Consider the multiplication operator version of the spectral theorem for bounded operators in a Hilbert space. Quoted from Wikipedia:
Let $A$ be a bounded self-adjoint operator on a Hilbert space $H$. Then there is a measure space $(X, \Sigma, \mu)$ and a real-valued essentially bounded measurable function $f$ on $X$ and a unitary operator $U:H \to L^2(X, \mu)$ such that
$U^\dagger T_f U = A$
where $T_f$ is the multiplication operator:
$T_f [\varphi](x) = f(x) \varphi(x)$
and $\|T_f\| = \|f\|_\infty$.
It is not much work to find an example where this measure space is not unique. For example, one can look at the sum of the left shift and right shift operators in $L^2(\mathbb{Z})$ and use Fourier series to see that the space $X$ can be any real interval.
My question is, using the boundedness of $A$, can we always find a measure space with finite measure satisfying this theorem? In particular, can we always find one such that $\mu(X)=1$?
Best Answer
For the separable setting, yes. The reference for what follows is Reed-Simon Vol. 1.
The theorem you quote as the spectral theorem is viewed as a quick corollary of the "real" spectral theorem, Reed-Simon Theorem VII.3 where $A$ is conjugated to a direct sum: $$ U:H\to \oplus_{n=1}^N L^2(\mathbb R,d\mu_n)\\ (UAU^*\psi)_n(\lambda)=\lambda\psi_n $$ where $N$ may be infinite. I.e. U conjugates $A$ to a direct sum of operators which are just multiplication by the independent variable on each of the (direct) summands. What you quote as the spectral theorem is then a quick corollary of the above (immediately below in the book), with the added property that the measure $\mu$ may be taken to be finite (and in their construction, it in fact satisfies $\mu(X)=1$).