An identity involving Radon-Nikodym derivatives The following result was stated without proof in [HAL] (note 4 to section 32 "Derivatives of Signed Measures", p. 136).

If $\mu_0$, $\mu_1$, and $\mu_2$ are finite measures, and if
$$
\begin{array}{lcl}
f_1 & = & \frac{d\mu_0}{d\left(\mu_0+\mu_1\right)} \\
f_2 & = & \frac{d\mu_0}{d\left(\mu_0+\mu_2\right)} \\
f & = & \frac{d\mu_0}{d\left(\mu_0+\mu_1+\mu_2\right)}
\end{array}
$$
then we have $\left(\mu_0+\mu_1+\mu_2\right)$-a.e.,
$$
f(x) = 
\begin{cases}
\frac{f_1(x)f_2(x)}{f_1(x)+f_2(x)-f_1(x)f_2(x)} &\mbox{if }f_1(x)f_2(x)\neq0,\\
0 &\mbox{if }f_1(x)=f_2(x)=0
\end{cases}
\space\space\mbox{(Alexander)}
$$

An intuitive, symbolic proof was suggested by postmortes in another thread:

Write
$$
\begin{array}{lcl}
d\mu_0 & = & f_1 d\left(\mu_0+\mu_1\right)\space\space\mbox{(Gary)} \\
d\mu_0 & = & f_2 d\left(\mu_0+\mu_2\right)\space\space\mbox{(Terrence)}\\
d\mu_0 & = & f d\left(\mu_0+\mu_1+\mu_2\right)\space\space\mbox{(Leon)}
\end{array}
$$
Multiply (Gary) and (Terrence) throughout by $f_2$ and $f_1$, respectively, to obtain
$$
\begin{array}{lcl}
f_2d\mu_0 & = & f_1f_2 d\left(\mu_0+\mu_1\right) \\
f_1d\mu_0 & = & f_1f_2 d\left(\mu_0+\mu_2\right)
\end{array}
$$
Add up to obtain
$$
(f_1+f_2)d\mu_0= 2f_1f_2d\mu_0 + f_1f_2d\mu_2 + f_1f_2d\mu_1
$$
Subtract $f_1f_2d\mu_0$ from both sides to obtain
$$
\left(f_1 + f_2 - f_1f_2\right)d\mu_0 = f_1f_2d\left(\mu_0 + \mu_1 +\mu_2\right)
$$
Finally divide both sides by $d\left(\mu_0+\mu_1+\mu_2\right)$ and use (Leon) to obtain (Alexander).

Unfortunately, i fail to see how to turn these symbolic manipulations into a rigorous argument. Any help will be welcome.
References
[HAL] Halmos, Paul Richard. "Measure Theory". Springer-Verlag, 1974.
 A: Everything is already more or less rigorous. The first step is just a reformulation of the definitions of the $f_i$. For example: $f_1$ is the density of $\mu_0$ with respect to $\mu_0 + \mu_1$, so for all measurable $A$ its measure $\mu_0(A)$ can be obtained by integrating $f_1$ with respect to $\mu_0 + \mu_1$ over $A$. »$f_1 \mathrm{d}(\mu_0 + \mu_1)$« is just a shorthand notation for the definition of a measure by the last sentence.
Then the next step follows from a theorem, which you can re-prove easily: if $\mu$ has density $f$ with respect to $\nu$ and $\lambda$ has density $g$ with respect to $\mu$, then $\lambda$ (which is $g \mathrm{d}\mu$ by what I've already said) has density $fg$ with respect to $\nu$ (i. e. $\mathrm{d}\lambda = fg \mathrm{d}\nu$).
The third step uses another theorem, which states that the sum of measures with density with respect to a common control measure has the sum of those densities as density with respect to the common measure. The right-hand sides just uses the definition of the sum of two measures.
The fourth step follows analogously, only that you should be careful not to subtract $\infty$ from $\infty$.
In the last step, you can substitute (Leon) first and get an equation of two measures with density with respect to $\nu := \mu_0 + \mu_1 + \mu_2$. At $\nu$-almost all points theses densities must coincide (otherwise integrate both densities over the set of points where equality does not hold to get a contradiction). Then it is just a matter of shuffling the terms around. Edit: For Martin's caveat: the comparison also gives that $f_1$ vanishes iff $f_2$ vanishes ($\nu$-almost everywhere).
