let $(X,d)$ be a metric space with the Borel $\sigma$-algebra. Let $\mu$ be a finite measure on $(X,\Sigma)$. Show that $\mu$ is regular in the following sense: for every $A\in \Sigma$ and for every $\epsilon>0$ there exist a closed set $C$ and an open set $U$ such that $C\subset E \subset U$ and $\mu(U-C)<\epsilon$.
Hint: show that this is true for every closed set $E\in \Sigma$ and then show that the collection of measurable sets for which this assertion holds is a $\sigma$-algebra.
I'm stuck on the first step. What I believe is the answer is to take $C=E$ and $U=\cup_{x\in E} B(x,\epsilon)$. Now I "feel" like $\mu(U-C)<\epsilon$ but I fail to see how to connect the idea of distance $\epsilon$ in a metric space to the idea of a measure. Any help would be appreciated.
If $C$ is a closed set then $C=\cap U_n$ where $U_n=\{x:d(x,C)<\frac 1 n\}$. [ $d(x,C)$ is defined as $\inf \{d(x,y):y\in C\}$]. Each $U_n$ is open and $U_{n+1} \subset U_n$. Hence $\mu (U_n) \to \mu (C)$ as $ n \to \infty$. From this you should be able to see that $\mu (C)=\inf \{\mu(U): C \subset U, U \text {open} \}$.
For the second part choose a closed set $D \subseteq E^{c}$ such that $\mu (E^{c}\setminus D) <\epsilon$. Then $U=D^{c}$ is open, $E \subseteq U$ and $\mu (U\setminus E)=\mu (U)-\mu (E)=[\mu (X)-\mu(D)]-[\mu(X) -\mu (E^{c})]=\mu (E^{c}\setminus D) <\epsilon$.