Do semi-continuous functions generate bounded Borel measurable functions as a $C^*$-algebra? This question is related to Question 2 of my previous posting.
Question. Let $\mu$ be a Radon measure on a compact Hausdorff space $\Omega$ and $L^{\infty}(\Omega,\mu)$ the set of essentially bounded Borel measurable functions on $\Omega$. Suppose that $S$ is the set of bounded lower or upper semi-continuous functions on $\Omega$. Does $S$ generate $L^{\infty}(\Omega,\mu)$ as a $C^*$-algebra?
It suffices to consider whether the indicator function of any Borel set is obtained from $S$ by algebraic operations and (essential-supremum) norm limits. If necessary, you may assume $\Omega$ to be second countable. Thank you.
 A: Any upper or lower semicontinuous function is continuous almost everywhere in the sense of Baire category (since it is a pointwise limit of a sequence of continuous functions, at least when $\Omega$ is compact metrizable). The algebra of Baire-a.e. continuous functions is itself a $C^\ast$-algebra.
So the answer to your question is no, once we show that there exists a function $f \in L^\infty$ that is not $\mu$-equivalent to a Baire-a.e. continuous function. For an example we may take any indicator function of a set $S$ such that both $\mathrm{supp}\ \mu\restriction S$ and $\mathrm{supp}\ \mu\restriction (\Omega \setminus S)$ equal $\Omega$.
A: No. The bounded Baire class one functions on $[0,1]$ are stable under uniform limits and hence constitute a C*-algebra. This C*-algebra contains every semicontinuous function on [0,1]. Every function in $L^\infty[0,1]$ is equal almost everywhere to a Baire class two function, but not a Baire class one function. (The previous version of my answer neglected this essential point.)
See http://www.encyclopediaofmath.org/index.php/Baire_classes
