I came up with a question while I was studying the Radon-Nikodym theorem.

Referring to Folland’s book page 90, the statement of Radon-Nikodym theorem is the following:

Theorem (Radon-Nikodym theorem). Let $\nu$ be a $\sigma$-finite measure and $\mu$ a $\sigma$-finite positive measure on $(X,\mathcal{M})$. If $\nu \ll \mu$, then there is an extended $\mu$-integrable function $f:X\to\mathbb{R}$ such that $d\rho = fd\mu$, and any two such functions are equal $\mu$-a.e.

Here, given a measure space $(X,\mathcal{M},\mu)$ with positive measure $\mu$, a measurable function $f: X \to [-\infty ,\infty]$ is called extended $\mu$-integrable if at least one of $\int f^+ d\mu$ and $\int f^- d\mu$ is finite.

I’ve seen other questions here concerning with generalizing the theorem in terms of “finite”-issue of $\mu$ and $\nu$. However, that is not the case I am curious of, so, as you can see, I am already considering the case when both $\mu$ and $\nu$ are $\sigma$-finite.

The question I have is: “can we can further generalize the theorem by weakening the “positive” condition of $\mu$?” In other words, can we say the Radon-Nikodym derivative $\frac{d\nu}{d\mu}$ still exists even if $\mu$ is signed $\sigma$-finite measure?

Thank you.

  • 1
    $\begingroup$ You can write both $\mu$ and $\nu$ in terms of $\lvert\mu\rvert$. Can you combine that to write $\nu$ as $h\,d\mu$ for some $h$? If not always, under which conditions? $\endgroup$ Commented Oct 12, 2017 at 12:24
  • $\begingroup$ Thank you for your reply. I think got your point: since $\nu \ll \mu$ Implies $\nu \ll |\mu|$, and always $\mu \ll |\mu|$ is satisfied, using the Radon-Nikodym theorem for positive-$\mu$ case, we have $d\nu = f d|\mu|$ and $d\mu = g d|\mu|$. In case of $g$ is non-vanishing a.e., then $h=f/g$ may be the desired function, i.e. $\nu = h d\mu$. $\endgroup$
    – Matholic
    Commented Oct 12, 2017 at 12:56
  • $\begingroup$ So far so good. What do you know about $g$? $\endgroup$ Commented Oct 12, 2017 at 12:59
  • $\begingroup$ Consider the set $B:=\{x \in X | g(x) = 0\}$. Then $\mu(B) = \int_B g d|\mu| = \int g\chi_B d|\mu|=0$, for $g\chi_B$ is identically zero on X, by the definition of $B$. Thus we conclude g is non-vanishing a.e, and the $h$ is well-defined a.e.! $\endgroup$
    – Matholic
    Commented Oct 12, 2017 at 13:26

1 Answer 1


Thanks to Daniel Fishcer, I got to the point and the logic is like the following:

Considering the total variation measure $|\mu|$, we have $\mu \ll |\mu|$ and thus $\nu \ll |\mu|$. Therefore we can apply the Radon-Nikodym theorem for positive case on two pairs of measures; $(\mu,|\mu|)$ and $(\nu, |\mu|)$ regarding $|\mu|$ as a positive measure. As a result, we have $d\nu = fd|\mu|$ and $d\mu = gd|\mu|$ for some extended $|\mu|$-measurable functions $f$ and $g$.

Now it seems that we have some sort of a relation between $d\nu$ and $d\mu$, which we desired,



$g=\frac{d\mu}{d|\mu|}\neq0$ a.e.

Fortunately, this comes out to be true! Indeed, defining a set $B=\{x \in X| g(x) = 0\}$, $\mu(B)=\int_B gd|\mu|=\int g\chi_B d|\mu|=0$, since $g \chi_B$ is identically zero on $X$ by the definition of $B$.

Thus, we can conclude $\frac{f}{g}$ is the desired Radon-Nikodym derivative, which the extended theorem asked for.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .