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 choice of $Z$ being made independently of $\delta$. Instead, let's choose $Z$ in a more refined way. Recall that $\pi^{-1}(B) \cap B_n \to \pi^{-1}(B)$ (mod $\nu$). Since $\nu(\pi^{-1}(B) \setminus B_n) \to 0$ as $n \to \infty$, we can choose $m$ large enough such that:

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

Let's take $Z = B$. This choice clearly satisfies $Z \subseteq B$ and $\mu(B \setminus Z) = 0 < \varepsilon$.

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

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

For any $E \in \mathcal{A}$, we have:

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

Therefore:

\begin{align*} \nu((\pi^{-1}(Z) \cap B_m) \cap \pi^{-1}(Z \cap E)) &= \nu(\pi^{-1}(Z \cap E) \cap B_m) \\ &= \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*}

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 $E \in \mathcal{A}$:

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

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 $$\nu\left( (\pi^{-1}(Z) \cap B_m) \cap \pi^{-1}(Z \cap E) \right) \ge (1-\delta) \cdot c \mu(Z \cap E)$$ for a suitable choice of $Z \subseteq B$ with $\mu(B \setminus Z) < \varepsilon$. As you correctly noted, this is equivalent to showing $$\delta \cdot c \mu(Z \cap E) \ge \nu(\pi^{-1}(Z \cap E) \setminus B_m).$$ 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, $$\nu(\pi^{-1}(Z \cap E) \setminus B_m) \le \nu(\pi^{-1}(B) \cap B_m^c) < \varepsilon \cdot c \mu(Z).$$ We want this to be less than or equal to $\delta \cdot c \mu(Z \cap E)$, i.e., we want $$\varepsilon \cdot c \mu(Z) \le \delta \cdot c \mu(Z \cap E).$$ However, this is not generally true, as $\mu(Z \cap E)$ can be arbitrarily small.

The issue arises from the choice of $Z$ being made independently of $\delta$. Instead, let's choose $Z$ in a more refined way. Recall that $\pi^{-1}(B) \cap B_n \to \pi^{-1}(B) \pmod{\nu}$. Let $A_n = \pi^{-1}(B) \setminus (\pi^{-1}(B) \cap B_n) = \pi^{-1}(B) \cap B_n^c$. Then $\nu(A_n) \to 0$ as $n \to \infty$.

Choose $m$ such that $\nu(A_m) < \delta c \mu(B)$. Now, define a function $f(x) = \nu(\pi^{-1}(x) \cap B_m^c)$. Then $\int_B f d\mu = \nu(\pi^{-1}(B) \cap B_m^c) < \delta c \mu(B)$. Let $Z = \{x \in B : f(x) < \delta c\}$. Then $$\int_B f d\mu = \int_Z f d\mu + \int_{B \setminus Z} f d\mu.$$ Also, $$\int_{B \setminus Z} f d\mu \ge \delta c \mu(B \setminus Z).$$ Thus, $$\delta c \mu(B) > \nu(\pi^{-1}(B) \cap B_m^c) \ge \int_{B \setminus Z} f d\mu \ge \delta c \mu(B \setminus Z),$$ which implies $\mu(B) > \mu(B \setminus Z)$. So, $\mu(Z) > 0$.

Now let $E \in \mathcal{A}$. We want to show $$\nu((\pi^{-1}(Z) \cap B_m) \cap \pi^{-1}(Z \cap E)) \ge (1 - \delta) c \mu(Z \cap E).$$ We have $$\nu(\pi^{-1}(Z \cap E) \setminus B_m) = \nu(\pi^{-1}(Z \cap E) \cap B_m^c) = \int_{Z \cap E} f d\mu < \int_{Z \cap E} \delta c d\mu = \delta c \mu(Z \cap E).$$ Therefore, \begin{align*}\nu((\pi^{-1}(Z) \cap B_m) \cap \pi^{-1}(Z \cap E)) &= \nu(\pi^{-1}(Z \cap E) \cap B_m) \\ &= \nu(\pi^{-1}(Z \cap E)) - \nu(\pi^{-1}(Z \cap E) \cap B_m^c) \\ &= c \mu(Z \cap E) - \nu(\pi^{-1}(Z \cap E) \cap B_m^c) \\ &> c \mu(Z \cap E) - \delta c \mu(Z \cap E) \\ &= (1 - \delta) c \mu(Z \cap E).\end{align*}

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 choice of $Z$ being made independently of $\delta$. Instead, let's choose $Z$ in a more refined way. Recall that $\pi^{-1}(B) \cap B_n \to \pi^{-1}(B)$ (mod $\nu$). Since $\nu(\pi^{-1}(B) \setminus B_n) \to 0$ as $n \to \infty$, we can choose $m$ large enough such that:

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

Let's take $Z = B$. This choice clearly satisfies $Z \subseteq B$ and $\mu(B \setminus Z) = 0 < \varepsilon$.

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

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

For any $E \in \mathcal{A}$, we have:

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

Therefore:

\begin{align*} \nu((\pi^{-1}(Z) \cap B_m) \cap \pi^{-1}(Z \cap E)) &= \nu(\pi^{-1}(Z \cap E) \cap B_m) \\ &= \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*}