$\mu $ ergodic $\iff \mathcal{O}^{+}_{\phi}(x)\cap A\neq\emptyset$ Let $\mathcal{A}$ be a $\sigma$-Algebra on $X$ and $\phi: X\rightarrow X$ measurable (i.e. $\phi^{-1}(A)\in\mathcal{A}$) .
Let $\mu:\mathcal{A}\rightarrow[0,1]$ be probability measure which is $\phi$-invariant (i.e. $\mu(\phi^{-1}(A))=\mu(A)\thinspace \forall A\in\mathcal{A}$). 

To prove:
$\mu$ is ergodic $\iff$$\forall A\in \mathcal{A}$ with $\mu(A)>0 \colon\qquad\mathcal{O}^{+}_{\phi}(x)\cap A\neq\emptyset$ for almost all $x\in X $
Has anyone an idea how to start both ways ? Maybe the fact that $\mathcal{O}^{+}_\phi(x)\subseteq X $ for an invariant $\phi$ could help. I've tried the facts that $\mu(\mathcal{O}^{+}_\phi(x)\cup A) = \mu(\mathcal{O}^{+}_\phi(x))+\mu(A)-\mu(\mathcal{O}^{+}_\phi(x)\cap A)$. and $\mathcal{O}^{+}_\phi(x)$ is countable  $\mu(\mathcal{O}^{+}_\phi(x))=0$ 
I'm thankful for any tips
 A: For all $A \in \mathcal{A}$, let $x \in \phi^{-n}(A)$ iff $\phi^n(x) \in F$. Then, $O_\phi(x) \cap A \neq \emptyset$ a.s. iff $\mu(\cup_{n=1}^\infty\phi^{-n}(A))=1$.
($\Rightarrow$) Suppose that $\mu$ is ergodic and let $\mu(A)>0$.
Let $F_n = \bigcup_{k=n}^\infty \phi^{-k}(A)$ and note that $F_{n+1} \subset F_n$ and $\phi^{-1}(F_n) = F_{n+1}$. By $\phi$-invariance, $\mu(F_n) = \mu(F_{n+1})$. Let $F = \bigcap_{n=1}^\infty F_n$ and note that $\phi^{-1}(F)=F$.  
By ergodicity, $\mu(F) = 0$ or $1$. But $F_n \downarrow F$ and $\mu(F_n) = \mu(F_1)$ imply that $\mu(F_1)=0$ or $1$.
To finish, then, it suffices to show that $\mu(F_1)>0$. Simply observe that $\phi^{-1}(A) \subset F_1$ and use $\phi$-invariance once more to get $\mu(F_1) \geq \mu(\phi^{-1}(A)) = \mu(A)>0$.
($\Leftarrow$) For the converse assume that $\mu$ is not ergodic and let $A$ be a witness to this fact. Then, $\phi^{-1}(A)=A$, and $0 < \mu(A)<1$. By induction $\phi^{-n}(A)=A$ for all $n$, hence $$\mu(\cup_{n=1}^\infty\phi^{-n}(A)) = \mu(A) < 1.$$
