Intuition behind Caratheodory's definition for measurable sets?

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I start saying that I know this question has been already proposed a thousand times on this forum BUT I couldn't find a direct answer so I try asking on my own.



What's the intuitive meaning of this definition of measurable set?



$cdot$ Let $X$ be a set and $lambda^*$ an outer measure. Then $X$ is said to be measurable if for each $Bin X lambda^*(B)=lambda^*(Bcap X) + lambda^*(Bcap X^C)$



Or equivalently if $lambda_*(X)=lambda^*(X)$.



I feel OK with the second definition because it looks kind of similar to the intuition in Riemann Jordan "measure", but why are the two equivalent?



Thanks.




measure-theory




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    up vote
    1
    down vote

    favorite












    I start saying that I know this question has been already proposed a thousand times on this forum BUT I couldn't find a direct answer so I try asking on my own.



    What's the intuitive meaning of this definition of measurable set?



    $cdot$ Let $X$ be a set and $lambda^*$ an outer measure. Then $X$ is said to be measurable if for each $Bin X lambda^*(B)=lambda^*(Bcap X) + lambda^*(Bcap X^C)$



    Or equivalently if $lambda_*(X)=lambda^*(X)$.



    I feel OK with the second definition because it looks kind of similar to the intuition in Riemann Jordan "measure", but why are the two equivalent?



    Thanks.




    measure-theory




    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I start saying that I know this question has been already proposed a thousand times on this forum BUT I couldn't find a direct answer so I try asking on my own.



      What's the intuitive meaning of this definition of measurable set?



      $cdot$ Let $X$ be a set and $lambda^*$ an outer measure. Then $X$ is said to be measurable if for each $Bin X lambda^*(B)=lambda^*(Bcap X) + lambda^*(Bcap X^C)$



      Or equivalently if $lambda_*(X)=lambda^*(X)$.



      I feel OK with the second definition because it looks kind of similar to the intuition in Riemann Jordan "measure", but why are the two equivalent?



      Thanks.




      measure-theory




      share|cite|improve this question













      I start saying that I know this question has been already proposed a thousand times on this forum BUT I couldn't find a direct answer so I try asking on my own.



      What's the intuitive meaning of this definition of measurable set?



      $cdot$ Let $X$ be a set and $lambda^*$ an outer measure. Then $X$ is said to be measurable if for each $Bin X lambda^*(B)=lambda^*(Bcap X) + lambda^*(Bcap X^C)$



      Or equivalently if $lambda_*(X)=lambda^*(X)$.



      I feel OK with the second definition because it looks kind of similar to the intuition in Riemann Jordan "measure", but why are the two equivalent?



      Thanks.





      measure-theory





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited Jul 25 at 13:52









      Looper

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      asked Jul 25 at 13:44









      Baffo rasta

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