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Thread: finite additivity vs countable additivity

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

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

    Can someone provide a counterexample? Thanks
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  2. #2
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    $\displaystyle \mu$ is a finitely additive measure?

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

    For counterexample, take an ultrafilter $\displaystyle \mathcal{U}$ containing Fréchet filter on $\displaystyle \mathbb{N}$ (we don't want $\displaystyle \mathcal{U}$ to be trivial) and for every $\displaystyle A \subseteq \mathbb{N}$ set
    $\displaystyle \mu(A) = \begin{cases} 1 & \mbox{if } A \in \mathcal{U} \\
    0 & \mbox{otherwise} \end{cases}$
    It is straightforward to verify that $\displaystyle \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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