Arveson says that an operator $A$ acting on a separable Hilbert space $H$ is $\textit{diagonalizable}:$
"If there is a (necessarily $\textit{separable}$) $\sigma$-finite measure space $(X, \mu)$, a function $f \in L^{\infty}(X,\mu)$, and a unitary operator $W:L^{2}(X,\mu) \mapsto H$ such that $WM_{f}=AW$",
where $M_f$ is multiplication by $f \in L^{\infty}(X,\mu)$.
Could someone elaborate on the notion of a separable $\sigma$-finite measure space and how it enters into the context of diagonalizable operators and the Spectral Theorem for (Normal) operators on a separable Hilbert space?
Saying that a measure space is separable is not standard terminology. Here it just means that $L^2(X)$ is separable, which is obvious since it is isomorphic to $H$.
If $A$ is diagonalizable then it is unitarily equivalent to the normal operator $M_f$, so it is normal.
To see the relation with the Spectral Theorem, given $\Delta\subset\sigma(A)$ Borel, define $$ F(\Delta)=WM_{1_{f^{-1}(\Delta)}}W^*. $$ Then one checks that $F$ is a spectral measure and that $$ A=\int_{\sigma(A)}\lambda\,dF(\lambda). $$