Writing a Hilbert C*-submodule of L2(X)L^2(X) as an integral sum over Hilbert subundles

In Ergodic Theory, some (though not all) presentations of compact extensions use Hilbert bundles.
Given $(X,\mathcal{B},\mu,T)$ and a sub $\sigma$-algebra $\mathcal{G}\subseteq\mathcal{A}$ one has an associated factor map $\pi:(X,\mu,T)\to (Y,\nu,T)$ (where $\pi(x)=\pi(x')$ iff ($x\in A\leftrightarrow x'\in A$) for every $A\in\mathcal{G}$) and a decomposition $\mu=\int\mu_yd\nu$ (where $\nu=\pi_*(\mu)$).

Then one can informally think of the space of functions in $L^2(X,\mu)$ as an “integral” over the spaces of functions $H_y:=L^2(X,\mu_y)$ for $\nu$-a.e. $y$. Moreover, the union $\coprod_yH_y$ defines a Hilbert bundle over $Y$ (of course, $Y$ a priori has no topological structure, so this is not strictly precise).
This is the approach taken in these notes by Yuri Lima and in Ch. 9 of Glasner’s book, for instance.

One then considers $L^\infty(Y)$-submodules (i.e. closed $L^\infty(Y,\nu)$-invariant subspaces) of $L^2(X,\mu)$. Now, here comes my problem. In proving many of the properties that one wants to prove for any such submodule $M$ (I won’t go into these details or into compact extensions themselves) the authors usually consider the collection of $\{M_y\}_{y\in Y}$, which are thought of as closed subspaces of the $H_y$ defined above (i.e. $\{M_y\}_{y\in Y}$ is a subbundle of the Hilbert bundle). If $f\in M$, then $f_y$ is in $M_y$, where $f_y$ is just $f$ “seen” as an element of $L^2(X,\mu_y)$ instead of $L^2(X,\mu)$ (warning: this is not the conditional expectation $\mu_y(f))$. For instance, if we fix topological models such that $X=Y\times U$ then $f_y=f(y,\cdot)$ The problem is that $f$ is not a function, it is an equivalence class of functions. Certainly, for a fixed $f\in L^2(Y\times U)$ the function $f_y=f(y,\cdot)$ is defined for $\nu$-a.e. $y$. But that doesn’t mean we can actually define a map $M\mapsto \{M_y\}_{\nu\text{-a.e.} y}$ for any closed subspace $M$, since such subspaces have uncountably many $f$‘s.

Question: Is there a way around this technical problem?

Answer

Attribution
Source : Link , Question Author : Moritz , Answer Author : Community