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 $\Omega$.

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 $\Omega$.

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). 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 $\Omega$.

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 $\Omega$.