I'm faced with the following situation: let $X$ be a Hausdorff compact space and let $\mu$ be a regular finite Borel measure on $X$. In classes, we proved that $C(X)$ (meaning all continuous complex functions on $X$, endowed with the sup norm) are weak*-dense in $L^\infty (X, \mu)$ using Luzin's theorem. However, in the earlier lecture we proved the following lemma:
Lemma: Let $X$ be a Banach space. Then the image of the inclusion $$\iota: X \to X^{**},\quad \iota (x) = (f \mapsto f(x))$$ is weak*-dense in $X^{**}$.
My question is: could we somehow avoid using Luzin's theorem and prove that $C(X)$ is weak*-dense in $L^\infty (X, \mu)$ using just the above lemma?
Actually, yes.
Let $M(X)$ be the space of complex regular Borel measures on $X$. By the Riesz-Markov-Kakutani theorem, $M(X)$ is the dual of $C(X)$ with the norm $|\nu|(X)$ where $|\nu|$ is the total variation of $\nu$.
$L_1(X,\mu)$ is isometrically embedded as a closed subspace in $M(X)$ via the identification $f\to fd\mu$. $|fd\mu|(X)=\int_{X}|f|d|\mu|$.
$L_\infty(X,\mu)$ is the dual of $L_1(X,\mu)$ via $\langle f,g\rangle=\int_{X} \overline{f}gd\mu$ for $f\in L_\infty(X,\mu),g\in L_1(X,\mu)$.
Now fix $f\in L_\infty(X,\mu)$. $f$ acts as a functional on $L_1(X,\mu)$. By Hahn-Banach, there is a functional $\alpha\in M(X)^*$ such that $\alpha\rvert_{L_1(X,\mu)}=f$.
Since $C(X)$ is weak-$*$ dense in $M(X)^*$, there is a net $f_\gamma\in C(X)$ such that $f_\gamma(\nu)\to \alpha(\nu)$ for all $\nu\in M(X)$. This means $$\int_{X} \overline{f_\gamma}d\nu\to\alpha(\nu)$$ for all $\nu\in M(X)$. In particular for $\nu=gd\mu$ with $g\in L_1(X,\mu)$ we have $$\int_{X}\overline{f_\gamma}gd\mu\to \alpha(gd\mu)=\int_{X}\overline{f}gd\mu$$ and this proves that $f_\gamma\to f$ in the weak-$*$ topology of $L_\infty(X,\mu)$ as the dual of $L_1(X,\mu)$.
Whether this avoids Lusin's theorem I can't tell because the assertions about the relations between $C(X),L_1(X,\mu),M(X),L_\infty(X,\mu)$ may need Lusin's theorem to prove.