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Definition of measure on an algebra violating algebra definition


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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







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$endgroup$





















    3












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$

















      3












      3








      3


      0



      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited 8 hours ago









      Henno Brandsma

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      asked 8 hours ago









      Red FloydRed Floyd

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          2 Answers
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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.






          share|cite|improve this answer









          $endgroup$























            1














            $begingroup$

            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).






            share|cite|improve this answer











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              2 Answers
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              2 Answers
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              active

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              active

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              active

              oldest

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              3














              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$




















                3














                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$


















                  3














                  3










                  3







                  $begingroup$

                  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.






                  share|cite|improve this answer









                  $endgroup$



                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 8 hours ago









                  Matt SamuelMatt Samuel

                  41.7k6 gold badges39 silver badges73 bronze badges




                  41.7k6 gold badges39 silver badges73 bronze badges




























                      1














                      $begingroup$

                      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).






                      share|cite|improve this answer











                      $endgroup$




















                        1














                        $begingroup$

                        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).






                        share|cite|improve this answer











                        $endgroup$


















                          1














                          1










                          1







                          $begingroup$

                          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).






                          share|cite|improve this answer











                          $endgroup$



                          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).







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 7 hours ago

























                          answered 8 hours ago









                          Henno BrandsmaHenno Brandsma

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