Question: 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.
$$UA\_kU^\*f=\varphi\_kf,\;\forall f\in H,\,k=1,2.$$

> 
> I look for a reference which contains the proof of the above theorem.
>  More precisely I hope to find a proof which show that $\mu$ can be taken semifinite: i.e. every set of infinite measure contains a subset
>  of positive finite measure. 
> 
> 
>

Answer: 
In the book "A course in abstract harmonic analysis" by "Gerald B. Folland"., we have the follwoing theorem

[![enter image description here](https://i.stack.imgur.com/XSeyf.png)](https://i.stack.imgur.com/XSeyf.png)

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[![enter image description here](https://i.stack.imgur.com/tUvxY.png)](https://i.stack.imgur.com/tUvxY.png)
Start with the normal operator $A\_1$ and its spectral resolution $A\_1 = \int \lambda dE\_1(\lambda)$. Let $x$ be a non-zero vector, and consider the closed subspace $\mathscr{A}\_1(x)$ generated by $\{ x,A\_1 x,A\_1^\*,A\_1^2 x,A\_1^{\*2}x\cdots\}$. Every element of $\mathscr{A}\_1(x)$ can be written as a limit of vectors of the form $p\_n(A\_1,A\_1^\*)x$ as $n\rightarrow\infty$, where $p\_n(\lambda,\overline{\lambda})$ is a polynomial in the two variables. Knowing the spectral resolution of $A\_1$ allows you to trade this limit for a function limit in $L^2(d\|E\_1(\lambda)x\|^2)$ because
$$
 \|p\_n(A\_1,A\_1^\*)x\|^2 = \int |p\_n(\lambda,\overline{\lambda})|^2d\|E\_1(\lambda)x\|^2.
$$
So there is a map $U\_x : \mathscr{A}\_1(x) \rightarrow L^2\_{\mu\_x}$ where $\mu\_x(S)=\|E\_1(S)x\|^2$, and
$$
 U\_x A\_1 x = \lambda U\_x x \\
 U\_x A\_1^\* x = \overline{\lambda}U\_x x.
$$
Because $A\_2$ commutes with $A\_1$, then $U\_x A\_2 U\_x^\*$ commutes with multiplication by all polynomials in $\lambda,\overline{\lambda}$ on $L^2\_{\mu\_x}$, which forces $U\_xA\_2U\_x^\*f=a\_2(\lambda)f$ for all $f\in L^2\_{\mu\_x}$ because the operator space of bounded multiplication operators on $L^2\_{\mu\_x}$ is maximally Abelian. Finally, there is a trick that I do not recall at the moment for piecing together mutually orthogonal spaces $U\_x\mathscr{A}\_1(x)=L^2\_{\mu\_x}$ together in order to write $A\_1$ as a multiplication operator by $a\_1\in L^{\infty}\_{\mu}$ on the composite $L^2\_\mu$ space; the corresponding representation for $A\_2$ as a multiplication follows as well. So this is a partial answer, but it gives you the basic building block coming from the spectral theorem.