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Definition of measure on an algebra violating algebra definition
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My textbook's definition of measure goes like this
A set function $mu$ defined on an algebra $mathscr A$ is called measure if:
a) $mu ( emptyset ) = 0 $
b) $mu$ is countably additive: i.e., if ${A_n}^infty_{n=1} $ is a countable collection of sets in $mathscr A$ such that (i) $A_n bigcap A_m = emptyset$ for $ n neq m$, and (ii) $ A = bigcup^infty_{n=1} A_n in mathscr A $, then $mu(A) = sum^infty_{n=1} mu(A_n$).
However, in the definition of algebra, the book defined it to be closed under finite unions. To cross-check, this maths stack exchange post confirms that an algebra and sigma algebra's difference is the fact that algebra may not be closed over infinite unions.
Therefore, $A$ in the definition of measure may not belong to the algebra which doesn't make sense as I think the set function $mu$ is $mathscr A rightarrow [0, infty] $ (closed braces on infinity as given in my book). Can someone please solve my confusion? I think it may be that the set function is defined from smallest sigma field containing $mathscr A$ then it would make sense but it is not apparent.
abstract-algebra measure-theory
edited 8 hours ago
Henno Brandsma
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asked 8 hours ago
Red FloydRed Floyd
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$endgroup$
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$begingroup$
My textbook's definition of measure goes like this
A set function $mu$ defined on an algebra $mathscr A$ is called measure if:
a) $mu ( emptyset ) = 0 $
b) $mu$ is countably additive: i.e., if ${A_n}^infty_{n=1} $ is a countable collection of sets in $mathscr A$ such that (i) $A_n bigcap A_m = emptyset$ for $ n neq m$, and (ii) $ A = bigcup^infty_{n=1} A_n in mathscr A $, then $mu(A) = sum^infty_{n=1} mu(A_n$).
However, in the definition of algebra, the book defined it to be closed under finite unions. To cross-check, this maths stack exchange post confirms that an algebra and sigma algebra's difference is the fact that algebra may not be closed over infinite unions.
Therefore, $A$ in the definition of measure may not belong to the algebra which doesn't make sense as I think the set function $mu$ is $mathscr A rightarrow [0, infty] $ (closed braces on infinity as given in my book). Can someone please solve my confusion? I think it may be that the set function is defined from smallest sigma field containing $mathscr A$ then it would make sense but it is not apparent.
abstract-algebra measure-theory
edited 8 hours ago
Henno Brandsma
132k4 gold badges53 silver badges137 bronze badges
asked 8 hours ago
Red FloydRed Floyd
3003 silver badges16 bronze badges
$endgroup$
add a comment
|
$begingroup$
My textbook's definition of measure goes like this
A set function $mu$ defined on an algebra $mathscr A$ is called measure if:
a) $mu ( emptyset ) = 0 $
b) $mu$ is countably additive: i.e., if ${A_n}^infty_{n=1} $ is a countable collection of sets in $mathscr A$ such that (i) $A_n bigcap A_m = emptyset$ for $ n neq m$, and (ii) $ A = bigcup^infty_{n=1} A_n in mathscr A $, then $mu(A) = sum^infty_{n=1} mu(A_n$).
However, in the definition of algebra, the book defined it to be closed under finite unions. To cross-check, this maths stack exchange post confirms that an algebra and sigma algebra's difference is the fact that algebra may not be closed over infinite unions.
Therefore, $A$ in the definition of measure may not belong to the algebra which doesn't make sense as I think the set function $mu$ is $mathscr A rightarrow [0, infty] $ (closed braces on infinity as given in my book). Can someone please solve my confusion? I think it may be that the set function is defined from smallest sigma field containing $mathscr A$ then it would make sense but it is not apparent.
abstract-algebra measure-theory
edited 8 hours ago
Henno Brandsma
132k4 gold badges53 silver badges137 bronze badges
asked 8 hours ago
Red FloydRed Floyd
3003 silver badges16 bronze badges
$endgroup$
My textbook's definition of measure goes like this
A set function $mu$ defined on an algebra $mathscr A$ is called measure if:
a) $mu ( emptyset ) = 0 $
b) $mu$ is countably additive: i.e., if ${A_n}^infty_{n=1} $ is a countable collection of sets in $mathscr A$ such that (i) $A_n bigcap A_m = emptyset$ for $ n neq m$, and (ii) $ A = bigcup^infty_{n=1} A_n in mathscr A $, then $mu(A) = sum^infty_{n=1} mu(A_n$).
However, in the definition of algebra, the book defined it to be closed under finite unions. To cross-check, this maths stack exchange post confirms that an algebra and sigma algebra's difference is the fact that algebra may not be closed over infinite unions.
Therefore, $A$ in the definition of measure may not belong to the algebra which doesn't make sense as I think the set function $mu$ is $mathscr A rightarrow [0, infty] $ (closed braces on infinity as given in my book). Can someone please solve my confusion? I think it may be that the set function is defined from smallest sigma field containing $mathscr A$ then it would make sense but it is not apparent.
abstract-algebra measure-theory
abstract-algebra measure-theory
edited 8 hours ago
Henno Brandsma
132k4 gold badges53 silver badges137 bronze badges
asked 8 hours ago
Red FloydRed Floyd
3003 silver badges16 bronze badges
edited 8 hours ago
Henno Brandsma
132k4 gold badges53 silver badges137 bronze badges
asked 8 hours ago
Red FloydRed Floyd
3003 silver badges16 bronze badges
edited 8 hours ago
Henno Brandsma
132k4 gold badges53 silver badges137 bronze badges
edited 8 hours ago
Henno Brandsma
132k4 gold badges53 silver badges137 bronze badges
edited 8 hours ago
Henno Brandsma
132k4 gold badges53 silver badges137 bronze badges
132k4 gold badges53 silver badges137 bronze badges
asked 8 hours ago
Red FloydRed Floyd
3003 silver badges16 bronze badges
asked 8 hours ago
Red FloydRed Floyd
3003 silver badges16 bronze badges
asked 8 hours ago
Red FloydRed Floyd
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3003 silver badges16 bronze badges
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The book explicitly specifies that this only applies if the infinite union happens to be in the algebra. If it isn't, then it doesn't assert that the measure of that set exists.
answered 8 hours ago
Matt SamuelMatt Samuel
41.7k6 gold badges39 silver badges73 bronze badges
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The condition b) says that if we happen to have a disjoint family of members of $mathcal{A}$ such that the union is also in $mathcal{A}$ (this is indeed not guaranteed to be the case (algebras are only required to be closed under finite unions), but it could be the case and the condition only applies when it does), in that case we know that $mu(A)$ is the sum of the $mu(A_n)$. If the union is not in $mathcal{A}$, too bad, and there is nothing to check in that case.
A simple example: let $X=mathbb{N}$ and $mathcal{A}$ is the algebra of finite and co-finite (complement is finite) subsets of $mathbb{N}$ and $mu(A)$ is $+infty$ for $A$ cofinite, and the number of elements of $A$ for $A$ finite. We only need to check the condition for situations like e.g. $A_n = {2n,2n+1}$ for $n=0,1,2,ldots$ where $bigcup_n A_n = Bbb N in mathcal{A}$ but not for situations like when $A_n = {2n}$ where the union is neither finite not co-finite (though it does hold then, but that's irrelevant for the definition).
answered 8 hours ago
Henno BrandsmaHenno Brandsma
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$begingroup$
The book explicitly specifies that this only applies if the infinite union happens to be in the algebra. If it isn't, then it doesn't assert that the measure of that set exists.
answered 8 hours ago
Matt SamuelMatt Samuel
41.7k6 gold badges39 silver badges73 bronze badges
$endgroup$
add a comment
|
$begingroup$
The book explicitly specifies that this only applies if the infinite union happens to be in the algebra. If it isn't, then it doesn't assert that the measure of that set exists.
answered 8 hours ago
Matt SamuelMatt Samuel
41.7k6 gold badges39 silver badges73 bronze badges
$endgroup$
add a comment
|
$begingroup$
The book explicitly specifies that this only applies if the infinite union happens to be in the algebra. If it isn't, then it doesn't assert that the measure of that set exists.
answered 8 hours ago
Matt SamuelMatt Samuel
41.7k6 gold badges39 silver badges73 bronze badges
$endgroup$
The book explicitly specifies that this only applies if the infinite union happens to be in the algebra. If it isn't, then it doesn't assert that the measure of that set exists.
answered 8 hours ago
Matt SamuelMatt Samuel
41.7k6 gold badges39 silver badges73 bronze badges
answered 8 hours ago
Matt SamuelMatt Samuel
41.7k6 gold badges39 silver badges73 bronze badges
answered 8 hours ago
Matt SamuelMatt Samuel
41.7k6 gold badges39 silver badges73 bronze badges
answered 8 hours ago
Matt SamuelMatt Samuel
41.7k6 gold badges39 silver badges73 bronze badges
41.7k6 gold badges39 silver badges73 bronze badges
add a comment
|
add a comment
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$begingroup$
The condition b) says that if we happen to have a disjoint family of members of $mathcal{A}$ such that the union is also in $mathcal{A}$ (this is indeed not guaranteed to be the case (algebras are only required to be closed under finite unions), but it could be the case and the condition only applies when it does), in that case we know that $mu(A)$ is the sum of the $mu(A_n)$. If the union is not in $mathcal{A}$, too bad, and there is nothing to check in that case.
A simple example: let $X=mathbb{N}$ and $mathcal{A}$ is the algebra of finite and co-finite (complement is finite) subsets of $mathbb{N}$ and $mu(A)$ is $+infty$ for $A$ cofinite, and the number of elements of $A$ for $A$ finite. We only need to check the condition for situations like e.g. $A_n = {2n,2n+1}$ for $n=0,1,2,ldots$ where $bigcup_n A_n = Bbb N in mathcal{A}$ but not for situations like when $A_n = {2n}$ where the union is neither finite not co-finite (though it does hold then, but that's irrelevant for the definition).
answered 8 hours ago
Henno BrandsmaHenno Brandsma
132k4 gold badges53 silver badges137 bronze badges
$endgroup$
add a comment
|
$begingroup$
The condition b) says that if we happen to have a disjoint family of members of $mathcal{A}$ such that the union is also in $mathcal{A}$ (this is indeed not guaranteed to be the case (algebras are only required to be closed under finite unions), but it could be the case and the condition only applies when it does), in that case we know that $mu(A)$ is the sum of the $mu(A_n)$. If the union is not in $mathcal{A}$, too bad, and there is nothing to check in that case.
A simple example: let $X=mathbb{N}$ and $mathcal{A}$ is the algebra of finite and co-finite (complement is finite) subsets of $mathbb{N}$ and $mu(A)$ is $+infty$ for $A$ cofinite, and the number of elements of $A$ for $A$ finite. We only need to check the condition for situations like e.g. $A_n = {2n,2n+1}$ for $n=0,1,2,ldots$ where $bigcup_n A_n = Bbb N in mathcal{A}$ but not for situations like when $A_n = {2n}$ where the union is neither finite not co-finite (though it does hold then, but that's irrelevant for the definition).
answered 8 hours ago
Henno BrandsmaHenno Brandsma
132k4 gold badges53 silver badges137 bronze badges
$endgroup$
add a comment
|
$begingroup$
The condition b) says that if we happen to have a disjoint family of members of $mathcal{A}$ such that the union is also in $mathcal{A}$ (this is indeed not guaranteed to be the case (algebras are only required to be closed under finite unions), but it could be the case and the condition only applies when it does), in that case we know that $mu(A)$ is the sum of the $mu(A_n)$. If the union is not in $mathcal{A}$, too bad, and there is nothing to check in that case.
A simple example: let $X=mathbb{N}$ and $mathcal{A}$ is the algebra of finite and co-finite (complement is finite) subsets of $mathbb{N}$ and $mu(A)$ is $+infty$ for $A$ cofinite, and the number of elements of $A$ for $A$ finite. We only need to check the condition for situations like e.g. $A_n = {2n,2n+1}$ for $n=0,1,2,ldots$ where $bigcup_n A_n = Bbb N in mathcal{A}$ but not for situations like when $A_n = {2n}$ where the union is neither finite not co-finite (though it does hold then, but that's irrelevant for the definition).
answered 8 hours ago
Henno BrandsmaHenno Brandsma
132k4 gold badges53 silver badges137 bronze badges
$endgroup$
The condition b) says that if we happen to have a disjoint family of members of $mathcal{A}$ such that the union is also in $mathcal{A}$ (this is indeed not guaranteed to be the case (algebras are only required to be closed under finite unions), but it could be the case and the condition only applies when it does), in that case we know that $mu(A)$ is the sum of the $mu(A_n)$. If the union is not in $mathcal{A}$, too bad, and there is nothing to check in that case.
A simple example: let $X=mathbb{N}$ and $mathcal{A}$ is the algebra of finite and co-finite (complement is finite) subsets of $mathbb{N}$ and $mu(A)$ is $+infty$ for $A$ cofinite, and the number of elements of $A$ for $A$ finite. We only need to check the condition for situations like e.g. $A_n = {2n,2n+1}$ for $n=0,1,2,ldots$ where $bigcup_n A_n = Bbb N in mathcal{A}$ but not for situations like when $A_n = {2n}$ where the union is neither finite not co-finite (though it does hold then, but that's irrelevant for the definition).
answered 8 hours ago
Henno BrandsmaHenno Brandsma
132k4 gold badges53 silver badges137 bronze badges
edited 7 hours ago
answered 8 hours ago
Henno BrandsmaHenno Brandsma
132k4 gold badges53 silver badges137 bronze badges
answered 8 hours ago
Henno BrandsmaHenno Brandsma
132k4 gold badges53 silver badges137 bronze badges
answered 8 hours ago
Henno BrandsmaHenno Brandsma
132k4 gold badges53 silver badges137 bronze badges
132k4 gold badges53 silver badges137 bronze badges
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