I don't know very much about spectral theory so probably the answer to my question has a basic reference which I would appreciate.

Let's say I have two self-adjoint operators on a Hilbert space and that they commute. I know that, since they generate a commutative $C^*$ algebra, on the one hand they can be viewed as continuous functions on some compact Hausdorff space. On the other hand, individually, they can be viewed as multiplication operators $L^2$ on some measure space.

Now, one (extremely basic) question I have is whether it can be arranged for them to be viewed as multiplication operators on the same measure space. I am guessing the answer is yes, but I would like a reference for it or a counterexample since I don't really know.

Now I am sort of assuming the answer to question 1 is yes, but here's question 2. Let's say I already have two multiplication operators in hand, $\phi_1 : X \to {\mathbb R}$ and $\phi_2 : X \to {\mathbb R}$ where $X$ has measure $\mu$. I would say that $\phi_1$ and $\phi_2$ are independent if the pushforward of $\mu$ by $\phi_1 \times \phi_2 : X \to {\mathbb R}^2$ is the product measure of the pushforwards of $\mu$ by the maps $\phi_1$ and $\phi_2$ individually. Notice that if I compose $\phi_1 \circ T$ and $\phi_2 \circ T$ with any measure-preserving map, the resulting maps are still independent since the pushed-forward measures do not change at all.

My second question is whether the notion of independence makes sense for such operators (or more generally for members of a commutative $C^*$ algebra) without having a measure at all. For example, is being independent an operator-theoretic property of $\phi_1$ and $\phi_2$?