Is this measure on extension of $sigma$-algebra well defined?












0












$begingroup$


Let $(X, xi, mu)$ be a measure space and define:
$$xi_mu := {A in mathcal{P} : exists B, C in xi: mu(C) = 0 text{ and } A ,triangle, Bsubseteq C}$$
I already showed that $xi_mu$ is a $sigma$-algebra and $xi subseteq xi_mu$. For $A in xi_mu$ with $B, C in xi$ as in the definition we define:



$$barmu(A) := mu(B)$$



Now I'll need to prove that this is even well defined. Let's consider two $B_1, B_2, C_1, C_2$ with $mu(C_1), mu(C_2) = 0$ and:



$$A ,triangle, B_1 subseteq C_1$$
$$A ,triangle, B_2 subseteq C_2$$



but I'm stuck showing $mu(B_1) = mu(B_2)$.










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    0












    $begingroup$


    Let $(X, xi, mu)$ be a measure space and define:
    $$xi_mu := {A in mathcal{P} : exists B, C in xi: mu(C) = 0 text{ and } A ,triangle, Bsubseteq C}$$
    I already showed that $xi_mu$ is a $sigma$-algebra and $xi subseteq xi_mu$. For $A in xi_mu$ with $B, C in xi$ as in the definition we define:



    $$barmu(A) := mu(B)$$



    Now I'll need to prove that this is even well defined. Let's consider two $B_1, B_2, C_1, C_2$ with $mu(C_1), mu(C_2) = 0$ and:



    $$A ,triangle, B_1 subseteq C_1$$
    $$A ,triangle, B_2 subseteq C_2$$



    but I'm stuck showing $mu(B_1) = mu(B_2)$.










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $(X, xi, mu)$ be a measure space and define:
      $$xi_mu := {A in mathcal{P} : exists B, C in xi: mu(C) = 0 text{ and } A ,triangle, Bsubseteq C}$$
      I already showed that $xi_mu$ is a $sigma$-algebra and $xi subseteq xi_mu$. For $A in xi_mu$ with $B, C in xi$ as in the definition we define:



      $$barmu(A) := mu(B)$$



      Now I'll need to prove that this is even well defined. Let's consider two $B_1, B_2, C_1, C_2$ with $mu(C_1), mu(C_2) = 0$ and:



      $$A ,triangle, B_1 subseteq C_1$$
      $$A ,triangle, B_2 subseteq C_2$$



      but I'm stuck showing $mu(B_1) = mu(B_2)$.










      share|cite|improve this question









      $endgroup$




      Let $(X, xi, mu)$ be a measure space and define:
      $$xi_mu := {A in mathcal{P} : exists B, C in xi: mu(C) = 0 text{ and } A ,triangle, Bsubseteq C}$$
      I already showed that $xi_mu$ is a $sigma$-algebra and $xi subseteq xi_mu$. For $A in xi_mu$ with $B, C in xi$ as in the definition we define:



      $$barmu(A) := mu(B)$$



      Now I'll need to prove that this is even well defined. Let's consider two $B_1, B_2, C_1, C_2$ with $mu(C_1), mu(C_2) = 0$ and:



      $$A ,triangle, B_1 subseteq C_1$$
      $$A ,triangle, B_2 subseteq C_2$$



      but I'm stuck showing $mu(B_1) = mu(B_2)$.







      measure-theory






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      share|cite|improve this question











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      asked Jan 7 at 14:52









      user7802048user7802048

      382211




      382211






















          1 Answer
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          $begingroup$

          Hint: First prove that
          $$
          B_1triangle B_2subset (Atriangle B_1)cup(Atriangle B_2)subset C_1cup C_2.
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Perfect. I concluded $mu(B_1) = mu(B_1 cup B_2) - mu(B_2 setminus B_1) = mu(B_1 cup B_2) - mu(B_1 setminus B_2) = mu(B_2)$. Your hint proves the second equality, right?
            $endgroup$
            – user7802048
            Jan 7 at 15:48










          • $begingroup$
            Sure :) I'm glad it helped.
            $endgroup$
            – Song
            Jan 7 at 15:49











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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









          1












          $begingroup$

          Hint: First prove that
          $$
          B_1triangle B_2subset (Atriangle B_1)cup(Atriangle B_2)subset C_1cup C_2.
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Perfect. I concluded $mu(B_1) = mu(B_1 cup B_2) - mu(B_2 setminus B_1) = mu(B_1 cup B_2) - mu(B_1 setminus B_2) = mu(B_2)$. Your hint proves the second equality, right?
            $endgroup$
            – user7802048
            Jan 7 at 15:48










          • $begingroup$
            Sure :) I'm glad it helped.
            $endgroup$
            – Song
            Jan 7 at 15:49
















          1












          $begingroup$

          Hint: First prove that
          $$
          B_1triangle B_2subset (Atriangle B_1)cup(Atriangle B_2)subset C_1cup C_2.
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Perfect. I concluded $mu(B_1) = mu(B_1 cup B_2) - mu(B_2 setminus B_1) = mu(B_1 cup B_2) - mu(B_1 setminus B_2) = mu(B_2)$. Your hint proves the second equality, right?
            $endgroup$
            – user7802048
            Jan 7 at 15:48










          • $begingroup$
            Sure :) I'm glad it helped.
            $endgroup$
            – Song
            Jan 7 at 15:49














          1












          1








          1





          $begingroup$

          Hint: First prove that
          $$
          B_1triangle B_2subset (Atriangle B_1)cup(Atriangle B_2)subset C_1cup C_2.
          $$






          share|cite|improve this answer









          $endgroup$



          Hint: First prove that
          $$
          B_1triangle B_2subset (Atriangle B_1)cup(Atriangle B_2)subset C_1cup C_2.
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 7 at 15:11









          SongSong

          13.6k633




          13.6k633












          • $begingroup$
            Perfect. I concluded $mu(B_1) = mu(B_1 cup B_2) - mu(B_2 setminus B_1) = mu(B_1 cup B_2) - mu(B_1 setminus B_2) = mu(B_2)$. Your hint proves the second equality, right?
            $endgroup$
            – user7802048
            Jan 7 at 15:48










          • $begingroup$
            Sure :) I'm glad it helped.
            $endgroup$
            – Song
            Jan 7 at 15:49


















          • $begingroup$
            Perfect. I concluded $mu(B_1) = mu(B_1 cup B_2) - mu(B_2 setminus B_1) = mu(B_1 cup B_2) - mu(B_1 setminus B_2) = mu(B_2)$. Your hint proves the second equality, right?
            $endgroup$
            – user7802048
            Jan 7 at 15:48










          • $begingroup$
            Sure :) I'm glad it helped.
            $endgroup$
            – Song
            Jan 7 at 15:49
















          $begingroup$
          Perfect. I concluded $mu(B_1) = mu(B_1 cup B_2) - mu(B_2 setminus B_1) = mu(B_1 cup B_2) - mu(B_1 setminus B_2) = mu(B_2)$. Your hint proves the second equality, right?
          $endgroup$
          – user7802048
          Jan 7 at 15:48




          $begingroup$
          Perfect. I concluded $mu(B_1) = mu(B_1 cup B_2) - mu(B_2 setminus B_1) = mu(B_1 cup B_2) - mu(B_1 setminus B_2) = mu(B_2)$. Your hint proves the second equality, right?
          $endgroup$
          – user7802048
          Jan 7 at 15:48












          $begingroup$
          Sure :) I'm glad it helped.
          $endgroup$
          – Song
          Jan 7 at 15:49




          $begingroup$
          Sure :) I'm glad it helped.
          $endgroup$
          – Song
          Jan 7 at 15:49


















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