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.
Best Answer
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 $n$.