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Math Help - Rondon Nikodym derivative

  1. #1
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    Rondon Nikodym derivative

    Let \mu,\nu be measures on (X,F) such that \nu << \mu and \lambda = \mu+\nu. Show that f= \frac{d\nu}{d\lambda} exists and satisfies 0 \leq f \leq 1 a.e \lambda.
    (Hint: \mu_1 << \mu_2 and \mu_2 << \mu_1 implies \frac{d\mu_1}{d\mu_2}=(\frac{d\mu_2}{d\mu_1})^{-1}.
    i do not have any idea how to do this. any help would be appreciated.
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  2. #2
    Senior Member Shanks's Avatar
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    since v<<u, \lambda=u+v,therefore v<<\lambda.
    the Radon Nikodym Theorem asserts that f=\frac{dv}{d\lambda}exist.
    similarly,since u<<\lambda, g=\frac{du}{d\lambda}exist.
    Therefore, d\lambda=du+dv=gd\lambda+fd\lambda
    therefore 1=g+f a.e.[\lambda]
    since g is nonnegative, which implies 0\geq f\geq 1, a.e.[\lambda].
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