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Let $\pi: (X,T) \to (Y,S)$ be a factor map of minimal topological dynamical systems, and let $U \subseteq X$ be open, non-empty.

Question: Does there exist a $T$-invariant, Borel probability measure $\mu$ on $X$, an $\eta > 0$, and a non-empty, open set $V \subseteq Y$ so that $\mathbb{E}(1_U \mid Y)$ is bounded from below by $\eta$ on $V$?

Here are the definitions I'm using. The maps $T$ and $S$ are homeomorphisms of the compact metric spaces $X$ and $Y$, respectively. The map $\pi$ is a continuous surjection of $X$ onto $Y$ that intertwines $T$ and $S$: $\pi \circ T = S \circ \pi$. That the system $(X,T)$ is minimal means $\emptyset$ and $X$ are the only closed, $T$-invariant subsets of $X$; minimality of the system $(Y,S)$ follows from that of $(X,T)$ by an easy exercise. A Borel probability measure $\mu$ on $X$ is $T$-invariant if for all Borel sets $B \subseteq X$, $\mu(T^{-1}B) = \mu(B)$. Given a Borel probability measure $\mu$ on $X$, the function $\mathbb{E}(1_U \mid Y)$ is the conditional expectation of $1_U$, the indicator function of $U$, with respect to the pre-image under $\pi$ of the Borel $\sigma$-algebra on $Y$. While technically a function in $L^\infty(X,\mu)$, it is interpreted in the question as being a function in $L^\infty(Y, \pi_*\mu)$, where $\pi_*\mu$ is the push-forward of $\mu$ by $\pi$.

The question can be rephrased in terms of the disintegration of $\mu$ with respect to $\pi$.

Question (rephrased): Does there exist a $T$-invariant, Borel probability measure $\mu$ on $X$ with the following property? Let $\nu = \pi_*\mu$, and let $\mu = \int \mu_y \ d\nu$ be the disintegration of $\mu$ with respect to $\pi$. There is an $\eta > 0$ and a non-empty, open set $V \subseteq Y$ such that for $\nu$-a.e. $y \in V$, $\mu_y(U) > \eta$.

I'm trying to understand how much mass the fiber measures $\mu_y$ give to the open set $U$. Here are a couple of blunt first attempts to answer this question.

Fact 1: There exists a non-empty, open set $V \subseteq Y$ so that for all $y \in V$, $\pi^{-1}(\{y\}) \cap U \neq \emptyset$. (More generally, factor maps of minimal systems are semiopen.)

Proof: Let $F \subseteq U$ be a closed set with non-empty interior. Because $(X,T)$ is minimal and $F$ has non-empty interior, there exists $\ell \in \mathbb{N}$ so that $F \cup T^{-1}F \cup \cdots \cup T^{-\ell}F = X$. Let $H = \pi F$, and note that $H$ is a closed subset of $Y$. Applying $\pi$ to both sides, we see that $H \cup S^{-1}H \cup \cdots \cup S^{-\ell}H = Y$. By the Baire Category Theorem, some $S^{-i} H$ has non-empty interior. Since $S$ is a homeomorphism, this implies that $H$, and hence $\pi U$, has non-empty interior. Now any non-empty, open subset $V$ of the interior of $\pi U$ has the desired property.$\Box$

Fact 2: For any $T$-invariant, Borel probability measure $\mu$ on $X$, there exists an $\eta > 0$ and sets $W \subseteq V \subseteq Y$ with $V$ non-empty, open and $W$ a ``$\pi_*\mu$-dense'' subset of $V$ (meaning that for all non-empty, open $V' \subseteq V$, $\pi_*\mu(V' \cap W) > 0$) such that $\mathbb{E}(1_U \mid Y)$ is bounded from below by $\eta$ on $W$.

Proof: Let $\mu$ be a $T$-invariant, Borel probability measure on $X$, and put $\nu = \pi_* \mu$. Since $(X,T)$ is minimal and $U$ is open, there exists $\ell \in \mathbb{N}$ so that $U \cup T^{-1}U \cup \cdots \cup T^{-\ell}U = X$. Put $\eta = (2 \ell)^{-1}$. Let $B = \{y \in Y \mid \mathbb{E}(1_U \mid Y)(y) > \eta\}$. The support of $B$ is $\text{supp}(B) = Y \setminus \cup \big\{Q \subseteq Y \mid \text{$Q$ open subset of $Y$ with $\nu(B \cap Q) = 0$}\big\}.$ We need only to show that $\text{supp}(B)$ has non-empty interior. (Indeed, if $V$ is non-empty, open and contained in the interior of $\text{supp}(B)$, then $W = B \cap V$ satisfies the conclusions of the fact.) Because $U \cup T^{-1}U \cup \cdots \cup T^{-\ell}U = X$, we have $B \cup S^{-1}B \cup \cdots \cup S^{-\ell}B \stackrel{\nu}{=} Y$ (up to a set of $\nu$ measure $0$). It follows that $\text{supp}(B) \cup S^{-1}\text{supp}(B) \cup \cdots \cup S^{-\ell}\text{supp}(B) = Y$. Since $\text{supp}(B)$ is closed, the Baire Category Theorem gives that some $S^{-i} \text{supp}(B)$ has non-empty interior. Since $S$ is a homeomorphism, this implies that $\text{supp}(B)$ has non-empty interior.$\Box$

Both of these facts depend on the Baire Category Theorem. Perhaps there's some analogue of the BCT in the measurable category that would help here. I would be interested in answers to the question in special cases, too. For example, what if we assume $(X,T)$ is uniquely ergodic or that the factor map $\pi$ is open?

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