# Spectral measure is sigma finite

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?

• you should probably say "on a Hilbert space $H$" – mathworker21 Oct 14 '18 at 23:32

If every real $$x\in[0,1]$$ is an eigenvalue, then the $$\mu$$ cannot be sigma finite.
• For instance, let $H = \ell^2([0,1])$, i.e. $L^2$ of $[0,1]$ equipped with the discrete $\sigma$-algebra and counting measure. Then define $(Af)(x) = x f(x)$. – Nate Eldredge Oct 15 '18 at 4:30