In Billingsley's Book(Convergence of probability measure), There is a question about interchanging role of inner and outer regular. In book, they defined as
$$P(A) = \sup_{F \subset A}P(F) = \inf_{A \subset G}P(G)$$ that $G$ implies open set and $F$ implies closed set.
In book question, they ask in what condition this role interchanged, i.e.
$$P(A) = \sup_{G \subset A}P(G) = \inf_{A \subset F}P(F)$$
I tried to prove this as the existence of $P(A) = \sup_{F \subset A}P(F) = \inf_{A \subset G}P(G)$(they proved firstly at the closed set, and showed the satisfying set is $\sigma$-algebra. ), but it failed because countable intersection of closed set is not closed.
Can you help me?
There is a largest open set contained in $A$, namely $A^{0}$ (the interior of $A$). So we get $P(A)=P(A^{0})$. Similarly the smallest closed set containing $A$ is it closure. So we get $P(A)=P(A^{0})=P(\overline A)$. These are equivalent to the fact that $P(\partial A)=0$.