TITLE: Radon-Nikodým (chain rule and other properties)
QUESTION [15 upvotes]: (1) For three $\sigma$-finite measures $\mu\ll\nu\ll\eta$ it is 
    $$
\frac{d\mu}{d\eta}=\frac{d\mu}{d\nu}\frac{d\nu}{d\eta}~~\eta-\text{a.s.}
$$
    (2) For two finite measures $\mu\sim\nu$ it is $\mu$-a.s. (and therefore $\nu$-a.s.)
    $$ 
0<\frac{d\mu}{d\nu}\frac{d\nu}{d\mu}<\infty~~~~~~~~~\text{and}~~~~~~~~~~~\frac{d\mu}{d\nu}=\left(\frac{d\nu}{d\mu}\right)^{-1}
$$


Hello!
Here's what I've already tried.
(1) From $\mu\ll\nu\ll\eta$ it follows $\mu\ll\eta$ and from this by Radon-Nikodým, that it exists a density $\frac{d\mu}{d\eta}$ of $\mu$ relating to $\eta$, that is $\eta-\text{ a.s.}$ unique.
Moreover, there is a second density of $\mu$ relating to $\eta$, getting by using $\mu\ll\nu$ and $\nu\ll\eta$ and again Radon-Nikodým: It exists a density $\frac{d\mu}{d\nu}$, that is $\nu-\text{ a.s.}$ unique and for which it is $\mu=\frac{d\mu}{d\nu}\nu$. Similarly, there exists a density $\frac{d\nu}{d\eta}$, that is $\eta-\text{ f.s.}$ unique and for which it is $\nu=\frac{d\nu}{d\eta}\eta$. Putting this together, it is
$$
\mu=\frac{d\mu}{d\nu}\nu=\frac{d\mu}{d\nu}\frac{d\nu}{d\eta}\eta.
$$
So another density of $\mu$ relating to $\eta$ is found.
But because, as said, by Radon-Nikodým the densitiy of $\mu$ relating to $\eta$ (named above $\frac{d\mu}{d\eta}$) is $\eta-\text{ a.s.}$ unique, it follows
$$ 
\frac{d\mu}{d\eta}=\frac{d\mu}{d\nu}\frac{d\nu}{d\eta}~\eta-\text{ a.s.}
$$
(2) From $\mu\sim\nu$ it follows by Radon-Nikodým:

$\mu=\frac{d\mu}{d\nu}\nu$, whereat $\frac{d\mu}{d\nu}$ is $\nu-\text{ a.s.}$ unique 
$\nu=\frac{d\nu}{d\mu}\mu$, whereat $\frac{d\nu}{d\mu}$ is $\mu-\text{ a.s.}$ unique

$$
\implies \nu=\frac{d\nu}{d\mu}\mu=\frac{d\nu}{d\mu}\frac{d\mu}{d\nu}\nu
$$
Additionally, it is $\nu\ll\nu$ and $\nu=f\nu$ with $f\equiv 1$. From the $\mu-\text{ a.s.}$-uniqueness of the density of $\nu$ relating to $\nu$ it follows 
$$
\frac{d\nu}{d\mu}\frac{d\mu}{d\nu}=1~\nu-\text{ a.s.}
$$
Finally it is
$$
\frac{d\nu}{d\mu}\frac{d\mu}{d\nu}=1~\nu-\text{ a.s.}\Leftrightarrow \frac{d\mu}{d\nu}=\left(\frac{d\nu}{d\mu}\right)^{-1}~\nu-\text{ a.s.}.
$$
It remains to show, that
$$
0<\frac{d\mu}{d\nu}\frac{d\nu}{d\mu}<\infty~\mu-\text{ a.s.}\text{ resp. }\nu-\text{ a.s}
$$
I was not able to show that yet.
Would be great to get some help.

Is my previous proof allright?
With greetings,
math12

REPLY [3 votes]: I think that your proof is fine. Regarding the missing inequality, you have already shown that 
$\frac{d\nu}{d\mu}\frac{d\mu}{d\nu}=1$ a.e., so of course you have 
$$
0<\frac{d\nu}{d\mu}\frac{d\mu}{d\nu}<\infty.
$$