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Unformatted text preview: Analysis Midterm 2 Write solutions in a neat and logical fashion, giving complete reasons for all steps. 1. Let (Ω , F ) be a measurable space on which λ, μ and ν are (finite) measures. Show that if ν μ and μ λ then ν λ ; further, carefully verify the relation dν dλ = dν dμ dμ dλ among Radon-Nikodym derivatives (which should be made precise). Solution To see that ν λ let C ∈ F with λ ( C ) = 0: as μ λ it follows that μ ( C ) = 0; as ν μ it follows that ν ( C ) = 0. Let (0 6 ) f (= dμ dλ ) ∈ L (Ω , F , λ ) and (0 6 ) g (= dν dμ ) ∈ L (Ω , F , μ ) be Radon- Nikodym derivatives as indicated, so that ν = μ g and μ = λ f . The indicated relation may be expressed by saying that the product gf is a Radon-Nikodym derivative of ν by λ : that is, ν = λ gf ; explicitly, if C ∈ F then ν ( C ) = Z C gfdλ. To justify this, ν ( C ) = Z C gdμ = Z C gdλ f = Z C gfdλ where Theorem 7.2 is used at the last step. Note that an R-N derivative is defined uniquely up to null sets.defined uniquely up to null sets....
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This note was uploaded on 11/09/2011 for the course MAA 5228 taught by Professor Robinson during the Fall '09 term at University of Florida.
- Fall '09