Real Analysis, Folland Problem 1.3.15 Measures 
Given a measure $\mu$ on $(X,M)$, define $\mu_0$ on $M$ by $$\mu_0(E) = \sup\{\mu(F): F\subset E \ \text{and} \ \mu(F) < \infty\}$$
a.) $\mu_0$ is a semifinite measure. It is called the semifinite part of $\mu$.
b.) If $\mu$ is semifinite, then $\mu = \mu_0$ (Use Exercise 14)
c.) There is a measure $\nu$ on $M$ (in general, not unique) which assumes only the values of $0$ and $\infty$ such that $\mu = \mu_0 + \nu$.

The proof of Exercise 14 can be found here
For a.) I believe we need to show that $\mu_0(E) = \infty$. Perhaps we can do this by proof of contradiction, similarly to what the proof in the link above does (actually it seems identical).
For b.) I am not sure how we can show $\mu = \mu_0$ I think maybe we have to show by construction that $\mu \leq \mu_0$ and $\mu \geq \mu_0$
For c.) I haven't any idea for.
Any suggestions is greatly appreciated.
 A: 
Given a measure $\mu$ on $(X,M)$, define $\mu_0$ on $M$ by $$\mu_0(E) = \sup\{\mu(F):  F\subseteq E \ \text{and} \ \mu(F) < \infty\}$$


a.) $\mu_0$ is a semifinite measure. It is called the semifinite part of $\mu$.


b.) If $\mu$ is semifinite, then $\mu = \mu_0$ (Use Exercise 14)


c.) There is a measure $\nu$ on $M$ (in general, not unique) which assumes only the values of $0$ and $\infty$ such that $\mu = \mu_0 + \nu$.

Proof:
a.) Let us prove $\mu_0$ is a semifinite measure.
i.) Since, for any $E\in M$, $\emptyset\subseteq E$ and $\mu(\emptyset)=0<+\infty$, we have that $0\in \{\mu(F):  F\subseteq E \ \text{and} \ \mu(F) < \infty\}$. So
$$\mu_0(E) = \sup\{\mu(F):  F\subseteq E \ \text{and} \ \mu(F) < \infty\}\geqslant 0$$
ii.) Since, if $F\subseteq \emptyset$, then $F= \emptyset$. So we have
$$\mu_0(\emptyset) = \sup\{\mu(F): F\subseteq \emptyset \ \text{and} \ \mu(F) < \infty\}= \sup\{0\}=0$$
iii.) Now, let $\{E_i\}_{i\in \mathbb{N}}$ be a family of disjoint sets in $M$. Let $E=\bigcup_{i\in \mathbb{N}} E_i$. For any set $F\in M$ such that $F\subseteq E$ and $\mu(F)<+\infty$, we have that  $\{F\cap E_i\}_{i\in \mathbb{N}}$ is a family of disjoint sets in $M$ and
$$F=F\cap E= \bigcup_{i\in \mathbb{N}}(F\cap E_i)$$
So,
$$\mu(F)=\sum_{i\in \mathbb{N}} \mu(F\cap E_i)$$
On the other hand, we have that, for all $i \in \mathbb{N}$, $F\cap E_i \subseteq E_i$ and $\mu(F\cap E_i)<+\infty$. So we have, for each $i \in \mathbb{N}$,
$$  F\cap E_i \in \{H :  H\subset E_i \ \text{and} \ \mu(H) < \infty\}$$
so, we have, for each $i \in \mathbb{N}$,
$$\mu(F\cap E_i)\leqslant \sup \{\mu(H):  H\subset E_i \ \text{and} \ \mu(H) < \infty\}=\mu_0(E_i)$$
Thus, we get
$$\mu(F)=\sum_{i\in \mathbb{N}} \mu(F\cap E_i)\leqslant \sum_{i\in \mathbb{N}}\mu_0(E_i)$$
So we have
$$\mu_0(E) = \sup\{\mu(F):  F\subset E \ \text{and} \ \mu(F) < \infty\}\leqslant\sum_{i\in \mathbb{N}}\mu_0(E_i)$$
So we have proved
$$\mu_0(E) \leqslant\sum_{i\in \mathbb{N}}\mu_0(E_i)\tag{1}$$
Now note that, if there is a $i_0 \in \mathbb{N}$ such that $\mu_0(E_{i_0})=+\infty$, then since $ E_{i_0} \subseteq E$, we have
$$\{F:  F\subset E_{i_0} \ \text{and} \ \mu(F) < \infty\} \subseteq 
\{F:  F\subset E \ \text{and} \ \mu(F) < \infty\}$$ then
$$+\infty = \mu_0(E_{i_0})=\sup \{\mu(F):  F\subset E_{i_0} \ \text{and} \ \mu(F) < \infty\} \leqslant  \\ \leqslant
\sup \{\mu(F) :  F\subset E \ \text{and} \ \mu(F) < \infty\}=\mu_0(E)$$
So $\mu_0(E)=+\infty$ and we have
$$\mu_0(E)=+\infty =\sum_{i\in \mathbb{N}}\mu_0(E_i) \tag{2}$$
Now suppose that for all $i \in \mathbb{N}$, $\mu_0(E_i)<+\infty$
Given any $\varepsilon>0$ then, by the definition of $\mu_0$, there is $\{F_i\}_{i\in \mathbb{N}}$ a family of sets, such that   $F_i \subseteq E_i$ and
$$\mu_0(E_i)-\frac{\varepsilon}{2^{i+1}}\leqslant \mu(F_i)\leqslant \mu_0(E_i)<+\infty$$
Since  $F_i \subseteq E_i$, we have that $\{F_i\}_{i\in \mathbb{N}}$ is a family of disjoint sets and $\bigcup_{i\in \mathbb{N}} F_i\subseteq \bigcup_{i\in \mathbb{N}} E_i=E$. So we get
\begin{align} 
\sum_{i\in \mathbb{N}} \mu_0(E_i)-\varepsilon &=\sum_{i\in \mathbb{N}} \left(\mu_0(E_i)-\frac{\varepsilon}{2^{i+1}}\right )\leqslant \sum_{i\in \mathbb{N}} \mu(F_i)=\mu\left(\bigcup_{i\in \mathbb{N}} F_i\right) = \\
&=\sup\left\{\mu\left(\bigcup_{i=0}^k F_i\right): k\in\mathbb{N} \right\} \leqslant \\
&\leqslant\sup\{\mu(F):  F\subseteq E \ \text{and} \ \mu(F) < \infty\}= \mu_0(E)
\end{align}
So, for any $\varepsilon>0$,
$$\sum_{i\in \mathbb{N}} \mu_0(E_i)-\varepsilon\leqslant \mu_0(E)$$
So
$$\sum_{i\in \mathbb{N}} \mu_0(E_i)\leqslant \mu_0(E)\tag{3}$$
From $(1)$, $(2)$ and $(3)$ we get
$$\mu_0(E)=\sum_{i\in \mathbb{N}} \mu_0(E_i)$$
So $\mu_0$ is a measure.
iv.) It is easy to prove that $\mu_0$ is semifinite. In fact, given any $E\in M$ such that
$$ \mu_0(E) = \sup\{\mu(F):  F\subseteq E \ \text{and} \ \mu(F) < \infty\}=+\infty$$
there is at least one $ F\subseteq E$ and  $\mu(F) < \infty$ such that $0<\mu(F)$, that is, $0< \mu(F) < \infty$.
b.) First, note that, for any $E\in M$, if $\mu(E)<+\infty$, then $\mu(E)=\mu_0(E)$.
Supose now that $\mu$ is semifinite. If $\mu(E)=+\infty$, we know (from Exercise 14) that for any $C\in \mathbb{R}$, $C>0$, there is $F\subseteq E$ such that $C<\mu(F)<+\infty$. So we have
$$ \mu_0(E) = \sup\{\mu(F):  F\subseteq E \ \text{and} \ \mu(F) < \infty\}=+\infty=\mu(E)$$
So, if $\mu$ is semifinite, then for all $E\in M$, $\mu(E)=\mu_0(E)$.
c.) Remember that, given any measure $\mu$ defined on $M$,  a set $E\in M$ is $\sigma$-finite (w.r.t. $\mu$) if $E$ is the countable union of sets $E_i$ such that $\mu(E_i)<+\infty$, for all $i\in \mathbb{N}$.
Let $\nu$ be a measure on $M$ defined by, for any $E\in M$,
\begin{align} 
\nu(E) = 0\phantom{\infty} &\;\textrm{ if $E$ is $\sigma$-finite} \\
\nu(E) = +\infty &\;\textrm{ if $E$ is not $\sigma$-finite} 
\end{align}
It is easy to see that $\nu$ is well defined (it is uniquely defined for all $E\in M$) and it is also easy to see that $\nu$ is a measure.
In fact, for any $E\in M$, $\nu(E)\geqslant 0$ and $\nu(\emptyset)=0$. Let $\{E_i\}_{i\in \mathbb{N}}$ be a family of disjoint sets in $M$. Let $E=\bigcup_{i\in \mathbb{N}} E_i$. If, for all $i\in \mathbb{N}$, $E_i$ is $\sigma$-finite, then $E$ is $\sigma$-finite
So $$\nu(E)=0=\sum_{i\in \mathbb{N}}\nu(E_i) $$
If there is  $i_0\in \mathbb{N}$, such that $E_{i_0}$ is not $\sigma$-finite, then $E$ is not $\sigma$-finite
So $$\nu(E)=+\infty=\nu(E_{i_0})\leqslant \sum_{i\in \mathbb{N}}\nu(E_i)  \leqslant +\infty $$
So $$\nu(E)=+\infty=\sum_{i\in \mathbb{N}}\nu(E_i)  $$
So we have proved $\nu$ is a measure.
Now, it is easy to see that, for any $E\in M$, if  $E$ is $\sigma$-finite, then $\mu_0(E)=\mu(E)$ and, since $\nu(E)=0$, we have
$$\mu(E)=\mu_0(E)+\nu(E)$$
On the other hand, if  $E$ is not  $\sigma$-finite, then $\mu(E)=+\infty$ and $\nu(E)=+\infty$. So we have
$$\mu(E)=\mu_0(E)+\nu(E)$$
So we have prove
$ \mu=\mu_0+\nu$.
Remark: We might be tempted to define $\nu$ by, for any $E\in M$,
\begin{align} 
\nu(E) = 0\phantom{\infty} &\;\textrm{ if $\mu(E)<+\infty$} \\
\nu(E) = +\infty &\;\textrm{ if $\mu(E)=+\infty$} 
\end{align}
The issue with such definition is that this $\nu$ may not be a (countable additive) measure. In fact, suppose there is $E\in M$ such that $E$ is $\sigma$-finite and $\mu(E)=+\infty$. Then there is $\{E_i\}_{i\in \mathbb{N}}$ a family of disjoint sets in $M$ such that, for all $i \in \mathbb{N}$, $\mu(E_i)<+\infty$ and  $E=\bigcup_{i\in \mathbb{N}} E_i$. We have then
$\nu(E)=+\infty$ and, for all $i \in \mathbb{N}$, $\nu(E_i)=0$. So,
$\nu(E)>\sum_{i \in \mathbb{N}}\nu(E_i)$.
Remark 2: The measure $\nu$ that satisfies the item c.) does not need to be unique. Here is another way to define a different $\nu$ (which we are going to call $\nu'$) that will also satisfy the item c.).
Remember that, given any measure $\mu$ defined on $M$,  a set $E\in M$ is semi-finite (w.r.t. $\mu$) if, for all $F \subseteq E$, if $\mu(F)= +\infty$, there is $G\subseteq F$ such that $0<\mu(G)< \infty$. Note that if $\mu(E)<\infty$, then $E$ is trivially semi-finite.
We will also use this lemma:

Let $\{E_i\}_{i\in \mathbb{N}}$ be a family of disjoint sets in $M$. Let $E=\bigcup_{i\in \mathbb{N}} E_i$. If, for all $i\in \mathbb{N}$, $E_i$ is semi-finite, then $E$ is semi-finite.

Proof: If $F \subseteq E$ and $\mu(F)= +\infty$, then there is $i\in \mathbb{N}$, such that $\mu(F \cap E_i) >0$. We have to possibilities:

*

*If $\mu(F \cap E_i) < \infty$, take $G=F \cap E_i$, then we have $G = F \cap E_i \subseteq F$ such that $0< \mu(G)< \infty$.

*If $\mu(F \cap E_i) = \infty$, since $E_i$ is semi-finite, there is $G \subseteq F \cap E_i \subseteq F$ such that $0< \mu(G)< \infty$.

So we proved that $E$ is semi-finite. $\square$
Let $\nu'$ be a measure on $M$ defined by, for any $E\in M$,
\begin{align} 
\nu'(E) = 0\phantom{\infty} &\;\textrm{ if $E$ is semi-finite} \\
\nu'(E) = +\infty &\;\textrm{ if $E$ is not semi-finite} 
\end{align}
It is easy to see that $\nu'$ is well defined (it is uniquely defined for all $E\in M$) and it is also easy to see that $\nu'$ is a measure.
In fact, for any $E\in M$, $\nu'(E)\geqslant 0$ and $\nu'(\emptyset)=0$.
Let $\{E_i\}_{i\in \mathbb{N}}$ be a family of disjoint sets in $M$. Let $E=\bigcup_{i\in \mathbb{N}} E_i$. If, for all $i\in \mathbb{N}$, $E_i$ is semi-finite, then, by the lemma, $E$ is semi-finite.
So
$$\nu'(E)=0=\sum_{i\in \mathbb{N}}\nu'(E_i) $$
If there is  $i_0\in \mathbb{N}$, such that $E_{i_0}$ is not semi-finite, then it is immediate that $E$ is not semi-finite.
So
$$\nu'(E)=+\infty=\nu'(E_{i_0})\leqslant \sum_{i\in \mathbb{N}}\nu'(E_i)  \leqslant +\infty $$
So
$$\nu'(E)=+\infty=\sum_{i\in \mathbb{N}}\nu'(E_i)  $$
So we have proved $\nu'$ is a measure.
Now, it is easy to see that, for any $E\in M$, if  $E$ is semi-finite, then $\mu_0(E)=\mu(E)$ and, since $\nu'(E)=0$, we have
$$\mu(E)=\mu_0(E)+\nu'(E)$$
On the other hand, if  $E$ is not  semi-finite, then $\mu(E)=+\infty$ and $\nu'(E)=+\infty$. So we have
$$\mu(E)=\mu_0(E)+\nu'(E)$$
So we have prove
$ \mu=\mu_0+\nu'$.
It is easy to see that $\nu'$ is, in general, different from the measure  $\nu$ that we defined   in  c.). In fact, if $E \in M$ is semi-finite but not $\sigma$-finite, we have $\nu'(E)=0 < +\infty = \nu(E)$.
A: For part a) you need to show that it is indeed a measure (usual axioms) and that if $\mu_0(E)=\infty$ then there is $F\subset E$ with finite measure (this follows by the definition of sup). For part (b) if $\mu$ is already semifinite and $\mu(E)=\infty$ by Exercise 4 we have that $\forall C>0:\exists F\subset E,F\in \mathfrak M:C<\mu(F)<\infty$ which yields that $\mu_0(E)=\infty$. If $\mu(E)<\infty$ then $E$ realizes the sup of $\mu_0(E)$ so $\mu_0(E)=\mu(E)$. Hence the two measures are the same.
For part (c), if $\mu$ is not semi-finite we have the problem that we might not have the sequence of sets $F_n\subset E$ such that their measure diverge to infinity when $\mu(E)=\infty$. We can define $\mu$ to be zero on all sets with finite measure and infinity otherwise, this is a measure on $\mathfrak M$. If $\mu(E)<\infty$ it is clear that $\mu(E)=\mu_0(E)+\nu(E)$. 
If $\mu(E)=\infty$ even if we have the sequence of sets we are not perturbing the value of the sup but we are just correcting the case $\mu(E)=\infty$ and $E$ does not contain a sequence of set with diverging measure.
