Let me define $T_1,T_2:[0,1)\rightarrow [0,1)$ . Let $\lambda_1, \lambda_2$ be measures on $[0,1)$ with total mass $1$ which are invariant for $T_1$ resp. $T_2$. Let $$T:[0,1)^2\rightarrow [0,1)^2; (x_1,x_2)\mapsto (T_1x_1,T_2x_2) $$ and $\lambda=\lambda_1\otimes \lambda_2$. I want to show that if $T$ is strong mixing for $\lambda$ then $T_i$ is strong mixing for $\lambda_i$.
I know that by definition I need to show that for all $B,B'\in \mathcal{B}([0,1))$ $\lambda_i(T_i^{-n}(B)\cap B')\rightarrow \lambda_i(B)\lambda_i(B')$. I thought that since sets of the form $[a,b)$ with $0\leq a<b\leq 1$ generates the borel-$\sigma$-algebra $\mathcal{B}([0,1))$, we can first choose $B=[a,b)$ and $B'=[c,d)$. But by assumption I only know that for all $A,A'\in \mathcal{B}([0,1)^2)$, $\lambda(T^{-n}(A)\cap A')\rightarrow \lambda(A)\lambda(A')$. Now also for $A$ and $A'$ one could take sets of the form $[u,v)\times[w,z)$, but I don't see how to write $B,B'$ as a set of the form $A,A'$ to apply the assumption. Because $[a,b)\times [0,1)$ and $[c,d)\times [0,1)$ would give a result really close to what I want, but $[a,b)\times [0,1)\neq B$.
Can someone help me further?
I will write $\Omega := [0, 1)$ for brevity.
For $B \in \mathcal{B}$, strong mixing of $T$ implies $$\lambda(T^{-n}(B \times \Omega) \cap (B' \times \Omega)) \to \lambda(B \times \Omega) \lambda(B' \times \Omega).$$ The right-hand side is equal to $\lambda_1(B) \lambda_1(B')$. The left-hand side is equal to \begin{align} \lambda(T^{-n}(B \times \Omega) \cap (B' \times \Omega)) &= \lambda((T_1^{-n}(B) \times T^{-n}(\Omega)) \cap (B' \times \Omega))\\ &= \lambda((T_1^{-n}(B) \times \Omega) \cap (B' \times \Omega))\\ &= \lambda((T_1^{-n}(B) \cap B') \times \Omega)\\ &= \lambda_1(T_1^{-n}(B) \cap B'). \end{align}