# 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.

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Why is the series in the end a finite sum ? The next term is going to be zero but what about higher order terms ? –  user69870 Mar 28 '13 at 11:35

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

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But we first choose $A$, and then $N$, so I don't really understand what you mean? Or do you pick $A = T^{-n} B$, then get an $N$ and use that one? –  Jonas Teuwen Mar 2 '11 at 20:38
@Jonas You are right. I will delete my post. –  Did Mar 3 '11 at 6:56
I actually think this is quite useful +1. –  Jonas Teuwen Mar 5 '11 at 23:23
Thanks. Nevertheless, this would probably be better as a comment, no? –  Did Jan 16 at 7:01

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: $(\mathcal{S},\Delta) \cong (\{\chi_A:A \in \mathcal{B} \},\| \cdot \|_{L^1})$.

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