Radon Nikodym Thm: extending to $\sigma$-finite case I am reading Bartle's "Elements of Integration". 
 Radon-Nikodym Thm: Let $\lambda,\mu$ be $\sigma$-finite measures on a measurable space $(X,\textbf{X})$ and say $\lambda \ll \mu$.  Then $\exists$ unique $\mu$-a.e. measurable $f:X \to \bar{\mathbb{R}}_{\geq 0}$ (that's my notation for the nonnegative extended reals) s.t. $\lambda(E)= \int_E f \, d \mu$, $\forall E \in \textbf{X}$.
From his pf, I'm comfortable with the case where $\lambda,\mu$ are finite (both existence and uniqueness). 
Here is how Bartle generalizes to the $\sigma$-finite case. 


*
*Let $X_1 \subseteq X_2 \subseteq\cdots$ be s.t. $X= \bigcup_{n=1}^\infty X_n$, each $\lambda(X_n),\mu(X_n)< \infty$.

* $\exists$ fns $h_n: X \to \bar{\mathbb{R}}_{\geq 0}$ s.t. $h_n(x)=0$ for $x \notin X_n$ and $$\lambda(E)= \int_E h_n d \mu, \forall E \subseteq X_n \text{ measurable} \tag1$$

* If $n \leq m$, then $X_n \subseteq X_m$ and $\int_E h_n d \mu = \int_E h_m d \mu$, $\forall E \in \textbf{X}$.  So $h_n 1_{X_n}= h_m 1_{X_n}$ $\mu$-a.e.

* Put $f_n:=\sup \{ h_1,\ldots,h_n\}$, so $(f_n)$ is a monotone seq.  Put $f:= \lim_{n \to \infty} f_n$.  Then $\lambda(E \cap X_n)= \int_E f_n \, d \mu$, so $\lambda(E)= \int_E f \, d \mu$, $\forall E \in \textbf{X}$, by the Monotone Conv Thm.


Here are my questions regarding the proof: 
a) (2nd bullet point) I can see $\lambda_n (E):= \lambda(E \cap X_n)$ and $\mu_n (E):= \mu(E \cap X_n)$ are finite measures s.t. $\lambda_n \ll \mu_n$.  Thus, $\exists$ fns $h_n$ s.t. $\lambda(E)= \int_E h_n d \mu_n$, $\forall E \subseteq X_n$ measurable.  But why can we replace $\mu_n$ with $\mu$ to get (1)?  This seems intuitive but how do I make a rigorous argument?  Perhaps my attempt to define $\lambda_n, \mu_n$ was not helpful but it was the only way I could think of to use the result for finite measures. 
b) (4th bullet pt) Why does he define $f_n$ at all?  Why not just use $h_n$ in place of $f_n$?  It seems the $(h_n)$ are already an incr seq: for instance, $h_1=h_2$ $\mu$-a.e. on $X_1$ but $h_1=0$ on $X_1^c$ (while we know $h_2 \geq$0) so $h_1 \leq h_2$ $\mu$-a.e. 
I appreciate any help as I am new to measure theory and trying to understand how arguments are made.
 A: Perhaps I do not understand your question, but it seems fairly straightforward to extend the theorem to the $\sigma $ -finite case:
$X=\bigcup_n X_n$ is a countable disjoint union of $\mu $-finite sets, $X_n$. For each integer $n$ and each measureable $E\subseteq X$, define $\mu_n=\mu (E\cap X_n)$. It's easy to see that $\mu_n$ is a measure. 
Then, by what you already know, for each integer $n$, there is a unique $f_n$ such that $d\lambda_n =f_nd\mu_n$, where each $ \lambda _n(E)=\lambda (E\cap X_n)$, and where we may take $f_n=0$ on $X_n^{c}$, because $\mu_n (X_n^{c})=0$.
Now it only remains to set $f=\sum_nf_n$ and $\lambda=\sum_n\lambda_n$ to obtain the result on $X$.
A: a) The Radon-Nikodym derivative $\frac{d\mu_n}{d\mu}=1_{X_n}$, because
$$
\mu_n(E)=\mu(E\cap X_n)=\int_E 1_{X_n}\,d\mu.
$$
Thus for any non-negative measurable function $f$, the identity
$$
\int_Xf\,d\mu_n=
\int_Xf\cdot 1_{X_n}\,d\mu=
\int_{X_{n}}f\,d\mu
$$
holds. Apply this idea when
$f=h_n\cdot1_{E}$, for any measurable set $E\subseteq X_n$.
b) I can not agree more: the $h_n\leq h_{n+1}$ a.e $[\mu]$, that is $f_n=h_n$ (a.e.). Thus the proof is working without introducing the sequence $f_n$.
