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The Wikipedia proof of Radon-Nikodym first shows for positive measures, then extends to real measures by applying the positive measure Radon-Nikodym theorem to the unique + and - (positive) measures obtained via Hahn-Jordan decomposition. But one crucial assumption of Radon-Nikodym is that the (positive) measures are absolutely continuous w.r.t. $\mu$.

My question is, if $\nu \ll \mu$ how do we know $\nu^+ \ll \mu$? Isn't it possible that $\nu(E) = 0$ but $\nu^+(E) = \nu^-(E) > 0$?

kaiwenw
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  • There exists a measurable set $A$ such that $\nu^{+}(E)=\nu (A\cap E)$ for every measurable set $E$. – Kavi Rama Murthy Nov 25 '20 at 23:40
  • @KaviRamaMurthy Can't $E$ have an $A$ and a $B$ component? For example, $A$ can be the positive reals, and $B$ be the negative reals, and $\nu^+ = \nu^- = \lambda$ on their respective sets. Then $\nu((-1, 1)) = 0$ but $\nu^+((-1, 1)) = \nu^+((0, 1)) = 1$ – kaiwenw Nov 25 '20 at 23:46
  • I think for signed measures, absolute continuity is defined in a stronger manner, basically requiring both $\nu^+<!<\mu$ and $\nu^-<!<\mu$. – Berci Nov 25 '20 at 23:53
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    $\mu$ is a positive measure. So $\mu (E)=0$ implies $\mu (A \cap E)=0$ so $\nu^{+}(E)=\nu (A \cap E)=0$. – Kavi Rama Murthy Nov 25 '20 at 23:54

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