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Math Help - Outer Measure

  1. #1
    Senior Member Sampras's Avatar
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    Outer Measure

    So one of the axioms of outer measure is the following:

    Countable subadditivity axiom: If  A = \bigcup_{n=1}^{\infty} A_n then  m^{*}A \leq \sum_{n=1}^{\infty} m^{*}A_n .

    Well it's not really an axiom since we want/have to prove it. But what is the point of using the "  \varepsilon/2^n trick" to prove this? E.g. we have the following:

     \sum_{k=1}^{\infty} |I_{kn}| < m^{*} A_n+ \frac{\varepsilon}{2^n} (1)

    This is analogous to  \inf S \leq x \Rightarrow x < \inf S + \varepsilon for some  \varepsilon >0 (with  x \in S ). But what is special about  \varepsilon/2^n ? Why not just use  \varepsilon ?
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  2. #2
    Moo
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    Hello,

    Because \sum_{n=1}^\infty \frac{\varepsilon}{2^n}=\varepsilon, wouldn't it ?
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  3. #3
    Senior Member Sampras's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    Because \sum_{n=1}^\infty \frac{\varepsilon}{2^n}=\varepsilon, wouldn't it ?

    yes, but the sum doesn't have to sum to  \varepsilon since it's arbitrary. I guess its convention.
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    Moo
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    Quote Originally Posted by Sampras View Post
    yes, but the sum doesn't have to sum to  \varepsilon since it's arbitrary. I guess its convention.
    Not really...

    As you're dealing with a limit, you want \varepsilon to appear. This one is arbitrary !

    Since you'll sum from 1 to infinity, if you keep \varepsilon, you'll sum something positive an infinite amount of times.
    Which is irrelevant for your problem.

    \varepsilon/2^n is arbitrary and > 0. It's a good candidate, because of this infinite sum.

    And you'll get \varepsilon in the end.


    You could also have taken \varepsilon/4^n or whatever you want. The main point is to get a converging series, arbitrary small.
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