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Recently, in a particular problem I was solving, I needed some kind of Radon-Nikodym theorem for measures where one of them is not necessarily absolutely continuous with respect to other. My colleague informed that he believes that this slightly more general version is true:

Let $\mu$ and $\rho$ be two $\sigma$-finite measures defined one the same measurable space. Then there exists a nonnegative measurable function $f$ such that $$\rho(A) = \int_A f\ d\mu + \rho(A\cap B),$$ where $B$ is some measurable set such that $\mu(B)=0$. However, I cannot prove it and thus convince myself that it is true (even in finite case), nor can I find a counterexample. Of course, my approach is to use standard Radon-Nikodym theorem, and with its help hypothesis trivially reduces to:

Let $\mu$ and $\rho$ be two $\sigma$-finite measures defined one the same measurable space. Then there exists a measurable set $B$ such that $\mu(B)=0$ and $\rho(\ \cdot\ \cap B^c)$ is absolutely continuous with respect to $\mu$.

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changed "absolutely convergent" to "absolutely continuous" – Gerald Edgar May 18 '10 at 15:47
up vote 6 down vote accepted

You looking for the Lebesgue's decomposition theorem.

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If you want a proof, it's theorem 3.8 of G. Folland, Real Analysis, or theorem 6.10 of W. Rudin, Real and Complex Analysis. – Nate Eldredge May 18 '10 at 16:15

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