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My question stems from the following problem.

Suppose $\mu$ is a $\sigma-$finite measure on $(X,M)$ and $\{E_n\}$ a sequence of measurable sets. Define $\nu$ on $M$ by $\nu(E)=\sum \mu(E\cap E_n)$. Find the Radon-Nikodym derivative $\frac{d\nu}{d\mu}$.

For all measurable set $E$, $\nu(E)=\sum \mu(E\cap E_n)=\sum \int \chi_{E\cap E_n} d\mu = \sum \int_E \chi_{E_n}d\mu=\int (\sum \chi_{E_n}) d\mu$, so that $\frac{d\nu}{d\mu}=\sum \chi_{E_n}$.

However, for the Radon-Nikodym derivative to exist, I need the fact that $\nu$ is $\sigma-$finite, but I can't show how this is possible. Is $\nu$ sigma finite even when the $E_n$'s are not disjoint? And how can I show this in that case? I would greatly appreciate any help.

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$\nu$ need not be $\sigma$-finite, even if $\mu$ is finite. For instance, let $\delta_0$ be the Dirac measure at $0$ on $\mathbb{R}$. Let $E_n=\{0\}$ for $n\in\mathbb{N}$. Then $\nu=\infty\cdot\delta_0$; that is, $$\nu(E)=\left\{\begin{array}{ll}\infty&\mbox{if }0\in E,\\ 0&\mbox{if }0\notin E.\end{array}\right.$$

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For the Radon–Nikodym theorem you don't need the measure $\nu$ to be $\sigma$-finite, only $\mu$.

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  • $\begingroup$ I learned the Radon-Nikodym theorem from Folland's text and it states that $\nu$ is a $\sigma$-finite signed measure and $\mu$ a $\sigma$-finite positive measure on $(X,M)$. Can you tell me where I can get your version? $\endgroup$ – nomadicmathematician Apr 14 '16 at 23:49
  • $\begingroup$ See for example Rudin's Real and Complex Analysis. In fact, the absolutely continuous part of the Lebesgue decomposition of a $\sigma$-finite measure has a Radon-Nikodym derivative. $\endgroup$ – John B Apr 14 '16 at 23:57
  • $\begingroup$ @takecare, Jonas is right. For the Radon–Nikodym theorem you don't need the measure $\nu$ to be $\sigma$-finite, only $\mu$. If $\nu$ is not σ-finite then the Radon–Nikodym is not necessarily finite valued. (See, for instance, Halmos, Measure Theory, chapter 6, section 31 remark 7). $\endgroup$ – Ramiro Apr 15 '16 at 1:47

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