Exercise 20 - Let $\mu^*$ be an outer measure on $X$, $M^*$ the $\sigma$-algebra of $\mu^*$-measurable sets, $\overline{\mu} = \mu^*|M^*$, and $\mu^+$ the outer measure induced by $\overline{\mu}$ as in (1.12) (with $\overline{\mu}$ and $M^*$ replacing $\mu_0$ and $\mathcal{A}$).
a.) If $E\subset X$, we have $\mu^*(E)\leq \mu^+(E)$ with equality iff there exists $A\in M^*$ with $A\supset E$ and $\mu^*(A) = \mu^*(E)$.
b.) If $\mu^*$ is induced from a premeasure, then $\mu^* = \mu^+$ (Use exercise 18a found [here][1])
c.) If $X = \{0,1\}$ there exists an outer measure $\mu^*$ on $X$ such that $\mu^* \neq \mu^+$
Proof
a.) [Your proof that $\mu^*(E)\leq \mu^+(E)$ is essentially correct, I include it here, with minor rewording, just for completeness]
Let $E\subset X$. Then
$$\mu^+(E) = \inf\{\sum_{j=1}^{\infty}\overline{\mu}(A_j): A_j\in M^*, E\subset \bigcup_{j=1}^{\infty}A_j\}$$
Note that for any sequence $\{A_j\}$, such that, for all $j$, $A_j \in M^*$ and $E\subset \bigcup_{1}^{\infty}A_j$, we have by monotonocity \begin{align*}
\mu^*(E) &\leq \mu^*\left(\bigcup_{j=1}^{\infty}A_j\right)\\
&\leq \sum_{j=1}^{\infty}\mu^*(A_j) \ \text{(subadditivity)} \\
&= \sum_{j=1}^{\infty}\overline{\mu}(A_j) \ (\text{since} \ \overline{\mu} = \mu^*|M^*)
\end{align*}
It follows that $\mu^*(E)\leq \mu^+(E)$.
Now let us prove that
$\mu^*(E)= \mu^+(E)$ iff there exists $A\in M^*$ with $A\supset E$ and $\mu^*(A) = \mu^*(E)$
($\Leftarrow$)If there exists an $A\in M^*$ such that $E\subset A$ and $\mu^*(A) = \mu^*(E)$, then let $\{C_j\}$ be defined as follow: $C_1=A$ and $C_j=\emptyset$, if $j>1$. Note that, for all $j$, $C_j\in M^*$ and $E\subset \bigcup_{1}^{\infty}C_j$. So we have
\begin{align} \mu^+(E)&= \inf\{\sum_{j=1}^{\infty}\overline{\mu}(A_j): A_j\in M^*, E\subset \bigcup_{j=1}^{\infty}A_j\} \leq \\
& \leq \sum_{j=1}^{\infty}\overline{\mu}(C_j)= \overline{\mu}(A) + \sum_{j=2}^{\infty}\overline{\mu}(\emptyset)= \overline{\mu}(A) = \mu^*(A) = \mu^*(E)\end{align}
So we have $\mu^+(E) \leq \mu^*(E)$ and so we can conclude $\mu^*(E)= \mu^+(E)$.
($\Rightarrow$) Suppose $\mu^*(E)= \mu^+(E)$. If $\mu^*(E)=+\infty$, then since $E\subset X$ we have that $\mu^*(X) = +\infty =\mu(E)$. So just take $A=X$.
Now suppose $\mu^*(E)<\infty$, since $\mu^*(E)= \mu^+(E)$, we have $\mu^+(E)<\infty$. Then from the definition of $\mu^+$, we have that, for each $n\in \mathbb{N}$, $n>0$ there is a sequence $\{A_{j,n}\}$, such that, for all $j$, $A_{j,n} \in M^*$ and $E\subset \bigcup_{j=1}^{\infty}A_{j,n}$ and
$$\mu^+(E)\leq \sum_{j=1}^{\infty}\overline{\mu}(A_{j,n}) \leq \mu^+(E)+\frac{1}{n}$$
Now since $A_{j,n} \in M^*$, we have that $\overline{\mu}(A_{j,n})=\mu^*(A_{j,n})$. And, since $ \mu^*(E)=\mu^+(E)$, So we have
$$\mu^*(E)\leq \sum_{j=1}^{\infty}\mu^*(A_{j,n}) \leq \mu^*(E)+\frac{1}{n}$$
Now, since $E\subset \bigcup_{j=1}^{\infty}A_{j,n}$, using the monotonicity and the sub-additivity, we get
$$\mu^*(E)\leq \mu^*\left(\bigcup_{j=1}^{\infty}A_{j,n}\right)\leq \sum_{j=1}^{\infty}\mu^*(A_{j,n}) \leq \mu^*(E)+\frac{1}{n}$$
Note, for all $n\in \mathbb{N}$, $n>0$
$$E\subset \bigcap_{m=1}^\infty \bigcup_{j=1}^{\infty}A_{j,m} \subset \bigcup_{j=1}^{\infty}A_{j,n}$$ So using monotonicity we have, for all $n\in \mathbb{N}$, $n>0$,
$$\mu^*(E)\leq \mu^*\left(\bigcap_{m=1}^\infty \bigcup_{j=1}^{\infty}A_{j,m}\right) \leq \mu^*\left(\bigcup_{j=1}^{\infty}A_{j,n}\right)\leq \sum_{j=1}^{\infty}\mu^*(A_{j,n}) \leq \mu^*(E)+\frac{1}{n}$$
So,
$$\mu^*(E)= \mu^*\left(\bigcap_{m=1}^\infty \bigcup_{j=1}^{\infty}A_{j,m}\right)$$
Note that, since for all $m\in \mathbb{N}$, $m>0$, and for all $j$, $A_{j,m}\in M^*$ we have that $\bigcap_{m=1}^\infty \bigcup_{j=1}^{\infty}A_{j,m} \in M^*$. So just take $A=\bigcap_{m=1}^\infty \bigcup_{j=1}^{\infty}A_{j,m}$, and we have $A\in M^*$, $E \subset A$ and $\mu^*(A) = \mu^*(E)$.
b.) Suppose $\mu^*$ is induced from a premeasure $\mu_0$ defined on an algebra $\mathcal{A}$, then we have
$$\mu^{*}(E) = \inf\left\{ \sum_{j=1}^{\infty} \mu_{0} ( A_{j} ) : A_{j} \in \mathcal{A}, E \subset \bigcup_{j=1}^{\infty} A_{j} \right \}$$
We are going to prove that, in this case, for every $E \subset X$, there exists $A\in M^*$ with $A\supset E$ and $\mu^*(A) = \mu^*(E)$. Then we can apply item a.) and conclude that $\mu^*=\mu^+$.
[ In fact we are going to apply the same kind of reasoning we used in the ($\Rightarrow$) part of item a.) ]
If $\mu^*(E)=+\infty$, then since $E\subset X$ we have that $\mu^*(X) = +\infty =\mu(E)$. So just take $A=X$.
Now suppose $\mu^*(E)<\infty$, Then we have that, for each $n\in \mathbb{N}$, $n>0$ there is a sequence $\{A_{j,n}\}$, such that, for all $j$, $A_{j,n} \in \mathcal{A} $ and $E\subset \bigcup_{j=1}^{\infty}A_{j,n}$ and
$$\mu^*(E)\leq \sum_{j=1}^{\infty}\mu_0(A_{j,n}) \leq \mu^*(E)+\frac{1}{n}$$
Now, since $E\subset \bigcup_{j=1}^{\infty}A_{j,n}$, using the monotonicity and the sub-additivity, we get
$$\mu^*(E)\leq \mu^*\left(\bigcup_{j=1}^{\infty}A_{j,n}\right)\leq \sum_{j=1}^{\infty}\mu^*(A_{j,n}) \leq \mu^*(E)+\frac{1}{n}$$
Note, for all $n\in \mathbb{N}$, $n>0$
$$E\subset \bigcap_{m=1}^\infty \bigcup_{j=1}^{\infty}A_{j,m} \subset \bigcup_{j=1}^{\infty}A_{j,n}$$ So using monotonicity we have, for all $n\in \mathbb{N}$, $n>0$,
$$\mu^*(E)\leq \mu^*\left(\bigcap_{m=1}^\infty \bigcup_{j=1}^{\infty}A_{j,m}\right) \leq \mu^*\left(\bigcup_{j=1}^{\infty}A_{j,n}\right)\leq \sum_{j=1}^{\infty}\mu^*(A_{j,n}) \leq \mu^*(E)+\frac{1}{n}$$
So,
$$\mu^*(E)= \mu^*\left(\bigcap_{m=1}^\infty \bigcup_{j=1}^{\infty}A_{j,m}\right)$$
Note that, since for all $m\in \mathbb{N}$, $m>0$, and for all $j$, $A_{j,m}\in \mathcal{A} \subset M^*$ we have that $\bigcap_{m=1}^\infty \bigcup_{j=1}^{\infty}A_{j,m} \in M^*$. So just take $A=\bigcap_{m=1}^\infty \bigcup_{j=1}^{\infty}A_{j,m}$, and we have $A\in M^*$, $E \subset A$ and $\mu^*(A) = \mu^*(E)$.
Now from item a.) we have that $\mu^*=\mu^+$.
c.) Let $X=\{0,1\}$ and let $\mu^*$ be an outer mesure defined on $P(X)$ by: $\mu^*(\emptyset)=0$, $\mu^*(\{1\})=\mu^*(\{2\})=1$ and $\mu^*(\{1,2\})=\frac{3}{2}$.
It is easy to check that $\mu^*$ is an outer measure.
It is also easy to check that the only $\mu^*$-measurable sets are $\emptyset$ and $X$. So $M^*=\{\emptyset, X\}$ and we have that $\overline{\mu}$ (the restriction of $\mu^*$ to $M^*$) is the measure such that $\overline{\mu}(\emptyset)=0$ and $\overline{\mu}(X)=\frac{3}{2}$.
As a consequence, $\mu^+$ (the outer measure induced by $\overline{\mu}$ ) is defined on $P(X)$ by: $\mu^+(\emptyset)=0$, $\mu^+(\{1\})=\mu^+(\{2\})=\mu^*(\{1,2\})=\frac{3}{2}$.
So $\mu^*\neq \mu^+$.
