Saturation of a measure Folland Problem 1.3.16 
Exercise 16 - Let $(X,M,\mu)$ be a measure space. A set $E\subset X$ is called locally measurable if $E\cap A\in M$ for all $A\in M$ such that $\mu(A) < \infty$. Let $\tilde{M}$ be the collection of all locally measurable sets. Clearly, $M\subset \tilde{M}$, if $M = \tilde{M}$, then $\mu$ is saturated. 
a.) If $\mu$ is $\sigma$-finite, then $\mu$ is saturated.

Proof - Suppose $\mu$ is $\sigma$-finite. Let $A\in\tilde{M}$, and let $X = \bigcup_{1}^{\infty}E_j$ where $E_j\in M$ and $\mu(E_j) < \infty$ for all $j$. We know $M\subset \tilde{M}$ what we want to show is that $\tilde{M}\subset M$. We can write $$A = A\cap X = A\cap \left(\bigcup_{1}^{\infty}E_j\right) = \bigcup_{1}^{\infty}E_j\cap A$$ Since $\mu(E_j)<\infty$ we have $E_j\cap A\in M$ for all $j$. Therefore, $\tilde{M}\subset M$, thus $\mu$ is saturated.

b.) $\tilde{M}$ is $\sigma$-algebra.

Proof - 
i.) $\emptyset\in M\subset \tilde{M}$, so $\emptyset\in \tilde{M}$.
ii.) Let $B\in \tilde{M}$. Take any $E\in M$ such that $\mu(E) < \infty$. Then $$E\setminus B = E\cap B^c = E\cap (E\cap B)^c$$
since $E\in M$ and $(E\cap B)\in M$ then $(E\cap (E\cap B)^c\in M$. Thus we have $B^c\in \tilde{M}$.
iii.) Let $\{B_j\}_{1}^{\infty}\in \tilde{M}$. Take any $E\in M$ with $\mu(E)< \infty$. Then, $$\left(\bigcup_{1}^{\infty}B_j\right)\cap E = \bigcup_{1}^{\infty}(B_j\cap E)\in M$$ 
so, by definition of $\tilde{M}$, $\bigcup_{1}^{\infty}B_j\in \tilde{M}$. Therefore $\tilde{M}$ us a $\sigma$-algebra.

c.) Define $\tilde{\mu}$ on $\tilde{M}$ by $\tilde{\mu}(E) = \mu(E)$ if $E\in M$ and $\tilde{\mu}(E) = \infty$ otherwise. Then $\tilde{\mu}$ is a saturated measure on $\tilde{M}$, called the saturation of $\mu$.

Step 1: Show that $\tilde{\mu}$ is a measure on $\tilde{M}$.
Proof - 
i.) $\tilde{\mu}(\emptyset) = \mu(\emptyset) = 0$.
ii.) Let $\{E_j\}_{1}^{\infty}\in \tilde{M}$ that is pairwise disjoint. Let $$E = \bigcup_{1}^{\infty}E_j$$
If $E \in M$ then \begin{align*} \tilde{\mu}(E) = \tilde{\mu}\left(\bigcup_{1}^{\infty}E_j\right) &= \mu\left(\bigcup_{1}^{\infty}E_j\right)\\
&= \sum_{1}^{\infty}\mu(E_j)\\
&= \sum_{1}^{\infty}\tilde{\mu}(E_j)
\end{align*}
If $E\notin M$ then $\tilde{\mu}(E) = \infty$... not sure where to go from here.
Step 2 - $\tilde{\mu}$ is saturated.
Proof - Let $E\subset X$ such that $E\cap A\in \tilde{M}$ when $\tilde{\mu}(A) < \infty$. Choose a $B\in M$ such that $\mu(B)<\infty$. Then, clearly $\tilde{\mu}(B)<\infty$ and $E\cap B\in \tilde{M}$. So, $E\cap B = (E\cap B)\cap B\in M$ so $E\in \tilde{M}$ it thus follows that $\tilde{\mu}$ is saturated.

d.) If $\mu$ is complete, so is $\tilde{\mu}$.

Proof - Suppose $\mu$ is complete. Let $A\subset X$ and suppose there is a $B\in \tilde{M}$ such that $A\subset B$ and $\mu(B) = 0$. Since $B\in\tilde{M}$ and $\mu(B) = 0$ then $\tilde{\mu}(B) < \infty$ and hence $B\in M$. This, since $A\subset B$ we have $A\in M$ by completeness of $\mu$. Therefore, $A\in \tilde{M}$ and $\tilde{\mu}$ is complete.

e.) Suppose that $\mu$ is semifinite. For $E\in\tilde{M}$, define $\underline{\mu}(E) = \sup\{\mu(A):A\in M, A\subset E\}$. Then $\underline{\mu}$ is a saturated measure on $\tilde{M}$ that extends $\mu$.

Step 1 - $\underline{\mu}$ is a measure.
Proof - 
i.) $\overline{\mu}(\emptyset) = \mu(\emptyset) = 0$
ii.) Let $\{E_j\}_{1}^{\infty}$ be a sequence of disjoint sets in $\tilde{M}$. Set, $$E = \bigcup_{1}^{\infty}E_j$$ then by definition of $\tilde{M}$ there is an $A\in M$ and $A\subset E$.
Case 1 - $\mu(A) < \infty$. Then $$\mu(A) = \mu\left(\bigcup_{1}^{\infty}E_j\cap A\right) = \sum_{1}^{\infty}\mu(E_j\cap A)\leq \sum_{1}^{\infty}\underline{\mu}(E_j)$$
Case 2 - $\mu(A) = \infty$. By semifiniteness, for all $C>0$ there exists a $F\subset A$ such that $F\in M$ and $\mu(F) = C$. Then by case 1, $\leq \sum_{1}^{\infty}\underline{\mu}(E_j) = \infty$. Therefore, $\mu(A) \leq \sum_{1}^{\infty}\underline{\mu}(E_j)$. Taking the supremum over all $A$ we have $$\underline{\mu}\left(\bigcup_{1}^{\infty}E_j\right)\leq \sum_{1}^{\infty}\underline{\mu}(E_j)$$ Now we need to show the reverse inequality. By the definition of supremum there exists a sequence $\{B_i\}_{1}^{\infty}\in M$ and $B_i\subset E_i$ for all $i$. Thus, $\underline{\mu}(E_i)\leq \mu(B_i) + \epsilon 2^{-i}$. Therefore, 
\begin{align*}
\sum_{1}^{\infty}\underline{\mu}(E_i) &\leq \sum_{1}^{\infty}\mu(B_i) + \epsilon\\
&= \mu\left(\bigcup_{1}^{\infty}B_i\right) + \epsilon \ \ \text{is this true because of case 1?}\\
&\leq \underline{\mu}\left(\bigcup_{1}^{\infty}E_i\right) + \epsilon
\end{align*}
Since this holds for all $\epsilon > 0$, we have $$\sum_{1}^{\infty}\underline{\mu}(E_j)\leq \underline{\mu}\left(\bigcup_{1}^{\infty}E_j\right)$$ Therefore, $$\underline{\mu}\left(\bigcup_{1}^{\infty}E_j\right) = \sum_{1}^{\infty}\underline{\mu}(E_j)$$
and hence $\underline{\mu}$ is a measure.
Step 2 - $\underline{\mu}$ is saturated.
Proof - Let $E\subset X$ be such that $E\cap A\in \tilde{M}$ when $\underline{\mu}(A)< \infty$. Take any $B\in M$ such that $\mu(B) < \infty$. Then $\underline{\mu}(B) < \infty$ so $E\cap B\in \tilde{M}$. Thus, $E\cap B = (E\cap B)\cap B\in M$ and $E\in\tilde{M}$, hence, $\underline{\mu}$ is saturated.
Step 3 - $\underline{\mu}$ is an extention of $\mu$.
Proof - Let $E\in M$. For any $A\in M$ such that $A\subset E$, we have by monotonicity that $\mu(A)\leq \mu(E)$. Since $\underline{\mu}(E)$ is the supremum over all such $A$, we must have that $\underline{\mu}(E)\leq \mu(E)$. OTOH.... not sure how to show the reverse inequality.

f.) Let $X_1$ and $X_2$ be disjoint uncountable sets, $X = X_1\cup X_2$, and $M$ the $\sigma$-alegbra of countable or co-countable sets in $X$. Let $\mu_0$ be counting measure on $\mathcal{P}(X_1)$ and define $\mu$ on $M$ by $\mu(E) = \mu_0(E\cap X_1)$. Then $\mu$ is a measure on $M$, $\tilde{M} = \mathcal{P}(X)$, and in the notation of parts (c) and (e), $\tilde{\mu}\neq \underline{\mu}$.

Step 1 - $\mu$ is a measure on $M$.
Proof - 
i.) $\mu(\emptyset) = \mu_0(\emptyset\cap X_1) = 0$
ii.) Let $\{E_j\}_{1}^{\infty}$ be a sequence of disjoint sets in $M$, then 
\begin{align*}
\mu\left(\bigcup_{1}^{\infty}E_j\right) &= \mu_0\left(\bigcup_{1}^{\infty}E_j\cap X_1\right)\\
&= \sum_{1}^{\infty}\mu_0(E_j\cap X_1)\\
&= \sum_{1}^{\infty}\mu(E_j) \ \ \ \text{is this true because} \ \mu_0 \ \text{is a counting measure?}
\end{align*}
Therefore $\mu$ is a measure on $M$.
Step 2 - $\tilde{M} = \mathcal{P}(X)$
Proof - 
Step 3 - $\tilde{\mu}\neq \underline{\mu}$
Proof - Take $y_1,y_2\in X_1$. Let $E = \{y_1,y_2\}\cup X_2$. Then $E\notin M$, so $\tilde{\mu}(E) = \infty$. However, $\underline{\mu}(E) = 2$.
I will re-edit my question and include these other proofs as I continue to do them. I am pretty sure my proof for $\tilde{\mu}$ is a measure is incorrect. But I am not really sure how to do it. Any suggestions on any of these is greatly appreciated.
 A: Items a and b are correct.  For the rest of the items, some need just minor improvements and some really need to be fixed. I tried to keep the proof as close to your proof as possible.

c.) Define $\tilde{\mu}$ on $\tilde{M}$ by $\tilde{\mu}(E) = \mu(E)$ if $E\in M$ and $\tilde{\mu}(E) = \infty$ otherwise. Then $\tilde{\mu}$ is a saturated measure on $\tilde{M}$, called the saturation of $\mu$.

Step 1: Show that $\tilde{\mu}$ is a measure on $\tilde{M}$.
Proof -
It is clear that $\tilde{\mu}$ is a non-negative well-defined function on $\tilde{M}$.
i.) $\tilde{\mu}(\emptyset) = \mu(\emptyset) = 0$.
ii.) Let $\{E_j\}_{1}^{\infty}\in \tilde{M}$ that is pairwise disjoint. Let $$E = \bigcup_{1}^{\infty}E_j$$
If $E \in M$, then we have two cases. First case: if, for all $j$, $E_j\in M$
then \begin{align*} \tilde{\mu}(E) = \tilde{\mu}\left(\bigcup_{1}^{\infty}E_j\right) &= \mu\left(\bigcup_{1}^{\infty}E_j\right) = \sum_{1}^{\infty}\mu(E_j)= \sum_{1}^{\infty}\tilde{\mu}(E_j)
\end{align*}
Second case: we begin remarking that if  $E \in M$ and $\mu(E)< \infty$ then, for all $j$, $E_j=E_j \cap E \in M$. So, if  $E \in M$ and, there is $j_0$ such that $E_{j_0}\notin M$ then $\mu(E)=\infty$, and we have
$$\tilde{\mu}(E) = \mu(E) = \infty = \tilde{\mu}(E_{j_0})\leqslant \sum_{1}^{\infty}\tilde{\mu}(E_j)\leqslant \infty$$  So we have
$$\tilde{\mu}(E) = \infty = \sum_{1}^{\infty}\tilde{\mu}(E_j)$$
If $E\notin M$ then $\tilde{\mu}(E) = \infty$. Since $E = \bigcup_{1}^{\infty}E_j$ and $M$ is a $\sigma$-algebra, there is $j_0$ such that $E_{j_0}\notin M$. So  $\tilde{\mu}(E_{j_0}) = \infty$. So we have
$$\tilde{\mu}(E) = \infty = \tilde{\mu}(E_{j_0})\leqslant \sum_{1}^{\infty}\tilde{\mu}(E_j)\leqslant \infty$$ So we have
$$\tilde{\mu}(E) = \infty = \sum_{1}^{\infty}\tilde{\mu}(E_j)$$
Step 2 - $\tilde{\mu}$ is saturated.
Proof - Let $E\subset X$ such that $E\cap A\in \tilde{M}$ when $\tilde{\mu}(A) < \infty$. For any $B\in M$ such that $\mu(B)<\infty$, we clearly have  $\tilde{\mu}(B)=\mu(B)<\infty$ and $E\cap B\in \tilde{M}$. Now, since $E\cap B\in \tilde{M}$ and $B\in M$ such that $\mu(B)<\infty$ we have $E\cap B = (E\cap B)\cap B\in M$. So we proved that, for any $B\in M$ such that $\mu(B)<\infty$, $E\cap B \in M$. So $E\in \tilde{M}$. It thus follows that $\tilde{\mu}$ is saturated.

d.) If $\mu$ is complete, so is $\tilde{\mu}$.

Proof - Suppose $\mu$ is complete. Let $A\subset X$ and suppose there is a $B\in \tilde{M}$ such that $A\subset B$ and $\tilde{\mu}(B) = 0$. Since $B\in \tilde{M}$  and $\tilde{\mu}(B) = 0 < \infty$ and hence $ B \in M$ and $\mu(B)=\tilde{\mu}(B) = 0$. Since $A\subset B$ we have $A\in M \subset \tilde{M}$ by completeness of $\mu$. Therefore, $A\in \tilde{M}$ and $\tilde{\mu}$ is complete.

e.) Suppose that $\mu$ is semifinite. For $E\in\tilde{M}$, define $\underline{\mu}(E) = \sup\{\mu(A):A\in M, A\subset E\}$. Then $\underline{\mu}$ is a saturated measure on $\tilde{M}$ that extends $\mu$.

Before proving the item e we prove a lemma

Lemma: Suppose that $\mu$ is semifinite. For $E\in\tilde{M}$,  $$\underline{\mu}(E) = \sup\{\mu(A):A\in M, A\subset E \textrm{ and } \mu(A)<\infty\}$$

Proof of the lemma:
We clearly have
$$\underline{\mu}(E)=\sup\{\mu(A):A\in M, A\subset E \} \geqslant \sup\{\mu(A):A\in M, A\subset E \textrm{ and } \mu(A)<\infty\}$$
Case 1: If, for all $A\in M$, $A\subset E$, we have  $\mu(A)<\infty$ then we have
$$\underline{\mu}(E)=\sup\{\mu(A):A\in M, A\subset E \} = \sup\{\mu(A):A\in M, A\subset E \textrm{ and } \mu(A)<\infty\}$$
Case 2: Now, suppose there is $A_0\in M$, $A_0\subset E$ such that $\mu(A_0)=\infty$. Then $\underline{\mu}(E)=\infty$ and, since $\mu$ is semifinite, we have that
$$\sup\{\mu(A):A\in M, A\subset A_0 \textrm{ and } \mu(A)<\infty\}=\mu(A_0)=\infty$$
So we have
$$\infty=\underline{\mu}(E)=\sup\{\mu(A):A\in M, A\subset E \} \geqslant \sup\{\mu(A):A\in M, A\subset E \textrm{ and } \mu(A)<\infty\}\geqslant \\ \geqslant \sup\{\mu(A):A\in M, A\subset A_0 \textrm{ and } \mu(A)<\infty\}=\infty$$
So we have
$$\underline{\mu}(E)=\infty= \sup\{\mu(A):A\in M, A\subset E \textrm{ and } \mu(A)<\infty\}$$
End of proof of the lemma
Step 1 - $\underline{\mu}$ is a measure.
Proof -
It is clear that $\tilde{\mu}$ is a non-negative well-defined function on $\tilde{M}$.
i.) $\underline{\mu}(\emptyset) = \mu(\emptyset) = 0$
ii.) Let $\{E_j\}_{1}^{\infty}$ be a sequence of disjoint sets in $\tilde{M}$. Set, $$E = \bigcup_{1}^{\infty}E_j$$
Then, by the lemma, we have:
\begin{align*}
\underline{\mu}\left (  \bigcup_{1}^{\infty}E_j   \right ) & = \sup\left\{\mu(A):A\in M, A\subset \bigcup_{1}^{\infty}E_j \textrm{ and } \mu(A)<\infty \right\}
\end{align*}
Since each $E_j \in \tilde{M}$, then, for each $A\in M$, $A\subset \bigcup_{1}^{\infty}E_j$ and $\mu(A)<\infty$, we have $E_j \cap A \in M$ and
$$\mu(A)=\sum_1^\infty\mu(E_j \cap A)$$
So we have
\begin{align*}
\underline{\mu}\left (  \bigcup_{1}^{\infty}E_j   \right ) & = \sup\left\{\mu(A):A\in M, A\subset \bigcup_{1}^{\infty}E_j \textrm{ and } \mu(A)<\infty \right\}=\\
& = \sup\left\{\sum_1^\infty\mu(A \cap E_j):A\in M, A\subset \bigcup_{1}^{\infty}E_j \textrm{ and } \mu(A)<\infty \right\}\leqslant \\
&\leqslant \sum_1^\infty\sup\left\{\mu(A \cap E_j):A\in M, A\subset \bigcup_{1}^{\infty}E_j \textrm{ and } \mu(A)<\infty \right\}\leqslant \\
&\leqslant\sum_1^\infty\sup\left\{\mu(B):B\in M, B\subset E_j \textrm{ and } \mu(B)<\infty \right\}= \\
&= \sum_1^\infty \underline{\mu}( E_j )
& \leqslant 
\end{align*}
So we proved
$$\underline{\mu}\left(\bigcup_{1}^{\infty}E_j\right)\leqslant \sum_{1}^{\infty}\underline{\mu}(E_j)$$
Now note that from the defintion of $\overline{\mu}$ we have
\begin{align*}
\sum_{1}^{\infty}\underline{\mu}(E_j)&= \sum_{1}^{\infty}\sup\left\{\mu(B_j):B_j\in M, B_j\subset E_j  \right\}= \\
&= \sup\left\{\sum_{1}^{\infty}\mu(B_j):B_j\in M, B_j\subset E_j  \right\}= \\
&= \sup\left\{\mu(\bigcup_{1}^{\infty}B_j):B_j\in M, B_j\subset E_j  \right\}\leqslant \\
&\leqslant \sup\left\{\mu(B):B\in M, B\subset \bigcup_{1}^{\infty} E_j  \right\}= \\
&=\underline{\mu}\left(\bigcup_{1}^{\infty}E_j\right)
\end{align*}
So we proved
$$\underline{\mu}\left(\bigcup_{1}^{\infty}E_j\right)= \sum_{1}^{\infty}\underline{\mu}(E_j)$$
and hence $\underline{\mu}$ is a measure.
Step 2 - $\underline{\mu}$ is saturated.
Proof - Let $E\subset X$ be such that $E\cap A\in \tilde{M}$ when $A \in \tilde{M}$ and $\underline{\mu}(A)< \infty$.
Take any $B\in M$ such that $\mu(B) < \infty$. Then $\underline{\mu}(B) \leqslant \mu(B) < \infty$ so $E\cap B\in \tilde{M}$. Thus, $E\cap B = (E\cap B)\cap B\in M$. So, we proved that, for any $B\in M$ such that $\mu(B) < \infty$, $E\cap B \in M$. So $E\in\tilde{M}$, hence, $\underline{\mu}$ is saturated.
Step 3 - $\underline{\mu}$ is an extention of $\mu$.
Proof - Let $E\in M$. Then
$$\underline{\mu}(E)=\sup\{\mu(A): A\in M, A\subset E \}=\mu(E)$$
So  $\underline{\mu}$ is an extention of $\mu$.

f.) Let $X_1$ and $X_2$ be disjoint uncountable sets, $X = X_1\cup X_2$, and $M$ the $\sigma$-alegbra of countable or co-countable sets in $X$. Let $\mu_0$ be counting measure on $\mathcal{P}(X_1)$ and define $\mu$ on $M$ by $\mu(E) = \mu_0(E\cap X_1)$. Then $\mu$ is a measure on $M$, $\tilde{M} = \mathcal{P}(X)$, and in the notation of parts (c) and (e), $\tilde{\mu}\neq \underline{\mu}$.

Step 1 - $\mu$ is a measure on $M$.
Proof -
i.) $\mu(\emptyset) = \mu_0(\emptyset\cap X_1) = 0$
ii.) Let $\{E_j\}_{1}^{\infty}$ be a sequence of disjoint sets in $M$, then
\begin{align*}
\mu\left(\bigcup_{1}^{\infty}E_j\right) &= \mu_0\left(\bigcup_{1}^{\infty}E_j\cap X_1\right)\\
&= \sum_{1}^{\infty}\mu_0(E_j\cap X_1)\\
&= \sum_{1}^{\infty}\mu(E_j) \ \ \ \text{is this true because} \ \mu_0 \ \text{is a counting measure?}
\end{align*}
Therefore $\mu$ is a measure on $M$.
Step 2 - $\tilde{M} = \mathcal{P}(X)$
Proof - Note that if $A \in M$ and $\mu(A)<\infty$ then $A \cap X_1$  is a finite set, so $A$ can not be co-countable. So $A$ is countable.
Given any set $E\subset X$, then for all  $A \in M$ and $\mu(A)<\infty$, $A$ is countable and so is $E \cap A$. So $E \cap A \in M$. So $\tilde{M}=\mathcal{P}(X)$.
Step 3 - $\tilde{\mu}\neq \underline{\mu}$
Proof - Take $y_1,y_2\in X_1$. Let $E = \{y_1,y_2\}\cup X_2$. Then $E\notin M$, so $\tilde{\mu}(E) = \infty$. However, $\underline{\mu}(E) = 2$.
