Why it mustn't be a sigma algebra?
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We have $(Omega, mathcal F, P)$, some probability space. Let $mathcal F_1$ - some sub-algebra of $mathcal F$ and $forall n$ define $mathcal F_{n+1}$ as class of sets, which received by countable intersecting or countable union from $mathcal F_n$. How to prove, that $cup_{n in mathbb N} mathcal F_n$ mustn't be even $sigma$-algebra? Does it exists some counterexample?
probability-theory measure-theory
asked Dec 3 '18 at 15:11
anykkanykk
665
$endgroup$
add a comment |
$begingroup$
We have $(Omega, mathcal F, P)$, some probability space. Let $mathcal F_1$ - some sub-algebra of $mathcal F$ and $forall n$ define $mathcal F_{n+1}$ as class of sets, which received by countable intersecting or countable union from $mathcal F_n$. How to prove, that $cup_{n in mathbb N} mathcal F_n$ mustn't be even $sigma$-algebra? Does it exists some counterexample?
probability-theory measure-theory
asked Dec 3 '18 at 15:11
anykkanykk
665
$endgroup$
add a comment |
$begingroup$
We have $(Omega, mathcal F, P)$, some probability space. Let $mathcal F_1$ - some sub-algebra of $mathcal F$ and $forall n$ define $mathcal F_{n+1}$ as class of sets, which received by countable intersecting or countable union from $mathcal F_n$. How to prove, that $cup_{n in mathbb N} mathcal F_n$ mustn't be even $sigma$-algebra? Does it exists some counterexample?
probability-theory measure-theory
asked Dec 3 '18 at 15:11
anykkanykk
665
$endgroup$
We have $(Omega, mathcal F, P)$, some probability space. Let $mathcal F_1$ - some sub-algebra of $mathcal F$ and $forall n$ define $mathcal F_{n+1}$ as class of sets, which received by countable intersecting or countable union from $mathcal F_n$. How to prove, that $cup_{n in mathbb N} mathcal F_n$ mustn't be even $sigma$-algebra? Does it exists some counterexample?
probability-theory measure-theory
probability-theory measure-theory
asked Dec 3 '18 at 15:11
anykkanykk
665
asked Dec 3 '18 at 15:11
anykkanykk
665
asked Dec 3 '18 at 15:11
anykkanykk
665
asked Dec 3 '18 at 15:11
anykkanykk
665
asked Dec 3 '18 at 15:11
anykkanykk
665
665
add a comment |
add a comment |
1 Answer
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$begingroup$
Ok so I can give you an example to this. Assuming that you already aware of this question which says that there exists a Borel set which is not obtained by finitely many applications of intersections/unions of closed or open sets.
Let $X$ be such a space. We let $F_1$ be the set of open and closed sets of $X$. This is an algebra.
Let $F$ be the union of all $F_n$, if by contradiction $F$ is an $sigma$-algebra it must contain all Borel sets (because the Borel is the minimal $sigma$-algebra which contains all open sets).
But if you believe to the answer in the linked question, there exists a Borel measurable set that is not a obtained from an open or closed sets by finitely many applications of intersection or union. Hence for this $X$ we have a counter-example.
answered Dec 3 '18 at 15:43
YankoYanko
7,0381629
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1 Answer
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1 Answer
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$begingroup$
Ok so I can give you an example to this. Assuming that you already aware of this question which says that there exists a Borel set which is not obtained by finitely many applications of intersections/unions of closed or open sets.
Let $X$ be such a space. We let $F_1$ be the set of open and closed sets of $X$. This is an algebra.
Let $F$ be the union of all $F_n$, if by contradiction $F$ is an $sigma$-algebra it must contain all Borel sets (because the Borel is the minimal $sigma$-algebra which contains all open sets).
But if you believe to the answer in the linked question, there exists a Borel measurable set that is not a obtained from an open or closed sets by finitely many applications of intersection or union. Hence for this $X$ we have a counter-example.
answered Dec 3 '18 at 15:43
YankoYanko
7,0381629
$endgroup$
add a comment |
$begingroup$
Ok so I can give you an example to this. Assuming that you already aware of this question which says that there exists a Borel set which is not obtained by finitely many applications of intersections/unions of closed or open sets.
Let $X$ be such a space. We let $F_1$ be the set of open and closed sets of $X$. This is an algebra.
Let $F$ be the union of all $F_n$, if by contradiction $F$ is an $sigma$-algebra it must contain all Borel sets (because the Borel is the minimal $sigma$-algebra which contains all open sets).
But if you believe to the answer in the linked question, there exists a Borel measurable set that is not a obtained from an open or closed sets by finitely many applications of intersection or union. Hence for this $X$ we have a counter-example.
answered Dec 3 '18 at 15:43
YankoYanko
7,0381629
$endgroup$
add a comment |
$begingroup$
Ok so I can give you an example to this. Assuming that you already aware of this question which says that there exists a Borel set which is not obtained by finitely many applications of intersections/unions of closed or open sets.
Let $X$ be such a space. We let $F_1$ be the set of open and closed sets of $X$. This is an algebra.
Let $F$ be the union of all $F_n$, if by contradiction $F$ is an $sigma$-algebra it must contain all Borel sets (because the Borel is the minimal $sigma$-algebra which contains all open sets).
But if you believe to the answer in the linked question, there exists a Borel measurable set that is not a obtained from an open or closed sets by finitely many applications of intersection or union. Hence for this $X$ we have a counter-example.
answered Dec 3 '18 at 15:43
YankoYanko
7,0381629
$endgroup$
Ok so I can give you an example to this. Assuming that you already aware of this question which says that there exists a Borel set which is not obtained by finitely many applications of intersections/unions of closed or open sets.
Let $X$ be such a space. We let $F_1$ be the set of open and closed sets of $X$. This is an algebra.
Let $F$ be the union of all $F_n$, if by contradiction $F$ is an $sigma$-algebra it must contain all Borel sets (because the Borel is the minimal $sigma$-algebra which contains all open sets).
But if you believe to the answer in the linked question, there exists a Borel measurable set that is not a obtained from an open or closed sets by finitely many applications of intersection or union. Hence for this $X$ we have a counter-example.
answered Dec 3 '18 at 15:43
YankoYanko
7,0381629
answered Dec 3 '18 at 15:43
YankoYanko
7,0381629
answered Dec 3 '18 at 15:43
YankoYanko
7,0381629
answered Dec 3 '18 at 15:43
YankoYanko
7,0381629
7,0381629
add a comment |
add a comment |
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