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

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 $0 < \mu(F) < \infty$.

If $H$ is a separable complex Hilbert space, then $(X,\mathcal{E},\mu)$ is a $\sigma$-finite measure space and so $\mu$ is semifinite.

existsa semifinite measure satisfying the given properties. Of course, there will always exist non-semifinite ones as well (take any such measure and if it's semifinite then consider a space with one additional point that has measure infinity). My suspicion is that you can prove there's a semifinite one by saying "let $\mu$ be a measure satisfying the conclusion of the spectral theorem; if it is not semifinite, let $\mu_0$ be its semifinite part and then $\mu_0$ satisfies." But I can't quite finish the proof of this. $\endgroup$ – Nate Eldredge Mar 9 '18 at 16:06