If $X$ is compact and $\mu$ is a positive Borel measure on $X$, then $\pi_\mu$ : $C(X) \rightarrow \mathcal{B}\left(L^2(\mu)\right)$ defined by $\pi_\mu(f)=M_f$ is a representation of $C(X)$
Question: Determine the kernel of $\pi_\mu$. If $X$ is metrizable, show that the measure $\mu$ can be chosen so that $\pi_\mu$ is injective.
My attempt:
I think, clearly, the set of a.e. zero functions is the kernel.
In the second part, I am getting that for any measure $\mu$, $\pi_\mu$ is injective:
$\pi_\mu(f)=\pi_\mu(g)$ for $f, g\in C(X)$
$\implies M_fh=M_gh \, \forall h \in L^2(\mu)$
$\implies fh=gh \, \forall h \in L^2(\mu)$
$\implies f=g$, putting $h$ as the identity function
Please let me know where I am making a mistake.
The mistake is that the equalities in your reasoning are up to a nullset. And while for Lebesgue measure nullsets are intuitively irrelevant, that's not the case for arbitrary measures.
Take $X=[0,1]$, and $\mu$ the Dirac delta. Then, since $\int_Xfd\mu=f(0)$ you have that $f$ and $g$ are equivalent in $L^2$ if and only if $f(0)=g(0)$. It follows that $$ \ker\pi_\mu=\{f\in C[0,1]:\ f(0)=0\}. $$
For arbitrary $\mu$, $f\in\ker\pi_\mu$ if and only if $fg=0$ $\mu$-a.e. for all $g$. Which is equivalent with $f=0$ $\mu$-a.e. That is, $$ \ker\pi_\mu=\{f\in C(X):\ \mu(|f|>0)=0\}. $$