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? $\endgroup$ – Daniel Fischer Oct 12 '17 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 Oct 12 '17 at 12:56
  • $\begingroup$ So far so good. What do you know about $g$? $\endgroup$ – Daniel Fischer Oct 12 '17 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 Oct 12 '17 at 13:26

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.


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.