Let $F$ be a finite set equipped the discrete topology. Let $X = F \times F \times ...$ be the countably infinite product space equipped with the product topology. Let $\mathcal A$ be any field of subsets of $X$ that contains the open subsets of $X$. Let $\mu$ be a finitely additive, finite measure with domain $\mathcal A$.

Suppose that $\mu$ has the following "clopen approximation property": For any $\epsilon > 0$ and any open subset $G$ of $X$, there is a clopen subset $C$ of $X$ such that $\mu(G \triangle C)< \epsilon$.

Is the clopen approximation property equivalent to the following "inner regularity property"? For every open subset $G$ of $X$, $\mu(G) = \sup\{\mu(C): C \subseteq G, C \text{ clopen}\}$.

Clearly inner regularity implies clopen approximation, but I am unable to see that the converse is true.

If $G$ is open, then it can be written as a countable union $G = C_1 \cup C_2 \cup ...$ of pairwise disjoint clopen sets. It seems reasonable to expect that if $G$ is approximable by some sequence of clopen sets, then it should be approximable from within by finite unions of its constiuent clopen sets. i.e. $\mu(\cup_{i=1}^n C_i) \to \mu(G)$ as $n \to \infty$. But, again, I am unable to see that this is the case.

If $G$ is approximable by some clopen sequence $B_n$, so that $\mu(G \triangle B_n) \to 0$, then $\mu(G - B_n) \to 0$ and $G - B_n$ is an open subset of $G$. If we could replace this sequence of open subsets of $G$ with a "similar" sequence of clopen subsets of $G$, maybe we could prove the result. I played around with sequences like $(C_1 \cup ... \cup C_n) - B_n$ (recall that $G = C_1 \cup C_2 \cup ...$), but didn't get anywhere.