Pairwise measurable derivatives imply measurability of combined derivative I've found the following simple claim in an article. Unfortunately, i don't understand the proof given there nor can i come up with an alternative proof of my own. Maybe math.stackexchange can give me a hand?
Claim ([HAL], a combination of Lemma 9 and Corollary 3) Let $S=\left(\Omega,\mathcal{A}\right)$ be a measurable space and let $\mathfrak{M}=\left\{\mu_0,\mu_1,\mu_2\right\}$ be a set of three finite measures on $S$. Suppose $\mathcal{T}$ is a sub-$\sigma$-algebra of $\mathcal{A}$ such that for any two measures $\mu,\nu\in\mathfrak{M}$, the Radon-Nikodym derivative $\frac{d\mu}{d\left(\mu+\nu\right)}$ is $\mathcal{T}/\mathfrak{B}$-measurable. Then $\frac{d\mu_0}{d\left(\mu_0+\mu_1+\mu_2\right)}$ is $\mathcal{T}/\mathfrak{B}$-measurable.
As an alternative to help in proving the claim, i'd appreciate help in understanding the proof given in the article that i reproduce below almost verbatim. The points that baffle me are:


*

*Why is $(*)$ true?

*In the last paragraph, why would $f$ need to be redefined?

*In the last paragraph, why is $f$ redefined on a set of $\mu_0$-measure $0$? Shouldn't it be redefined on a set of $\left(\mu_0+\mu_1+\mu_3\right)$-measure $0$, since $f$ is unique up to a set of $\left(\mu_0+\mu_1+\mu_3\right)$-measure $0$?

*Why is the redefined $f$ $\mathcal{T}/\mathfrak{B}$-measurable?


Proof
Define
$$
\begin{array}{lcl}
f_1 & := & \frac{d\mu_0}{d\left(\mu_0+\mu_1\right)} \\
f_2 & := & \frac{d\mu_0}{d\left(\mu_0+\mu_2\right)}
\end{array}
$$
Since $d\mu_0=f_1d\left(\mu_0+\mu_1\right)=f_2d\left(\mu_0+\mu_2\right)$, we have $f_1d\mu_0=f_1f_2d\left(\mu_0+\mu_2\right)$ and $f_2d\mu_0=f_1f_2d\left(\mu_0+\mu_1\right)$, so that
$$\left(f_1+f_2-f_1f_2\right)d\mu_0=f_1f_2d\left(\mu_0+\mu_1+\mu_2\right) \tag{*}$$
If we define
$$f:=\frac{d\mu_0}{d\left(\mu_0+\mu_1+\mu_2\right)}$$
then it follows that
$$\left(f_1+f_2+f_1f_2\right)f=f_1f_2\space\left[\mu_0+\mu_1+\mu_2\right]-\mathrm{a.e.}$$
Since $0\leq f_1\leq1$ and $0\leq f_2\leq1$, the equation $f_1+f_2+f_1f_2=0$ is equivalent to $f_1=f_2=0$. Since $\mu_0\left(\left\{x:f_1(x)=f_2(x)=0\right\}\right)=0$, it follows that $f$ may be redefined, if necessary, to be $0$ on the set $\left\{x:f_1(x)=f_2(x)=0\right\}$ without affecting the relation $d\mu_0=fd\left(\mu_0+\mu_1+\mu_2\right)$; since outside this set $f=f_1f_2/\left(f_1+f_2-f_1f_2\right)$, the proof is complete.
References
[HAL] Halmos, Paul R., Savage, L. J. "Application of the Radon-Nikodym Theorem to the Theory of Sufficient Statistics". Annals of Mathematical Statistics, 20, 225-241 (1949)
 A: I can answer some of these points, but from your question I can't see what $\cal{B}$ is, so I'm afraid I've got no answer for your last question.  In order then:
(*) holds because your measures are finite and because of the relationships defined:
$$f_1d\mu_0 = f_1f_2d(\mu_0 + \mu_2)$$ and $$f_2d\mu_0 = f_1f_2d(\mu_0+ \mu_1)$$ so adding these together gives:
$$(f_1+f_2)d\mu_0= 2f_1f_2d\mu_0 + f_1f_2\mu_2 + f_1f_2\mu_1$$ and so:
$$(f_1 + f_2 - f_1f_2)d\mu_0 = f_1f_2d(\mu_0 + \mu_1 +\mu_2)$$
For your second question and third questions, the aim is to be able to divide by the $(f_1 + f_2 -f_1f_2)$ term. So because we know that $(f_1+f_2-f_1f_2) = 0$ iff $f_1 = 0 = f_2$ (with respect to the $d\mu_0$ measure) we may have to redefine $f$ to be zero on that set of points (so that we're not trying to divide by 0)
For your third question look back at (*): $(f_1+f_2-f_1f_2)$ is measured by $d\mu_0$, not $d(\mu_0 +\mu_1+\mu_2)$.
EDIT: ah, I should have surmised that $\cal B$ was Borel really....
That follows from everything that you're using in this theorem being Borel-measurable: if you have two measurable functions $f$ and $g$ then $f+g$, $f/g$ and $kf$ (for $k$ real) are all measurable functions.  Since $f_1$ and $f_2$ are given as Radón-Nikodym derivatives they are certainly measurable, and so $f_1f_2/(f_1 + f_2 -f_1f_2)$ is measurable being a permitted combination of measurable functions.
A: I have found the following unproven proposition in a book by one of the authors of the article that resolves all my questions and can be seen as a concise proof of the claim that was the focus of my original post. I reproduce it here.

If $\mu_0$, $\mu_1$, and $\mu_2$ are totally finite measures, and if $d\mu_0=f_1d\left(\mu_0+\mu_1\right)=f_2d\left(\mu_0+\mu_2\right)=fd\left(\mu_0+\mu_1+\mu_2\right)$, then we have, almost everywhere with respect to $\mu_0+\mu_1+\mu_2$,
$$
f(x) = 
\begin{cases}
\frac{f_1(x)f_2(x)}{f_1(x)+f_2(x)-f_1(x)f_2(x)} &\mbox{if }f_1(x)f_2(x)\neq0,\\
0 &\mbox{if }f_1(x)=f_2(x)=0
\end{cases}
$$
[HAL] Note 4 to section 32 "Derivatives of Signed Measures", p. 136

Consult the thread An identity involving Radon-Nikodym derivatives for a rigorous proof.
References
[HAL] Halmos, Paul Richard. "Measure Theory". Springer-Verlag, 1974.
