Spectral theorem and semifinite measure when $H$ is not separable 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 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.
$$UA_kU^*f=\varphi_kf,\;\forall f\in H,\,k=1,2.$$

Why $\mu$ can be taken  semifinite? i.e. for each $E \in \mathcal{E}$ with $\mu(E) = \infty$ , there exists $F \subset E$ and $F \in \mathcal{E}$ and $0 < \mu(F) < \infty$.

If $\mu(E) = \infty$ for all $E\neq\varnothing$, then $L^2(X,\mu)=\{0\}$. If we replace $\mu$ by its semifinite part $\mu_1$ given in (1) and defined as
$$
\mu_{1}(E) = \sup\{\mu(A) \mid A \subseteq E \text{ measurable, } \mu(A)< \infty\} \quad \text{for }E \in \mathcal{E},
$$

it is possible to show that
   $$L^2(\mu)"=" L^2(\mu_1)?$$

 A: I believe the answer is yes. First,


If $\mu_1(E)<\infty$ then there exists $A\subset E$ with $\mu(A)=\mu_1(E)$. (Hence $\mu_1(E\setminus A)=0$.)


By definition of $\mu_1$ there exist $A_n\subset E$ with $\sup_n\mu(A_n)=\mu_1(E)$ and $A_n\subset A_{n+1}$. Let $A=\bigcup_n A_n$. Then $\mu(A)\ge\sup_n\mu(A_n)=\mu_1(E)$, while the definition of $\mu_1$ shows that $\mu(A)\le\mu_1(E)$, since $A\subset E$ and $\mu(A)=\lim_n\mu(A_n)<\infty$.
And now $\mu_1(E)<\infty$ shows that  $\mu(A)<\infty$, hence $\mu_1(A)=\mu(A)=\mu_1(E)<\infty$, so $\mu_1(E\setminus A)=0$.
Hence


If $\phi\ge0$ is simple and $\int \phi d\mu_1<\infty$ then $\phi=\psi$ $\mu_1$-almost-everywhere, where $\int \psi d\mu=\int\phi d\mu_1$.


And hence by the definition of the integral of a non-negative function


If $f\ge0$ is measurable then $f=g$ $\mu_1$-almost-everywhere, where $\int g d\mu=\int f d\mu_1$.


So if $f\in L^2(\mu_1)$ then $f=g$ $\mu_1$-almost-everywhere, with $\int|g|^2d\mu=\int|f|^2d\mu_1$. On the other hand $L^2(\mu)\subset L^2(\mu_1)$, since $\mu_1\le\mu$.
