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If u is a nonnegative, additive function, then u is countably subadditive

  1. Mar 4, 2012 #1
    I'm trying to prove the following:

    [​IMG]

    I ran into a roadblock at the end. I can't use the assumption that [itex]\mu[\itex] is additive because we don't know that [itex](\cup{A_k}) \cap A_{j + 1} = \emptyset[\itex].

    We do know that [itex]\mu(\cup_{k=1}^jA_k) + \mu(A_{j + 1} \leq \sum_{k=1}^{j+1}\mu(A_k)[\itex].
     
  2. jcsd
  3. Mar 8, 2012 #2

    morphism

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    You don't really need to worry about the intersection stuff. It's enough to note that a nonnegative additive function will be (finitely) subadditive.
     
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