Stronger than strong-mixing I have the following exercise:
"Show that if a measure-preserving system $(X, \mathcal B, \mu, T)$ has the property that for any $A,B \in \mathcal B$ there exists $N$ such that
$$\mu(A \cap T^{-n} B) = \mu(A)\mu(B)$$
for all $n \geq N$, then $\mu(A) = 0$ or $1$ for all $A \in \mathcal B$"
Now the back of the book states that I should fix $B$ with $0 < \mu(B) < 1$ and then find $A$ using the Baire Category Theorem. Edit: I'm now pretty sure that this "$B$" is what "$A$" is in the required result.
Edit: This stopped being homework so I removed the tag. Any approach would be nice. I have some idea where I approximate $A$ with $T^{-n} B^C$ where the $n$ will be an increasing sequence and then taking the $\limsup$ of the sequence. I'm not sure if it is correct. I will add it later on.
My attempt after @Did's comment: "proof":
First pick $B$ with $0 < \mu(B) < 1$. Then set $A_0 = B^C$ and
determine the smallest $N_0$ such that
$$\mu(A_0 \cap T^{-N_0} B) = \mu(A_0) \mu(B)$$
Continue like this and set
$$A_k = T^{-N_{k - 1}} B^C$$
Now we note that the $N_k$ are a strictly increasing sequence, since
suppose not, say $N_{k} \leq N_{k - 1}$ then
$$\mu \left ( T^{-N_{k - 1}} B^C \cap T^{-N_{k - 1}} B \right ) = 0 \neq
    \mu(B^C) \mu(B) > 0$$
Set $A = \limsup_n A_n$, then note that
\begin{align}
\sum_n \mu(A_n) = \sum_n \mu(B^C) = \infty
\end{align}
So $\mu(A) = 1$, by the Borel-Cantelli lemma. Well, not yet, because
we are also required to show that the events are independent, so it is
sufficient to show that $\mu(A_{k + 1} \cap A_k) = \mu(A_{k + 1
})\mu(A_k)$
We know that $\mu(T^{N_k} B^C \cap T^{N_{k + 1}} B) = \mu(B^C)\mu(B)$. So
does a similar result now hold if we replace $B$ with $B^C$ in the
second part?
Note:
\begin{align}
\mu(A \cap T^{-M} B^C) &= \mu(A \setminus (T^{-M} B \cap A))\\\
&= \mu(A) - \mu(A)\mu(B) \\\
&= \mu(A) - \mu(A \cap T^{-M} B)\\\
&= \mu(A)\mu(B^C)
\end{align}
which is what was required.
For this $A$ and $B$ we can find an $M$ and a $k$ such that $N_k \leq
M < N_{k + 1}$. Now note that $\limsup_n A \cap T^{-M} B = \limsup_n
(A \cap T^{-M} B)$.
Further,
$$\sum_n \mu(A_n \cap T^{-N_{k +1}}) = \mu(A_0 \cap T^{-N_{k + 1}}) +
\ldots + \mu(A_{k + 1} \cap T^{N_{k + 1}}) < \infty$$
So again by the Borel-Cantelli Lemma we have 
$\mu(\limsup_n A_n \cap T^{-M} B) = 0$.
Thus we get
$$\mu(A) \mu(B) = \mu(B) = \mu(A \cap T^{-M} B) = 0$$
which is a contradiction since $\mu(B) > 0$. So, such $B$'s
violate the condition.
Added: Actually the metric on the space of events $d(A,B) = \mu(A \Delta B)$ can work together with Baire's Category Theorem.
 A: let $\mathcal{B}$ be the space of events, then $\mathcal{S}= \mathcal{B}$ (mod $\mu$) equipped with the metric $d(A,B)=\mu(A\Delta B)$ is a complete metric space, (where if $(A_n)$ is a koshi sequence, then for any subsequence $(A_{n_k})$ with $\sum_k \mu(A_{n_k}\Delta A_{n_{k+1}})<\infty$, $A_n\to \limsup_kA_{n_k}$).
let $B \in \mathcal{S}$ such that $0<\mu(B)<1$, and define 
$\Lambda_N = \{A \in \mathcal{S}:\mu(A\cap T^{-n}B)=\mu(A)\mu(B)\space \space(\forall n\geq N) \}$
we shall prove that $\Lambda_N$ is closed and meagre for all $N \in \mathbb{N}$:
if $(A_m) \subset \Lambda_N$, and $A_m \to A$, then for any $C \in \mathcal{S}$, $A_m\cap C \to A \cap C$, hence for any $n \geq N$ $\mu(A \cap T^{-n}B)=\lim_m\mu(A_n \cap T^{-n}B)=\lim_m \mu(A_m) \mu(B)=\mu(A) \mu(B)$, thus $A \in \Lambda_N$, so $\Lambda_N$ is closed.      (observe that $\mu (\lim_m A_m)= \lim_m \mu(A_m)$ in $(\mathcal{S},\Delta)$) 
for any $A \in \Lambda_N, \mu (A) \neq 1$ and any $\epsilon >0$, take $C \subset A^c \cap T^{-N}B^c$ such that $0<\mu(C)< \epsilon$, then $\mu((A\sqcup C) \Delta A)= \mu (C)< \epsilon$.
(observe that $\mu(A^{c} \cap T^{-N}B^c)= \mu(A^c) \mu(B^c) \neq0$, and since there exist k and  a sequence $N_1 ... N_k$ such that $\mu((A^{c} \cap T^{-N}B^c) \cap T^{-N_1}B \cap ... \cap T^{-N_k}B)= \mu(A^c) \mu(B^c) \mu(B)^k< \epsilon$, we can find such $C$)
but $\mu((A \sqcup C) \cap T^{-N}B)=\mu(A  \cap T^{-N}B)=\mu(A) \mu(B) \neq \mu(A \sqcup C)\mu(B)$, hence $A\sqcup C \notin \Lambda_N$, so $\Lambda_N$ is meagre.
now, by baire category theorem, $\bigcup_{N \in \mathbb{N}} \Lambda_N \subsetneq \mathcal{S}$, hence there must be some $A \in \mathcal{S} - (\bigcup_{N \in \mathbb{N}} \Lambda_N)$ such that for any $N$, there exists $n \geq N$ for which $\mu(A \cap T^{-n}B) \neq \mu(A) \mu(B)$.
Remark: $(\mathcal{S},\Delta) \cong (\{\chi_A:A \in \mathcal{B} \},\| \cdot \|_{L^1})$. 
A: Hint: what happens if $A=T^{-N}B$?
