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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $n$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.