In the exercise 4.21(i) in the book Measures, Integrals, and Martingales,
Let $X$ be a metric space and $\mu$ be a finite measure on the Borel sets $\mathcal{B} = \mathcal{B}(X)$ and denote the open sets by $\mathcal{O}$ and the closed sets by $\mathcal{F}$. Define a family of sets $$ \Sigma = \{A \subseteq X: \forall \epsilon > 0, \exists U \in \mathcal{O}, F \in \mathcal{F} \ \text{ s.t } F \subset A \subset U, \mu(U \setminus F) < \epsilon\} $$ (i) Show that $A \in \Sigma \Longrightarrow A^c \in \Sigma$ and that $\mathcal{F} \subseteq \Sigma$
I'm able to show the first part of the question, but couldn't show $\mathcal{F} \subseteq \Sigma$. Find this exercise in the solution manual, I got the following:
Denote $d$ is the metric on $X$ and $B_{1/n}(0)$ for the open ball center at $0$ with radius $1/n$. As in the solution of Problem 3.14(ii) we see that $U_n ∶= F +_{1/n}(0)$ is a sequence of open sets such that $U_n \downarrow F$. Because of the continuity of measures we get $\mu(_n \setminus F) \rightarrow 0$ and since $\mathcal{F} \ni \subset F \subset U_n \in \mathcal{O}$, this means that $\mathcal{F} \subseteq \Sigma$.
I have trouble showing $U_n$ is an open set so any hints are appreciated. Here, $$ A + B = \{a + b \vert a \in A, b \in B\} $$ Note. The problem 3.14(ii) that the solution above mentioned is:
(ii) If $\mathcal{F}$ is stable w.r.t complements, i.e $F \in \mathcal{F} \Longrightarrow F^c \in \mathcal{F}$, then so is $m(\mathcal{F})$
where, in the context of problem 3.14, $\mathcal{F}$ is any family of sets and $m(\mathcal{F})$ is the minimal monotone class generated by $\mathcal{F}$. I read the solution of this, which just follows standard procedure, and I think this has nothing to do with the exercise, but if it is, I'd love to know too.
$$ F+B_{1/n}(0)=\bigcup_{x\in F}B_{1/n}(x), $$ and any union of open sets is itself an open set.