Omega$.
b) $sigma{A_i cap B_j}=sigma{{A_i} cup {B_j}}=sigma{sigma{A_i} cup sigma{B_j}}$
I already managed to proof a) and the inclusions $sigma{A_i cap B_j}subseteqsigma{{A_i} cup {B_j}}$,$sigma{{A_i} cup {B_j}}subseteqsigma{sigma{A_i} cup sigma{B_j}}$,
But I have problems to prove:
$sigma{sigma{A_i} cup sigma{B_j}}subseteqsigma{A_i cap B_j}.$
Note: here, $sigma { A_i }$ refers to the $sigma$-algebra generated by ${ A_i }$ in $Omega$; ditto for $sigma{ B_j }$, $sigma { A_i cap B_j }$, etc.
Please, someone help me with this.
probability probability-theory measure-theory
probability probability-theory measure-theory
edited Dec 6 '18 at 7:09
Brahadeesh
6,46942363
asked Dec 5 '18 at 19:09
Israel BarquínIsrael Barquín
276
edited Dec 6 '18 at 7:09
Brahadeesh
6,46942363
asked Dec 5 '18 at 19:09
Israel BarquínIsrael Barquín
276
edited Dec 6 '18 at 7:09
Brahadeesh
6,46942363
edited Dec 6 '18 at 7:09
Brahadeesh
6,46942363
edited Dec 6 '18 at 7:09
Brahadeesh
6,46942363
6,46942363
asked Dec 5 '18 at 19:09
Israel BarquínIsrael Barquín
276
asked Dec 5 '18 at 19:09
Israel BarquínIsrael Barquín
276
asked Dec 5 '18 at 19:09
Israel BarquínIsrael Barquín
276
276
$begingroup$
Hi and welcome to Math.SE. I don't know what you are meaning with $sigma$. Could you provide a little more context?
$endgroup$
– Daniele Tampieri
Dec 5 '18 at 19:16
1
$begingroup$
@DanieleTampieri: Sigma algebra.
$endgroup$
– Alex R.
Dec 5 '18 at 19:17
add a comment |
$begingroup$
Hi and welcome to Math.SE. I don't know what you are meaning with $sigma$. Could you provide a little more context?
$endgroup$
– Daniele Tampieri
Dec 5 '18 at 19:16
1
$begingroup$
@DanieleTampieri: Sigma algebra.
$endgroup$
– Alex R.
Dec 5 '18 at 19:17
$begingroup$
Hi and welcome to Math.SE. I don't know what you are meaning with $sigma$. Could you provide a little more context?
$endgroup$
– Daniele Tampieri
Dec 5 '18 at 19:16
$begingroup$
Hi and welcome to Math.SE. I don't know what you are meaning with $sigma$. Could you provide a little more context?
$endgroup$
– Daniele Tampieri
Dec 5 '18 at 19:16
1
1
$begingroup$
@DanieleTampieri: Sigma algebra.
$endgroup$
– Alex R.
Dec 5 '18 at 19:17
$begingroup$
@DanieleTampieri: Sigma algebra.
$endgroup$
– Alex R.
Dec 5 '18 at 19:17
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: Prove that for every $mathcal{A} subset mathcal{P}(Omega)$, where $mathcal{P}(Omega)$ is the power set of $Omega$, one has $$sigma{ sigma { mathcal{A} } } = sigma{ mathcal{A} }.$$
answered Dec 6 '18 at 7:06
BrahadeeshBrahadeesh
6,46942363
$endgroup$
$begingroup$
Oh, but this would be used to proved the whole equality in another way, right?
$endgroup$
– Israel Barquín
Dec 6 '18 at 17:39
$begingroup$
@JBr Yes, you would use this to show that each of the two containments you have proved also hold in the other direction, thereby proving that they are all equal.
$endgroup$
– Brahadeesh
Dec 6 '18 at 17:46
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Hint: Prove that for every $mathcal{A} subset mathcal{P}(Omega)$, where $mathcal{P}(Omega)$ is the power set of $Omega$, one has $$sigma{ sigma { mathcal{A} } } = sigma{ mathcal{A} }.$$
answered Dec 6 '18 at 7:06
BrahadeeshBrahadeesh
6,46942363
$endgroup$
$begingroup$
Oh, but this would be used to proved the whole equality in another way, right?
$endgroup$
– Israel Barquín
Dec 6 '18 at 17:39
$begingroup$
@JBr Yes, you would use this to show that each of the two containments you have proved also hold in the other direction, thereby proving that they are all equal.
$endgroup$
– Brahadeesh
Dec 6 '18 at 17:46
add a comment |
$begingroup$
Hint: Prove that for every $mathcal{A} subset mathcal{P}(Omega)$, where $mathcal{P}(Omega)$ is the power set of $Omega$, one has $$sigma{ sigma { mathcal{A} } } = sigma{ mathcal{A} }.$$
answered Dec 6 '18 at 7:06
BrahadeeshBrahadeesh
6,46942363
$endgroup$
$begingroup$
Oh, but this would be used to proved the whole equality in another way, right?
$endgroup$
– Israel Barquín
Dec 6 '18 at 17:39
$begingroup$
@JBr Yes, you would use this to show that each of the two containments you have proved also hold in the other direction, thereby proving that they are all equal.
$endgroup$
– Brahadeesh
Dec 6 '18 at 17:46
add a comment |
$begingroup$
Hint: Prove that for every $mathcal{A} subset mathcal{P}(Omega)$, where $mathcal{P}(Omega)$ is the power set of $Omega$, one has $$sigma{ sigma { mathcal{A} } } = sigma{ mathcal{A} }.$$
answered Dec 6 '18 at 7:06
BrahadeeshBrahadeesh
6,46942363
$endgroup$
Hint: Prove that for every $mathcal{A} subset mathcal{P}(Omega)$, where $mathcal{P}(Omega)$ is the power set of $Omega$, one has $$sigma{ sigma { mathcal{A} } } = sigma{ mathcal{A} }.$$
answered Dec 6 '18 at 7:06
BrahadeeshBrahadeesh
6,46942363
answered Dec 6 '18 at 7:06
BrahadeeshBrahadeesh
6,46942363
answered Dec 6 '18 at 7:06
BrahadeeshBrahadeesh
6,46942363
answered Dec 6 '18 at 7:06
BrahadeeshBrahadeesh
6,46942363
6,46942363
$begingroup$
Oh, but this would be used to proved the whole equality in another way, right?
$endgroup$
– Israel Barquín
Dec 6 '18 at 17:39
$begingroup$
@JBr Yes, you would use this to show that each of the two containments you have proved also hold in the other direction, thereby proving that they are all equal.
$endgroup$
– Brahadeesh
Dec 6 '18 at 17:46
add a comment |
$begingroup$
Oh, but this would be used to proved the whole equality in another way, right?
$endgroup$
– Israel Barquín
Dec 6 '18 at 17:39
$begingroup$
@JBr Yes, you would use this to show that each of the two containments you have proved also hold in the other direction, thereby proving that they are all equal.
$endgroup$
– Brahadeesh
Dec 6 '18 at 17:46
$begingroup$
Oh, but this would be used to proved the whole equality in another way, right?
$endgroup$
– Israel Barquín
Dec 6 '18 at 17:39
$begingroup$
Oh, but this would be used to proved the whole equality in another way, right?
$endgroup$
– Israel Barquín
Dec 6 '18 at 17:39
$begingroup$
@JBr Yes, you would use this to show that each of the two containments you have proved also hold in the other direction, thereby proving that they are all equal.
$endgroup$
– Brahadeesh
Dec 6 '18 at 17:46
$begingroup$
@JBr Yes, you would use this to show that each of the two containments you have proved also hold in the other direction, thereby proving that they are all equal.
$endgroup$
– Brahadeesh
Dec 6 '18 at 17:46
add a comment |
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$begingroup$
Hi and welcome to Math.SE. I don't know what you are meaning with $sigma$. Could you provide a little more context?
$endgroup$
– Daniele Tampieri
Dec 5 '18 at 19:16
1
$begingroup$
@DanieleTampieri: Sigma algebra.
$endgroup$
– Alex R.
Dec 5 '18 at 19:17