Let $A_{1},...,A_{n}$ be densely defined self-adjoint operators on a separable Hilbert space $\mathscr{H}$. Suppose these have a common dense domain $D\subset \mathscr{H}$ and satisfy commutation relations: $$A_{i}A_{j}+A_{j}A_{i} = 0 \quad \forall i,j = 1,...,n$$$$A_{i}A_{j}-A_{j}A_{i} = 0 \quad \forall i,j = 1,...,n$$ By the spectral theorem, there exists a measure space $(\Omega, \mathcal{F},\mu)$, a unitary operator $U: \mathscr{H} \to L^{2}(\Omega,\mu)$ and complex-valued functions $g_{1},...,g_{n}$ on $\Omega$ such that $A_{i}$ is unitarily equivalent to a multiplication by $g_{i}$.
On Mathematical Aspects of Quantum Field Theory, the authors state the following, without proof (I am adapting the notations here). Suppose the set of operators $A_{1},...,A_{n}$ form a complete system, in the sense that any other operator $B$ which commutes with all $A_{1},...,A_{n}$ is of the form $B = f(A_{1},...,A_{n})$, for some Borel measurable function $f$. Then, the measure space $(\Omega,\mathcal{F},\mu)$ is unique and can be taken to be $\Omega = \mathbb{R}^{n}$ with the Borel $\sigma$-algebra and Borel measure.
I don't see how to prove this result, and also I haven't found this result anywhere I looked. Any help or reference is appreciated.