Dgdrxrt

Problem 3.3.14 - If $nu$ is an arbitrary signed measure and $mu$ is a $sigma$-finite measure on $(X,M)$ such that $null mu$, there exists an extended $mu$-integrable function $f:Xrightarrow [-infty,infty]$ such that $dnu = fdmu$. Hints:

a.) It suffices to assume that $mu$ is finite and $nu$ is positive.

b.) With these assumptions, there exists an $Ein M$ 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 $Fcap E = emptyset$, 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 $$L = supF: F textis sigma-textfinite for nu$$
Now lets take a sequence $E_n_1^infty$ that are $sigma$-finite with respect to $nu$ and $mu(E_n)rightarrow L$ as $nrightarrow infty$. Now let $$E = bigcup_1^inftyE_n$$ 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 $Fcap E = emptyset$ well since $nullmu$ 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 $mu(Fcup E) > mu(E)$ then this implies that $Ecup F$ 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 $f:Xrightarrow [-infty,infty]$ such that $dnu = fdmu$.

I am not sure if this is completely correct, any suggestions is greatly appreciated.

@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 $null mu$, there exists an extended $mu$-integrable function $f:Xrightarrow [-infty,infty]$ such that $dnu = fdmu$. Hints:

a.) It suffices to assume that $mu$ is finite and $nu$ is positive.

b.) With these assumptions, there exists an $Ein M$ 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 $Fcap E = emptyset$, 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 $X=Pcup N$, $Pcap N= emptyset$ and, for all measurable set $A$,
$$nu^+(A)=nu(Acap P) quad nu^-(A)=-nu(Acap N)$$

For each $n$, $nu^+|_X_nll mu|_X_n$ and $nu^-|_X_n ll mu|_X_n$.
If we can prove for those cases, that there are extended $mu$-integrable functions $f_n:Xrightarrow [0,infty]$ such that $dnu^+|_X_n = f_ndmu|_X_n$ and extended $mu$-integrable functions $g_n:Xrightarrow [0,infty]$ such that $dnu^-|_X_n = g_ndmu|_X_n$.

It is easy to see that $g_n = 0$ $mu$-a.e. on $Pcap X_n$ and $f_n = 0$ $mu$-a.e. on $Ncap X_n$. 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 $h:Xrightarrow [-infty,infty]$ and we also have, for any $A$ measurable set,
beginalign*
nu(A) &= sum_nnu(Acap X_ncap P) + sum_nnu(Acap X_ncap N)= \
&= sum_nnu^+(Acap X_n) - sum_nnu^-(Acap X_n)= \
&= sum_n int_Acap X_n f_n dmu - sum_n int_Acap X_n g_n dmu= \
&= int_A sum_n f_n dmu - int_A sum_n g_n dmu= \
&= int_A sum_n (f_n - g_n) dmu = \
& =int_A sum_n h_n dmu = int_A h dmu
endalign*
So we have proved that, for any $A$ measurable set,
$$nu(A) = int_A h dmu$$
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 $int h^+ dmu <infty$ or $int h^- dmu <infty$ 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 $null mu$, there exists an extended measurable function $f:Xrightarrow [0,infty]$ such that $dnu = fdmu$.

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
$$L = supmu(F) : F textis a sigmatext-finite set for nu$$
Now lets take a sequence $E_n_1^infty$ that are $sigma$-finite with respect to $nu$ and $mu(E_n)rightarrow L$ as $nrightarrow infty$. Now let $$E = bigcup_1^inftyE_n$$ 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 $nu|_Ell mu|_E$ we apply the Radon-Nikodym theorem, and so there is $f_1:Erightarrow [0,+infty)$ such that
$$nu(Acap E) =int_Acap E f_1 dmu tag1$$

Now, given any measurable set $F$ such that $Fcap E = emptyset$, suppose that $mu(F)>0$ and $nu(F) <infty$. Then $Ecup F$ is $sigma$-finite for $nu$, (that is $Ecup F in S$) and
$$mu(Ecup F) =mu(E) +mu( F) > mu(E)=L = sup S$$
Contradiction.

So, for any measurable set $F$ such that $Fcap E = emptyset$,
we have two cases:

1. 
   $mu(F) =0$ then, since $nu ll mu$, $nu(F)=0$.


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

So, for any measurable set $F$ such that $Fcap E = emptyset$,

$$ nu(F) = int_F +infty chi_E^c dmu $$

That means, for any measurable set $A$,

$$ nu(Acap E^c) = int_Acap E^c +infty chi_E^c dmu tag2$$

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 $A$,
$$nu(A)= nu(Acap E)+ nu(Acap E^c)= int_Acap E f_1 dmu +int_Acap E^c +infty chi_E^c dmu= int_A f dmu$$