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Prove that the Extension Theorem is Complete.
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Prove that the probability triple $(Omega, mathcalM, mathbbP^*) $ constructed from the Extension Theorem is complete.
Specifically, this problem asks us to show that, given any $A subseteq Omega$, such that $A in mathcalM$ and $ mathbbP^* (A)=0$, and any $B subseteq A$, then $$ mathbbP^*(B cap E) + mathbbP^*(B^c cap E) = mathbbP^*(E).$$
The proof of this theorem posted below seemed fairly straightforward, but there are two issues (one of which has been previously asked by another user, without answer), which make me doubt it.
The text I am reading writes that, from this result, it follows $B$ has outer measure $0$, i.e.:
$$B in mathcalM implies mathbbP^*(B)=0,$$ but doesn’t that result follow regardless of whether $B$ is in $mathcalM$ or not?
The other issue is that the following proof does not seem to make use of the fact that $$A in mathcalM.$$
In fact, it seems we could rewrite the definition of a complete probability triple as follows:
Definition: If $A subseteq Omega$, such that $$mathbbP^*(A)=0,$$ then $$S(A) in mathcalM,$$ where $S(A)$ is the set of all subsets of $A$.
Proof:
By the monitonicity of outer measure, for any $E subseteq Omega$, $$mathbbP^*(B cap E) leq mathbbP^*(A cap E).$$
Further, since $(A cap E) subseteq A$, $$mathbbP^*(B cap E)=0. qquad (1)$$
Clearly, $(B^c cap E) subseteq E$, which implies $$mathbbP^*(B^c cap E) leq mathbbP^*(E).qquad (2)$$
Putting $(1)$ and $(2)$ together $$mathbbP^*(B cap E) + mathbbP^*(B^c cap E) leq mathbbP^*(E).$$
Since outer measure is also subadditive: $$ mathbbP^*(E) leq
mathbbP^*(B cap E) + mathbbP^*(B^c cap E).$$
Thus $mathbbP^*$ is additive on the union $B cap E$ and $B^c cap E$, for all subsets $E$ of $Omega$, from which we conclude $B in mathcalM$.
$square$
probability-theory measure-theory proof-verification
asked Jul 24 at 7:28
Moed Pol Bollo
19318
add a comment |Â
up vote
0
down vote
favorite
Prove that the probability triple $(Omega, mathcalM, mathbbP^*) $ constructed from the Extension Theorem is complete.
Specifically, this problem asks us to show that, given any $A subseteq Omega$, such that $A in mathcalM$ and $ mathbbP^* (A)=0$, and any $B subseteq A$, then $$ mathbbP^*(B cap E) + mathbbP^*(B^c cap E) = mathbbP^*(E).$$
The proof of this theorem posted below seemed fairly straightforward, but there are two issues (one of which has been previously asked by another user, without answer), which make me doubt it.
The text I am reading writes that, from this result, it follows $B$ has outer measure $0$, i.e.:
$$B in mathcalM implies mathbbP^*(B)=0,$$ but doesn’t that result follow regardless of whether $B$ is in $mathcalM$ or not?
The other issue is that the following proof does not seem to make use of the fact that $$A in mathcalM.$$
In fact, it seems we could rewrite the definition of a complete probability triple as follows:
Definition: If $A subseteq Omega$, such that $$mathbbP^*(A)=0,$$ then $$S(A) in mathcalM,$$ where $S(A)$ is the set of all subsets of $A$.
Proof:
By the monitonicity of outer measure, for any $E subseteq Omega$, $$mathbbP^*(B cap E) leq mathbbP^*(A cap E).$$
Further, since $(A cap E) subseteq A$, $$mathbbP^*(B cap E)=0. qquad (1)$$
Clearly, $(B^c cap E) subseteq E$, which implies $$mathbbP^*(B^c cap E) leq mathbbP^*(E).qquad (2)$$
Putting $(1)$ and $(2)$ together $$mathbbP^*(B cap E) + mathbbP^*(B^c cap E) leq mathbbP^*(E).$$
Since outer measure is also subadditive: $$ mathbbP^*(E) leq
mathbbP^*(B cap E) + mathbbP^*(B^c cap E).$$
Thus $mathbbP^*$ is additive on the union $B cap E$ and $B^c cap E$, for all subsets $E$ of $Omega$, from which we conclude $B in mathcalM$.
$square$
probability-theory measure-theory proof-verification
asked Jul 24 at 7:28
Moed Pol Bollo
19318
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Prove that the probability triple $(Omega, mathcalM, mathbbP^*) $ constructed from the Extension Theorem is complete.
Specifically, this problem asks us to show that, given any $A subseteq Omega$, such that $A in mathcalM$ and $ mathbbP^* (A)=0$, and any $B subseteq A$, then $$ mathbbP^*(B cap E) + mathbbP^*(B^c cap E) = mathbbP^*(E).$$
The proof of this theorem posted below seemed fairly straightforward, but there are two issues (one of which has been previously asked by another user, without answer), which make me doubt it.
The text I am reading writes that, from this result, it follows $B$ has outer measure $0$, i.e.:
$$B in mathcalM implies mathbbP^*(B)=0,$$ but doesn’t that result follow regardless of whether $B$ is in $mathcalM$ or not?
The other issue is that the following proof does not seem to make use of the fact that $$A in mathcalM.$$
In fact, it seems we could rewrite the definition of a complete probability triple as follows:
Definition: If $A subseteq Omega$, such that $$mathbbP^*(A)=0,$$ then $$S(A) in mathcalM,$$ where $S(A)$ is the set of all subsets of $A$.
Proof:
By the monitonicity of outer measure, for any $E subseteq Omega$, $$mathbbP^*(B cap E) leq mathbbP^*(A cap E).$$
Further, since $(A cap E) subseteq A$, $$mathbbP^*(B cap E)=0. qquad (1)$$
Clearly, $(B^c cap E) subseteq E$, which implies $$mathbbP^*(B^c cap E) leq mathbbP^*(E).qquad (2)$$
Putting $(1)$ and $(2)$ together $$mathbbP^*(B cap E) + mathbbP^*(B^c cap E) leq mathbbP^*(E).$$
Since outer measure is also subadditive: $$ mathbbP^*(E) leq
mathbbP^*(B cap E) + mathbbP^*(B^c cap E).$$
Thus $mathbbP^*$ is additive on the union $B cap E$ and $B^c cap E$, for all subsets $E$ of $Omega$, from which we conclude $B in mathcalM$.
$square$
probability-theory measure-theory proof-verification
asked Jul 24 at 7:28
Moed Pol Bollo
19318
Prove that the probability triple $(Omega, mathcalM, mathbbP^*) $ constructed from the Extension Theorem is complete.
Specifically, this problem asks us to show that, given any $A subseteq Omega$, such that $A in mathcalM$ and $ mathbbP^* (A)=0$, and any $B subseteq A$, then $$ mathbbP^*(B cap E) + mathbbP^*(B^c cap E) = mathbbP^*(E).$$
The proof of this theorem posted below seemed fairly straightforward, but there are two issues (one of which has been previously asked by another user, without answer), which make me doubt it.
The text I am reading writes that, from this result, it follows $B$ has outer measure $0$, i.e.:
$$B in mathcalM implies mathbbP^*(B)=0,$$ but doesn’t that result follow regardless of whether $B$ is in $mathcalM$ or not?
The other issue is that the following proof does not seem to make use of the fact that $$A in mathcalM.$$
In fact, it seems we could rewrite the definition of a complete probability triple as follows:
Definition: If $A subseteq Omega$, such that $$mathbbP^*(A)=0,$$ then $$S(A) in mathcalM,$$ where $S(A)$ is the set of all subsets of $A$.
Proof:
By the monitonicity of outer measure, for any $E subseteq Omega$, $$mathbbP^*(B cap E) leq mathbbP^*(A cap E).$$
Further, since $(A cap E) subseteq A$, $$mathbbP^*(B cap E)=0. qquad (1)$$
Clearly, $(B^c cap E) subseteq E$, which implies $$mathbbP^*(B^c cap E) leq mathbbP^*(E).qquad (2)$$
Putting $(1)$ and $(2)$ together $$mathbbP^*(B cap E) + mathbbP^*(B^c cap E) leq mathbbP^*(E).$$
Since outer measure is also subadditive: $$ mathbbP^*(E) leq
mathbbP^*(B cap E) + mathbbP^*(B^c cap E).$$
Thus $mathbbP^*$ is additive on the union $B cap E$ and $B^c cap E$, for all subsets $E$ of $Omega$, from which we conclude $B in mathcalM$.
$square$
probability-theory measure-theory proof-verification
asked Jul 24 at 7:28
Moed Pol Bollo
19318
edited Jul 24 at 7:51
asked Jul 24 at 7:28
Moed Pol Bollo
19318
asked Jul 24 at 7:28
Moed Pol Bollo
19318
asked Jul 24 at 7:28
Moed Pol Bollo
19318
19318
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
Any set $A$ with $P^*(A)=0$ is measurable. The condition $A in mathcal M$ is not necessary.
answered Jul 24 at 7:38
Kavi Rama Murthy
20.2k2829
Okay, well that’s good. Is there any reason why it was included in the question, then?
– Moed Pol Bollo
Jul 24 at 7:50
1
They were just trying state completeness of $P^*$ restricted to the class of measurable sets. This restriction is the main object of interest in analysis. Outer measure is only a tool in the study of measures.
– Kavi Rama Murthy
Jul 24 at 7:57
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Any set $A$ with $P^*(A)=0$ is measurable. The condition $A in mathcal M$ is not necessary.
answered Jul 24 at 7:38
Kavi Rama Murthy
20.2k2829
Okay, well that’s good. Is there any reason why it was included in the question, then?
– Moed Pol Bollo
Jul 24 at 7:50
1
They were just trying state completeness of $P^*$ restricted to the class of measurable sets. This restriction is the main object of interest in analysis. Outer measure is only a tool in the study of measures.
– Kavi Rama Murthy
Jul 24 at 7:57
add a comment |Â
up vote
1
down vote
Any set $A$ with $P^*(A)=0$ is measurable. The condition $A in mathcal M$ is not necessary.
answered Jul 24 at 7:38
Kavi Rama Murthy
20.2k2829
Okay, well that’s good. Is there any reason why it was included in the question, then?
– Moed Pol Bollo
Jul 24 at 7:50
1
They were just trying state completeness of $P^*$ restricted to the class of measurable sets. This restriction is the main object of interest in analysis. Outer measure is only a tool in the study of measures.
– Kavi Rama Murthy
Jul 24 at 7:57
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Any set $A$ with $P^*(A)=0$ is measurable. The condition $A in mathcal M$ is not necessary.
answered Jul 24 at 7:38
Kavi Rama Murthy
20.2k2829
Any set $A$ with $P^*(A)=0$ is measurable. The condition $A in mathcal M$ is not necessary.
answered Jul 24 at 7:38
Kavi Rama Murthy
20.2k2829
answered Jul 24 at 7:38
Kavi Rama Murthy
20.2k2829
answered Jul 24 at 7:38
Kavi Rama Murthy
20.2k2829
answered Jul 24 at 7:38
Kavi Rama Murthy
20.2k2829
20.2k2829
Okay, well that’s good. Is there any reason why it was included in the question, then?
– Moed Pol Bollo
Jul 24 at 7:50
1
They were just trying state completeness of $P^*$ restricted to the class of measurable sets. This restriction is the main object of interest in analysis. Outer measure is only a tool in the study of measures.
– Kavi Rama Murthy
Jul 24 at 7:57
add a comment |Â
Okay, well that’s good. Is there any reason why it was included in the question, then?
– Moed Pol Bollo
Jul 24 at 7:50
1
They were just trying state completeness of $P^*$ restricted to the class of measurable sets. This restriction is the main object of interest in analysis. Outer measure is only a tool in the study of measures.
– Kavi Rama Murthy
Jul 24 at 7:57
Okay, well that’s good. Is there any reason why it was included in the question, then?
– Moed Pol Bollo
Jul 24 at 7:50
Okay, well that’s good. Is there any reason why it was included in the question, then?
– Moed Pol Bollo
Jul 24 at 7:50
1
1
They were just trying state completeness of $P^*$ restricted to the class of measurable sets. This restriction is the main object of interest in analysis. Outer measure is only a tool in the study of measures.
– Kavi Rama Murthy
Jul 24 at 7:57
They were just trying state completeness of $P^*$ restricted to the class of measurable sets. This restriction is the main object of interest in analysis. Outer measure is only a tool in the study of measures.
– Kavi Rama Murthy
Jul 24 at 7:57
add a comment |Â
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