[proofplan]
We first verify that the Koopman operator is a unitary operator on the given separable Hilbert space. The Hahn-Hellinger theorem for unitary operators then supplies a countable orthogonal decomposition into cyclic reducing subspaces, each represented as multiplication by the coordinate function on an $L^2$ space over the unit circle. Renaming the abstract cyclic vectors and measures gives the stated Koopman decomposition, and finite decompositions are padded by zero summands.
[/proofplan]
[step:Verify that the Koopman operator is unitary]
The Koopman operator is the map
\begin{align*}
U_T:H&\to H\\
f&\mapsto f\circ T.
\end{align*}
Since the system is measure-preserving and invertible modulo $\mu$-null sets, the inverse Koopman operator is
\begin{align*}
U_{T^{-1}}:H&\to H\\
f&\mapsto f\circ T^{-1}.
\end{align*}
For every $f\in H$, the measure-preserving property of $T$ gives
\begin{align*}
\|U_Tf\|_H^2
&=\int_X |f(Tx)|^2\,d\mu(x)\\
&=\int_X |f(y)|^2\,d\mu(y)\\
&=\|f\|_H^2.
\end{align*}
Thus $U_T$ is an isometry. Since $U_{T^{-1}}U_T$ and $U_TU_{T^{-1}}$ are both the identity operator on $H$, $U_T$ is surjective. Hence $U_T$ is unitary.
[guided]
We need the hypotheses of the Hahn-Hellinger theorem. The operator under consideration is
\begin{align*}
U_T:H&\to H\\
f&\mapsto f\circ T.
\end{align*}
Because the measure-preserving system is invertible modulo $\mu$-null sets, $T^{-1}$ also defines a measure-preserving transformation modulo null sets, hence
\begin{align*}
U_{T^{-1}}:H&\to H\\
f&\mapsto f\circ T^{-1}
\end{align*}
is well defined.
Fix $f\in H$. Since $T$ preserves $\mu$, integration after composition with $T$ preserves the $L^2$ norm:
\begin{align*}
\|U_Tf\|_H^2
&=\int_X |f(Tx)|^2\,d\mu(x)\\
&=\int_X |f(y)|^2\,d\mu(y)\\
&=\|f\|_H^2.
\end{align*}
Therefore $U_T$ is an isometry. Invertibility gives surjectivity: for every $f\in H$,
\begin{align*}
U_{T^{-1}}U_Tf=f
\qquad\text{and}\qquad
U_TU_{T^{-1}}f=f
\end{align*}
as elements of $H$, because the compositions agree with the identity modulo $\mu$-null sets. Thus $U_T$ is unitary. The separability of $H=L^2(X,\mu)$ is part of the theorem statement.
[/guided]
[/step]
[step:Apply Hahn-Hellinger to the unitary Koopman operator]
Let
\begin{align*}
\mathbb{T}:=\{z\in\mathbb{C}:|z|=1\},
\end{align*}
and let $\mathcal{B}(\mathbb{T})$ be its Borel $\sigma$-algebra. By the Hahn-Hellinger theorem for unitary operators, applied to the unitary operator $U_T:H\to H$, there are an index set $I$, either finite or countably infinite, vectors $e_i\in H$, and finite positive Borel measures $\nu_i:\mathcal{B}(\mathbb{T})\to[0,\infty)$ such that, with
\begin{align*}
K_i:=\overline{\operatorname{span}_{\mathbb{C}}\{U_T^n e_i:n\in\mathbb{Z}\}}^{\,H},
\end{align*}
one has
\begin{align*}
H=\bigoplus_{i\in I}K_i.
\end{align*}
Moreover, defining
\begin{align*}
\zeta:\mathbb{T}&\to\mathbb{C}\\
z&\mapsto z
\end{align*}
and
\begin{align*}
M_{\zeta,i}:L^2(\mathbb{T},\nu_i)&\to L^2(\mathbb{T},\nu_i)\\
\varphi&\mapsto \zeta\varphi,
\end{align*}
there is a unitary map
\begin{align*}
W_i:K_i&\to L^2(\mathbb{T},\nu_i)
\end{align*}
such that
\begin{align*}
W_i(U_T|_{K_i})=M_{\zeta,i}W_i.
\end{align*}
The measures may be chosen in Hahn-Hellinger order: $\nu_i\gg\nu_{i+1}$ whenever both indices occur.
[guided]
The Hahn-Hellinger theorem for unitary operators requires a unitary operator on a separable Hilbert space. The previous step proves unitarity of $U_T$, and separability of $H$ is assumed. Hence the theorem applies.
Let
\begin{align*}
\mathbb{T}:=\{z\in\mathbb{C}:|z|=1\}
\end{align*}
with Borel $\sigma$-algebra $\mathcal{B}(\mathbb{T})$. Hahn-Hellinger gives cyclic vectors $e_i\in H$ and finite positive Borel measures
\begin{align*}
\nu_i:\mathcal{B}(\mathbb{T})\to[0,\infty)
\end{align*}
indexed by a finite or countable set $I$. For each $i\in I$, the generated cyclic subspace is
\begin{align*}
K_i:=\overline{\operatorname{span}_{\mathbb{C}}\{U_T^n e_i:n\in\mathbb{Z}\}}^{\,H},
\end{align*}
and these subspaces are mutually orthogonal with
\begin{align*}
H=\bigoplus_{i\in I}K_i.
\end{align*}
The theorem also identifies each summand with a multiplication model. Define
\begin{align*}
\zeta:\mathbb{T}&\to\mathbb{C}\\
z&\mapsto z
\end{align*}
and
\begin{align*}
M_{\zeta,i}:L^2(\mathbb{T},\nu_i)&\to L^2(\mathbb{T},\nu_i)\\
\varphi&\mapsto \zeta\varphi.
\end{align*}
Since $|\zeta(z)|=1$ for every $z\in\mathbb{T}$, this multiplication operator is unitary. Hahn-Hellinger supplies a unitary map
\begin{align*}
W_i:K_i&\to L^2(\mathbb{T},\nu_i)
\end{align*}
with
\begin{align*}
W_i(U_T|_{K_i})=M_{\zeta,i}W_i.
\end{align*}
Finally, Hahn-Hellinger orders the measures by absolute continuity, meaning $\nu_i\gg\nu_{i+1}$ whenever both indices occur.
[/guided]
[/step]
[step:Rename the cyclic data and obtain the stated decomposition]
For each $i\in I$, define
\begin{align*}
f_i:=e_i\in H
\qquad\text{and}\qquad
\sigma_i:=\nu_i:\mathcal{B}(\mathbb{T})\to[0,\infty).
\end{align*}
Then
\begin{align*}
H_{f_i}:=\overline{\operatorname{span}_{\mathbb{C}}\{U_T^n f_i:n\in\mathbb{Z}\}}^{\,H}
\end{align*}
equals $K_i$, so
\begin{align*}
H=\bigoplus_{i\in I}H_{f_i}.
\end{align*}
The measure $\sigma_i$ is finite and positive, and $U_T|_{H_{f_i}}$ is unitarily equivalent to $M_{\zeta,i}$ on $L^2(\mathbb{T},\sigma_i)$. The Hahn-Hellinger ordering gives
\begin{align*}
\sigma_i\gg\sigma_{i+1}
\end{align*}
whenever both indices occur.
If $I=\{1,\dots,N\}$ is finite, define $f_i:=0\in H$ and $\sigma_i:=0$ for every $i>N$. Then $H_{f_i}=\{0\}$ for $i>N$, the orthogonal direct sum is unchanged, and the absolute-continuity chain continues because every finite positive measure dominates the zero measure. Thus the required sequence $f_1,f_2,\dots$ and finite positive measures $\sigma_1,\sigma_2,\dots$ exist.
[/step]
