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I'm studying the spectral Theorem (for bounded self-adjoint operators) by myself and I'm following Nik Weaver's nice book. Let me first introduce some notations first.

Notations: If $\mathcal{H}$ is a Hilbert space, $\mathcal{B}(\mathcal{H})$ is the (Banach space) of all bounded linear operators $A: \mathcal{H} \to \mathcal{H}$. If $A \in \mathcal{B}(\mathcal{H})$, $\mbox{sp}(A)$ is the spectrum of $A$.

Now, let $(X, \mathcal{F},\mu)$ be a $\sigma$-finite measure space. A measurable Hilbert bundle over $X$ is a disjoint union: $$\mathcal{X} = \bigcup_{n\in \mathbb{N}}(X_{n}\times \mathcal{H}_{n}) $$ where $\{X_{n}\}_{n\in \mathbb{N}}$ is a measurable partition of $X$ and, for each $0 \le n \le \infty$, $\mathcal{H}_{n}$ is a Hilbert space with dimension $n$.

Finally, $f: X \to \mathcal{H}$ is weakly-measurable if the function $x \mapsto \langle f(x),v\rangle$ is measurable for every $v \in \mathcal{H}$. We denote $L^{2}(X;\mathcal{H})$ the set of all weakly measurable functions $f: X \to \mathcal{H}$ such that: $$||f|| := \int_{x}||f(x)||^{2}d\mu(x) < +\infty $$ modulo functions which are zero almost everywhere. This is a Hibert space with inner product: $$\langle f,g\rangle := \int_{x}\langle f(x),g(x)\rangle d\mu(x) $$ If $f \in L^{2}(X;\mathcal{H})$, $M_{f}$ is the operator multiplication by $f$. Also, $L^{2}(X;\mathcal{X}) := \bigoplus_{n\in \mathbb{N}}L^{2}(\mathcal{X}_{n};\mathcal{H}_{n})$.

Now, the statement of the spectral Theorem in this reference is as follows.

Theorem: Let $\mathcal{B}(\mathcal{H})$ be self-adjoint. Then there exits a probability measure $\mu$ on $\mbox{sp}(A)$, a measurable Hilbert bundle $\mathcal{X}$ over $\mbox{sp}(A)$ and an isometric isomorphism $U: L^{2}(\mbox{sp}(A);\mathcal{X}) \to \mathcal{H}$ such that $A = UM_{x}U^{-1}$.

However, I'm more interested in another version of this Theorem, which is stated in Dimock's book and goes like (with adapted notation)

Theorem: Let $A \in \mathcal{B}(\mathcal{H})$ be self-adjoint. Then, there exists a measure space $(\mathcal{M},\mathcal{\Omega},\mu)$, a bounded measurable function $\tau: \mathcal{M}\to \mathbb{R}$ and a unitary operator $U: \mathcal{H}\to L^{2}(\mathcal{M},\mu)$ such that $A = UM_{\tau}U^{-1}$.

Question: How can I obtain Dimock's version of the spectral Theorem from Weaver's version of it?

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Let $\mathcal{M}$ be a disjoint union consisting of $n$ copies of $X_n$ for each $n$. The given measure on $\mbox{sp}(A)$ restricts to a measure on $X_n$ and thus induces a measure on $\mathcal{M}$. There is then an isomorphism $L^2(\mathcal{M})\cong L^2(\mbox{sp}(A);\mathcal{X})$: if you pick an orthonormal basis for each $\mathcal{H}_n$, then $L^2(X_n;\mathcal{H}_n)$ is just a direct sum of $n$ copies of $L^2(X_n)$, and when you take the direct sum of these over all $n$ you get $L^2(\mathcal{M})$. This isomorphism $L^2(\mathcal{M})\cong L^2(\mbox{sp}(A);\mathcal{X})$ turns multiplication by $x$ on $\mbox{sp}(A)$ to multiplication by the function $\tau$ on $\mathcal{M}$ which is given by the inclusion function $X_n\to\mathbb{R}$ on each copy of each $X_n$.

(Alternatively, without directly using Weaver's version, Dimock's version follows by using the same proof as Weaver does but using his Theorem 3.4.2 instead of Corollary 3.4.3. Weaver himself comments on this (since it applies in the non-separable case as well) at the top of page 62.)

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  • $\begingroup$ Amazing answer! Thank you so much. I've read Weaver's comments on page 62 and I was pretty sure my doubt was related to it, but I was not 100% sure I was in the right path. Now I know! Thanks again! $\endgroup$
    – Idontgetit
    Aug 23, 2020 at 2:32
  • $\begingroup$ Incidentally, Weaver's proof of Corollary 3.4.3 is wrong (specifically, Proposition 2.4.7 does not justify identifying $L^2(X;\mathcal{X})$ with $\bigoplus L^2(S_k,\mu|_{S_k})$). The result is correct but requires a more difficult proof, and is part of what is known as the Hahn-Hellinger theorem. $\endgroup$ Aug 23, 2020 at 2:59
  • $\begingroup$ would you know another reference where I could check a correct proof of this result? $\endgroup$
    – Idontgetit
    Aug 23, 2020 at 3:39
  • $\begingroup$ I don't know a reference for the full proof. You can find a more general theory (which applies even in the non-separable case) in the final chapter of Halmos's nice little book Introduction to Hilbert Space and the Theory of Spectral Multiplicity. Roughly speaking, this gives a "measurable Hilbert bundle" on the spectrum except that instead of having disjoint sets $X_n$ with respect to a single measure $\mu$, you have a family of pairwise orthogonal measures (and instead of indexing by $n\in\mathbb{N}\cup\{\infty\}$, it can be indexed by arbitrary cardinalities). $\endgroup$ Aug 23, 2020 at 3:49
  • $\begingroup$ In the separable case, this family of pairwise orthogonal measures is countable and each measure is $\sigma$-finite, and so you can use the Lebesgue decomposition theorem to realize them as the restrictions of a single measure to disjoint subsets of the spectrum, thus obtaining a measurable Hilbert bundle in Weaver's sense. $\endgroup$ Aug 23, 2020 at 3:51

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