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Let $(X,\mathcal M, \mu)$ be a measured space where $\mu$ is a positive measure. Let $\lambda$ be a complex measure on $(X,\mathcal M)$. When $\mu$ is sigma-finite, the Radon-Nikodym theorem provides a decomposition of $\lambda$ in a sum of an absolutely continuous measure wrt $\mu$ plus a singular measure wrt $\mu$.

Question. Is that statement still true without the sigma-finiteness of $\mu$? It does seems to be so easy for the case of the Hausdorff measures of dimension $d<n$ in $\mathbb R^n$.

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    $\begingroup$ One may take $\mathbb{R}$ with the Lebesgue measure (or some other ac measure if you insist on finite measure) and the counting measure of the Borel sets (which is not $\sigma$-finite). Then the ac measure is obviously ac with respect to the counting measure (having counting measure zero means just empty set), but it is easy to show that the RN derivative in this case (if exists) should be just $0$. PS - usually the decomposition you've mentioned is called the Lebesgue decomposition. $\endgroup$ – Asaf Jan 6 '17 at 14:29
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The measure spaces for which all of the nice theorems about $\sigma$-finite measures (except those about product measures, which require other generalizations) generalize are called localizable measure spaces. There are a number of equivalent definitions of localizable measure spaces; perhaps the simplest states that a positive measure $\mu$ is localizable if it is semifinite and has essential suprema, and a complex measure is localizable if its total variation is.

A measure is semifinite if whenever $\mu(E) = \infty$ there exists a measurable $F \subseteq E$ such that $0 < \mu(F) < \infty$. A measure has essential suprema if for every family $\mathcal{E}$ of measurable sets there exists a measurable set $H$ such that

  • $E \setminus H$ is null for every $E \in \mathcal{E}$, and
  • if $G$ is measurable and $E \setminus G$ is negligible for every $E \in \mathcal{E}$, then $H \setminus G$ is negligible.

The most commonly used equivalent definition of a localizable measure space is that real-valued $L^\infty(\mu)$ is Dedekind-complete as a lattice. Essentially all measures that arise in applications are localizable, e.g. all regular Borel measures on locally compact Hausdorff spaces are localizable.

The theorem that you would like to generalize is normally called the Lebesgue Decomposition Theorem rather than the Radon-Nikodym Theorem, which is about the representation of absolutely continuous measures in terms of integration against functions. You are missing the hypothesis that $\lambda$ is also $\sigma$-finite, or at least that $\mu + \lambda$ is $\sigma$-finite.

If $\mu$ and $\nu$ are measures on the same measurable space such that $\mu + \nu$ is localizable, then there exists a decomposition $\mu = \mu_c + \mu_s$ where $\mu_c$ is absolutely continuous with respect to $\nu$ and $\mu_s$ is singular. For a simple proof, see Decomposition and Representation Theorems in Measure Theory by Kelley.

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  • $\begingroup$ What is the proper setting for product measures? The version in Fremlin doesn't seem like a big improvement over the traditional version. $\endgroup$ – arsmath Sep 8 '18 at 19:23
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One can formulate the Radon-Nikodym so that it applies to all measure spaces without any restriction. Let $(X,\Sigma,\nu)$ be a measure space and $\nu$ be a real valued signed measure on $\Sigma$.

Then $\nu$ is absolutely continuous with respect to $\mu$ if and only if for all $\epsilon>0$ there exists $\delta>0$ such that for all $A\in\Sigma$, $\mu(A)\leq\delta$ implies $|\nu(A)|\leq\epsilon$.

We say that $\nu$ is truly continuous with respect to $\mu$ if and only for all $\epsilon>0$ there exists $\delta>0$ and $F\in\Sigma$ with $\mu(F)<\infty$ such that for all $A\in\Sigma$, $\mu(A\cap F)\leq\delta$ implies $|\nu(A)|\leq\epsilon$.

If $(X,\Sigma,\nu)$ is $\sigma$-finite, then a real-valued signed measure $\nu$ on $\Sigma$ is truly continuous with respect of $\mu$ if and only if it is absolutely continuous with respect to $\mu$.

Theorem: Let $(X,\Sigma,\nu)$ be a measure space and $\nu$ be a real valued signed measure on $\Sigma$. Then $\nu$ is truly continuous with respect to $\mu$ if and only if there exists a measurable function $f:\Sigma\to \mathbb{R}$ such that $$\nu(A)=\int_A f~\text{d}\mu$$ for all $A\in\Sigma$.

This formulation of the Radon-Nikodym is from the second volume of David Fremlin's treatise on measure theory. Fremlin doesn't assume $\nu$ to be countably additive, which requires additional distinctions. There is also a treatment in Section 5.4 of the book Measure and Integration by Dietmar Salamon (preprint available here), which also contains a useful characterization by Heinz König of being truly continuous in terms of absolute continuity and an inner regularity property

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