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Math Help - finite additivity vs countable additivity

  1. #1
    Junior Member
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    finite additivity vs countable additivity

    A lecturer made a passing reference to this... he said that finite additivity plus continuity is equivalent to countable additivity... could someone offer a proof of this or else direct me to one?

    I don't understand why the continuity criterion is necessary

    If F_i\cap F_j = \emptyset \ \forall i\neq j \Rightarrow \mu(\bigcup_{i=1}^NF_i) = \sum_{i=1}^N\mu(F_i)

    Then don't we get countable additivity straight away just by induction?

    Can someone provide a counterexample? Thanks
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  2. #2
    Member
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    Aug 2009
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    \mu is a finitely additive measure?

    What induction already gave you is that "two-set additivity" A \cap B = 0  \rightarrow \mu(A \cup B) = \mu(A)+\mu(B) implies finite additivity, it can't give you more (e.g. sigma-additivity).

    For counterexample, take an ultrafilter \mathcal{U} containing Fréchet filter on \mathbb{N} (we don't want \mathcal{U} to be trivial) and for every A \subseteq \mathbb{N} set
    \mu(A) = \begin{cases} 1  & \mbox{if } A \in \mathcal{U} \\ <br />
0 & \mbox{otherwise} \end{cases}
    It is straightforward to verify that \mu is a finitely additive measure but is not sigma additive.

    I don't quite understand the remark about continuity though.
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