The spectral theorem states that for a self-adjoint linear operator $A$ with $\left\|A\right\|=1$ on a Hilbert space, there is a measure space $(X,\mathcal M,\mu)$ so that $A$ is unitarily equivalent to a multiplication operator on $L^2(I,d\mu)$ where $I:=[-1,1]$. If $H$ is separable, then $\mu$ can be taken to be a finite measure.
Now my question is: If $H$ is not separable, can we always take $\mu$ to be $\sigma$-finite?
If every real $x\in[0,1]$ is an eigenvalue, then the $\mu$ cannot be sigma finite.