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Consider the probability space $(X,\mathcal{B},\mu)$ where $X$ is a compact metric space and $\mu(A) > 0$ for all $A$ non-empty. Let $T:X \to X$ be a continuous ergodic measure-preserving transformation.

By a theorem in my course, if $A,B \in \mathcal{B}$ are open and non-empty, then we have that $$ \lim_{n \to \infty} \frac{1}{n} \sum_{k = 0}^{n - 1} \mu(T^{-k}A \cap B) = \mu(A)\mu(B) > 0 $$ thus there exists $m \in \mathbb{N}$ such that $\mu(T^{-m}A \cap B) > 0$ and so $T^{-m} A \cap B \neq \emptyset$. Now I am trying to show that $T$ is transitive therefore need that there exists $l \in \mathbb{N}$ such that $T^{l} A \cap B \neq \emptyset$. This is close to what I have derived so far but I am stuck trying to understand how to get to the final result.

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You have that for a $m$, $\mu(T^{-m}A \cap B) > 0$.

By hypothesis on your measure a set has zero measure if and only if it is empty or to say it an other way a set has non-zero measure if and only if it is non empty.

As the set $T^{-m}A \cap B$ has strictly positive measure then it is non empty that is $T^{-m}A \cap B \ne \emptyset$. Then taking $T^m(T^{-m}A \cap B)=A \cap T^m(B)$, you find that $T^{m}B \cap A \ne \emptyset$.

Changing the role of $B$ and $A$, you find what you want.

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