Real Analysis, Problem 3.2.14 The Radon Nikodym Theorem 
Problem 3.3.14 - If $\nu$ is an arbitrary signed measure and $\mu$ is a $\sigma$-finite measure on $(X,M)$ such that $\nu\ll \mu$, there exists an extended $\mu$-integrable function $f:X\rightarrow [-\infty,\infty]$ such that $d\nu = fd\mu$. Hints:
a.) It suffices to assume that $\mu$ is finite and $\nu$ is positive.
b.) With these assumptions, there exists an $E\in M$ that is $\sigma$-finite for $\nu$ such that $\mu(E)\geq \mu(F)$ for all sets $F$ that are $\sigma$-finite for $\nu$.
c.) The Radon-Nikodym theorem applies on $E$. If $F\cap E = \emptyset$, then either $\nu(F) = \mu(F) = 0$ or $\mu(F) > 0$ and $|\nu(F)| = \infty$.

Attempted proof - Consider $\mu(E)$ where $E$ is a $\sigma$-finite set for $\nu$. Since we have that $\mu$ is finite then clearly $\mu(E)$ must be bounded which implies it has a supremum. We will define the supremum as $$L = \sup\{F: F \ \text{is} \  \sigma-\text{finite for} \ \nu\}$$
Now lets take a sequence $\{E_n\}_{1}^{\infty}$ that are $\sigma$-finite with respect to $\nu$ and $\mu(E_n)\rightarrow L$ as $n\rightarrow \infty$. Now let $$E = \bigcup_{1}^{\infty}E_n$$ then $\mu(E) \leq L$ since it is taken to be a countable union of $\sigma$-finite sets. On the other hand, $\mu(E)\geq \mu(E_n)$ for all $n$ and since $\mu(E_n)\rightarrow L$ then $\mu(E) = L$. Now since $\mu(E) = L$ then clearly by definition of $L$ we have that $\mu(E) > \mu(F)$ for all sets $F$ that are $\sigma$-finite for $\nu$. Now we apply the Radon-Nikodym theorem: so, suppose that $F\cap E = \emptyset$ well since $\nu\ll\mu$ then by definition $\nu(F) = \mu(F) = 0$. If $\mu(F) > 0$ then for sake of contradiction suppose $|\nu(F)|\neq \infty$. Then since $\mu$ is finite $\mu(F\cup E) > \mu(E)$ then this implies that $E\cup F$ is not $\sigma$-finite since a $\sigma$-finite set with a union of a finite set is $\sigma$-finite. Therefore $|\nu(F)| = \infty$.
Thus I believe we can refer to Radon-Nikidym theorem to conclude that there exists an extended $\mu$-integrable function $f:X\rightarrow [-\infty,\infty]$ such that $d\nu = fd\mu$.
I am not sure if this is completely correct, any suggestions is greatly appreciated.
 A: @Wolfy, here is a detailed proof, following Folland's hints.

Problem 3.3.14 - If $\nu$ is an arbitrary signed measure and $\mu$ is a $\sigma$-finite measure on $(X,M)$ such that $\nu\ll \mu$, there exists an extended $\mu$-integrable function $f:X\rightarrow [-\infty,\infty]$ such that $d\nu = fd\mu$. Hints:
a.) It suffices to assume that $\mu$ is finite and $\nu$ is positive.
b.) With these assumptions, there exists an $E\in M$ that is $\sigma$-finite for $\nu$ such that $\mu(E)\geq \mu(F)$ for all sets $F$ that are $\sigma$-finite for $\nu$.
c.) The Radon-Nikodym theorem applies on $E$. If $F\cap E = \emptyset$, then either $\nu(F) = \mu(F) = 0$ or $\mu(F) > 0$ and $|\nu(F)| = \infty$.

Proof - Let us work on hint a.).
Since $\mu$ is a $\sigma$-finite measure on $(X,M)$, there is a countable family $\{X_n\}_n$ of disjoint measurable subsets of $X$ such that, fo all $n$, $\mu(X_n)<\infty$ and $X=\bigcup_n X_n$.
Now, let $P$ and $N$ be a Hahn decomposition of $X$ for $\nu$. Then $X=P\cup N$, $P\cap N= \emptyset$ and, for all measurable set $A$,
$$\nu^+(A)=\nu(A\cap P) \quad \nu^-(A)=-\nu(A\cap N)$$
For each $n$,  $\nu^+|_{X_n}\ll \mu|_{X_n}$ and $\nu^-|_{X_n} \ll \mu|_{X_n}$.
If we can prove for those cases, that there are extended $\mu$-integrable functions $f_n:X\rightarrow [0,\infty]$ such that $d\nu^+|_{X_n} = f_nd\mu|_{X_n}$ and extended $\mu$-integrable functions $g_n:X\rightarrow [0,\infty]$ such that $d\nu^-|_{X_n} = g_nd\mu|_{X_n}$.
It is easy to see that $g_n = 0$ $\mu$-a.e. on $P\cap X_n$ and $f_n = 0$ $\mu$-a.e. on $N\cap X_n$. So, $h_n=f_n-g_n$ is well defined $\mu$-a.e.. (I means we will not have $\infty -\infty$).
Since $\{X_n\}_n$ is a countable family  of disjoint measurable subsets of $X$, we have that
$h= \sum_n h_n$ is a measurable function and $h:X\rightarrow [-\infty,\infty]$ and we also have, for any $A$ measurable set,
\begin{align*} 
\nu(A) &= \sum_n\nu(A\cap X_n\cap P) + \sum_n\nu(A\cap X_n\cap N)= \\
&= \sum_n\nu^+(A\cap X_n) - \sum_n\nu^-(A\cap X_n)= \\
&= \sum_n \int_{A\cap X_n} f_n d\mu - \sum_n \int_{A\cap X_n} g_n d\mu= \\
&=  \int_{A} \sum_n f_n d\mu - \int_{A} \sum_n  g_n d\mu= \\
&=  \int_{A} \sum_n (f_n - g_n) d\mu = \\
& =\int_{A}  \sum_n  h_n d\mu =  \int_{A}  h d\mu
\end{align*}
So we have proved that, for any $A$ measurable set,
$$\nu(A) = \int_{A}  h d\mu$$
Since $\nu$ is a signed measure, it does not attain either $+\infty$ or $-\infty$. It follows then that $h$ is not only measurable but it an extended $\mu$-integrable function. (In other word, either $\int h^+ d\mu <\infty$ or  $\int h^- d\mu <\infty$ or both).
Now, to complete the proof, all we need is to prove that

If $\nu$ is a (positive) measure and $\mu$ is a finite measure on $(X,M)$ such that $\nu\ll \mu$, there exists an extended measurable function $f:X\rightarrow [0,\infty]$ such that $d\nu = fd\mu$.

Consider the set
$$S = \{\mu(F) : \textrm{ $F$ is a $\sigma$-finite set for $\nu$ } \}$$
Since we have that $\mu$ is finite then clearly $S$ must be bounded which implies it has a finite supremum. We will define the supremum as
$$L = \sup\{\mu(F) : F \ \text{is a} \  \sigma\text{-finite set for} \ \nu\}$$
Now lets take a sequence $\{E_n\}_{1}^{\infty}$ that are $\sigma$-finite with respect to $\nu$ and $\mu(E_n)\rightarrow L$ as $n\rightarrow \infty$. Now let $$E = \bigcup_{1}^{\infty}E_n$$ then $\mu(E) \leq L$ since $E$ is a $\sigma$-finite set (being  a countable union of $\sigma$-finite sets). On the other hand, $\mu(E)\geq \mu(E_n)$ for all $n$ and since $\mu(E_n)\rightarrow L$ then $\mu(E) = L$. Now since $\mu(E) = L$ then clearly by definition of $L$ we have that $\mu(E) > \mu(F)$ for all sets $F$ that are $\sigma$-finite for $\nu$.
So we used hint b.). Now let us work on hint c.).
Now, since $\nu|_E$ is $\sigma$-finite and $\nu|_E\ll \mu|_E$ we apply the Radon-Nikodym theorem, and so there is $f_1:E\rightarrow [0,+\infty)$ such that
$$\nu(A\cap E) =\int_{A\cap E} f_1 d\mu  \tag{1}$$
Now, given any measurable set $F$ such that  $F\cap E = \emptyset$, suppose that $\mu(F)>0$ and $\nu(F) <\infty$.  Then $E\cup F$ is $\sigma$-finite for $\nu$, (that is $E\cup F \in S$) and
$$\mu(E\cup F) =\mu(E) +\mu( F) > \mu(E)=L = \sup S$$
Contradiction.
So, for any measurable set $F$ such that  $F\cap E = \emptyset$,
we have two cases:

*

*$\mu(F) =0$ then, since $\nu \ll \mu$, $\nu(F)=0$.

*if $\mu(F)>0$ then $\nu(F)=+\infty$.

So, for any measurable set $F$ such that  $F\cap E = \emptyset$,
$$ \nu(F) = \int_F +\infty \chi_{E^c} d\mu $$
That means, for any measurable set $A$,
$$ \nu(A\cap E^c) = \int_{A\cap E^c} +\infty \chi_{E^c} d\mu \tag{2}$$
Let $f:X \rightarrow [0,+\infty]$ be defined as $f(x)=f_1(x)$ for $x \in E$ and $f(x)=+\infty$ for $x \notin E$. From $(1)$ and $(2)$, we have, for any measurable set $A$,
$$\nu(A)= \nu(A\cap E)+ \nu(A\cap E^c)= \int_{A\cap E} f_1 d\mu +\int_{A\cap E^c} +\infty \chi_{E^c} d\mu= \int_A f d\mu$$
