$A=\{E \in \Sigma : \forall\epsilon>0\,\exists C\subset E\subset U,\,C\text{ closed , $U$ open, and }\mu(U-C)<\epsilon\}$ $(X,\Sigma,\mu)$ is a measure space, $X$ is metric space and $\mu(X) = 1$.
I need to prove that the collection
$$
A=\{E \in \Sigma : \forall\epsilon>0\,\exists C\subset E\subset U,\,C\text{ closed , $U$ open, and }\mu(U-C)<\epsilon\}
$$
is a $\sigma-$ algebra. 
To show $\emptyset \in A$ and that $A$ is closed under complements i did. 
How can I prove $A$ is closed under countable unions? 
Thanks for helping .
 A: I think you can prove that $A$ is closed under taking complements and finite unions by yourself, right? Now, let $E_n\in A$, $n\in\mathbb N$, and let $\varepsilon > 0$. Put $E := \bigcup_n E_n$. First, construct pairwise disjoint $E_n'\in A$ such that $E = \bigcup_n E_n'$ by taking complements and finite unions of the $E_n$'s. Then $\sum_n\mu(E_n') = \mu(E)\le\mu(X) = 1$. Hence, there exists $N\in\mathbb N$ such that $\sum_{n\ge N}\mu(E_n') < \varepsilon/4$. Now, choose $C'$ and $U'$ such that $C'\subset\bigcup_{n<N}E_n\subset U'$ and $\mu(U'-C') < \varepsilon/2$. Also choose $U_n\supset E_n'$ such that $\mu(U_n-E_n') < 2^{-n-2}\varepsilon$ for $n\ge N$. Now, put $C := C'$ and $U := U'\cup\bigcup_{n\ge N}U_n$. Then $C\subset E\subset U$, $C$ is closed, $U$ is open, and
\begin{align*}
\mu(U-C)
&= \mu\left((U'-C')\cup\bigcup_{n\ge N}U_n\setminus C'\right) < \frac\varepsilon 2 + \mu\left(\bigcup_{n\ge N}U_n\setminus C'\right)\\
&\le \frac\varepsilon 2 + \mu\left(\bigcup_{n\ge N}U_n\right)\\
&= \frac\varepsilon 2 + \mu\left(\bigcup_{n\ge N}U_n-\bigcup_{m\ge N}E_m'\right) + \mu\left(\bigcup_{m\ge N}E_m'\right)\\
&\le \frac\varepsilon 2 + \mu\left(\bigcup_{n\ge N}(U_n-E_n')\right) + \frac\varepsilon 4\\
&\le \frac{3\varepsilon}{4} + \sum_{n\ge N}\mu(U_n-E_n')\,\le\,\varepsilon.
\end{align*}
So, the "trick" is not to choose the union of all $C_n$'s (which might be non-closed) but only a finite union which is already sufficiently "close" to $E$.
