This question comes from my attempt to understand theorem $7.21$ in E-W. This concerns the dichotomy between relatively weak-mixing extensions and compact extensions. I cannot understand the proof as I do not understand these concepts. I will sketch my understanding of the proof until the part I'm certain I don't understand.
All systems are taken to be ergodic, invertible, and Borel.
Definition 1: An extension $\psi:(X,\mathscr{B},\mu,T) \to (Y,\mathscr{A},\nu,S)$ is said to be relatively weak-mixing if the system $(X\times X,\mathscr{B}\otimes\mathscr{B},\bar{\mu},T\times T)$ is ergodic. Here $\bar{\mu}$ is the relatively independent join over $Y$(See E-W definition $6.15$).
Definition 2: Again, let $\psi:(X,\mathscr{B},\mu,T) \to (Y,\mathscr{A},\nu,S)$ be an extension. Let $\mu^{\psi^{-1}\mathscr{A}}_x$ be the conditional measures on $(X,\mathscr{B})$. A function $f\in L^2(X,\mathscr{B},\mu)$ is almost-periodic $(AP)$ with respect to $Y$ if given $\varepsilon>0$, there exist $g_1,\dots,g_r\in L^2(X,\mathscr{B},\mu)$ such that $\forall n \in \mathbb{Z}$, $\nu$-a.e $y$, $$\min_{i}\|U_T^nf-g_i\|_{L^2(\mu_y^{\mathscr{A}})}<\varepsilon.$$ Here $U_Tf:=f\circ T$ and $\mu_y^{\mathscr{A}}$ are the measures on $(X,\mathscr{B})$ obtained by completing the conditional measure diagram: $\require{AMScd}$ \begin{CD} X @>\mu^{\psi^{-1}\mathscr{A}}_.>> M\left(\bar{X}\right)\\ @V{\psi}VV @| \\ Y @>\mu^{\mathscr{A}}_.>> M\left(\bar{X}\right) \end{CD}
Definition 3: The extension $\psi:(X,\mathscr{B},\mu,T) \to (Y,\mathscr{A},\nu,S)$ is said to be a compact extension if the set of $AP$ (with respect to $Y$) functions in $L^2(X,\mathscr{B},\mu)$ is dense.
Theorem 7.21: If the extension $\psi:(X,\mathscr{B},\mu,T) \to (Y,\mathscr{A},\nu,S)$ is not relatively weak-mixing, there exists a non-trivial intermediate factor $X\to X^* \to Y$ with the property that $X^* \to Y$ is a compact extension.
Sketch of proof: (For details see E-W page $200$ onwards) By the hypothesis, there exists a non-constant $H \in \mathscr{L}^\infty(X\times X,\bar{\mu})$ invariant under $T\times T$ ($\mathscr{L}$ denotes the fact that $H$ is bounded and not just essentially-bounded). For $\phi \in L^2(X,\mathscr{B},\mu)$, define $$H*\phi(x) = \int H(x,x')\phi(x')d\mu^{\psi^{-1}\mathscr{A}}_x(x').$$
One can show this formula defines a bounded operator $L^2(X,\mathscr{B},\mu)\to L^2(X,\mathscr{B},\mu).$ And, a priori, this formula defines a compact operator $L^2(X,\mathscr{B},\mu^{\psi^{-1}\mathscr{A}}_x) \to L^2(X,\mathscr{B},\mu^{\psi^{-1}\mathscr{A}}_x)$.
Using this and the additional fact that $U_T(H*\phi)=H*(U_T\phi)$, one shows that $\lbrace H*\phi : \phi \in \mathscr{L}^{\infty}(X,\mathscr{B},\mu)\rbrace$ is an $AP$ (wrt. $Y$) family in $L^{\infty}(X,\mathscr{B},\mu)$. Moreover, one can show that this family contains functions which are not $\psi^{-1}\mathscr{A}$-measureable.
Then if we define $\mathscr{F}=\lbrace f\in L^{\infty}(X,\mathscr{B},\mu): f \text{ is } AP \text{ wrt. } Y\rbrace $ and $\mathscr{B}^*$ to be the smallest $\sigma$-algebra making the functions in $\mathscr{F}$ measurable, one can show that $\mathscr{F} \subset L^2(X,\mathscr{B}^*,\mu)$ is dense.
Here's the part I don't understand: It is then claimed that the $T$-invariant $\sigma$-algebra $\mathscr{B}^* (\subset \mathscr{B}$) gives rise to the intermediate, non-trivial, compact extension. I assume that this is done by completing the following system of extensions from $M\left(\bar{X}\right)$ to $Y$: \begin{CD} X @>\psi>> Y\\ @V\mu^{\mathscr{B}^*}_.VV \\ M\left(\bar{X}\right) \end{CD}
The $T$-invariance of $\mathscr{B}^*$ shows that $\mu^{\mathscr{B}^*}_.$ gives an extension. And I guess the density of $\mathscr{F}$ will be used to show $M\left( \bar{X}\right)$ is a compact extension of $Y$.
Question: To complete the above diagram, don't we need to show that $\psi^{-1}\mathscr{A}\subset \mathscr{B}^*$? Is this obvious? It is not explicitly addressed in the text.
Thanks for reading, all help is appreciated.
Edit: As John Griesmer points out, $\psi^{-1}\mathscr{A}\subset\mathscr{B}^*$ would follow if one could show every characteristic function $1_{\psi^{-1}A}$ on $X$ was AP with respect to Y. I don't quite see why such a statement should be true though.
Take the function $1_{A}\circ\psi$. I want to find functions $g_i\in L^2(X,\mathscr{B},\mu)$ and estimate $$\int|1_{A}(\psi(T^nx))-g_i(x)|^2d\mu_{x'}^{\psi^{-1}\mathscr{A}}(x) = \int|1_{A}(S^n(\psi x))-g_i(x)|^2d\mu_{x'}^{\psi^{-1}\mathscr{A}}(x)$$
as $x'$ varies in $X$. Now if the $g_i$'s happened to be functions on $Y$, I could rewrite the last integral as $$\int|1_{A}(S^ny)-g_i(y)|^2d\nu_{\psi x'}^{\mathscr{A}}(x) = E\left(|U_S^n1_A-g_i|^2:\mathscr{A},\nu\right)(\psi(x')) = |U^n_S1_A - g_i|^2(\psi(x'))$$ by the properties of the conditional expectation operator. So I guess, I'm left trying to play around with the equation $$\min_i|U_s^n1_A(y) - g_i(y)|^2 < \varepsilon, $$ and hope that it holds for almost every $y$ and for all $n$. Am I on the right track? what would be sensible choices for the $g_i$? Apologies if I'm missing something obvious; I only understand these things formally and have no intuition for what's actually going on.
You are correct: you need to show $\psi^{-1}\mathscr A \subset \mathscr B^*$ (modulo null sets). Brief outline: if $A$ is a measurable subset of $Y$, then it should follow quickly from the definition that the characteristic function of $\psi^{−1}(A)$ is $AP$ wrt $Y$ (I don't have their exact definition in front of me, so I won't try to fill in the details here). Then the definition of $\mathscr F$ and $\mathscr B^*$ imply $A\in \mathscr B^*$.