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This question comes from this question. The answer therein missed an argument that the Radon-Nikodym derivative is real-valued a.e. Without this, the proof in that answer has flaw because either the sum of $f_n$ is not equal to $f$ (a.e.) or the measure corresponds to $+\infty$ is not finite. The following is a complete formulation of my question.

On an arbitrary measurable space $(E,\mathcal{E})$, $\mu\ll\nu$ and $\nu$ is a finite measure. Let $p$ denote the Radon-Nikodym derivative $d\mu/d\nu$. Show that $p$ is real-valued $\nu$-almost everywhere.

I can find no way to exclude the case that $\nu(\{x\in E|p(x)=+\infty\})=0$. Can you please help me show that this measure is zero? Thanks a lot.

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  • $\begingroup$ To the best of my knowledge the answer given in that link is wrong because the argument is circular. I think $\sigma$ finiteness of $\mu$ is required to assert then existence of a real valued measurable Radon Nikodym derivative. But may be I don't know the most general form of the theorem. $\endgroup$ Oct 31 '20 at 11:55
  • $\begingroup$ @Kavi Rama Murthy: As shown in that link, the ultimate goal is to show $\Sigma$-finiteness of $\mu$. But we know $\sigma$-finiteness implies $\Sigma$-finiteness per se, so the conditions of absolute continuity and finiteness of $\nu$ would be useless. So I think we cannot assume $\mu$ is $\sigma$-finite. $\endgroup$ Oct 31 '20 at 12:03
  • $\begingroup$ The answer to the linked question is indeed wrong! I have provided a counter-exmaple. Please visit that page again. $\endgroup$ Oct 31 '20 at 12:19
  • $\begingroup$ @DavidC.Ullrich It is assumed that $\nu$ is finite. No finiteness /sigma finiteness condition on $\mu$ $\endgroup$ Oct 31 '20 at 12:20
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    $\begingroup$ Obvious counterexample (or not, depending on exactly what the hypotheses mean): let $\mu$ be more or less anything and define $\nu(E)=0$ if $\mu(E)=0$, $\nu(E)=\infty$ if $\mu(E)>0$. Then $\nu(E)=\int_E p\,d\mu$ where $p=+\infty$. $\endgroup$ Oct 31 '20 at 12:36
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Application of Radon Nikodym Theorem requires sigma finiteness of $\mu$. Under this condition we can argue as follows: Let $\mu(A_n) <\infty$ and $X =\cup_n A_n$. Then $\int_{A_n} fd\nu <\infty$ so $f <\infty$ a.e. $[\nu]$ on on $A_n$. This true for each $n$ and hence $f <\infty$ almost everywhere $[\nu]$.

A counter-example: Let $X=\mathbb N$, $\mu (\emptyset)=0$ and $\mu (E)=\infty$ for all non-empty subsets of $X$. Let $\nu (E)=\sum_{n \in E} \frac 1 {2^{n}}$. Then $\nu$ is a finite measure and $\mu <<\nu$. If there exist $f$ such that $\mu(E)=\int_Efd\nu$ for all $E$ then $\infty=\mu(\{n\})=\int_{\{n\}}fd\nu=f(n) \frac 1 {2^n}$, so $f(n)=\infty$ for all $n$.

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  • $\begingroup$ Radon Nikodym Theorem only requires $\sigma$-finiteness of $\nu$ (the "source" measure, or integrator), but does not require $\sigma$-finiteness of $\mu$ (the indefinite integral). A source of this claim is here: math.stackexchange.com/questions/1760977/… $\endgroup$ Nov 1 '20 at 4:15
  • $\begingroup$ @user5280911 It is clearly stated in the anser to that question that Wikipedia has a mistake. It is not enough to assume that $\nu$ is sigma a finite. $\endgroup$ Nov 1 '20 at 4:42
  • $\begingroup$ I'm confused. Wikipedia is mistaken in that it assumes both measure to be $\sigma$-finite (I checked it just now). But the answer says that it suffices only for the source measure to be so. The answer also gives a counterexample in which the indefinite integral is not $\sigma$-finite. $\endgroup$ Nov 1 '20 at 4:53
  • $\begingroup$ @user5280911 I have added a simple counter-example to your staement. $\endgroup$ Nov 1 '20 at 5:02
  • $\begingroup$ I see it. Thank you. So we should not assume the $\sigma$-finiteness (which leads to both the real-valuedness a.e. and $\Sigma$-finiteness). Now, my question is solved. However, as answers in "Is σ -finiteness unnecessary for Radon Nikodym theorem?" mentioned, if $f$ (RN derivative) is allowed to take $+\infty$ value, $\sigma$-finiteness for the indefinite integral is not needed (hope my understanding is correct). So, the question of "absolutely continuous with respect to a finite measure, then Σ-finite" remains open. $\endgroup$ Nov 1 '20 at 5:14

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