I write you, because I have a problem to show two equivalences. But before writing them down, I give you all the definitions we had as background:
(I) The Quadruple $(\Omega,\mathcal{A},\mu,T)$ is called dynamical system, if $(\Omega,\mathcal{A},\mu)$ is a probability space and $ T\colon\Omega\to\Omega$ a measurable Transformation with the property $\mu\circ T^{-1}=\mu$.
(II) A set $A\in\mathcal{A}$ is called $T$-invariant, if $T^{-1}(A)=A $ a.s. (i.e. $T^{-1}(A)\Delta A$ is a null set).
(III) The system $(\Omega,\mathcal{A},\mu,T)$ is called ergodic, if for all T-invariant sets $A\in\mathcal{A}$ it is $\mu(A)\in\left\{0,1\right\}$.
Now the exercise is to show that the following two statements (1) and (2) are equivalent.
(1) $(\Omega,\mathcal{A},\mu,T)$ is ergodic
(2) $\forall B\in\mathcal{A}: T^{-1}B\subset B\implies \mu(B)\in\left\{0,1\right\}$
I think I managed it to prove $(1)\implies (2)$:
Consider any $B\in\mathcal{A}$ with $T^{-1}B\subset B$. It is $\mu(T^{-1}B)=\mu(B)$. Because of $T^{-1}B\subset B$ it is $T^{-1}B\Delta B=B\setminus T^{-1}(B)$. And because $\mu(T^{-1}(B))=\mu(B)<\infty$ it is $$ \mu(T^{-1}B\Delta B)=\mu(B\setminus T^{-1}B)=\mu(B)-\mu(T^{-1}B)=\mu(B)-\mu(B)=0. $$ From this it follows that $B$ is $T$-invariant and so because of the ergodicity it is $\mu(B)\in\left\{0,1\right\}$.
I think, that should be right?
But I am not able to prove $(2)\implies (1)$. I've tried very long time and now I cannot have a clear thought about it. So I would be very thankful if anybody could tell me how to prove that. I am perplexed...
Your reasoning is correct. For the part $(2)\Rightarrow (1)$, notice that if $\mu(A\Delta T^{-1}(A))=0$, then there is $A'$ invariant and $N$ of measure $0$ such that $A=A'\cup N$.