Understanding Radon Nikodym proof in the book Measure Theory from Donald L.Cohn

Question

I'm reading the book Measure Theory (second edition) from Donald L.Cohn and
have a problem in understanding his proof of the $\sigma$-finite case of his
proof of the Radon-Nikodym theorem. In the book he first proves the case for
finite measures $\mu$, $\nu$ and then proceed to prove it for $\mu$ and $\nu$
as $\sigma$-finite measures on the measurable space $\langle X, \mathcal{A}
\rangle$. For this he considers a disjoint sequence $\{ B_n \}_{n \in
\mathbb{N}_0}$ of measurable sets such that $\mu (B_n)$ and $\nu (B_n)$ are
finite. So far I can follow the proof, however then he proceeds to use the
finite case to find a $g_n : X \rightarrow [0, \infty)$ such that $\nu (A) =
\int_A g_n d \mu$ for every measurable subset of $B_n$ and it is this which is
not clear to me. If you want to apply the finite case you must have a
measurable space with finite measures on it. So I thought of using the
measurable space $\langle B_n, \mathcal{A}_n \rangle$ where $\mathcal{A}_n$={$B_n \bigcap A : A \in \mathcal{A}$} and instead of $\mu, \nu$ using the measures $\mu_n$, $\nu_n$ (the restrictions of $\mu, \nu$ to $\mathcal{A}_n$). Using the finite case we have
then the existence of a measurable function $g_n : X \rightarrow [0, \infty)$
so that $\nu (A) = \nu_n (A) = \int_A g_n d \mu_n$. Is this the correct way
to proceed and if so how would I prove that $\int_A g_n d \mu_n = \int_A g_n d
\mu$ and find a $g : X \rightarrow [0, \infty)$ such that $\int_A g
d \mu = \nu (A)$.

Answer 1

I think Cohn's $g_n$ is the same as yours. It is defined (and measurable) on $B_n$ by: for every measurable $A\subset B_n,$ $\nu(A)=\int_Ag_nd\mu.$

Next, simply define $g$ on $X$ by: its restriction to $B_n$ is $g_n.$ Since the $B_n$'s are disjoint (and there union is $X,$ which you forgot to mention), there is no problem.

Edit1: I just looked online at Cohn's book p.124 and this is exactly what he does(!).

Edit2: Note that your $\mathcal A_n$ is simply the set of measurable subsets of $B_n,$ and your $\mu_n,\nu_n$ are the restrictions of $\mu,\nu$ to it, so that for any measurable function $g$ and any $A\in\mathcal A_n,$ $\int_A gd\mu_n=\int_Agd\mu.$