For a measure space ( X , E , μ ) we introduced the complete measure space E μ defined as: E μ := { A ∈ P ( X ) | ∃ B , C ∈ E with A △ B ⊂ C and μ ( C ) = 0 }and we define the corresponding complete measure as μ ¯ 1 ( A ) = μ ( B )On the internet I've found exclusively an alternative definition, for which we define N := { N ⊂ X | ∃ M ∈ E mit μ ( M ) = 0 und N ⊂ M }and then the complete measure, I'll name it E ¯ is: E ¯ := { D ∪ N | D ∈ E , N ∈ N }and a measure μ ¯ 2 ( D ∪ N ) = μ ( D )I find it much easier to work with the alternative definition from the internet, and I was wondering what advantages the one from my professor would have. And secondly, if both are complete measure spaces with respect to E , is also E μ = E ¯ and μ 1 = μ 2 ? How could I start proving it?

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For a measure space   (  X  ,      E    ,  μ  ) we introduced the complete measure space             E        μ   defined as:            E        μ    :=  {  A  ∈      P    (  X  )         |       ∃  B  ,  C  ∈      E     with   A  △  B  ⊂  C   and   μ  (  C  )  =  0  }and we define the corresponding complete measure as            μ      ¯        1    (  A  )  =  μ  (  B  )On the internet I&#039;ve found exclusively an alternative definition, for which we define      N    :=  {  N  ⊂  X         |       ∃  M  ∈      E     mit   μ  (  M  )  =  0   und   N  ⊂  M  }and then the complete measure, I&#039;ll name it             E        ¯   is:            E        ¯    :=  {  D  ∪  N         |       D  ∈      E    ,  N  ∈      N    }and a measure            μ      ¯        2    (  D  ∪  N  )  =  μ  (  D  )I find it much easier to work with the alternative definition from the internet, and I was wondering what advantages the one from my professor would have. And secondly, if both are complete measure spaces with respect to       E  , is also             E        μ    =            E        ¯   and       μ    1    =      μ    2  ? How could I start proving it?

robegarj · Accepted Answer

Yes, these are the same.Given   D  ∪  N with   N  ⊆  M, put   B  =  D and   C  =  M. Then obviously             μ      ¯        1    (  D  ∪  N  )  =  μ  (  B  ) equals             μ      ¯        2    (  D  ∪  N  )  =  μ  (  D  ).On the other hand, given   A with associated   B and   C, put   D  =  B  ∖  C,   N  =  A  ∩  C, and   M  =  C. Then             μ      ¯        2    (  A  )  =  μ  (  D  ) equals             μ      ¯        1    (  A  )  =  μ  (  B  ) since   μ  (  C  )  =  0.

Damon Stokes · Accepted Answer

thank you!

For a measure space ( X , <mrow class="MJX-TeXAtom-ORD"> <mi class="MJX-tex-caligr

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