Let $\mu$ be a Borel measure on $\mathbb{R}$. By the Lebesgue decomposition theorem, there exists measures $\mu_\text{pp}$, $\mu_\text{ac}$ and $\mu_\text{sing}$ such that $\mu = \mu_\text{pp}+\mu_\text{ac}+\mu_\text{sing}$, where $\mu_\text{pp}$ is a pure point measure, $\mu_\text{ac}$ is absolutely continuous with respect to the Lebesgue measure $m$ on $\mathbb{R}$ and $\mu_\text{sing}$ is singular with respect to the latter.
The following decomposition is a consequence of the above: $$L^{2}(\mathbb{R},d\mu) = L^{2}(\mathbb{R},d\mu_\text{pp})\oplus L^{2}(\mathbb{R},d\mu_\text{ac})\oplus L^{2}(\mathbb{R},d\mu_\text{sing}). \tag{1}\label{1}$$
A generalization of \eqref{1} is: $$\mathscr{H} = \mathscr{H}_\text{pp}\oplus \mathscr{H}_\text{ac}\oplus \mathscr{H}_\text{sing}\tag{2}\label{2}$$ for a given separable Hilbert space $\mathscr{H}$, where: $$\tag{3}\label{3}\begin{align} &\mathscr{H}_\text{pp} = \{\psi: \text{$\mu_{\psi}$ is pure point}\} \\ &\mathscr{H}_\text{ac} = \{\psi: \text{$\mu_{\psi}$ is absolutely continuous w.r.t. Lebesgue measure}\} \\ &\mathscr{H}_\text{sing} = \{\psi : \text{$\mu_{\psi}$ is singular w.r.t. Lebesgue measure}\}.\end{align}$$
Now, from the Lebesgue Decomposition Theorem, there exist disjoint Borel sets $M_\text{pp}$ (which is countable), $M_\text{ac}$ and $M_\text{sing}$ such that $\mathbb{R} = M_\text{pp}\cup M_\text{ac}\cup M_\text{sing}$ and : $$\mu_\text{pp}(\Omega) = \mu(\Omega\cap M_\text{pp}), \quad \mu_\text{ac}(\Omega) = \mu(\Omega \cap M_\text{ac}) \quad \text{and} \quad \mu_\text{sing}(\Omega) = \mu(\Omega\cap M_\text{sing}). \tag{4}\label{4}$$
Suppose $A$ is a fixed self-adjoint operator on $\mathscr{H}$. From the Borel functional calculus, one can define projection-valued measures $P_{\Omega} \equiv \chi_{\Omega}(A)$ for every Borel set $\Omega$. These are projection operators. Consider $P_{M_\text{pp}}$, $P_{M_\text{ac}}$ and $P_{M_\text{sing}}$ and, for each of these sets (denote it generically by $M_{*}$ where $*$ means either $\text{pp}$, $\text{ac}$ or $\text{sing}$) $\psi_{*} := P_{M_{*}}\psi$, for a given vector $\psi \in \mathscr{H}$. This allows us to define subspaces $\mathscr{H}_{*} := P_{M_{*}}(\mathscr{H})$, i.e. the range of $P_{M_{*}}$. Each one of these subspaces is closed because if $\psi_{n}$ is a sequence of elements of $\mathscr{H}_{*}$ such that $\psi_{n} \to \psi$, write $\psi_{n} = P_{M_{*}}\varphi_{n}$ for some $\varphi_{n}$. We get: $$P_{M_{*}}\psi = P_{M_{*}}\lim_{n\to \infty}\psi_{n} = \lim_{n\to \infty}P_{M_{*}}P_{M_{*}}\varphi_{n} = \lim_{n\to \infty}P_{M_{*}}\varphi_{n} = \psi \tag{5}\label{5}$$ since $P_{M^{*}}$ is a projection operator. Hence $\psi = P_{M_{*}}\psi $ and $\psi \in P_{M_{*}}(\mathscr{H})$. Moreover each $\psi \in \mathscr{H}$ can be written as a sum: $$\psi = \psi_\text{pp}+\psi_\text{ac}+\psi_\text{sing}.$$ Since the sets $M_\text{pp}$, $M_\text{ac}$ and $M_\text{sing}$ are disjoint, this is an orthogonal decomposition and we end up with: $$\mathscr{H} = \mathscr{H}_\text{pp}\oplus \mathscr{H}_\text{ac}\oplus \mathscr{H}_\text{sing}.$$
Finally, each $P_{M_{*}}(\mathscr{H})$ coincides with the corresponding closed subspace $\mathscr{H}_{*}$ defined by \eqref{3}. In particular, if $\psi$ is such that $\mu_{\psi}$ is pure point then, by \eqref{4}: $$\mu_{\psi}(\Omega) = \langle \psi, P_{\Omega}\psi\rangle = \mu_{\psi,\text{pp}}(\Omega) = \langle \psi, P_{\Omega}P_{M_\text{pp}}\psi\rangle \Rightarrow \langle P_{\Omega}\psi, (I-P_{M_\text{pp}})\psi\rangle = 0.$$ Taking $\Omega = \mathbb{R}\setminus M_\text{pp}$ one concludes that $\psi = P_{M_\text{pp}}\psi$. The converse is easy.
Now, my question is: is there something wrong with this proof? I worked it on it based on my previous question on MSE, and I find nothing wrong with it. However, when reading Mathematical Methods in Quantum Mechanics by Teschl, the procedure of considering projections $P_{M_{*}}$ is only used to prove decomposition \eqref{1}, and there is a more elaborated reasoning behind decomposition \eqref{2}, where the author uses spectral basis and unitary transformations from $L^{2}$ to $\mathscr{H}$. In short, I was under the impression that I cannot use the above reasoning to prove the general decomposition \eqref{2}, although I see nothing wrong with it. So, is there anything wrong at all? Or maybe some limitation?
