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 jdinatale Mar4-12 03:03 PM

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

I'm trying to prove the following:

http://i3.photobucket.com/albums/y89...ale/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].

 morphism Mar8-12 11:36 PM

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

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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