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

  • Thread starter jdinatale
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  • #1
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I'm trying to prove the following:

fsdfsd.png


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].
 

Answers and Replies

  • #2
morphism
Science Advisor
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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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