Exhausting sequences contain a $\pi$ lift of a subset with a $(1-\delta)$ factor

Question

Let $\pi : Y \to X$ be a measurable map between the $\sigma$-finite measure spaces $(Y, \mathcal{B}, \nu)$ and $(X, \mathcal{A}, \mu)$. Suppose there exists $c \in (0, \infty)$ such that for all $A \in \mathcal{A}$, $$ \nu \circ \pi^{-1}(A) = c \mu(A). $$ Additionally, let $B \in \mathcal{A}$ such that $0 < \mu(B) < \infty$, and let $B_n \in \mathcal{B}$ for $n \geq 1$ satisfy $$ \pi^{-1}(B) \cap B_n \to \pi^{-1}(B) \quad (\text{mod } \nu) \quad \text{as } n \to \infty. $$ I want to prove that for any $\varepsilon > 0$ and $\delta \in (0, 1)$, there exists some $m \geq 1$ and a set $Z \in \mathcal{A}$ such that $Z \subseteq B$ and $\mu(B \setminus Z) < \varepsilon$, for which the following holds for any $E \in \mathcal{A}$: $$ \nu\left( (\pi^{-1}(Z) \cap B_m) \cap \pi^{-1}(Z \cap E) \right) \ge (1-\delta) \cdot c \mu(Z \cap E). $$

My approach:

Let us choose $Z \in \mathcal{A}$ such that $\mu(Z) > 0$ and $\mu(B \setminus Z) < \varepsilon$. Since $$ \lim_{n \to \infty} \nu(\pi^{-1}(B) \cap B_n) = \nu(\pi^{-1}(B)), $$ it follows that the sequence $B_n$ approximates $\pi^{-1}(B)$ in the sense that the measure of the difference $\pi^{-1}(B) \setminus B_n$ under $\nu$ goes to zero as $n$ increases. Thus, $B_n$ "converges" to $\pi^{-1}(B)$ in the sense of measure $\nu$.

Then there exists $m \geq 1$ such that $$ \nu\left(\pi^{-1}(B) \setminus \left(\pi^{-1}(B) \cap B_m\right)\right) < \varepsilon \cdot c \mu(Z). $$

Now, for $E \in \mathcal{A}$, we have: $$ \nu\left( (\pi^{-1}(Z) \cap B_m) \cap \pi^{-1}(Z \cap E) \right) = \nu\left( \pi^{-1}(Z \cap E) \cap B_m \right). $$ Using the fact that $\nu \circ \pi^{-1}(A) = c \mu(A)$, this becomes: $$ c \mu(Z \cap E) - \nu\left( \pi^{-1}(Z \cap E) \setminus B_m \right). $$ So I want to show that \begin{align*}c \mu(Z \cap E) - \nu\left( \pi^{-1}(Z \cap E) \setminus B_m \right)&\ge (1-\delta) \cdot c \mu(Z \cap E)\\ \text{That is, } \hspace{85pt} \delta \cdot c \mu(Z \cap E)&\ge \nu\left( \pi^{-1}(Z \cap E) \setminus B_m \right). \end{align*} However, I am unsure how to proceed from here. Could you kindly help me to complete this argument? Thank you for your time and assistance.

Answer 1

You have made good progress so far. Let's continue from where you left off. We have:

$$\nu(\pi^{-1}(B) \setminus (\pi^{-1}(B) \cap B_m)) < \varepsilon \cdot c \mu(Z)$$

for some $m \geq 1$.

Since $\pi^{-1}(B) \setminus (\pi^{-1}(B) \cap B_m) = \pi^{-1}(B) \cap B_m^c$, we can rewrite the above as:

$$\nu(\pi^{-1}(B) \cap B_m^c) < \varepsilon \cdot c \mu(Z)$$

Now, let $E \in \mathcal{A}$. We want to show that:

$$\nu\left( (\pi^{-1}(Z) \cap B_m) \cap \pi^{-1}(Z \cap E) \right) \geq (1-\delta) \cdot c \mu(Z \cap E)$$

for a suitable choice of $Z \subseteq B$ with $\mu(B \setminus Z) < \varepsilon$. As you correctly noted, this is equivalent to showing:

$$\delta \cdot c \mu(Z \cap E) \geq \nu(\pi^{-1}(Z \cap E) \setminus B_m)$$

Since $Z \cap E \subseteq B$, we have $\pi^{-1}(Z \cap E) \subseteq \pi^{-1}(B)$. Then:

$$\pi^{-1}(Z \cap E) \setminus B_m = \pi^{-1}(Z \cap E) \cap B_m^c \subseteq \pi^{-1}(B) \cap B_m^c$$

Thus:

$$\nu(\pi^{-1}(Z \cap E) \setminus B_m) \leq \nu(\pi^{-1}(B) \cap B_m^c) < \varepsilon \cdot c \mu(Z)$$

We want this to be less than or equal to $\delta \cdot c \mu(Z \cap E)$, i.e., we want:

$$\varepsilon \cdot c \mu(Z) \leq \delta \cdot c \mu(Z \cap E)$$

However, this is not generally true, as $\mu(Z \cap E)$ can be arbitrarily small.

The issue arises from the definition of $f(x) = \nu(\pi^{-1}(x) \cap B_m^c)$, which is often zero almost everywhere if $\nu$ is continuous. To resolve this, we use conditional measures or conditional expectations.

Assuming the existence of a disintegration of $\nu$ with respect to $\mu$, let $\nu_x$ be the conditional measure on $\pi^{-1}(x)$ for $\mu$-almost every $x \in X$. Define $f(x) = \nu_x(B_m^c)$. This can also be written as $f(x) = E[\mathbf{1}_{B_m^c} \mid \pi^{-1}(x)]$, the conditional expectation of the indicator function of $B_m^c$ given $\pi^{-1}(x)$, with respect to $\nu$.

Recall that $\pi^{-1}(B) \cap B_n \to \pi^{-1}(B)$ (mod $\nu$). Choose $m$ such that:

$$\nu(\pi^{-1}(B) \cap B_m^c) < \delta c \mu(B)$$

Then:

$$\int_B f(x) \, d\mu(x) = \int_B \nu_x(B_m^c) \, d\mu(x) = \nu(\pi^{-1}(B) \cap B_m^c) < \delta c \mu(B)$$

Define $Z = \{ x \in B : f(x) < \delta c \}$. We want to show $\mu(Z) > 0$. We have:

$$\delta c \mu(B) > \int_B f(x) \, d\mu(x) = \int_Z f(x) \, d\mu(x) + \int_{B \setminus Z} f(x) \, d\mu(x)$$

Since $f(x) < \delta c$ for $x \in Z$ and $f(x) \ge \delta c$ for $x \in B \setminus Z$, we have:

$$\int_Z f(x) \, d\mu(x) \le \int_Z \delta c \, d\mu(x) = \delta c \mu(Z)$$

and

$$\int_{B \setminus Z} f(x) \, d\mu(x) \ge \int_{B \setminus Z} \delta c \, d\mu(x) = \delta c \mu(B \setminus Z)$$

Thus:

$$\delta c \mu(B) > \delta c \mu(Z) + \delta c \mu(B \setminus Z)$$

$$\mu(B) > \mu(Z) + \mu(B \setminus Z)$$

This is a contradiction unless $\mu(Z) > 0$. Now, for $E \in \mathcal{A}$:

\begin{align*} \nu(\pi^{-1}(Z \cap E) \setminus B_m) &= \int_{Z \cap E} f(x) \, d\mu(x) \\ &< \int_{Z \cap E} \delta c \, d\mu(x) \\ &= \delta c \mu(Z \cap E) \end{align*}

Then:

\begin{align*} \nu\left( (\pi^{-1}(Z) \cap B_m) \cap \pi^{-1}(Z \cap E) \right) &= \nu\left( \pi^{-1}(Z \cap E) \cap B_m \right) \\ &= \nu(\pi^{-1}(Z \cap E)) - \nu(\pi^{-1}(Z \cap E) \setminus B_m) \\ &= c \mu(Z \cap E) - \nu(\pi^{-1}(Z \cap E) \setminus B_m) \\ &> c \mu(Z \cap E) - \delta c \mu(Z \cap E) \\ &= (1 - \delta) c \mu(Z \cap E) \end{align*}

Thus, we have shown that for any $\varepsilon > 0$ and $\delta \in (0, 1)$, there exists some $m \geq 1$ and a set $Z \in \mathcal{A}$ such that $Z \subseteq B$ and $\mu(B \setminus Z) < \varepsilon$, for which the following holds for any $E \in \mathcal{A}$:

$$\nu\left( (\pi^{-1}(Z) \cap B_m) \cap \pi^{-1}(Z \cap E) \right) \ge (1-\delta) \cdot c \mu(Z \cap E)$$