Finiteness of the Measure Space appearing in Spectral Theorem of Bounded Operators

Question

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 $\mu(X)=1$?

Answer 1

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 $\mu(X)=1$).

Answer 2

I'll refer to E.B. Davies' Spectral Theory and Differential Operators, and I'll limit myself to separable Hilbert spaces in case what I said in my comment above isn't true (I could have easily misinterpreted what Davies said) and you need to modify the statement so much that you lose the finiteness of the measure.

Given a self-adjoint operator $A$ (it can be unbounded) on $H$, you can decompose $H$ as a countable sum of cyclic subspaces $L_n$ for $A$, on each of which you can apply the Riesz-Markov theorem on a well-chosen linear functional to get a finite measure $\mu_n$ such that on $L_n$ the theorem is true for $X = \sigma(A)$ and $f$ is just the identity function. Then, you "gather" all these copies of $\sigma(A)$ and those $\mu_n$. Hence at the end $X$ can be chosen as $\sigma(A) \times \mathbb{N}$ and $f$ as the function $f : (s,n) \mapsto s$ (still the identity function in terms of $s$), with the (finite!) measure being $\mu_n$ on each $\sigma(A) \times \{n\}$. I don't know what happens in the general non-separable case.

I should note that that passage is pretty self-contained if you skip directly to Them 2.3.1 and accept that theorem (the existence and uniquness of a functional calculus on the continuous functions vanishing at infinity) as true, and even moreso if you only want to consider bounded operators and not general unbounded operators, in which case you can also ignore domain issues for the most part. Another possible reference is Reed and Simon's Functional Analysis, Chapter VII I believe.