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Math Help - A finite additive measure that is continuous from below is a measure

  1. #1
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    A finite additive measure that is continuous from below is a measure

    A finite additive measure  \mu is a measure iff it is continuous from above. [sorry, I made a mistake in the title, should be "from above"]

    I finished the proof from finite additive to continuous from above, but stuck on the other one...

    Proof so far.

    Suppose that  \mu : \mathbb {M} \rightarrow [0, \infty ] is continuous from above, that is, if  \{ E_j \} ^ \infty _{j=1} \subset \mathbb {M} with  E_{j+1} \subset E_j , then I have  \lim _{j \rightarrow \infty } \mu (E_j)= \mu ( \bigcap ^ \infty _{j=1} E_j ) .

    I will show that  \mu is a measure.

    Let  \{ E_j \} ^n _{j=1} \subset \mathbb {M} be disjoint sets.

    Claim:  \mu ( \bigcup ^n _{j=1} E_j) = \sum _{j=1}^n \mu (E_j)

    Let F_n = \bigcup ^ \infty _{j=1}E_j \ \bigcup ^n _{i=1} E_i, note that F_j \subset F_{j+1} and  \bigcup ^n_{j=1}F_j

    Now, since  \mu is continuous from above, I have  \lim _{n \rightarrow \infty } \mu (F_n) = \mu ( \bigcup ^ \infty _{j=1} F_j )

    I'm defining the sequence as such base on how I work the other direction, but I can't seem to be able to break the left hand side into a sum of  \mu (E_1) + . . . + \mu (E_n) .

    Any hints? Thank you
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  2. #2
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    Hello,

    Suppose that  \mu : \mathbb {M} \rightarrow [0, \infty ] is continuous from above, that is, if  \{ E_j \} ^ \infty _{j=1} \subset \mathbb {M} with  E_{j+1} \subset E_j , then I have  \lim _{j \rightarrow \infty } \mu (E_j)= \mu ( \bigcap ^ \infty _{j=1} E_j ) .
    This limit holds only if at least one of the E_j has a finite measure. If they're all of infinite measure, I think it should be easy to finish your proof by using (*) below. So now assume we can use this formula.

    I will show that  \mu is a measure.
    Since \mu is finite-additive, the first axiom of a measure :
    \mu(\emptyset)=0
    is verified.
    And the positivity too.

    Let \{ E_j \} ^n _{j=1} \subset \mathbb {M} be disjoint sets.

    Claim:  \mu ( \bigcup ^n _{j=1} E_j) = \sum _{j=1}^n \mu (E_j)
    And we want to prove that for any pairwise disjoint sequence \{F_j\}, we have :
    \mu\left(\bigcup_{j=1}^\infty F_j\right)=\sum_{j=1}^\infty \mu(F_j) (*)


    Let F_n = \bigcup ^ \infty _{j=1}E_j \backslash \bigcup ^n _{i=1} E_i, note that F_j \subset F_{j+1} and  \bigcup ^n_{j=1}F_j
    The backslash is not \ which gives a space in latex, but \backslash

    Now, since  \mu is continuous from above, I have  \lim _{n \rightarrow \infty } \mu (F_n) = \mu ( \bigcup ^ \infty _{j=1} F_j )
    I guess you're being confused with below and above...
    The above continuity deals with an intersection, not an union, and with F_{j+1}\subset F_j (which you actually have)


    ---------------------------------
    To sum up... (not too formally) :
    We know that E_{j+1}\subset E_j, \lim_{j\to\infty} \mu(E_j)=\mu\left(\bigcap_{j=1}^\infty E_j\right) (continuous from above)
    We know that E_j pairwise disjoint, \mu\left(\bigcup_{j=1}^n E_j\right)=\sum_{j=1}^n \mu(E_j) (finite additivity)

    We want to prove that E_j pairwise disjoint, \mu\left(\bigcup_{j=1}^\infty E_j\right)=\sum_{j=1}^\infty \mu(E_j)


    With the information we have, this is equivalent to proving \mu\left(\bigcup_{j=1}^\infty E_j\right)=\lim_{n\to\infty} \mu\left(\bigcup_{j=1}^n E_j\right)

    Then I recommend you read this : http://www.mathhelpforum.com/math-he...ection-ei.html (the second post).
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  3. #3
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    So base on the F_n that I define, I have the following:

    1.  \lim _{j \rightarrow \infty } \mu (F_j) = \mu ( \bigcap _{j=1} ^ \infty F_j)

    2.  \bigcup ^ \infty _{j=1} E_j = \bigcap _{j=1} ^ \infty F_j

    3.  F_j = \bigcup ^ \infty _{j=1} E_j \backslash \bigcup _{i=1} ^n E_i

    So I have:  \mu ( \bigcup ^ \infty _{j=1} E_j ) = \mu ( \bigcap _{j=1} ^ \infty F_j ) =  \lim _{j \rightarrow \infty } \mu (F_j) = \lim _{ j \rightarrow \infty } ( \bigcup ^ \infty _{j=1} E_j \backslash \bigcup _{i=1} ^n E_i )


    What I'm trying to do is get the last term to be  \sum _{j=1}^ \infty \mu (E_j) , but I'm suspecting that the F_n that I define is not really help...

    Will this get me to the right track? Thanks.
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  4. #4
    Moo
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    Okay, I confused some signs, so the reasoning I was pointing you at was not correct... Sorry

    Indeed you defined badly F_n. I believe I wrote it above

    Let's start it again.


    ------------------------------
    We know that E_{j+1}\subset E_j, \lim_{j\to\infty} \mu(E_j)=\mu\left(\bigcap_{j=1}^\infty E_j\right) (continuous from above)
    We know that E_j pairwise disjoint, \mu\left(\bigcup_{j=1}^n E_j\right)=\sum_{j=1}^n \mu(E_j) (finite additivity)

    We want to prove that E_j pairwise disjoint, \mu\left(\bigcup_{j=1}^\infty E_j\right)=\sum_{j=1}^\infty \mu(E_j)


    With the information we have, this is equivalent to proving \mu\left(\bigcup_{j=1}^\infty E_j\right)=\lim_{n\to\infty} \mu\left(\bigcup_{j=1}^n E_j\right), where the Ej are pairwise disjoint << this is important. It uses the finite additivity. Do you understand this point ?


    Then, assuming the measure of the whole space is finite (if it's infinite, it may work too but it ain't no fun)...

    Let's take a sequence Ej of pairwise disjoint measurable sets.

    Define F_n=\bigcap_{j=1}^n \{E\backslash E_j\}
    - why did I choose the intersection ? Because we need a decreasing sequence for using the continuity from above.
    - why did I take the complement ? Because it makes no sense taking the intersection of disjoint sets ! And because of de Moivre's formula, which will help us transform the intersection into an union.

    So by the continuity from above applied to the sequence F_n, we have \lim_{n\to\infty} \mu\left(\bigcap_{j=1}^n \{E\backslash E_j\}\right)=\mu\left(\bigcap_{j=1}^\infty\{E\back  slash E_j\}\right)

    \lim_{n\to\infty} \mu(E)-\mu\left(\bigcup_{j=1}^n E_j\right)=\mu(E)-\mu\left(\bigcup_{j=1}^\infty E_j\right)

    and finally \lim_{n\to\infty}\mu\left(\bigcup_{j=1}^n E_j\right)=\mu\left(\bigcup_{j=1}^\infty E_j\right)


    Does it look somehow clear ?
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