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I'm curious as to whether the converse of the Radon-Nikodym theorem holds:

Converse Theorem: let $\mu.\nu$ be measures on $(\Omega,\Sigma)$ such that $$\nu : A\mapsto \int_A f \ d\mu$$ for some measurable function $f$. Then

  1. $\nu\ll\mu$, and
  2. $\mu$ and $\nu$ are $\sigma$-finite.

Proof: the first part is easy, as $$\mu(A) = 0\implies \int_A f \ d\mu = 0.$$


Does the second part, regarding $\sigma$-finiteness, hold?

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1 Answer 1

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It does not. Take any measure $\mu$. Then

$$\mu(A)=\int_A~\mathrm{d}\mu.$$

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  • $\begingroup$ Oh! Well that was an easy counterexample. $\endgroup$
    – Sam
    Aug 14, 2023 at 17:51

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