Let $K$ be a compact Hausdorff space. For a bounded Borel measurable function $f$ on $K$, we define $\phi\_{f}\in C(K)^{\*\*}$ by $\phi\_{f}(\mu)=\int\_{K}fd\mu$ for all $\mu\in C(K)^{\*}$. It is easy to see that $\|\phi\_{f}\|=\|f\|\_{\infty}$. Thus the space of all bounded Borel measurable functions on $K$ can be considered as a closed subspace of $C(K)^{\*\*}$. We let $Z$ be the norm closed subspace of $C(K)^{\*\*}$ generated by $\{\chi\_{U}:U$ open $F\_{\sigma}$ sets in $K\}$. Given a sequence $(U\_{n})\_{n}$ of disjoint open $F\_{\sigma}$ sets in $K$ and a bounded scalar sequence $(a\_{n})\_{n}$, we define $g(t)=a\_{n}$ for $t\in U\_{n}$ and $g(t)=0$ for $t\notin \cup\_{n}U\_{n}.$ Then $g$ is a bounded Borel measurable function. It seems clear that $g$ belongs to $Z$. But I am not sure.


Question. Does $g$ belong to $Z$? 



A closed subspace generated by open $F_{\sigma}$ sets of $K$ in $C(K)^{**}$
Let $\epsilon > 0$. Since $a\_n$ is a bounded sequence, by compactness we can find a finite set of scalars $b\_1, \dots, b\_m$ such that for every $a\_n$ there exists a $b\_{k\_n}$ with $|a\_n - b\_{k\_n}| \le \epsilon$. Now consider the function $h = \sum\_{n=1}^\infty b\_{k\_n} \chi\_{U\_n}$, that is, $h(t) = b\_{k\_n}$ for $t \in U\_n$. Clearly $\|\phi\_g - \phi\_h\| = \|g-h\|\_{\infty} \le \epsilon$. Now set $V\_k = \bigcup\_{k\_n = k} U\_n$. Then $V\_k$ is an open $F\_\sigma$ since it is a countable union of open $F\_\sigma$s, and we can write $h = \sum\_{k=1}^m b\_k \chi\_{V\_k}$. Thus $h \in Z$, and since $\epsilon$ was arbitrary and $Z$ is closed, we have $g \in Z$ also.


