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.

Hint: what happens if $A=T^{-N}B$?
