Are Cat. II sets measurable (for a given outer measure)
$begingroup$
Given a complete metric space $X$. Let $mu^*(E)=0$ if $E$ is of Cat. I, $mu^*(E)=1$ if $E$ is of Cat. II.
I have proved that $mu^*$ is an outer measure.
What are the $mu^*$-measurable sets?
I've proved that all Cat. I sets are measurable. But I found no way to prove/disprove that the Cat. II sets are measurable.
Proof of All Cat. I sets are measurable:
Suppose $A$ is Cat. I, then the Carathéodory condition becomes $mu^*(S)gemu^*(Scap A)+mu^*(Scap A^C)=0+mu^*(Scap A^C)$ which is always true (subsets of Cat I are Cat I sets), thus Cat. I sets are measurable.
general-topology outer-measure
asked Dec 11 '18 at 9:56
PeptideChainPeptideChain
452311
$endgroup$
add a comment |
$begingroup$
Given a complete metric space $X$. Let $mu^*(E)=0$ if $E$ is of Cat. I, $mu^*(E)=1$ if $E$ is of Cat. II.
I have proved that $mu^*$ is an outer measure.
What are the $mu^*$-measurable sets?
I've proved that all Cat. I sets are measurable. But I found no way to prove/disprove that the Cat. II sets are measurable.
Proof of All Cat. I sets are measurable:
Suppose $A$ is Cat. I, then the Carathéodory condition becomes $mu^*(S)gemu^*(Scap A)+mu^*(Scap A^C)=0+mu^*(Scap A^C)$ which is always true (subsets of Cat I are Cat I sets), thus Cat. I sets are measurable.
general-topology outer-measure
asked Dec 11 '18 at 9:56
PeptideChainPeptideChain
452311
$endgroup$
$begingroup$
Are open and dense subsets measurable? Maybe also post your proof that first category subsets are, too.
$endgroup$
– Henno Brandsma
Dec 11 '18 at 22:17
add a comment |
$begingroup$
Given a complete metric space $X$. Let $mu^*(E)=0$ if $E$ is of Cat. I, $mu^*(E)=1$ if $E$ is of Cat. II.
I have proved that $mu^*$ is an outer measure.
What are the $mu^*$-measurable sets?
I've proved that all Cat. I sets are measurable. But I found no way to prove/disprove that the Cat. II sets are measurable.
Proof of All Cat. I sets are measurable:
Suppose $A$ is Cat. I, then the Carathéodory condition becomes $mu^*(S)gemu^*(Scap A)+mu^*(Scap A^C)=0+mu^*(Scap A^C)$ which is always true (subsets of Cat I are Cat I sets), thus Cat. I sets are measurable.
general-topology outer-measure
asked Dec 11 '18 at 9:56
PeptideChainPeptideChain
452311
$endgroup$
Given a complete metric space $X$. Let $mu^*(E)=0$ if $E$ is of Cat. I, $mu^*(E)=1$ if $E$ is of Cat. II.
I have proved that $mu^*$ is an outer measure.
What are the $mu^*$-measurable sets?
I've proved that all Cat. I sets are measurable. But I found no way to prove/disprove that the Cat. II sets are measurable.
Proof of All Cat. I sets are measurable:
Suppose $A$ is Cat. I, then the Carathéodory condition becomes $mu^*(S)gemu^*(Scap A)+mu^*(Scap A^C)=0+mu^*(Scap A^C)$ which is always true (subsets of Cat I are Cat I sets), thus Cat. I sets are measurable.
general-topology outer-measure
general-topology outer-measure
asked Dec 11 '18 at 9:56
PeptideChainPeptideChain
452311
asked Dec 11 '18 at 9:56
PeptideChainPeptideChain
452311
edited Dec 12 '18 at 7:11
PeptideChain
asked Dec 11 '18 at 9:56
PeptideChainPeptideChain
452311
asked Dec 11 '18 at 9:56
PeptideChainPeptideChain
452311
asked Dec 11 '18 at 9:56
PeptideChainPeptideChain
452311
452311
$begingroup$
Are open and dense subsets measurable? Maybe also post your proof that first category subsets are, too.
$endgroup$
– Henno Brandsma
Dec 11 '18 at 22:17
add a comment |
$begingroup$
Are open and dense subsets measurable? Maybe also post your proof that first category subsets are, too.
$endgroup$
– Henno Brandsma
Dec 11 '18 at 22:17
$begingroup$
Are open and dense subsets measurable? Maybe also post your proof that first category subsets are, too.
$endgroup$
– Henno Brandsma
Dec 11 '18 at 22:17
$begingroup$
Are open and dense subsets measurable? Maybe also post your proof that first category subsets are, too.
$endgroup$
– Henno Brandsma
Dec 11 '18 at 22:17
add a comment |
1 Answer
1
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votes
$begingroup$
There are 4 types of sets in a general space $X$ (a set is first category ($C^I$) or if not it's second category ($C^{II}$) :
$A$ $C^I$ and $A^complement$ $C^I$.
$A$ $C^I$ and $A^complement$ $C^{II}$.
$A$ $C^{II}$ and $A^complement$ $C^I$.
$A$ $C^{II}$ and $A^complement$ $C^{II}$.
In a complete space type 1 does not exist, as this would imply $X$ is first category.
You've already shown that type 2 is measurable.
Type 3. is exactly the complement of sets of type 2, so also measurable.
(also easy to show directly, by the same argument as for 2).
Type 4. is not measurable as we can take $S=X$ and get a contradiction $1ge 2$ in the Carathéodory condition.
So the measurable sets are those such that $A$ or its complement is first category (but not both as that is impossible in this situation).
answered Dec 12 '18 at 22:49
Henno BrandsmaHenno Brandsma
108k347114
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
There are 4 types of sets in a general space $X$ (a set is first category ($C^I$) or if not it's second category ($C^{II}$) :
$A$ $C^I$ and $A^complement$ $C^I$.
$A$ $C^I$ and $A^complement$ $C^{II}$.
$A$ $C^{II}$ and $A^complement$ $C^I$.
$A$ $C^{II}$ and $A^complement$ $C^{II}$.
In a complete space type 1 does not exist, as this would imply $X$ is first category.
You've already shown that type 2 is measurable.
Type 3. is exactly the complement of sets of type 2, so also measurable.
(also easy to show directly, by the same argument as for 2).
Type 4. is not measurable as we can take $S=X$ and get a contradiction $1ge 2$ in the Carathéodory condition.
So the measurable sets are those such that $A$ or its complement is first category (but not both as that is impossible in this situation).
answered Dec 12 '18 at 22:49
Henno BrandsmaHenno Brandsma
108k347114
$endgroup$
add a comment |
$begingroup$
There are 4 types of sets in a general space $X$ (a set is first category ($C^I$) or if not it's second category ($C^{II}$) :
$A$ $C^I$ and $A^complement$ $C^I$.
$A$ $C^I$ and $A^complement$ $C^{II}$.
$A$ $C^{II}$ and $A^complement$ $C^I$.
$A$ $C^{II}$ and $A^complement$ $C^{II}$.
In a complete space type 1 does not exist, as this would imply $X$ is first category.
You've already shown that type 2 is measurable.
Type 3. is exactly the complement of sets of type 2, so also measurable.
(also easy to show directly, by the same argument as for 2).
Type 4. is not measurable as we can take $S=X$ and get a contradiction $1ge 2$ in the Carathéodory condition.
So the measurable sets are those such that $A$ or its complement is first category (but not both as that is impossible in this situation).
answered Dec 12 '18 at 22:49
Henno BrandsmaHenno Brandsma
108k347114
$endgroup$
add a comment |
$begingroup$
There are 4 types of sets in a general space $X$ (a set is first category ($C^I$) or if not it's second category ($C^{II}$) :
$A$ $C^I$ and $A^complement$ $C^I$.
$A$ $C^I$ and $A^complement$ $C^{II}$.
$A$ $C^{II}$ and $A^complement$ $C^I$.
$A$ $C^{II}$ and $A^complement$ $C^{II}$.
In a complete space type 1 does not exist, as this would imply $X$ is first category.
You've already shown that type 2 is measurable.
Type 3. is exactly the complement of sets of type 2, so also measurable.
(also easy to show directly, by the same argument as for 2).
Type 4. is not measurable as we can take $S=X$ and get a contradiction $1ge 2$ in the Carathéodory condition.
So the measurable sets are those such that $A$ or its complement is first category (but not both as that is impossible in this situation).
answered Dec 12 '18 at 22:49
Henno BrandsmaHenno Brandsma
108k347114
$endgroup$
There are 4 types of sets in a general space $X$ (a set is first category ($C^I$) or if not it's second category ($C^{II}$) :
$A$ $C^I$ and $A^complement$ $C^I$.
$A$ $C^I$ and $A^complement$ $C^{II}$.
$A$ $C^{II}$ and $A^complement$ $C^I$.
$A$ $C^{II}$ and $A^complement$ $C^{II}$.
In a complete space type 1 does not exist, as this would imply $X$ is first category.
You've already shown that type 2 is measurable.
Type 3. is exactly the complement of sets of type 2, so also measurable.
(also easy to show directly, by the same argument as for 2).
Type 4. is not measurable as we can take $S=X$ and get a contradiction $1ge 2$ in the Carathéodory condition.
So the measurable sets are those such that $A$ or its complement is first category (but not both as that is impossible in this situation).
answered Dec 12 '18 at 22:49
Henno BrandsmaHenno Brandsma
108k347114
answered Dec 12 '18 at 22:49
Henno BrandsmaHenno Brandsma
108k347114
answered Dec 12 '18 at 22:49
Henno BrandsmaHenno Brandsma
108k347114
answered Dec 12 '18 at 22:49
Henno BrandsmaHenno Brandsma
108k347114
108k347114
add a comment |
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$begingroup$
Are open and dense subsets measurable? Maybe also post your proof that first category subsets are, too.
$endgroup$
– Henno Brandsma
Dec 11 '18 at 22:17