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Let $H$ be a complex Hilbert space. We recall the following theorem:

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 an 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. $$UA_kU^*f=\varphi_kf,\;\forall f\in H,\,k=1,2.$$

If $H$ is separable, why $\mu$ can be taken $\sigma$-finite?

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    $\begingroup$ Do you know how to do this for one operator? If so, then deal with the second operator as an operator that commutes with multiplication operators. $\endgroup$ – DisintegratingByParts Mar 28 '18 at 21:30
  • $\begingroup$ @DisintegratingByParts Unfortunately I don't find a proof even for one operator. $\endgroup$ – Student Mar 29 '18 at 14:32
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    $\begingroup$ Do you know the traditional spectral theorem using spectral measures? $\endgroup$ – DisintegratingByParts Mar 29 '18 at 17:32
  • $\begingroup$ @DisintegratingByParts Yes $\endgroup$ – Student Mar 29 '18 at 17:39
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If you know the traditional spectral theorem $A = \int_{\sigma}\lambda dE(\lambda)$, where $\sigma$ is the spectrum of $A$ and $E$ is the spectral measure, then you can start by fixing $x \in H$ that is non-zero, and consider the closure $H_x$ of $\{ x,Ax,A^2x,\cdots \}$. Note that if $p$ is a polynomial, $$ \|p(A)x\|^2 = \int_{\sigma}|p(\lambda)|^2 d\mu_x(\lambda), $$ where $\mu_x(S) = \langle E(S)x,x\rangle$. And $Ap(A)x = (\lambda p(\lambda))|_{\lambda=A}x$, which shows how the operator $A$ is transformed to multiplication by $\lambda$ on $H_x$. Every $y \in H_x$ can be written as a limit $$ y = \lim_n p_n(A)x, $$ where $\{ p_n \}$ is a sequence of polynomials. This is how you end up with $p \in L^2(\sigma,d\mu_x)$ such that $y = p(A)x$. So, at least on $H_x$, you can see that $U : L^2(\sigma,d\mu_x)\rightarrow H_x$ is a unitary map such that $$ A = U^{-1}M_{\lambda}U. $$ Then you have to glue together mutually orthogonal spaces $H_x$ to obtain the full representation.

Once you have such a representation, you are left to consider a second operator on an $L^2$ space that commutes with multiplication by $\lambda$, which ends up implying that the second operator is also a multiplication operator. If $H$ is separable, you end up being able to keep the full representation on a sigma-finite measure space because you only have to glue a countable number of $H_x$ representations together, each of which is represented using a finite measure.

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