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Math Help - Outer measure and measurable sets.

  1. #1
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    Outer measure and measurable sets.

    If  \mu is an outer measure on X and  \{ A_n \} ^ \infty _{n=1} is a sequence of disjointed  \mu measurable sets, prove that  \mu ( E \cap ( \bigcup ^ \infty _{n=1} A_n )) = \sum ^ \infty _{n=1} \mu (E \cap A_n ) \ \ \ \forall E \subset X

    Proof so far.

    Since  \mu is an outer measure on  X , we know that:

    1.  \mu (A) \leq \mu (B) if  A \subset B

    2.  \mu ( \bigcup _{n=1} ^ \infty A_n ) \leq \sum ^ \infty _{n=1} \mu (A_n)

    And since   \{ A_n \} ^ \infty _{n=1} is  \mu -measurable, then for each n, we have   \mu (E) = \mu (E \cap A_n )+ \mu (E \cap A_n ^c ) \ \ \ \forall E \subset X

    Now,  \mu (E \cap ( \bigcup _{n=1} ^ \infty A_n)) \leq \sum _{n=1} ^ \infty \mu (E \cap A_n ) since  \mu is an outer measure.

    So what is left to prove is that  \mu (E \cap ( \bigcup _{n=1} ^ \infty A_n)) \geq \sum _{n=1} ^ \infty \mu (E \cap A_n )

    So I have  \sum _{n=1} ^ \infty \mu (E \cap A_n ) = \mu (E \cap A_1)+ \mu (E \cap A_2) + . . .

    And I know I will have to use the measurable property, but I don't seem to be making any progress here...

    Any hints? Thank you.
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  2. #2
    Junior Member qspeechc's Avatar
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    Quote Originally Posted by tttcomrader View Post
    If  \mu is an outer measure on X and  \{ A_n \} ^ \infty _{n=1} is a sequence of disjointed  \mu measurable sets, prove that  \mu ( E \cap ( \bigcup ^ \infty _{n=1} A_n )) = \sum ^ \infty _{n=1} \mu (E \cap A_n ) \ \ \ \forall E \subset X

    Proof so far.

    Since  \mu is an outer measure on  X , we know that:

    1.  \mu (A) \leq \mu (B) if  A \subset B

    2.  \mu ( \bigcup _{n=1} ^ \infty A_n ) \leq \sum ^ \infty _{n=1} \mu (A_n)

    And since   \{ A_n \} ^ \infty _{n=1} is  \mu -measurable, then for each n, we have   \mu (E) = \mu (E \cap A_n )+ \mu (E \cap A_n ^c ) \ \ \ \forall E \subset X

    Now,  \mu (E \cap ( \bigcup _{n=1} ^ \infty A_n)) \leq \sum _{n=1} ^ \infty \mu (E \cap A_n ) since  \mu is an outer measure.

    So what is left to prove is that  \mu (E \cap ( \bigcup _{n=1} ^ \infty A_n)) \geq \sum _{n=1} ^ \infty \mu (E \cap A_n )

    So I have  \sum _{n=1} ^ \infty \mu (E \cap A_n ) = \mu (E \cap A_1)+ \mu (E \cap A_2) + . . .

    And I know I will have to use the measurable property, but I don't seem to be making any progress here...

    Any hints? Thank you.
    Look at
     \mu ( E \cap ( \bigcup ^k _{n=1} A_n )) = \sum ^k _{n=1} \mu (E \cap A_n ) \ \ \ \forall E \subset X
    Do you know that this is true (for all natural numbers k)?
    If you don't, then try prove it.
    If you do, then you will notice this is a sequence of real numbers indexed by k. This sequence is bounded above (by what?). Use some results about sequences of real numbers to get the other inequality.
    I hope I have given you enough hints.
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