# Is Radon-Nikodym derivative with respect to a finite measure real-valued a.e.?

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.

• 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. Oct 31 '20 at 11:55
• @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. Oct 31 '20 at 12:03
• The answer to the linked question is indeed wrong! I have provided a counter-exmaple. Please visit that page again. Oct 31 '20 at 12:19
• @DavidC.Ullrich It is assumed that $\nu$ is finite. No finiteness /sigma finiteness condition on $\mu$ Oct 31 '20 at 12:20
• 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 $p=+\infty$. Oct 31 '20 at 12:36

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 $$n$$.
• 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/… Nov 1 '20 at 4:15
• @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. Nov 1 '20 at 4:42
• 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. Nov 1 '20 at 4:53
• 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. Nov 1 '20 at 5:14