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Suppose $\sigma$ and $\mu$ are two positive measures on the complex unit circle $\mathbb{T}$ such that $\sigma\ll\mu$ and denote $\phi(\lambda)=d\sigma/d\mu(\lambda)$. Suppose that the set $\{\lambda\in\mathbb{T}/\phi(\lambda)=0\}$ has $\mu$-measure 0. Does this imply that $\mu\ll\sigma$ as well and $d\mu/d\sigma=\phi^{-1}$? If not, what could we demand from the function $\phi$ so $\mu\ll\sigma$ would be true?

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Do you work with finite measures? –  Davide Giraudo Oct 2 '12 at 11:14
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up vote 2 down vote accepted

Yes: If $\sigma(A) = 0$ then $\int_A \phi\,d\mu = 0$, so $\mu(\lbrace\phi>0\rbrace\cap A)=0$. If $\mu(\lbrace\phi=0\rbrace)=0$ then this implies $$\mu(A)= \mu(\lbrace\phi>0\rbrace\cap A) + \mu(\lbrace\phi=0\rbrace\cap A) = 0,$$ so $\mu\ll\sigma$. The relation between the derivatives follows from $$\sigma(A) = \int_A\frac{d\sigma}{d\mu}\,d\mu = \int_A\frac{d\sigma}{d\mu}\frac{d\mu}{d\sigma}\,d\sigma.$$

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