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Proving some properties on a complete measure space

  1. Mar 3, 2014 #1
    The problem statement, all variables and given/known data.

    A space ##(X,\Sigma, \mu)## is a complete measure space if given ## Z \in \Sigma## such that ##\mu(Z)=0##, for every ##Y \subset Z##, we have ##Y \in \Sigma##. In this case, prove that

    a) If ##Z_1 \in \Sigma##, ##Z_1ΔZ_2 \in \Sigma## and ##\mu(Z_1ΔZ_2)=0##, then ##Z_2 \in \Sigma##.

    b) If ##E_1, E_2 \in \Sigma## and ##\mu(E_1ΔE_2)=0##, then ##\mu(E_1)=\mu(E_2)##.



    The attempt at a solution.

    For a), ##Z_2\setminus Z_1 \subset Z_1ΔZ_2##, by definition of complete measure space, we have that ##Z_2\setminus Z_1 \in \Sigma##. If I could prove that ##Z_1\cap Z_2 \in \Sigma##, then, I could conclude that ##Z_2=(Z_2\setminus Z_1) \cup (Z_1 \cap Z_2) \in \Sigma##. I got stuck trying to prove the last part.

    For b), I think I got it but I would like to check if my solution is correct:

    First notice that since we are in a measure space and ##E_1, E_2 \in \Sigma \implies {E_1}^c, {E_2}^c \in \Sigma \implies E_1 \cap E_2=({E_1}^c \cup {E_2}^c)^c \in \Sigma##. Moreover, since it is a complete measure space, ##E_1\setminus E_2, E_2 \setminus E_1 \in \Sigma##. Having said that, we can write ##E_1=(E_1 \setminus E_2) \cup (E_1 \cap E_2)##, the same goes for ##E_2##. As ##\mu## is a measure, we have that ##\mu(E_1)=\mu(E_1 \setminus E_2) + \mu(E_1 \cap E_2)##, and ##\mu(E_2)=\mu(E_2 \setminus E_1) + \mu(E_1 \cap E_2)##.

    Again, since ##\mu## is a measure, ##0\leq \mu(E_1 \setminus E_2),\mu(E_2 \setminus E_1)\leq \mu(E_1ΔE_2)=0 \implies \mu(E_1 \setminus E_2)=0=\mu(E_2 \setminus E_1)##.

    But then, ##\mu(E_1)=0+\mu(E_1 \cap E_2)=\mu(E_2)##, which was what we wanted to prove.


    I would appreciate some help for part a)
     
  2. jcsd
  3. Mar 3, 2014 #2

    Dick

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    ##Z_1## and ##Z_1 \setminus Z_2## are in ##\Sigma##. ##(Z_1 \cap Z_2) = Z_1 \setminus (Z_1\setminus Z_2)##. The measureable sets form a sigma algebra, right?
     
  4. Mar 3, 2014 #3
    Oh, I see, as you've said ##(Z_1 \cap Z_2) = Z_1 \setminus (Z_1\setminus Z_2)##, and, ##Z_1 \setminus (Z_1\setminus Z_2)=Z_1 \cap (Z_1 \setminus Z_2)^c##, since a ##σ-## algebra is closed under countable intersection, one can deduce from here that ##Z_1 \cap Z_2 \in \Sigma## (for example, I can set ##\bigcap_{i \in \mathbb N} E_i## with ##E_1=Z_1, E_2=(Z_1 \setminus Z_2)^c## and ##E_i=A## for ##i\geq 2##). In a similar way, using the fact that a ##σ-##algebra is closed under countable union, it is proved that ##Z_2=(Z_2 \setminus Z_1) \cup (Z_1 \cap Z_2) \in \Sigma##

    Thanks!
     
  5. Mar 3, 2014 #4

    Dick

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    You're welcome! And you can just say that if C and D are in the sigma algebra then ##C \cap D## is in the sigma algebra. You don't have to spell out the whole countable intersection. Two is countable.
     
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