Inner measure and premeasures on an algebra 
Let $\mu$ be a finite premeasure on an algebra $\mathcal{A} \subseteq \mathcal{P}(X)$ and let $\mu^{*}$ be the induced outer measure on X. We define the inner measure $\mu_{*}$ by $$ \mu_{*}\left(E\right)=\mu\left(X\right)-\mu^{*}\left(E^{c}\right) $$
Let $F\subseteq X$ and let $\mathcal{B}$ be the algebra generated by $\mathcal{A}$ and F. Show that $$ \mathcal{B}=\left\{ \left(A\cap F\right)\cup\left(A'\cap F^{c}\right)\thinspace:\thinspace A,A'\in\mathcal{A}\right\}  $$
And that $$ \begin{cases}
\overline{\mu}\left(B\right)=\mu^{*}\left(B\cap F\right)+\mu_{*}\left(B\cap F^{c}\right)\\
\underline{\mu}\left(B\right)=\mu_{*}\left(B\cap F\right)+\mu^{*}\left(B\cap F^{c}\right)
\end{cases} $$
Are premeasures over $\mathcal{B}$.

I am trying to prove the above and have no clue where to start. As a hint I proved that $F\subseteq X $ is $\mu^{*} $ measurable if and only if $\mu_{*}(F)=\mu^{*}(F)$.
Any help would be appreciated.
Thanks in advance.
 A: 
Let $\mathcal{A} \subseteq \mathcal{P}(X)$ be an algebra.
Let $F\subseteq X$ and let $\mathcal{B}$ be the algebra generated by $\mathcal{A}$ and F. Show that $$ \mathcal{B}=\left\{ \left(A\cap F\right)\cup\left(A'\cap F^{c}\right)\thinspace:\thinspace A,A'\in\mathcal{A}\right\}  $$

Proof: Let $\mathcal{B}$ be the algebra generated by $\mathcal{A}$ and $F$ and let
$\mathcal{C}=\left\{ \left(A\cap F\right)\cup\left(A'\cap F^{c}\right)\thinspace:\thinspace A,A'\in\mathcal{A}\right\}  $.
It is immediate that:

*

*$\mathcal{C} \subseteq \mathcal{B}$.


*for all $A\in \mathcal{A}$, $ A= \left(A\cap F\right)\cup\left(A\cap F^{c}\right)\in \mathcal{C}$.
So $\mathcal{A} \subseteq \mathcal{C}$.


*Since $X, \emptyset \in \mathcal{A}$, we have that $F=(X\cap F) \cup (\emptyset \cap F^c) \in \mathcal{C}$.
So, if we prove that $\mathcal{C}$ is an algebra, it follows immediately that $\mathcal{C}=\mathcal{B}$ and the proof is comnplete.
Let us prove that $\mathcal{C}$ is an algebra. First note that, since $\emptyset \in \mathcal{A}$, using 2 above, we have that $\emptyset \in \mathcal{C}$.
Second, given any $C\in \mathcal{C}$, there are  $A,A'\in\mathcal{A}$, such that $C= 
\left(A\cap F\right)\cup\left(A'\cap F^{c}\right)$. So
\begin{align*} 
C^c&=\left (\left(A\cap F\right)\cup\left(A'\cap F^{c}\right)\right ) ^c =\\ 
&=\left(A^c\cup F^c\right)\cap\left(A'^c\cup F\right) = \\
&=\left(A^c \cap A'^c \right) \cup \left ( F^c \cap A'^c \right) \cup \left(A^c \cap  F\right) = \\ 
& = \left ( F^c \cap A'^c \right) \cup \left(A^c \cap  F\right) = \\ 
& = \left(A^c \cap  F\right) \cup \left ( A'^c \cap F^c \right)  
\end{align*}
Since $A^c,A'^c\in\mathcal{A}$, we have that $C^c \in \mathcal{C}$. So, $\mathcal{C}$ is closed by complement.
Third. for $i \in \{ 1,..., n\}$, Let $C_i\in \mathcal{C}$, then there are  $A_i,A'_i\in\mathcal{A}$, such that $C_i= 
\left(A_i\cap F\right)\cup\left(A'_i\cap F^{c}\right)$. So
\begin{align*} 
\bigcup_{i=1}^n C_i&= \bigcup_{i=1}^n \left (\left(A_i\cap F\right)\cup\left(A'_i\cap F^{c}\right) \right) = \\
&= \left (  \bigcup_{i=1}^n\left(A_i\cap F\right)\right) \cup \left (  \bigcup_{i=1}^n\left(A'_i\cap F^{c}\right)\right) = \\
&= \left (   \left(\bigcup_{i=1}^n A_i \right)\cap F\right) \cup \left ( \left( \bigcup_{i=1}^n A'_i\right)\cap F^{c}\right) 
\end{align*}
Since $ \bigcup_{i=1}^n A_i,  \bigcup_{i=1}^n A'_i \in \mathcal{A}$, we have that $\bigcup_{i=1}^n C_i \in \mathcal{C}$. So $\mathcal{C}$ is closed under finite unions.
So $\mathcal{C}$ is an algebra. So $\mathcal{C}= \mathcal{B}$, that means
$$ \mathcal{B}=\left\{ \left(A\cap F\right)\cup\left(A'\cap F^{c}\right)\thinspace:\thinspace A,A'\in\mathcal{A}\right\}  $$
Now let us prove that

$$ \begin{cases}
\overline{\mu}\left(B\right)=\mu^{*}\left(B\cap F\right)+\mu_{*}\left(B\cap F^{c}\right)\\
\underline{\mu}\left(B\right)=\mu_{*}\left(B\cap F\right)+\mu^{*}\left(B\cap F^{c}\right)
\end{cases} $$
are premeasures over $\mathcal{B}$.

To prove that, we will first prove that  $\nu_1(B) = \mu^{*}\left(B\cap F\right)$ and $\nu_2(B) = \mu_{*}\left(B\cap F^c\right)$ are premeasures over $\mathcal{B}$.
It is clear $\nu_1(\emptyset)= \mu^*(\emptyset \cap F) =0$. Now let $\{B_i\}_{i \in \Bbb N}$ be a family of disjoint sets in $\mathcal{B}$. Then, for each $i$ there are $A_i$ and $A'_i$ in $\mathcal{A}$, such that $B_i=(A_i \cap F) \cup (A'_i \cap F^c)$. So $B_i\cap  F= A_i \cap F$. Now let us define $D_0=A_0$ and, for all $i\in \Bbb N $, $D_{i+1}=A_{i+1} \setminus \bigcup_{j=0}^i D_j$. Note $\{D_i\}_{i \in \Bbb N}$ is a family of disjoint sets in $\mathcal{A}$. Now, let us prove that, for all $i\in \Bbb N$, $B_i\cap  F= D_i \cap F$. First note that that $B_0 \cap F = D_0 \cap F$. Now, given any $i\in \Bbb N$, suppose we know  for all $j$, $0\leqslant j \leqslant i$ ,   $B_j \cap F = D_j \cap F$. Then,
$$D_{i+1} \cap F= (A_{i+1} \cap F) \setminus \bigcup_{j=0}^i (D_j \cap F)=(B_{i+1} \cap F)\setminus \bigcup_{j=0}^i (B_j \cap F)=B_{i+1} \cap F$$
where the last equality is true because $\{B_i\}_{i \in \Bbb N}$ be a family of disjoint sets.
Now, since $\mu^{*}$ is the outer measure on $X$ induced by a  finite premeasure $\mu$ on an algebra $\mathcal{A}$, we know that there is a measurable cover $C$ of $\left ( \bigcup_i B_i \right) \cap F $. In particular $C$ is $\mu^*$-measurable, $\left ( \bigcup_i B_i \right) \cap F \subseteq C$ and $\mu^*(\left ( \bigcup_i B_i \right) \cap F )= \mu^*(C)$.
Note that, for all $i$, $D_i$ is in $\mathcal{A}$, so  $D_i$  and $D_i \cap C$ are $\mu^*$-measurable. Moreover, if $i\neq j$,
$ (D_i \cap C) \cap (D_j \cap C)= \emptyset$. So
\begin{align*}
\mu^*\left( \left(\bigcup_i B_i\right) \cap F \right)
&= \mu^*\left( \bigcup_i (B_i \cap F) \right) 
 \leqslant \sum_i \mu^*(B_i \cap F) = \\
& = \sum_i \mu^*(D_i \cap F)   
\leqslant  \sum_i \mu^*(D_i \cap C) =
\mu^*\left(\bigcup_i (D_i \cap C)  \right) = \\ 
& =  \mu^*\left( \left(\bigcup_i D_i\right) \cap C \right) 
\leqslant \mu^*(C) = \mu^* \left (\left ( \bigcup_i B_i \right) \cap F \right )
\end{align*}
So
$$ \nu_1 \left (\bigcup_i B_i  \right) = \mu^*\left( \left(\bigcup_i B_i\right) \cap F \right)= \sum_i \mu^*(B_i \cap F) = \sum_i \nu_1(B_i)$$
So $\nu_1$ is a pre-measure over $\mathcal{B}$.
Now, let us prove that $\nu_2(B) = \mu_{*}\left(B\cap F^c\right)$ is a premeasure over $\mathcal{B}$.
It is clear $\nu_2(\emptyset)= \mu^*(\emptyset \cap F) =0$. Now let $\{B_i\}_{i \in \Bbb N}$ be a family of disjoint sets in $\mathcal{B}$. Then, for each $i$ there are $A_i$ and $A'_i$ in $\mathcal{A}$, such that $B_i=(A_i \cap F) \cup (A'_i \cap F^c)$. So $B_i\cap  F^c= A'_i \cap F$. Now let us define $D'_0=A'_0$ and, for all $i\in \Bbb N $, $D'_{i+1}=A'_{i+1} \setminus \bigcup_{j=0}^i D'_j$. Note $\{D'_i\}_{i \in \Bbb N}$ is a family of disjoint sets in $\mathcal{A}$. Now, let us prove that, for all $i\in \Bbb N$, $B_i\cap  F^c= D'_i \cap F^c$. First note that that $B_0 \cap F^c = D'_0 \cap F^c$. Now, given any $i\in \Bbb N$, suppose we know  for all $j$, $0\leqslant j \leqslant i$ ,   $B_j \cap F^c = D'_j \cap F^c$. Then,
$$D'_{i+1} \cap F^c= (A'_{i+1} \cap F^c) \setminus \bigcup_{j=0}^i (D'_j \cap F^c)=(B_{i+1} \cap F^c)\setminus \bigcup_{j=0}^i (B_j \cap F^c)=B_{i+1} \cap F^c$$
where the last equality is true because $\{B_i\}_{i \in \Bbb N}$ be a family of disjoint sets.
Now, since $\mu^{*}$ is the outer measure on $X$ induced by a  finite premeasure $\mu$ on an algebra $\mathcal{A}$, we know that there is a measurable kernel $K$ of $\left ( \bigcup_i B_i \right) \cap F^c $. In particular $K$ is $\mu^*$-measurable, $K \subseteq \left ( \bigcup_i B_i \right) \cap F^c = \left ( \bigcup_i D'_i \right) \cap F^c $ and $\mu_*(\left ( \bigcup_i B_i \right) \cap F^c )= \mu_*(K)$.
Note that, for all $i$, $D'_i$ is in $\mathcal{A}$, so  $D'_i$  and $D'_i \cap K$ are $\mu^*$-measurable. Moreover, if $i\neq j$,
$ (D'_i \cap K) \cap (D'_j \cap K)= \emptyset$. So
\begin{align*}
\mu_*\left( \left(\bigcup_i B_i\right) \cap F^c \right)
&= \mu_*\left( \bigcup_i (B_i \cap F^c) \right) 
 \geqslant \sum_i \mu_*(B_i \cap F^c) = \\
& = \sum_i \mu_*(D'_i \cap F^c)   
\geqslant  \sum_i \mu_*(D'_i \cap K) =
\mu_*\left(\bigcup_i (D'_i \cap K)  \right) = \\ 
& =  \mu_*\left( \left(\bigcup_i D'_i\right) \cap K \right) 
= \mu_*(K) = \mu_* \left (\left ( \bigcup_i B_i \right) \cap F^c \right )
\end{align*}
So
$$ \nu_2 \left (\bigcup_i B_i  \right) = \mu_*\left( \left(\bigcup_i B_i\right) \cap F^c \right)= \sum_i \mu_*(B_i \cap F^F) = \sum_i \nu_2(B_i)$$
So $\nu_2$ is a pre-measure over $\mathcal{B}$.
Just reversing the roles of $F$ and $F^c$, we prove that that  $\nu_3(B) = \mu^{*}\left(B\cap F^c\right)$ and $\nu_4(B) = \mu_{*}\left(B\cap F\right)$ are premeasures over $\mathcal{B}$.
It follows immediately that
$$ \begin{cases}
\overline{\mu}\left(B\right)=\mu^{*}\left(B\cap F\right)+\mu_{*}\left(B\cap F^{c}\right)\\
\underline{\mu}\left(B\right)=\mu_{*}\left(B\cap F\right)+\mu^{*}\left(B\cap F^{c}\right)
\end{cases} $$
are premeasures over $\mathcal{B}$.
