In some articles that I was reading the fact that if two self-adjoint operators $A,B$ commute then this imples the existence of a third self-adjoint operator $C$ such that $f(C) = \Phi_C(f) = A$ and $g(C) = \Phi_C(g)= B$ for given borel measurable functions $f,g: \mathbb{C} \rightarrow \mathbb{C}$ where $\Phi_C$ is the Gelfand homomorphism associated with $C$ generating a functional calculus by the spectral theorem.
This result is cited as apparently proven in von Neumann's article "Uber Funktionen Von Funktionaloperatoren" of 1931 and in Theorem 6 of M.A. Neumark (Naimark) "Operatorenalgebren im Hilbertschen Raum" in Sowjetische Arbeiten zur Funktional analisys of 1954, the first reference is in german, I believe is the first proof of this fact and is very difficult to translante for me since I don't speak german, and the second is virtually impossible to locate and by the title is also in german.
Does this simply follow from the fact that we can write A and B as: $$A = \int_\mathbb{R} \chi_{\sigma(A)}(\lambda)\,\,dP_C(\lambda) \,\,\,\,\,\,\,\,\,\,\,\,\, B = \int_\mathbb{R} \chi_{\sigma(B)}(\lambda)\,\,dP_C(\lambda) $$
And since:
$$C = \int_\mathbb{R} \lambda \,dP_C(\lambda)$$
Then, are $f= \chi_{\sigma(A)}$ and $g = \chi_{\sigma(B)}$ ? And how does one show that $A$ and $B$ share not only the same eigenvectors (easy to show since they commute) but the same spectral measure (or are representable with the same) ?
What are the difficulties of this fact for unbounded operators and does this fact admit a even simpler demonstration for projectors instead of arbitrary self-adjoint operators $A$ and $B$?
I know that this is a elementar question about a classical theorem, but I can't seem to find a proof of this fact expressed in this basic way.