Example of measure for some algebra











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Can you give an example of a finitely additive measure given on $mathcal A$, not
having a countably additive extension to a $sigma$-algebra generated by $mathcal A$ for some algebra $mathcal A$ over $mathbb N$?










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    Can you give an example of a finitely additive measure given on $mathcal A$, not
    having a countably additive extension to a $sigma$-algebra generated by $mathcal A$ for some algebra $mathcal A$ over $mathbb N$?










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      favorite









      up vote
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      down vote

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      Can you give an example of a finitely additive measure given on $mathcal A$, not
      having a countably additive extension to a $sigma$-algebra generated by $mathcal A$ for some algebra $mathcal A$ over $mathbb N$?










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      Can you give an example of a finitely additive measure given on $mathcal A$, not
      having a countably additive extension to a $sigma$-algebra generated by $mathcal A$ for some algebra $mathcal A$ over $mathbb N$?







      probability-theory measure-theory






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      asked Nov 20 at 10:12









      Lisa

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          On $mathbb N$ with the sigma algebra of all subsets let $mu (A)=0$ if $A$ is a finite set and $infty$ if it is an infinite set. Then $mu (mathbb N)=infty$ and $sum _n mu({n})=0$ so $mu$ is not countably additive. It is trivial to check that $mu$ is finitely additive.






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          • Thank you, it helps me
            – Lisa
            Nov 20 at 10:34











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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          On $mathbb N$ with the sigma algebra of all subsets let $mu (A)=0$ if $A$ is a finite set and $infty$ if it is an infinite set. Then $mu (mathbb N)=infty$ and $sum _n mu({n})=0$ so $mu$ is not countably additive. It is trivial to check that $mu$ is finitely additive.






          share|cite|improve this answer





















          • Thank you, it helps me
            – Lisa
            Nov 20 at 10:34















          up vote
          2
          down vote



          accepted










          On $mathbb N$ with the sigma algebra of all subsets let $mu (A)=0$ if $A$ is a finite set and $infty$ if it is an infinite set. Then $mu (mathbb N)=infty$ and $sum _n mu({n})=0$ so $mu$ is not countably additive. It is trivial to check that $mu$ is finitely additive.






          share|cite|improve this answer





















          • Thank you, it helps me
            – Lisa
            Nov 20 at 10:34













          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          On $mathbb N$ with the sigma algebra of all subsets let $mu (A)=0$ if $A$ is a finite set and $infty$ if it is an infinite set. Then $mu (mathbb N)=infty$ and $sum _n mu({n})=0$ so $mu$ is not countably additive. It is trivial to check that $mu$ is finitely additive.






          share|cite|improve this answer












          On $mathbb N$ with the sigma algebra of all subsets let $mu (A)=0$ if $A$ is a finite set and $infty$ if it is an infinite set. Then $mu (mathbb N)=infty$ and $sum _n mu({n})=0$ so $mu$ is not countably additive. It is trivial to check that $mu$ is finitely additive.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 20 at 10:16









          Kavi Rama Murthy

          44.7k31852




          44.7k31852












          • Thank you, it helps me
            – Lisa
            Nov 20 at 10:34


















          • Thank you, it helps me
            – Lisa
            Nov 20 at 10:34
















          Thank you, it helps me
          – Lisa
          Nov 20 at 10:34




          Thank you, it helps me
          – Lisa
          Nov 20 at 10:34


















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