About unions of $sigma$-algebra being sigma algebras
$begingroup$
Let $Omega$ be a set and $mathcal{A}$ and $mathcal{B}$ be two sigma-algebras on $Omega$. Put $$mathcal{F}={Acap B:Ainmathcal{A};text{and};Binmathcal{B}}.$$
I have two question which seem intuitively true, but I am unable to prove them, since I am not a mathematician, but an engineer with an interest in probability theory:
- Is it true that the sigma-algebra generated by $mathcal{F}$ equals the sigma-algebra generated by $mathcal{Acup B}$, i.e. do we have $$sigma(mathcal{F})=sigma(mathcal{A}cupmathcal{B})?$$
- Does $mathcal{F}$ satisfy the property $$F,Ginmathcal{F}implies Fcap Ginmathcal{F}?$$
probability-theory measure-theory elementary-set-theory
edited Jan 8 at 12:46
Davide Giraudo
128k17156268
asked Aug 16 '13 at 20:37
quincequince
62
$endgroup$
add a comment |
$begingroup$
Let $Omega$ be a set and $mathcal{A}$ and $mathcal{B}$ be two sigma-algebras on $Omega$. Put $$mathcal{F}={Acap B:Ainmathcal{A};text{and};Binmathcal{B}}.$$
I have two question which seem intuitively true, but I am unable to prove them, since I am not a mathematician, but an engineer with an interest in probability theory:
- Is it true that the sigma-algebra generated by $mathcal{F}$ equals the sigma-algebra generated by $mathcal{Acup B}$, i.e. do we have $$sigma(mathcal{F})=sigma(mathcal{A}cupmathcal{B})?$$
- Does $mathcal{F}$ satisfy the property $$F,Ginmathcal{F}implies Fcap Ginmathcal{F}?$$
probability-theory measure-theory elementary-set-theory
edited Jan 8 at 12:46
Davide Giraudo
128k17156268
asked Aug 16 '13 at 20:37
quincequince
62
$endgroup$
add a comment |
$begingroup$
Let $Omega$ be a set and $mathcal{A}$ and $mathcal{B}$ be two sigma-algebras on $Omega$. Put $$mathcal{F}={Acap B:Ainmathcal{A};text{and};Binmathcal{B}}.$$
I have two question which seem intuitively true, but I am unable to prove them, since I am not a mathematician, but an engineer with an interest in probability theory:
- Is it true that the sigma-algebra generated by $mathcal{F}$ equals the sigma-algebra generated by $mathcal{Acup B}$, i.e. do we have $$sigma(mathcal{F})=sigma(mathcal{A}cupmathcal{B})?$$
- Does $mathcal{F}$ satisfy the property $$F,Ginmathcal{F}implies Fcap Ginmathcal{F}?$$
probability-theory measure-theory elementary-set-theory
edited Jan 8 at 12:46
Davide Giraudo
128k17156268
asked Aug 16 '13 at 20:37
quincequince
62
$endgroup$
Let $Omega$ be a set and $mathcal{A}$ and $mathcal{B}$ be two sigma-algebras on $Omega$. Put $$mathcal{F}={Acap B:Ainmathcal{A};text{and};Binmathcal{B}}.$$
I have two question which seem intuitively true, but I am unable to prove them, since I am not a mathematician, but an engineer with an interest in probability theory:
- Is it true that the sigma-algebra generated by $mathcal{F}$ equals the sigma-algebra generated by $mathcal{Acup B}$, i.e. do we have $$sigma(mathcal{F})=sigma(mathcal{A}cupmathcal{B})?$$
- Does $mathcal{F}$ satisfy the property $$F,Ginmathcal{F}implies Fcap Ginmathcal{F}?$$
probability-theory measure-theory elementary-set-theory
probability-theory measure-theory elementary-set-theory
edited Jan 8 at 12:46
Davide Giraudo
128k17156268
asked Aug 16 '13 at 20:37
quincequince
62
edited Jan 8 at 12:46
Davide Giraudo
128k17156268
asked Aug 16 '13 at 20:37
quincequince
62
edited Jan 8 at 12:46
Davide Giraudo
128k17156268
edited Jan 8 at 12:46
Davide Giraudo
128k17156268
edited Jan 8 at 12:46
Davide Giraudo
128k17156268
128k17156268
asked Aug 16 '13 at 20:37
quincequince
62
asked Aug 16 '13 at 20:37
quincequince
62
asked Aug 16 '13 at 20:37
quincequince
62
62
add a comment |
add a comment |
1 Answer
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$begingroup$
$mathcal F$ certainly contains $mathcal A$ and $mathcal B$, hence $sigma(mathcal F)supset sigma(mathcal Acupmathcal B)$. If $Fin mathcal F$, then $F=Acap B$ for some $Ainmathcal A$ and $Binmathcal B$. Sets of this form belong to the $sigma$-algebra generated by $mathcal Acupmathcal B$, as finite intersection of elements of $mathcal Acupmathcal B$.
Yes, since $mathcal A$ and $mathcal B$ are stable under finite intersections: write $F=Acap B$, $G=A'cap B'$, with $A,A'inmathcal A$ and $B,B'inmathcal B$. Then $Fcap G=underbrace{Acap A'}_{inmathcal A}cap underbrace{Bcap B'}_{inmathcal B}$.
answered Aug 16 '13 at 20:43
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
$begingroup$
@0xbadf00d Yes. I have edited.
$endgroup$
– Davide Giraudo
Jan 8 at 12:46
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$mathcal F$ certainly contains $mathcal A$ and $mathcal B$, hence $sigma(mathcal F)supset sigma(mathcal Acupmathcal B)$. If $Fin mathcal F$, then $F=Acap B$ for some $Ainmathcal A$ and $Binmathcal B$. Sets of this form belong to the $sigma$-algebra generated by $mathcal Acupmathcal B$, as finite intersection of elements of $mathcal Acupmathcal B$.
Yes, since $mathcal A$ and $mathcal B$ are stable under finite intersections: write $F=Acap B$, $G=A'cap B'$, with $A,A'inmathcal A$ and $B,B'inmathcal B$. Then $Fcap G=underbrace{Acap A'}_{inmathcal A}cap underbrace{Bcap B'}_{inmathcal B}$.
answered Aug 16 '13 at 20:43
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
$begingroup$
@0xbadf00d Yes. I have edited.
$endgroup$
– Davide Giraudo
Jan 8 at 12:46
add a comment |
$begingroup$
$mathcal F$ certainly contains $mathcal A$ and $mathcal B$, hence $sigma(mathcal F)supset sigma(mathcal Acupmathcal B)$. If $Fin mathcal F$, then $F=Acap B$ for some $Ainmathcal A$ and $Binmathcal B$. Sets of this form belong to the $sigma$-algebra generated by $mathcal Acupmathcal B$, as finite intersection of elements of $mathcal Acupmathcal B$.
Yes, since $mathcal A$ and $mathcal B$ are stable under finite intersections: write $F=Acap B$, $G=A'cap B'$, with $A,A'inmathcal A$ and $B,B'inmathcal B$. Then $Fcap G=underbrace{Acap A'}_{inmathcal A}cap underbrace{Bcap B'}_{inmathcal B}$.
answered Aug 16 '13 at 20:43
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
$begingroup$
@0xbadf00d Yes. I have edited.
$endgroup$
– Davide Giraudo
Jan 8 at 12:46
add a comment |
$begingroup$
$mathcal F$ certainly contains $mathcal A$ and $mathcal B$, hence $sigma(mathcal F)supset sigma(mathcal Acupmathcal B)$. If $Fin mathcal F$, then $F=Acap B$ for some $Ainmathcal A$ and $Binmathcal B$. Sets of this form belong to the $sigma$-algebra generated by $mathcal Acupmathcal B$, as finite intersection of elements of $mathcal Acupmathcal B$.
Yes, since $mathcal A$ and $mathcal B$ are stable under finite intersections: write $F=Acap B$, $G=A'cap B'$, with $A,A'inmathcal A$ and $B,B'inmathcal B$. Then $Fcap G=underbrace{Acap A'}_{inmathcal A}cap underbrace{Bcap B'}_{inmathcal B}$.
answered Aug 16 '13 at 20:43
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
$mathcal F$ certainly contains $mathcal A$ and $mathcal B$, hence $sigma(mathcal F)supset sigma(mathcal Acupmathcal B)$. If $Fin mathcal F$, then $F=Acap B$ for some $Ainmathcal A$ and $Binmathcal B$. Sets of this form belong to the $sigma$-algebra generated by $mathcal Acupmathcal B$, as finite intersection of elements of $mathcal Acupmathcal B$.
Yes, since $mathcal A$ and $mathcal B$ are stable under finite intersections: write $F=Acap B$, $G=A'cap B'$, with $A,A'inmathcal A$ and $B,B'inmathcal B$. Then $Fcap G=underbrace{Acap A'}_{inmathcal A}cap underbrace{Bcap B'}_{inmathcal B}$.
answered Aug 16 '13 at 20:43
Davide GiraudoDavide Giraudo
128k17156268
edited Jan 8 at 12:45
answered Aug 16 '13 at 20:43
Davide GiraudoDavide Giraudo
128k17156268
answered Aug 16 '13 at 20:43
Davide GiraudoDavide Giraudo
128k17156268
answered Aug 16 '13 at 20:43
Davide GiraudoDavide Giraudo
128k17156268
128k17156268
$begingroup$
@0xbadf00d Yes. I have edited.
$endgroup$
– Davide Giraudo
Jan 8 at 12:46
add a comment |
$begingroup$
@0xbadf00d Yes. I have edited.
$endgroup$
– Davide Giraudo
Jan 8 at 12:46
$begingroup$
@0xbadf00d Yes. I have edited.
$endgroup$
– Davide Giraudo
Jan 8 at 12:46
$begingroup$
@0xbadf00d Yes. I have edited.
$endgroup$
– Davide Giraudo
Jan 8 at 12:46
add a comment |
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