Let $(\Omega,\mathcal{A},\mu,T)$ be a dynamic system in meaasure theory. Show that the following statements are equivalent:
(1) $(\Omega,\mathcal{A},\mu,T)$ is ergodic
(2) $\forall A\in\mathcal{A}: T^{-1}(A)\subset A\implies\mu(A)\in\left\{0,1\right\}$
(3) $\forall A\in\mathcal{A}: \mu(A)>0\implies \mu(\bigcup_{k=1}^{\infty} T^{-k}A)=1$
I know it that way:
A dynamic system in measure theory consists of a measurable space $(\Omega,\mathcal{A})$, a measurable function $T\colon\Omega\to\Omega$ and a non-singular measure $\mu$ (i.e. for $A\in\mathcal{A}$ it is $\mu(A)0=0$ if and only if $\mu(T^{-1}(A))=0$). A measure $\mu$ is especially non-singular if it is invariant, i.e. for every $A\in\mathcal{A}$ it is $\mu(T^{-1}(A))=\mu(A)$.
A dynamic system is called ergodic if: $A\in\mathcal{A}: T^{-1}(A)\subset A\implies \mu(A)=0\text{ or }\mu(A^C)=0$
But then $(1)\implies (2)$ is trivial, isn't it?
Do you know a definition of "ergodic dynamic system" which does not use (2)?
First, I'll recall the broadest definition of ergodicity.
Let $(\Omega, \mathcal{A})$ be a measurable space. Let $T : \Omega \to \Omega$ be a measurable transformation. Let $\mu$ be a nonnegative, non-zero, $\sigma$-finite measure on $\Omega$.
We say that the dynamical system $(\Omega, \mathcal{A}, \mu, T)$ is non-singular if, for any Borel set $A$, we have $\mu (T^{-1}A) = 0 \Leftrightarrow \mu (A) = 0$. By Radon-Nikodym theorem, a system is non-singular if and only if $T_* \mu \ll \mu$, that is, if $T$ does not send some mass to a subset of measure zero.
We say that the dynamical system $(\Omega, \mathcal{A}, \mu, T)$ is measure-preserving if, for any Borel set $A$, we have $\mu (T^{-1}A) = \mu (A)$. That is, a system preserves the measure if and only if $T_* \mu = \mu$.
We say that the dynamical system $(\Omega, \mathcal{A}, \mu, T)$ is ergodic if it is non-singular and, for any Borel set $A$ such that $T^{-1}A \subset A$, we have $\mu (A) = 0$ or $\mu (A^c) = 0$. Ergodicity does not require that the measure be preserved, just that it be non-singular.
A system which is ergodic is, in some sense, irreducible: you cannot split it into two non-trivial (for the measure $\mu$) sub-systems which are invariant under $T$. For instance, the transformation $x \mapsto x+1/2 [1]$ on $[0, 1)$ is not ergodic with respect to the Lebesgue measure, because the subsets $[1, 1/4) \cup [1/2, 3/4)$ and $[1/4, 1/2) \cup [3/4, 1)$ are invariant and of positive measure.
Ergodicity can also be viewed functionally: the dynamical system $(\Omega, \mathcal{A}, \mu, T)$ is ergodic if and only if, for any measurable function $f: \Omega \to \mathbb{R}$, if $f = f \circ T$ almost everywhere for $\mu$, then there exists a constant $C$ such that $f = C$ almost everywhere for $\mu$.
In the following, I'll assume that $\mu$ is a probability measure, because ergodic theory with infinite measures is something very specific.
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The fact that $(1) \Leftrightarrow (2)$ is obvious, since $\mu (A) = 1 \Leftrightarrow \mu(A^c) = 0$ for any Borel subset $A$.
Let us prove that $(2) \Rightarrow (3)$, which you have already done more or less properly in the comments. We don't need to use the contraposition. Assume that the system is ergodic. Let $A$ be a Borel subset, with $\mu (A) > 0$. Let $B := \bigcup_{k=1}^{+ \infty} T^{-k} A$. Since the system is non-singular, $\mu (T^{-1}A) > 0$. But $T^{-1} A \subset B$, so $\mu(B)>0$. In addition,
$$T^{-1} B = T^{-1} \bigcup_{k=1}^{+ \infty} T^{-k} A = \bigcup_{k=1}^{+ \infty} T^{-(k+1)} A = \bigcup_{k=2}^{+ \infty} T^{-k} A \subset B.$$
By the property $(2)$, we have $\mu (B) = 1$.
Now, I'll give you some hints to prove that $(3) \Rightarrow (2)$. This time, I would use the contrapositive. That is, assume that there is a Borel subset $A$ with $T^{-1}A \subset A$ and $\mu (A) \in (0,1)$, and try to prove that $\mu (\bigcup_{k=1}^{+ \infty} T^{-k} A) < 1$. As a middle point, try to prove that $\mu (A^c \cap \bigcup_{k=1}^{+ \infty} T^{-k} A) = 0$.
Edit : Proof of $(3) \implies (2)$.
I'll use the contrapositive : it is enough to prove that $\not (2) \implies \not (3)$. Assume that $(2)$ is false. Then I can find a measurable subset $A$ such that $\mu (A) \in (0, 1)$ and $T^{-1} A \subset A$.
By induction, we show that $T^{-n} A \subset T^{-1}A$ for all $n \geq 1$. This is obviously true for $n = 1$. If it is true for some $n$, then $T^{-(n+1)} A = T^{-n} T^{-1} A \subset T^{-n}A \subset T^{-1} A$, as $ T^{-1} A \subset A$. Hence,
$$T^{-1} A \subset B := \bigcup_{k=1}^{+ \infty} T^{-k} A \subset T^{-1} A,$$
so $B = T^{-1} A$. In addition, $\mu (B) > 0$, as the measure $\mu$ is non-singular, and $B \subset A$, so $\mu (B) \leq \mu (A) < 1$. This contradicts $(3)$.