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Uniqueness of Finite Additive Measure
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I know that for two measures defined on the Borel set of real line $mu_1$ and $mu_2$, if $mu_1((a,b))=mu_2((a,b))$ for all real numbers a and b, then $mu_1=mu_2$.
Curious on whether similar conditions hold for finite-additive, but not countably additive measures. If $v_1(A)$, $v_2(A)$ are two finitely-additive measures and $v_1(A)$ and $v_2(A)$ are equal for all $A in beta$, in which $beta$ is a collection of Borel sets. What is the 'smallest' $beta$ to ensure that these two measures are equal? You can assume $mu(Omega)=1$ as non-standard probability theory if that's necessary.
If it takes a long story to answer, a book/paper name and author is also welcomed. Many thanks for that.
probability measure-theory
asked Jul 18 at 20:31
failedstatistician
165
add a comment |Â
up vote
3
down vote
favorite
I know that for two measures defined on the Borel set of real line $mu_1$ and $mu_2$, if $mu_1((a,b))=mu_2((a,b))$ for all real numbers a and b, then $mu_1=mu_2$.
Curious on whether similar conditions hold for finite-additive, but not countably additive measures. If $v_1(A)$, $v_2(A)$ are two finitely-additive measures and $v_1(A)$ and $v_2(A)$ are equal for all $A in beta$, in which $beta$ is a collection of Borel sets. What is the 'smallest' $beta$ to ensure that these two measures are equal? You can assume $mu(Omega)=1$ as non-standard probability theory if that's necessary.
If it takes a long story to answer, a book/paper name and author is also welcomed. Many thanks for that.
probability measure-theory
asked Jul 18 at 20:31
failedstatistician
165
1
I doubt there exists any interesting such set $beta$.
– Eric Wofsey
Jul 18 at 20:35
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I know that for two measures defined on the Borel set of real line $mu_1$ and $mu_2$, if $mu_1((a,b))=mu_2((a,b))$ for all real numbers a and b, then $mu_1=mu_2$.
Curious on whether similar conditions hold for finite-additive, but not countably additive measures. If $v_1(A)$, $v_2(A)$ are two finitely-additive measures and $v_1(A)$ and $v_2(A)$ are equal for all $A in beta$, in which $beta$ is a collection of Borel sets. What is the 'smallest' $beta$ to ensure that these two measures are equal? You can assume $mu(Omega)=1$ as non-standard probability theory if that's necessary.
If it takes a long story to answer, a book/paper name and author is also welcomed. Many thanks for that.
probability measure-theory
asked Jul 18 at 20:31
failedstatistician
165
I know that for two measures defined on the Borel set of real line $mu_1$ and $mu_2$, if $mu_1((a,b))=mu_2((a,b))$ for all real numbers a and b, then $mu_1=mu_2$.
Curious on whether similar conditions hold for finite-additive, but not countably additive measures. If $v_1(A)$, $v_2(A)$ are two finitely-additive measures and $v_1(A)$ and $v_2(A)$ are equal for all $A in beta$, in which $beta$ is a collection of Borel sets. What is the 'smallest' $beta$ to ensure that these two measures are equal? You can assume $mu(Omega)=1$ as non-standard probability theory if that's necessary.
If it takes a long story to answer, a book/paper name and author is also welcomed. Many thanks for that.
probability measure-theory
asked Jul 18 at 20:31
failedstatistician
165
asked Jul 18 at 20:31
failedstatistician
165
asked Jul 18 at 20:31
failedstatistician
165
asked Jul 18 at 20:31
failedstatistician
165
165
1
I doubt there exists any interesting such set $beta$.
– Eric Wofsey
Jul 18 at 20:35
add a comment |Â
1
I doubt there exists any interesting such set $beta$.
– Eric Wofsey
Jul 18 at 20:35
1
1
I doubt there exists any interesting such set $beta$.
– Eric Wofsey
Jul 18 at 20:35
I doubt there exists any interesting such set $beta$.
– Eric Wofsey
Jul 18 at 20:35
add a comment |Â
1 Answer
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2
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Finitely additive measures are in isometric correspondence with positive functionals over $L^infty(Omega)$. Since $L^infty(Omega) = C(X)$, where $X$ is the compact Hausdorff topological space given by the Gel'fand spectrum of $L^infty(Omega)$, we have that the finitely additive measures are the positive cone of $C(X)^* = M(X)$, where $M(X)$ are the Borel measures over $X$. The set $beta$ must generate the whole $sigma$-algebra of $X$, for example this happens when $beta$ is a base for the topology of $X$.
Example: When $Omega = mathbbN$, you have that $X = beta mathbbN$, the Stone-Cech compactification of $mathbbN$ and you can take $beta$ (confusing notation) to be $ mathbbN cap V : V in mathcalV subset mathcalP(beta mathbb N) $, where $mathcal V$ is a base for the topology of $beta mathbb N$.
answered Jul 19 at 15:05
Adrián González-Pérez
54138
I have virtually no knowledge on topology beyond metric space and I am totally lost...thanks anyway though.
– failedstatistician
Jul 22 at 7:09
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Finitely additive measures are in isometric correspondence with positive functionals over $L^infty(Omega)$. Since $L^infty(Omega) = C(X)$, where $X$ is the compact Hausdorff topological space given by the Gel'fand spectrum of $L^infty(Omega)$, we have that the finitely additive measures are the positive cone of $C(X)^* = M(X)$, where $M(X)$ are the Borel measures over $X$. The set $beta$ must generate the whole $sigma$-algebra of $X$, for example this happens when $beta$ is a base for the topology of $X$.
Example: When $Omega = mathbbN$, you have that $X = beta mathbbN$, the Stone-Cech compactification of $mathbbN$ and you can take $beta$ (confusing notation) to be $ mathbbN cap V : V in mathcalV subset mathcalP(beta mathbb N) $, where $mathcal V$ is a base for the topology of $beta mathbb N$.
answered Jul 19 at 15:05
Adrián González-Pérez
54138
I have virtually no knowledge on topology beyond metric space and I am totally lost...thanks anyway though.
– failedstatistician
Jul 22 at 7:09
add a comment |Â
up vote
2
down vote
Finitely additive measures are in isometric correspondence with positive functionals over $L^infty(Omega)$. Since $L^infty(Omega) = C(X)$, where $X$ is the compact Hausdorff topological space given by the Gel'fand spectrum of $L^infty(Omega)$, we have that the finitely additive measures are the positive cone of $C(X)^* = M(X)$, where $M(X)$ are the Borel measures over $X$. The set $beta$ must generate the whole $sigma$-algebra of $X$, for example this happens when $beta$ is a base for the topology of $X$.
Example: When $Omega = mathbbN$, you have that $X = beta mathbbN$, the Stone-Cech compactification of $mathbbN$ and you can take $beta$ (confusing notation) to be $ mathbbN cap V : V in mathcalV subset mathcalP(beta mathbb N) $, where $mathcal V$ is a base for the topology of $beta mathbb N$.
answered Jul 19 at 15:05
Adrián González-Pérez
54138
I have virtually no knowledge on topology beyond metric space and I am totally lost...thanks anyway though.
– failedstatistician
Jul 22 at 7:09
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Finitely additive measures are in isometric correspondence with positive functionals over $L^infty(Omega)$. Since $L^infty(Omega) = C(X)$, where $X$ is the compact Hausdorff topological space given by the Gel'fand spectrum of $L^infty(Omega)$, we have that the finitely additive measures are the positive cone of $C(X)^* = M(X)$, where $M(X)$ are the Borel measures over $X$. The set $beta$ must generate the whole $sigma$-algebra of $X$, for example this happens when $beta$ is a base for the topology of $X$.
Example: When $Omega = mathbbN$, you have that $X = beta mathbbN$, the Stone-Cech compactification of $mathbbN$ and you can take $beta$ (confusing notation) to be $ mathbbN cap V : V in mathcalV subset mathcalP(beta mathbb N) $, where $mathcal V$ is a base for the topology of $beta mathbb N$.
answered Jul 19 at 15:05
Adrián González-Pérez
54138
Finitely additive measures are in isometric correspondence with positive functionals over $L^infty(Omega)$. Since $L^infty(Omega) = C(X)$, where $X$ is the compact Hausdorff topological space given by the Gel'fand spectrum of $L^infty(Omega)$, we have that the finitely additive measures are the positive cone of $C(X)^* = M(X)$, where $M(X)$ are the Borel measures over $X$. The set $beta$ must generate the whole $sigma$-algebra of $X$, for example this happens when $beta$ is a base for the topology of $X$.
Example: When $Omega = mathbbN$, you have that $X = beta mathbbN$, the Stone-Cech compactification of $mathbbN$ and you can take $beta$ (confusing notation) to be $ mathbbN cap V : V in mathcalV subset mathcalP(beta mathbb N) $, where $mathcal V$ is a base for the topology of $beta mathbb N$.
answered Jul 19 at 15:05
Adrián González-Pérez
54138
edited Jul 22 at 17:09
answered Jul 19 at 15:05
Adrián González-Pérez
54138
answered Jul 19 at 15:05
Adrián González-Pérez
54138
answered Jul 19 at 15:05
Adrián González-Pérez
54138
54138
I have virtually no knowledge on topology beyond metric space and I am totally lost...thanks anyway though.
– failedstatistician
Jul 22 at 7:09
add a comment |Â
I have virtually no knowledge on topology beyond metric space and I am totally lost...thanks anyway though.
– failedstatistician
Jul 22 at 7:09
I have virtually no knowledge on topology beyond metric space and I am totally lost...thanks anyway though.
– failedstatistician
Jul 22 at 7:09
I have virtually no knowledge on topology beyond metric space and I am totally lost...thanks anyway though.
– failedstatistician
Jul 22 at 7:09
add a comment |Â
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1
I doubt there exists any interesting such set $beta$.
– Eric Wofsey
Jul 18 at 20:35