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

For each $n$, let $\mu_n(E)=\mu\left(E \cap B_n\right) \quad$ and $\quad \nu_n(E)=\nu\left(E \cap B_n\right) \quad$ for any $E \in \mathcal{M}$

Then $\mu_n$ and $\nu_n$ are finite measures. Moreover, since $\nu$ is absolutely continuous with respect to $\mu, \nu_n$ is absolutely continuous with respect to $\mu_n$. Hence for each positive integer $n$, there is a non-negative measurable function $f_n$ on $X$ such that $$ \nu_n(E)=\int_E f_n d \mu_n \quad \text { for all } E \in \mathcal{A} $$

Note that $\mu_n(E)=\mu\left(E \cap A_n\right)=\int_E \chi_{B_n} d \mu$. so $$ \begin{aligned} \nu(E) & =\sum_{n=1}^{\infty} \nu_n(E)=\sum_{n=1}^{\infty} \int_E f_n d \mu_n \\ & =\sum_{n=1}^{\infty} \int_E f_n \chi_{B_n} d \mu=\int_E \sum_{n=1}^{\infty} f_n \chi_{A_n} d \mu \end{aligned} $$

Define $f=\sum_{n=1}^{\infty} f_n \chi_{B_n}$. Then $f$ has the desired property. Hence the proof.

I would like to remark why $\int_E f_n d \mu_n =\int_E f_n \chi_{B_n} d \mu$

Suppose $f$ is a simple function, then it is easy to check. Now suppose $f$ is non-negative function, one way to define the integral of it is through the MCT. $\int_E f d \mu_n =\lim_{i}\int_E f_i d \mu_n=\lim_{i}\int_E f_i \chi_{B_n} d \mu=\int_E f \chi_{B_n} d \mu$. That is enough for our proof since we have $f_n \geq 0$

Answer 2

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 $A\in\mathcal A_n,$ $\int_A gd\mu_n=\int_Agd\mu.$