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Thread: m(E_1 U E_2)+m(E_1 ^ E_2)=mE_1+mE_2

  1. #1
    Junior Member Dark Sun's Avatar
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    m(E_1 U E_2)+m(E_1 ^ E_2)=mE_1+mE_2

    Hello everyone. I've been hitting my head against the chalk-board trying to figure this one out.

    Suppose $\displaystyle E_1$ and $\displaystyle E_2$ are Lebesgue measurable. Then we want to show that:

    $\displaystyle m(E_1\cup E_2)+m(E_1\cap E_2)=mE_1+mE_2$.

    Recall: $\displaystyle mA=\mbox{inf }_{A\subset \cup I_n} \sum \ell (I_n)$

    Also Recall: E is measurable means for any set $\displaystyle A, \; mA=m(A\cap E)+m(A\cap \widetilde E)$.

    Any help on this problem would be greatly appreciated!
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  2. #2
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    It might help to look at it this way:
    $\displaystyle m(E_1\cup E_2)=mE_1+mE_2-m(E_1\cap E_2)
    $

    Then use the measurability condition for the measurable set $\displaystyle E=E_1\setminus (E_1\cap E_2)$ on $\displaystyle A=E_1\cup E_2$
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  3. #3
    Moo
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    What are A and In ?
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  4. #4
    Junior Member Dark Sun's Avatar
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    Thumbs up

    Thanks! This proof is fairly self-explanatory, although it is missing a critical piece, and I cannot find the lemma anywhere in any textbook, and I am having difficulty proving it.

    Assume that $\displaystyle E_1$ and $\displaystyle E_2$ are measurable. w.t.s. $\displaystyle m(E_1\cup E_2)+m(E_1\cap E_2)=mE_1+mE_2$.

    Set $\displaystyle E=E_1\sim (E_1\cap E_2)$ and $\displaystyle A=E_1\cup E_2$. Clearly, $\displaystyle E$ is measurable.

    Now, $\displaystyle m(E_1\cup E_2)=m((E_1\cup E_2)\cap E)+m((E_1\cup E_2)\cap \widetilde E)=$

    $\displaystyle mE+mE_2=mE_1-m(E_1\cap E_2)+mE_2,$ and;

    $\displaystyle m(E_1\cup E_2)+m(E_1\cap E_2)=mE_1+mE_2$

    Now, I'm having trouble with $\displaystyle mE=mE_1-m(E_1\cap E_2)$, this seems to be the last road block to tackle.

    P.S.: A is any set for which m is defined. And, $\displaystyle \bigcup In$ is a covering of A.
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  5. #5
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    But of course since $\displaystyle E_1=E\cup (E_1\cap E_2)$ and $\displaystyle E, (E_1\cap E_2)$ are disjoint,

    $\displaystyle mE+m(E_1\cap E_2)=m(E_1)
    $ by additivity of a measure.
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  6. #6
    Junior Member Dark Sun's Avatar
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    Thumbs up

    You're a genius! Thanks a million!
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