# 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.

• Quick thoughts: $\mu =\cup \mu_n$, is well defined since $\mu_n =\mu _m$ if $m\le n$. You do need to check that it's a measure. The fact that $h_n=h_m$ whenever $m\le n$ does not imply that $h_n$ are increasing. – Matematleta Jun 20 '16 at 2:52
• @Chilango, what does $\cup \mu_n$ mean? Also, $\mu$ is a given $\sigma$-finite measure, so I don't see why we have to define it and prove it's a measure. Also, I'm going to edit my post to address why I think $h_n$ are incr a.e. – Jason Jun 20 '16 at 23:55
• Once you know Radon Nikodym for finite measures, you know it works for positive measures (by considering $\mu = \lim_n \mu_n$ where $\mu_n(E)$ is non-decreasing and finite, i.e. by monotone convergence), and for those kind of signed measures that can be written as $\mu = \mu_++\mu_-$ What do you want to say more ? – reuns Jun 21 '16 at 0:22
• @Jason I have tried to give a more complete answer. – Matematleta Jun 21 '16 at 0:38

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 $X$.
• Thx for elaborating. So if $E \subseteq \textbf{X}$, then $\lambda(E)= \sum_{n=1}^\infty \lambda_n (E)$ = $\sum_{n=1}^\infty \int_E f_n d \mu_n \overset{?}{=} \sum_{n=1}^\infty \int_E f_n d \mu$. The last step is intuitive, but is there a simple way to argue it rigorously? I'd like to avoid, if possible, to say it holds for indicator fns, so it holds for simple fns, so it holds for nonnegative fns, so it holds for general fns. – Jason Jun 21 '16 at 1:17