Induced sytem ergodic implies normal sytem ergodic Okay, we consider a measure preserving system $(X, \mathcal F, \mu, T)$ and let $A \in\mathcal F$ be such that $\mu(A) > 0$ and $\mu ( \cup ^{\infty}_{n=1} T^{-n}A) = 1 $. 
Now I want to show that if the induced system $(A,\mathcal F \cap A,\mu_A,T_A)$ is ergodic, then the system $(X,\mathcal F,\mu,T)$ is ergodic. 
My begin: 
Let $C \in F \cap A$ such that $T_A^{-1}C=C$. 
Because $T_A$ is ergodic, I know that $\mu_A(C)=0$ or $1$, equivalently $\mu(C)=0$ or $\mu(C)=\mu(A)$. 
...
I think my conclusion must be $\mu(A)=\mu(A)^2$ and $\mu(A) > 0$, thus $\mu(A)=1$, so $T$ is ergodic. 
Can someone help me? 
 A: In order to check ergodicity of the system $(X,\mathcal F, \mu,T)$, we have to show that for each subset $C$ such that $\mu(T^{-1}C\Delta C)=0$, we have $\mu(C)\in \{0,1\}$. To this aim, we define $C':=C\cap A'$, where $A'=\bigcup_{n\geqslant 1}T^{-n}A$. 
We show that 
$$\tag{$\star$}\mu\left(T^{-1}(C')\Delta C'\right)=0.$$ 
Indeed, we have 
$$\mu\left(T^{-1}(C')\setminus C'\right)\leqslant 
\mu(T^{-1}C\cap T^{-1}A'\cap C^c)+\mu(T^{-1}C\cap T^{-1}A'\cap (A')^c).$$
Since $\mu(T^{-1}C\setminus C)=0$ and $\mu(A'^c)=1$, we get 
$$\tag{1}\mu\left(T^{-1}(C')\setminus C'\right)=0.$$
Next, notice that
$$\mu\left(C'\setminus T^{-1}(C')\right)\leqslant \mu(A'\cap C\cap T^{-1}(A'^c))+\mu(A'\cap C\cap T^{-1}(A')\cap (T^{-1}C)^c).$$
Using the equality $\mu(T^{-1}(A'^c))=0$, we get 
$$\tag{2}\mu\left(C'\setminus T^{-1}(C')\right)=0.$$
Then $(\star)$ follows from (1) and (2).
Since the dynamical system $(A,A\cap\mathcal F,\mu_A,T_A)$ is ergodic and $C'$ belongs to $A\cap\mathcal F$, we have by $\star$, $\mu(C')\in \{0,1\}$. Using the fact that $\mu(A')=1$, we conclude that $\mu(C)\in \{0,1\}$.
