I'm following Reed & Simon's section on spectral theorem for bounded operators and there is a Corollary there which has a very quick undetailed proof and I'm working out the details. But I got some questions about it. For completeness, let me just state two previous results that are used in the proof of the mentioned corollary.
Lemma 1: Let $A$ be a bounded self-adjoint operator on a separable Hilbert space $\mathcal{H}$. Then there is a direct sum decomposition $\mathcal{H} = \bigoplus_{n=1}^{N}\mathcal{H}_{n}$ with $N=1,2,...$ or $\infty$ so that:
(a) $A$ leaves $\mathcal{H}_{n}$ invariant, that is, $\psi \in \mathcal{H}_{n}$ implies $A\psi_{n}\in \mathcal{H}_{n}$
(b) For each $n$, there is a $\phi_{n}\in \mathcal{H}_{n}$ which is cyclic for $A|_{\mathcal{H}_{n}}$, i.e. $\mathcal{H}_{n} = \overline{\{f(A)\phi_{n}| \hspace{0.1cm} f \in C(\sigma(A))\}}$. Here $C(\sigma(A))$ is the set of all continuous functions defined on the spectrum $\sigma(A)$ of $A$.
Lemma 2: Let $A$ be bounded and self-adjoint on a separable Hilbert space $\mathcal{H}$. Then, there exist measures $\{\mu_{n}\}_{n=1}^{N}$ ($N=1,2,...$ or $\infty$) on $\sigma(A)$ and a unitary operator $U: \mathcal{H}\to \bigoplus_{n=1}^{N}L^{2}(\mathbb{R},d\mu_{n})$ so that: \begin{eqnarray} (UAU^{-1}\psi)_{n}(\lambda) = \lambda \psi_{n}(\lambda) \tag{1}\label{1} \end{eqnarray} where we write an element $\psi \in \bigoplus_{n=1}^{N}L^{2}(\mathbb{R},d\mu_{n})$ as an $N$-uple $(\psi_{1}(\lambda),...,\psi_{N}(\lambda))$.
Now, let's go to the result I want to prove.
Corollary: Let $A$ be bounded and self-adjoint on a separable Hilbert space $\mathcal{H}$. Then there exists a finite measure space $(\mathcal{M},\mu)$, a bounded function $F$ on $\mathcal{M}$ and a unitary map $U: \mathcal{H}\to L^{2}(\mathcal{M},\mu)$ so that: \begin{eqnarray} (UAU^{-1}f)(m) = F(m)f(m) \tag{2}\label{2} \end{eqnarray}
Sketch of my proof: First, let us use Lemma 1 to decompose $\mathcal{H}=\bigoplus_{n=1}^{N}\mathcal{H}_{n}$ for some $N$. For each $n$, there exists a vector $\phi_{n} \in \mathcal{H}_{n}$ which is cyclic for $A|_{\mathcal{H}_{n}}$. In particular, we can take each $\phi_{n}$ such that $||\phi_{n}|| = 2^{-n}$. Now, if $\mathbb{R}_{n} := \mathbb{R}\times \{n\}$, let us define $\mathcal{M}$ as the disjoint union $\mathcal{M} := \bigoplus_{n=1}^{N}\mathbb{R}_{n}$ and define $\mu$ on $\mathcal{M}$ such that its restriction to the $n$-th copy $\mathbb{R}_{n}$ is $\mu_{n}$, where $\mu_{n}$ are the measures given by Lemma 2. Now, let $L^{2}(\mathcal{M},d\mu)$ be the set of all functions $f: \mathcal{M}\to \mathbb{C}$ such that $||f||_{L^{2}(\mathcal{M},d\mu)} := \int |f(m)|^{2}d\mu < +\infty$. Then, [I think] we have $L^{2}(\mathcal{M},d\mu) \cong \bigoplus_{n=1}^{N}L^{2}(\mathbb{R},d\mu_{n})$, so each $f \in L^{2}(\mathcal{M},d\mu)$ can be viewed as an $N$-uple $f = (f_{1},...,f_{n})$ of functions $f_{k} \in L^{2}(\mathbb{R},d\mu_{k})$. Thus, by Lemma 2, $U: \mathcal{H} \to \bigoplus_{n=1}^{N}L^{2}(\mathcal{M},d\mu_{n}) \cong L^{2}(\mathcal{M},d\mu)$ is unitary and (\ref{1}) implies: \begin{eqnarray} (UAU^{-1}f)_{n}(m) = mf_{n}(m) \tag{3}\label{3} \end{eqnarray} Let us prove that $(\mathcal{M},\mu)$ is finite. Using the fact that each $\mu_{n}$ is the spectral measure of $\phi_{n}$, we have: $$\mu_{n}(\mathbb{R}) = \int_{\mathbb{R}}1d\mu_{n}(\lambda) = \langle \phi_{n}, \phi_{n}\rangle = ||\phi_{n}||^{2} = 2^{-2n}$$ thus $\mu(\mathcal{M}) = \sum_{n=1}^{N}\mu_{n}(\mathbb{R} = \sum_{n=1}^{N}2^{-2n} <+\infty$.
Question 1: Is my reasoning correct so far?
Question 2: How can I prove (\ref{2}) from the above reasoning? I mean, as far as I understand, a function $f:\mathcal{M}\to \mathbb{C}$ is such that it takes a pair $(x,n)$ and associated $f(x,n) = f_{n}(x)$. So I think I must use this to write (\ref{3}) in an appropriate way? But this is where I get confused because I wrote $f = (f_{1},...,f_{N})$ but the result (\ref{2}) is not component-wise, so it seems that this decomposition is somewhat useless. How to fix it?