# Continuous functional calculus of multiplication operator in $L_2$

I would like to calculate the continuous calculus of the multiplication operator by an essentially bounded function $\varphi : X \rightarrow \mathbb{R}$ in $L_2 (X, \mu)$, where $\left( X, \mu \right)$ - finite measure space. But I have one problem. I know that $\sigma \left( M_{\varphi} \right) = \mathrm{ess.im} \left( \varphi \right)$, but to define $\gamma: C (\sigma) \rightarrow B \left( L_2 (X, \mu)\right)$ we need the domain of the function $f \in C (\sigma)$ to be $\mathrm{im} (\varphi)$.

I know, that if $$\forall A \subset X \quad \varphi(A) \cap \mathrm{ess.im} \left( \varphi \right) = \varnothing \implies \mu(A)=0$$ but I don't know how to prove this. Thank's for the help!

By definition, $$\sigma:=\operatorname{ess im}(\varphi)=\left\{x:\forall U^{open}\ni x,\mu(\varphi^{-1}(U))>0\right\}$$
and this is a compact set. Let $U_n$ be a sequence of open sets in $\mathbb{C}$ such that $$\mathbb{C}\setminus\sigma=\bigcup_n U_n.$$ Then for all $n$ we have $\mu(\varphi^{-1}(U_n))=0$, so using countability, $\mu(\varphi^{-1}(\mathbb{C}\setminus\sigma))=0$. In other words, $\varphi^{-1}(\sigma)$ is conull (has full measure) in $X$.
Therefore, if $f\in C(\sigma)$, the function $f\circ\varphi$ is defined a.e. on $X$, namely on the set $\varphi^{-1}(\sigma)$. If you substitute $\varphi$ by another function $\varphi'$ which coincides a.e. with $\varphi$, then $f\circ\varphi$ will coincide a.e. with $f\circ\varphi'$. (that is to say that the class of $f\circ\varphi$ in $L^\infty(X,\mu)$ depends only on the class of $\varphi$ in $L^\infty(X,\mu)$.)
The continuous calculus is given by $f(M_\varphi)=M_{f\circ\varphi}$. Check this on polynomials and extend by Weierstrass.