I am new to Ergodic Theory and I am trying to prove the following theorem:
Let $X$ be a metric space and $B$ the Borel $\sigma$-algebra. Let $(X, B, \mu)$ be a probability space where $\mu(U) > 0$ for every open set $U$, and let $f:X \to X$ be a surjective, ergodic measure preserving transformation. Then almost every point of $X$ has a dense orbit.
I am looking for any kind of hint on how to prove this, but if you could give me a hint on how to continue my approach below, that would be extremely helpful. Thank you very much!
Partial Solution:
Let $(B_n)$ be a base for $X$. The set of points which have a dense orbit is $$\{x \in X| \ \ \forall n \ge 1 \ \exists m \ge 0: \ f^m(x) \in B_n \}$$ which can be written as $$\bigcap_{n \ge 1} \bigcup_{m \ge 0}f^{-m}(B_n).$$
We want to show that this set has measure $1$. To do this, it is sufficient to show that $\bigcup_{m \ge 0}f^{-m}(B_n)$ has measure $1$ for each $n$. However, I am not sure how to proceed further.