Show that $\mu^{*} (A) = \mu (A)$ if $\mu$ is countably additive Let X be a set, let $\mathcal{A}$ be an algebra of subsets of X, and let $\mu$ be a fintiely additive measure on $\mathcal{A}$. For each subset $A$ of $X$ let $\mu^{*}(A)$  be the infimum of of the set of sums $\sum_{k = 1}^{\infty} \mu(A_{k})$, where $\{A_{k}\}$ ranges over the sequences of sets in $\mathcal{A}$ for which $A \subset \cup_{k = 1}^{\infty} A_{k}$.
Show that if $\mu$ is countably additive ( in the sense that $\mu(\cup
_{k} A_{k}) = \sum_{k}\mu(A_{k})$ holds whener $\{ A_{k} \}$ is a sequence of disjoint sets in $\mathcal{A}$ for which $\cup_{k} A_{k}$ belongs to $\mathcal{A}$), then each $A$ in $\mathcal{A}$ satisfies $\mu(A) = \mu^{*}(A)$.
We know that $\mu^{*}$ is an outer measureand that each set in $\mathcal{A}$ is $\mu^{*}$-measurable.
My incomplete try:
First, for the sequence $A_{1} = A$ and $A_{k} = \emptyset$ for $k \geq 2$ we have
$\mu^{*}(A) \leq \sum_{k} \mu(A_{k}) = \mu(A)$. 
So we now want $\mu(A) \leq \mu^{*}(A)$
Let $A\in \mathcal{A}$ and let $\{A_{k}\}$ be a sequence in $\mathcal{A}$ such that $A \subset \cup_{k} A_{k}$ and $\mu^{*}(\cup_{k} A_{k}) \leq \mu^{*}(A) + \epsilon$ for an $\epsilon > 0$. 
By defining $A'_{n} = \cap_{k}^{n}A_{k}$, we get a disjoint sequence $\{A'_{n}\}$ such that $\cup_{n} A'_{n} = \cup_{k}A_{k}$ and $\mu*(\cup_{n} A'_{n}$) = $\mu^{*}(\cup_{k}A_{k})$, so we can assume that $\{A_{k} \}$ is disjoint from beginning.
Let $B \in \mathcal{A}$ (all sets in $\mathcal{A}$ are measurable).
Then $\mu(A) = \mu(A \cap B ) + \mu(A \cap B^{C})$ since $A,B\in \mathcal{A}$ and $\mu$ is a measure on $\mathcal{A}$.
$\mu(A \cap B ) + \mu(A \cap B^{C}) \leq \sum_{k} \mu(A_{k}\cap B) + \mu(A_{k} \cap B^{C})$ 
since $\{A_{k} \cap B \}$ is a covering sequence of $A\cap B$ and similary $\{A_{k} \cap B^{C} \}$ covers $A \cap B$.
$ \sum_{k} \mu(A_{k}\cap B) + \mu(A_{k} \cap B^{C}) = \sum_{k} \mu(A_{k}) $, 
since $\mu$ is a measure and $ A_{k} \cap B , A_{k} \cap B^{C} \in \mathcal{A}$ for all $k$.
If $\cup_{k} A_{k} \in \mathcal{A}$, then 
$  \sum_{k} \mu(A_{k}) = \mu(\cup_{k} A_{k})  \leq \mu^{*}(A) + \epsilon$
and we have the wanted inequality.
However, as I understand it, the key problem is that $\mathcal{A}$ is an algebra so we do not know if we for any $A \in \mathcal{A}$ can find a disjoint sequence $\{A_{k} \}$ such that $A \subset \cup_{k} A_{k}$? 
Or can we create such a sequence by saying that " Let $\{C_{k}\}$ be sequence where each $C_{k}$ consists of one element of $A$." Then $C_{k}$ would be disjoint and $\cup_{k} C_{k} = A \in \mathcal{A}$. 
Any feedback is appreciated!
/ Erik
 A: Let $E \in \mathcal{A}.$ Using trivial covering $E=E \cup \emptyset \cup \emptyset \cup \dots$ we conclude that $\mu^*(E) \le \mu(E).$
On other way, if $\displaystyle E \subset \bigcup_{n=1}^{\infty}E_n$ is one covering of set $E$ with sets from $\mathcal{A}$, then $$E=E \cap \bigcup_{n=1}^{\infty} E_n = \bigcup_{n=1}^{\infty} (E \cap E_n),$$ with $E \cap E_n \in \mathcal{A}$ for all $n \in \mathbb{N},$ and then from countable subadditivity of measure $\mu$ we find $$\sum_{n=1}^{\infty} \mu(E_n) \ge \sum_{n=1}^{\infty} \mu(E \cap E_n) \ge \mu(E).$$ Using fact that covering $E \subset \bigcup_{n=1}^{\infty}E_n$ was arbitrary, we find $\mu^* (E) \ge \mu(E)$, so we proved our statement.
A: Ok, so instead of editing I try giving an answer myself.
Let $A\in \mathcal{A}$ and let $\{A_{k}\}$ be a sequence in $\mathcal{A}$ such that $A \subset \cup_{k} A_{k}$ and $\mu^{*}(\cup_{k} A_{k}) \leq \mu^{*}(A) + \epsilon$ for an $\epsilon > 0$. 
By defining $A'_{n} = A_{n} \setminus \cup_{k = 1}^{n-1} A_{k}$, we get a disjoint sequence $\{A'_{n}\}$ such that $\cup_{n} A'_{n} = \cup_{k}A_{k}$ and $\mu^*(\cup_{n} A'_{n}$) = $\mu^{*}(\cup_{k}A_{k})$. Then since $\{ A'_{k} \}$ are disjoint, we can create a new disjoint sequence $\{ C_{k} \}$ where $C_{k} = A'_{k} \cap A $. Then $\cup_{k} C_{k} = A \in \mathcal{A}$. 
Using the countable additivity of $\mu$ we get
$ \mu(A) = \mu(\cup_{k} C_{k}) = \sum_{k} \mu (C_{k}) \leq \sum_{k} \mu(A_{k}) \leq \mu^*(A) + \epsilon $.
Since $\epsilon $ was arbitrary we have the wanted inequality and $\mu(A) = \mu^*(A)$ whenever $A\in \mathcal{A}$. 
