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Math Help - Two Real Analysis Q's

  1. #1
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    Two Real Analysis Q's

    Hi!

    How would one show the following:

    1. If \mu(X)<\infty, \mu is a measure iff \mu is continuous from above (i.e. if \{E_j\}_1^\infty \subset M\subset P(X), E_1\supset E_2\supset \ldots and \mu(E_1)<\infty then \mu(\cub_1^{\infty}E_j)=lim_{j\rightarrow\infty}(E  _j)).

    ( \Rightarrow) I managed this direction, it's the ( \Leftarrow) that I need help with.

    2. If \mu is semifinite and \mu(E)=\infty, for C>0 there exists F\subset E with C<\mu(F)<\infty.

    Ok, so by def of semifinite we know that ther exists an F'\subset E such that 0<\mu(F')<\infty. Could I just set F=F'?
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  2. #2
    Senior Member roninpro's Avatar
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    On (1), it seems reasonable to check the conditions in the definition of a measure. However, I'm not exactly sure how to help you, because I don't know what kinds of assumptions you are putting on the function \mu. Are you taking \mu(X)<\infty? Is \mu assumed to be nonnegative?

    On (2), you're not quite done. You need to show that given C>0, you can produce F so that its measure is greater than C. Try using finite / countable additivity with the set F' that you found.
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  3. #3
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    On (1): Yes, I'm assuming that \mu is nonnegative and that \mu(X)=\infty not <\infty.
    (X,M) is the meausrable space

    On (2): I'm not quite sure what you mean. In order to use finite/countable additivity I need disjoint sets, and in order to form these I need at least two sets... however I only have F'.
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  4. #4
    Senior Member roninpro's Avatar
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    Quote Originally Posted by mgarson View Post
    On (1): Yes, I'm assuming that \mu is nonnegative and that \mu(X)=\infty not <\infty.
    (X,M) is the meausrable space
    The first thing that we can try to show is that the measure of the empty set is zero. But I don't see how that follows only from the condition in the problem. In particular, I don't see how to eliminate the possibility that \mu(\emptyset)=\infty.

    Quote Originally Posted by mgarson View Post
    On (2): I'm not quite sure what you mean. In order to use finite/countable additivity I need disjoint sets, and in order to form these I need at least two sets... however I only have F'.
    You can try setting that first set to be F_1. Then consider X\backslash F_1. That will be a set with infinite measure, so there exists a subset F_2 of finite measure. Then consider X\backslash (F_1\bigcup F_2). That will be a set of infinite measure, so there exists a subset F_3 of finite measure. Keep iterating that procedure to get a sequence of (disjoint) sets. The real work is to show that \sum \mu(F_n) does not converge (which I have not worked out).
    Last edited by roninpro; January 29th 2011 at 01:34 PM.
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