given $\epsilon >0$ , there is an $A\in \mathfrak{a}$ with $\overline \mu (A-E)+\overline \mu (E-A)<\epsilon$ Let $\mu$ be a measure on an algebra $\mathfrak{a}$ and $\overline \mu $ the extension of it given by the Caratheodory process. Let $E$ be measurable with respect to $\overline \mu $ and $\overline \mu (E)<\infty$  . Then given $\epsilon >0$ , there is an $ A\in \mathfrak{a}$ with $\overline \mu (A-E)+\overline \mu (E-A)<\epsilon$.

Here ${\overline\mu}(A):=inf\{\Sigma_{k=1}^{\infty}\mu(A_k): A\subseteq A_k, A_k\in \mathfrak{a}, \forall k\}$ and $E$ has the following property: $\forall A \subseteq X$, $\overline\mu(A)=\overline\mu(A\cap E)+\overline\mu(A\cap E^c)$ 
Attempt:
Given $\epsilon >0$, I choose a countable cover $(A_k)_k$ of $E$ by sets in $\mathfrak{a}$ such that $\Sigma\mu(A_k)\leq \overline\mu(E)+\epsilon$. Then I want to claim that $\bigcup A_k $ is the desired set in $\mathfrak{a}$ but I couldn't do it since an algebra is not closed under countable union. Thanks for your helps!
 A: Let $\epsilon>0$. Choose a sequence of sets $\left\{A_k\right\}_{k=1}^\infty\subseteq\mathfrak{a}$ such that $E\subseteq\bigcup_{k=1}^\infty A_k$ and $\sum_{k=1}^\infty\mu(A_k)<\overline{\mu}(E)+\epsilon/2$. In particular, the series $\sum_{k=1}^\infty\mu(A_k)$ converges, so there exists $N>0$ such that $\sum_{k=N+1}^\infty\mu(A_k)<\epsilon/2$.
Let $A=\bigcup_{k=1}^NA_k\in\mathfrak{a}$. Let's show that $\mu(A-E)+\mu(E-A)<\epsilon$. Note that
\begin{align*}
\overline{\mu}(A-E)=\overline{\mu}(\bigcup_{k=1}^NA_k-E)\leq\overline{\mu}(\bigcup_{k=1}^\infty A_k-E)
\end{align*}
Since $E\subseteq\bigcup_{k=1}^\infty A_k$ and $\overline{\mu}$ is a measure (in a certain $\sigma$-algebra containing $\mathfrak{a}$ and $E$), then
\begin{align*}
\overline{\mu}(\bigcup_{k=1}^\infty A_k-E)&=\overline{\mu}(\bigcup_{k=1}^\infty A_k)-\overline{\mu}(E)\leq\sum_{k=1}^\infty\overline{\mu}(A_k)-\overline{\mu}(E)=\sum_{k=1}^\infty\mu(A_k)-\overline{\mu}(E)\\
&<\epsilon/2
\end{align*}
so $\overline{\mu}(A-E)<\epsilon/2$. For the term $\mu(E-A)$, note that, since $E\subseteq\bigcup_{k=1}^\infty A_k$, then
\begin{align*}
\overline{\mu}(E-A)&\leq\overline{\mu}((\bigcup_{k=1}^\infty A_k)-(\bigcup_{k=1}^NA_k))\leq\overline{\mu}(\bigcup_{k=N+1}^\infty A_k)\leq\sum_{k=N+1}^\infty\overline{\mu}(A_k)=\sum_{k=N+1}^\infty\mu(A_k)\\
&<\epsilon/2.
\end{align*}
