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I am looking for a case in which Lebesgue decomposition of a measure fails.

More precisely, I am looking for some measure space $(X,\mathscr{M})$ , a positive measure $\mu$ on the $\sigma$-algebra $\mathscr{M}$ and some positive bounded measure $\lambda$ on $\mathscr{M}$ (that is $\lambda$ is a positive measure s.t. $\lambda(X)<\infty$), such that there exists no decomposition of the form \begin{equation} \lambda = \lambda_{a} + \lambda_{s}, \end{equation} where $\lambda_{a}$ is a positive measure absolutely continous with respect to $\mu$, and $\lambda_{s}$ is a positive measure, with $\lambda_{s}$ and $\mu$ mutually singular.

Since the Lebesgue decomposition Theorem is stated under the hypothesis that $\mu$ is $\sigma$-finite, I suppose that such a counterxample exists (obviously $\mu$ must be not $\sigma$-finite in this case), but I could not find it.

Thank you very much in advance for your help.

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I discovered that the counterexample I was looking for cannot exist. Actually, the hypothesis that $\mu$ is $\sigma$-finite can be removed. See Theorem 4.3.1 in Cohn, Measure Theory (by the way this is a wonderful book on measure theory!).

Thank you very much for your attention.

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  • $\begingroup$ Wait, the measure must be "finite signed, complex or sigma-finite positive". It must be sufficient to find a measure which does not relate to any of these cases. $\endgroup$ Commented Nov 25, 2016 at 9:04
  • $\begingroup$ Please, read carefully the question: I required $\lambda$ to a finite positive measure! $\endgroup$ Commented Dec 16, 2016 at 16:18
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The counting measure can not be decomposed with respect to the Lebesgue Measure. Here is a proof: Show that the counting measure has no Lebesgue decomposition.

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  • $\begingroup$ Please read carefully my question: I required $\lambda$ to a finite positive measure!! $\endgroup$ Commented Dec 16, 2016 at 16:20

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