I'm reading out of Operator theoretic aspects of ergodic theory, and specifically looking at pages 380 and 381, and specifically at Remark 18.14. The authors have already proven the existence of a maximal spectral type for normal operators on a separable Hilbert space. This means that given an operator $T$ on a separable Hilbert space $\mathscr{H}$, there exists a positive bounded Borel measure $\mu_\mathrm{max}$ on $K = \sigma(T)$ such that (1) $\mu_y \ll \mu_\mathrm{max}$ for all $y \in \mathscr{H}$, and (2) if $0 \leq \mu \ll \mu_\mathrm{max}$, then there exists $y \in \mathscr{H}$ such that $\mu_y = \mu$. Further, this $\mu_\mathrm{max}$ is unique up to equivalence, earning it the name maximal spectral type.
Now, in Remark 18.14, they mention that this result can be sharpened. Specifically, they say that there exists a (possibly finite) sequence $(x_n)$ in $\mathscr{H}$ such that $\mu_{x_1} \gg \mu_{x_2} \gg \cdots$, and $\mathscr{H} = \bigoplus_n Z(x_n)$, where $Z(x) : = \overline{\operatorname{span}} \left\{ T^k x : k \geq 0 \right\} \subseteq \mathscr{H}$. One can then recover the maximal spectral type as $\mu_{x_1}$.
If I'm understanding it correctly, what I want to do is take $\mu_{x_1}$ to be the maximal spectral type of $T$ on $\mathscr{H}$, then find $\mu_{x_2}$ as the maximal spectral type of $T \vert_{\mathscr{H} \ominus Z(x_1)}$, and continue by finding $x_n$ so that $\mu_{x_n}$ is the MST of $T$ restricted to $\mathscr{H} \ominus \left( Z(x_1) \oplus \cdots \oplus Z(x_{n - 1}) \right)$.
My hang-up here is that it's not clear to me why $\mathscr{H} = \bigoplus_n Z(x_n)$. I understand that $\mathscr{H}$ is separable, so it can't hold an uncountable number of nonzero orthogonal subspaces. But it seems to me possible that perhaps $\bigoplus_n Z(x_n)$ is only a proper subspace of $\mathscr{H}$, and I need to do some kind of transfinite recursion method so that I can choose $x_\omega$ by finding the MST of $T \vert_{\mathscr{H} \ominus \bigoplus_{n = 1}^\infty Z(x_n)}$, and continue this way. How do I know that $\mathscr{H} = \bigoplus_n Z(x_n)$?
EDIT: I've resolved my issue. Looking at another book, their discussion of the MST mentioned that given a fixed vector $z \in \mathscr{H}$, one could specifically choose a vector $x$ such that $\mu_x = \mu_\mathrm{max}$ and $z \in Z(x)$. To answer my question, we can fix an orthonormal basis $\{e_n\}_{n = 1}^\infty$ and just make sure that we always choose $x_n$ so that $e_n \in \bigoplus_{i = 1}^n Z(x_i)$. If we keep this schedule, then we'll have that $\mathscr{H} = \bigoplus_{n = 1}^\infty Z(x_n)$.