I'm studying right now spectral theory of unbounded self-adjoint operators. A corollary of spectral theorem states the following: let $H$ be a (separable) Hilbert space and $(D_T, T)$ a self-adjoint operator on H. There exists a unique map $$ L^\infty(\mathbb{R};\mathbb{C}) \longmapsto L(H),\ f \longrightarrow f(T) $$ continuos with norm $\le 1$ with the following properties:
(i )This is a ring homomorphism
(ii) $f(T)^* = \overline{f}(T)$ and $f \ge 0$ iff $f(T) \ge 0$
(iii) If $f_n$ converges pointwise to $f \in L^\infty (\mathbb{R};\mathbb{C}$) and $\sup_n ||f_n||_\infty < +\infty $ then $f_n(T) \mapsto f(T)$ strongly.
The author starts the proof setting $H=L^2_\mu(X)$ with $(X, \mu)$ is a measure space with $\mu < \infty$ and $(D_T, T) = (D_g, M_g)$ where $D_g := \{ \phi \in L^2_\mu | g\phi \in L^2_\mu \}$, $M_g \phi = g \phi$. Then he set
$f(T) := M_{f \circ g}$ for every $f \in L^\infty$.
My question is: why $f(T)$ is well defined? I haven't understood why if $f_1$, $f_2$ are in the same equivalence class $f \in L^\infty(\mathbb{R}; \mathbb{C})$ (remember we're using the Lebesgue measure now) then $M_{f_1 \circ g} = M_{f_2 \circ g}$, that is $f_1 \circ g = f_2 \circ g\ \mu$ a.e.
I'm attending a master degree in Math, and I read this on lecture notes of a Spectral Theory course made in ETH, Zurich. This is the corollary 4.43.