Let $H$ be a complex Hilbert space (not necessary separable).


**Spectral Theorem:** Let $A\_1$ and $A\_2$ be two commuting normal operators, then there exists a measure space $(X,\mathcal{E},\mu)$,
two functions $\varphi\_1,\varphi\_2\in L^\infty(\mu)$ and a unitary operator $U:H\longrightarrow L^2(\mu)$, such that each $A\_k$ is unitarily equivalent to multiplication by $\varphi\_k$, $k=1,2$. i.e.
$$UA\_kU^\*f=\varphi\_kf,\;\forall f\in H,\,k=1,2.$$



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> Is $\mu$ semifinite? i.e. for each $E \in \mathcal{E}$ with $\mu(E) = \infty$ , there exists $F \subset E$ and $F \in \mathcal{E}$ and $0 < \mu(F) < \infty$.
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If $H$ is a separable complex Hilbert space, then $(X,\mathcal{E},\mu)$ is a $\sigma$-finite measure space and so $\mu$ is semifinite.



Semifinite measure and spectral theorem
The answer is yes. Let me explain why.


Let $v \in E$ be any nonzero vector. Then define $E\_0$ to be the closure of the set of vectors of the form $p(A\_1,A\_1^\*, A\_2, A\_2^\*)v$ where $p$ is a complex polynomial. By restricting $p$ to have rational coefficients you can see that $E\_0$ is separable. Now the point is that $A\_1E\_0 \subseteq E\_0$ and $A\_1E\_0^\perp \subseteq E\_0^\perp$, and similarly for $A\_1^\*$, $A\_2$, and $A\_2^\*$. I.e., $E\_0$ is a "reducing" subspace. Now you can Zornicate and get a family of mutually orthogonal separable subspaces $E\_\alpha$, each of which is reducing for $A\_1$ and $A\_2$, and which span all of $E$. Apply the spectral theorem separately on each $E\_\alpha$.


This shows you something even stronger than semifiniteness --- it shows you that $\mu$ is *localizable*: the measure space can be decomposed into a disjoint family of finite measure subsets, such that the measure of any set is the sum of the measures of its intersections with the members of this family.


